I have struck on the following:
We can define 0^0 =0 or 0^0 = 1 = depending on the scale of 0 we use in first 0 and exponent ( based on various orders of infinitesimals ). if we denote 0 as dx, dy, than:
1) dx^dy = 0 if dx is infinitely smaller than dy, but dx=0, dy=0 dx+dy=0, dx/dy = 0 or
2) dx^dy=1 if dx is infinitely greater than dy, but dx=0, dy=0, dx+dy=0, dx/dy=infinity.
3) The only problem arises when dx=dy=0 and dx+dy=0, dx/dy=finite.
Then the problem 0^0 is solved axiomatically to fit other math, by defining 0^0=1.
but let us leave this case open.
if we look at h(0)= 0^0^0^0^............. infinite times:
and take partial exponents:
if 0^0 = 1, 0^(0^0)=0, 0^(0^(0^0))=0^(0^1)=0^0=1,
so we get oscillating values 1,0,1,0,1,0,1,0........
if 0^0=0, 0^(0^0)=0^0 = 0, 0^(0^(0^0))=0^(0^0) =0^0 =0
We get value 0 all the time.
4) So it might be correct to suppose that value of 0^0 when 0/0=finite value lies somewhere between these 2 cases.
But how to evaluate h(0) when 0^0=1?
5) If we look at sum 1-1+1-1+1-1+1..............we see that partial sums develop like 0,1,0,1 .....similar to case h(0) when 0^0=1.
We know that Euler summation gives values to those series =1/2. We might expect that something similar can work in case h(0) when 0^0=1.
6) but there is a difference:
we can look at 1-1+1-1+1 ....... as sum of
1+1+1+1+1.........= + infinity
and -1-1-1-1-1- = - infinity
So value 1/2 is actually attributed to sum -infinity+infinity= 1/2.
7) In case with tetration, we can look at :
0^1^0^1^0^1.............. = 0, 0^1=0
0^-1^-1^-1^-1.........= infinity , 0^-1=infinity
0^(-1)^0^-1^0^-1^0.........= either infinity or 0 or 1 depending on 0/0 and 0^0.
0^-1 = infinity.
minus infinity is never as result in exponentation of 0, so we are looking at value that would describe multiplication
0^(-1) ^-1^-1^........* 0^(1)^1^1^...... = 0^0^0^0^.......which is something like infinity*0 so could be finite.
8) the conjecture is, that there could exist a value atributable to the h(0) when 0^0=1 as a value of 0,1,0,1,0,1.............and that this value also gives the correct answer to 0^0 when 0/0 = 1.
h(0) when 0^0=0 is 0.
9) This value seems not to be 1/2. What could it be? pi/2? pi/4? 1/e? i/2? sqrt(i)? something else?
The definition of h(z) by - W(-ln(0)/ln(0) does not seem very helpful here, but perhaps to ln(0) can be formally attributed some value as well?
QUOTE (Ivars+Nov 18 2007, 02:18 PM)
We can define 0^0 =0 or 0^0 = 1 = depending on the scale of 0 we use in first 0 and exponent ( based on various orders of infinitesimals ).
Zero is zero is zero is zero. There's no 'scale of zero'. 0=5*0 doesn't mean that 0/0 = 5, that one of the zeros is a scale of 5 larger than the other, zero is zero.
0^0 is undefined. You can consider things like the limit as x->0 of x^x, which is 1, but that doesn't mean that 0^0 = 1.
Zero is zero is zero is zero. There's no 'scale of zero'. 0=5*0 doesn't mean that 0/0 = 5, that one of the zeros is a scale of 5 larger than the other, zero is zero.
0^0 is undefined. You can consider things like the limit as x->0 of x^x, which is 1, but that doesn't mean that 0^0 = 1.
QUOTE (Ivars+Nov 18 2007, 02:18 PM)
if we denote 0 as dx, dy
Yes, let's define 0 by a non-zero quantity. Genius! 
QUOTE (Ivars+Nov 18 2007, 02:18 PM)
1) dx^dy = 0 if dx is infinitely smaller than dy, but dx=0, dy=0 dx+dy=0, dx/dy = 0 or
2) dx^dy=1 if dx is infinitely greater than dy, but dx=0, dy=0, dx+dy=0, dx/dy=infinity.
3) The only problem arises when dx=dy=0 and dx+dy=0, dx/dy=finite.
2) dx^dy=1 if dx is infinitely greater than dy, but dx=0, dy=0, dx+dy=0, dx/dy=infinity.
3) The only problem arises when dx=dy=0 and dx+dy=0, dx/dy=finite.
Limits are something you never bothered to look are, aren't they?
QUOTE (Ivars+Nov 18 2007, 02:18 PM)
5) If we look at sum 1-1+1-1+1-1+1
Which is undefined under standard summations.
QUOTE (Ivars+Nov 18 2007, 02:18 PM)
1+1+1+1+1.........= + infinity
and -1-1-1-1-1- = - infinity
So value 1/2 is actually attributed to sum -infinity+infinity= 1/2.
and -1-1-1-1-1- = - infinity
So value 1/2 is actually attributed to sum -infinity+infinity= 1/2.
If you'd ever bothered to learn anything about convergent series, you'd know that your logic is wrong. You cannot reorder a convergent series which is not absolutely convergent. Otherwise -1=0=1 = -1+1-1+1....
Yet again, you obviously waste a lot of your time avoiding reading and learning any maths and just assume whatever you can pull out your backside is even close to valid. Why do you bother to talk about such things as Euler summations when it's clear you've made no attempt to learn about series summations and convergence?
Yet again, you obviously waste a lot of your time avoiding reading and learning any maths and just assume whatever you can pull out your backside is even close to valid. Why do you bother to talk about such things as Euler summations when it's clear you've made no attempt to learn about series summations and convergence?
QUOTE (AlphaNumeric+Nov 18 2007, 01:28 PM)
0^0 is undefined. You can consider things like the limit as x->0 of x^x, which is 1, but that doesn't mean that 0^0 = 1.
Yes, let's define 0 by a non-zero quantity. Genius!
Limits are something you never bothered to look are, aren't they?
hej Alphanumeric
So what is the answer? For 0^0^0^............... = h(0)? Undefined again? As is 0^0? but 0!=1 - why? what is ln 0 -> - infinity or undefined?
I hope I will survive without studying limits and with belief that there are many 0-es, infinities, even ONES. And relations between them can be studied.
To be honest, I have strong dislike of limits based math - every time i had encountered it - and I like the approach of Euler who deals with various degrees of infinities and infinitesimals, and divergent series elegantly. He considered dx =0 not as a limit, but as identity. And he did write a/dx=a/0=infinity and infinity*0 = a -> finite value. And it worked fine. Now, if things work, and efficiently, then what is the problem to use them again?
You will again twist his quotes, of course, but that is your problem. Read his books yourself and do not ask me to quote them for You to prove I have read them. You never bothered to read Euler, did You? May be You have not read Einstein or Planck either?
Yes, let's define 0 by a non-zero quantity. Genius!
Limits are something you never bothered to look are, aren't they?
hej Alphanumeric
So what is the answer? For 0^0^0^............... = h(0)? Undefined again? As is 0^0? but 0!=1 - why? what is ln 0 -> - infinity or undefined?
I hope I will survive without studying limits and with belief that there are many 0-es, infinities, even ONES. And relations between them can be studied.
To be honest, I have strong dislike of limits based math - every time i had encountered it - and I like the approach of Euler who deals with various degrees of infinities and infinitesimals, and divergent series elegantly. He considered dx =0 not as a limit, but as identity. And he did write a/dx=a/0=infinity and infinity*0 = a -> finite value. And it worked fine. Now, if things work, and efficiently, then what is the problem to use them again?
You will again twist his quotes, of course, but that is your problem. Read his books yourself and do not ask me to quote them for You to prove I have read them. You never bothered to read Euler, did You? May be You have not read Einstein or Planck either?
QUOTE (Ivars+Nov 18 2007, 09:26 PM)
I hope I will survive without studying limits and with belief that there are many 0-es, infinities, even ONES. And relations between them can be studied.
Then you will not be doing math, and probably nothing you do will be at all rigorous, and most likely not even self-consistent.
Then you will not be doing math, and probably nothing you do will be at all rigorous, and most likely not even self-consistent.
QUOTE (Ivars+Nov 18 2007, 10:26 PM)
So what is the answer? For 0^0^0^............... = h(0)? Undefined again?
Yes. You also fail to realise that exponentiation is not associative. a^(b^c) is different from (a^
^c and even if 0^0 was a valid quantity, you cannot write 0^0^0 without brackets because (0^0)^0 would not be the same as 0^(0^0) is all likelihood.
But you don't realise this, do you?
Yes. You also fail to realise that exponentiation is not associative. a^(b^c) is different from (a^
^c and even if 0^0 was a valid quantity, you cannot write 0^0^0 without brackets because (0^0)^0 would not be the same as 0^(0^0) is all likelihood.But you don't realise this, do you?
QUOTE (Ivars+Nov 18 2007, 10:26 PM)
but 0!=1
Well if you define 'factorial' by Euler's integral of the second kind, you find that 0!=1.
But then you didn't know this, did you? You have talked about such an integral before, it's the Gamma function! So much for you having studied that.
But then you didn't know this, did you? You have talked about such an integral before, it's the Gamma function! So much for you having studied that.
QUOTE (Ivars+Nov 18 2007, 10:26 PM)
I hope I will survive without studying limits
I doubt a meteorite will strike you down if you don't bother to read about limits but you'll never get very far if you don't.
QUOTE (Ivars+Nov 18 2007, 10:26 PM)
To be honest, I have strong dislike of limits based math
Derivatives and integrals are not something you like then. Well that's all your vortex crap out the window. That's the Gamma and Zeta functions you talk so much about out the window.
Basically, you're trying to make excuses for your ignorance.
Basically, you're trying to make excuses for your ignorance.
QUOTE (Ivars+Nov 18 2007, 10:26 PM)
nd I like the approach of Euler who deals with various degrees of infinities and infinitesimals
And as you quoted from Wikipedia, much of Euler's methods are not considered rigorous by todays standards because it wasn't until 100+ years after his death that real rigour came into it.
But you aren't bothered even about Euler's notion of rigour.
But you aren't bothered even about Euler's notion of rigour.
QUOTE (Ivars+Nov 18 2007, 10:26 PM)
Now, if things work, and efficiently, then what is the problem to use them again?
The problem is that you have no idea if things work. Again and again I've had to correct you on claims you've made which are complete nonsense.
You thought your method for solving y''+5y'+6y = exp(3x) was 'working' but it wasn't. 3 minutes with a highschool textbook would have told you otherwise.
You thought your method for solving y''+5y'+6y = exp(3x) was 'working' but it wasn't. 3 minutes with a highschool textbook would have told you otherwise.
QUOTE (Ivars+Nov 18 2007, 10:26 PM)
You will again twist his quotes, of course, but that is your problem.
The quotes you provided never divided through by zero. That was all you.
QUOTE (Ivars+Nov 18 2007, 10:26 PM)
ead his books yourself and do not ask me to quote them for You to prove I have read them. You never bothered to read Euler, did You? May be You have not read Einstein or Planck either?
Well considering he didn't write in English and generally neither did Einstein or Planck, no I've not. I've read the works of people who have put their work into English and who have the 100~250 years of hindsight since those people to develop their ideas. Reading the original works of famous people isn't always the best thing. 'Freethis' claimed to be reading Maxwell's original work, despite it being very contrived by todays standard because the elegant notation now used didn't exist in Maxwell's day. But if you want to talk about books I've read on various subjects, here's a few :
A First Course in Mathematical Analysis
Mathematical Methods for Physics and Engineering
A First Course in GR
Superstring theory (both volumes!)
String theory (both volumes)
Lie algebras in particle physics
Advanced GR
Supersymmetric Gauge Field Theory and String Theory
Supersymmetry and Supergravity
Asymptotic Expansions
An Introduction to QFT
Ideals, Varieties and Algorithms
And that's not even close to an exhaustive list. And that's on top of the 4 years of Cambridge lecture notes I have dotted about my room covering analysis, complex analysis (including the Gamma and Zeta functions), geometry, vector calculus, quantum mechanics, special and general relativity, field theory, string theory, supersymmetry and differential geometry.
How much have you read there Chuckles? Not much. You're reduced to directly quoting Euler and still not understanding what he's saying. You knew you'd fail if you tried to read a well known standard text on analysis like the book by Burkill I mentioned, so you jump to an obscure text and just fall back on "If you didn't read his original work, you aren't working on the same stuff as me". What's the matter, child's work too hard for you? But then we know that.
A First Course in Mathematical Analysis
Mathematical Methods for Physics and Engineering
A First Course in GR
Superstring theory (both volumes!)
String theory (both volumes)
Lie algebras in particle physics
Advanced GR
Supersymmetric Gauge Field Theory and String Theory
Supersymmetry and Supergravity
Asymptotic Expansions
An Introduction to QFT
Ideals, Varieties and Algorithms
And that's not even close to an exhaustive list. And that's on top of the 4 years of Cambridge lecture notes I have dotted about my room covering analysis, complex analysis (including the Gamma and Zeta functions), geometry, vector calculus, quantum mechanics, special and general relativity, field theory, string theory, supersymmetry and differential geometry.
How much have you read there Chuckles? Not much. You're reduced to directly quoting Euler and still not understanding what he's saying. You knew you'd fail if you tried to read a well known standard text on analysis like the book by Burkill I mentioned, so you jump to an obscure text and just fall back on "If you didn't read his original work, you aren't working on the same stuff as me". What's the matter, child's work too hard for you? But then we know that.
The series (1-1+1-1+1-...) can be expressed as a geometric progression,
u=(1+i^2+i^4+i^6+...+i^2k)
u=1+i^2(u)=1-u
u=1/2
I don't understand what Ivars is trying to show with 0^0
u=(1+i^2+i^4+i^6+...+i^2k)
u=1+i^2(u)=1-u
u=1/2
I don't understand what Ivars is trying to show with 0^0
the geometric progression S=1+z+z^2+...=1/(1-z) is only defined on the open unit disc centered at 0 as there is an unremoveable singularity at z=1. so z=-1 doesn't work as an example.
whatever ivars is saying, if it involves 1=0, he can show whatever he wants.
whatever ivars is saying, if it involves 1=0, he can show whatever he wants.
QUOTE (AlphaNumeric+Nov 18 2007, 11:06 PM)
Well considering he didn't write in English and generally neither did Einstein or Planck, no I've not. I've read the works of people who have put their work into English and who have the 100~250 years of hindsight since those people to develop their ideas. Reading the original works of famous people isn't always the best thing. 'Freethis' claimed to be reading Maxwell's original work, despite it being very contrived by todays standard because the elegant notation now used didn't exist in Maxwell's day. But if you want to talk about books I've read on various subjects, here's a few :
A First Course in Mathematical Analysis
Mathematical Methods for Physics and Engineering
A First Course in GR
Superstring theory (both volumes!)
String theory (both volumes)
Lie algebras in particle physics
Advanced GR
Supersymmetric Gauge Field Theory and String Theory
Supersymmetry and Supergravity
Asymptotic Expansions
An Introduction to QFT
Ideals, Varieties and Algorithms
And that's not even close to an exhaustive list. And that's on top of the 4 years of Cambridge lecture notes I have dotted about my room covering analysis, complex analysis (including the Gamma and Zeta functions), geometry, vector calculus, quantum mechanics, special and general relativity, field theory, string theory, supersymmetry and differential geometry.
Hej AlphaNumeric,
I do not doubt that You have read it all. I call it Readers Digest.It is very efficient way of reading books. Luckily that is not allowed for Philosphy or Literature students.
A First Course in Mathematical Analysis
Mathematical Methods for Physics and Engineering
A First Course in GR
Superstring theory (both volumes!)
String theory (both volumes)
Lie algebras in particle physics
Advanced GR
Supersymmetric Gauge Field Theory and String Theory
Supersymmetry and Supergravity
Asymptotic Expansions
An Introduction to QFT
Ideals, Varieties and Algorithms
And that's not even close to an exhaustive list. And that's on top of the 4 years of Cambridge lecture notes I have dotted about my room covering analysis, complex analysis (including the Gamma and Zeta functions), geometry, vector calculus, quantum mechanics, special and general relativity, field theory, string theory, supersymmetry and differential geometry.
Hej AlphaNumeric,
I do not doubt that You have read it all. I call it Readers Digest.It is very efficient way of reading books. Luckily that is not allowed for Philosphy or Literature students.
QUOTE (NeoDevin+Nov 18 2007, 10:01 PM)
Then you will not be doing math, and probably nothing you do will be at all rigorous, and most likely not even self-consistent.
Hej NeoDevin
It is not rigorous that matters, but results. Self consistency can be brought in later- If results are true, (as in Euler case) , self consistency will be found,.
Hej NeoDevin
It is not rigorous that matters, but results. Self consistency can be brought in later- If results are true, (as in Euler case) , self consistency will be found,.
QUOTE (phyti+Nov 19 2007, 04:17 AM)
The series (1-1+1-1+1-...) can be expressed as a geometric progression,
u=(1+i^2+i^4+i^6+...+i^2k)
u=1+i^2(u)=1-u
u=1/2
I don't understand what Ivars is trying to show with 0^0
Hej phyti
I am trying to show that limits are limits-to the knowledge.
u=(1+i^2+i^4+i^6+...+i^2k)
u=1+i^2(u)=1-u
u=1/2
I don't understand what Ivars is trying to show with 0^0
Hej phyti
I am trying to show that limits are limits-to the knowledge.
QUOTE (Ivars+Nov 19 2007, 10:07 AM)
I call it Readers Digest.
I call it physics and mathematics.
I call what you read.... nothing.
I call it physics and mathematics.
I call what you read.... nothing.
QUOTE (Ivars+Nov 19 2007, 10:07 AM)
It is very efficient way of reading books.
Ah, so someone has pointed out that they've read books, after you challenged them and now you don't believe them.
I've demonstrated on here many times that I'm familiar with such topics, I can do such topics and I even once posted a picture of a bunch of those books because someone else didn't believe me.
Of course we can both claim to have read whatever we like, but what it comes down to is what we actually show we understand and are capable of. You've failed to understand even a single paragraph of the book you claim to be reading. I'm in my 2nd year of a theoretical physics PhD and have a degree and masters in maths so a great many of those books, even if I hadn't read them, are material I'm familiar with and I've proven that.
I've demonstrated on here many times that I'm familiar with such topics, I can do such topics and I even once posted a picture of a bunch of those books because someone else didn't believe me.
Of course we can both claim to have read whatever we like, but what it comes down to is what we actually show we understand and are capable of. You've failed to understand even a single paragraph of the book you claim to be reading. I'm in my 2nd year of a theoretical physics PhD and have a degree and masters in maths so a great many of those books, even if I hadn't read them, are material I'm familiar with and I've proven that.
QUOTE (Ivars+Nov 19 2007, 10:07 AM)
Luckily that is not allowed for Philosphy or Literature students.
It's not allowed for physics students to just say "I read it" when they didn't. Why? Because I have to demonstrate working understanding of such material every day. Now you, you don't. You aren't tested, you aren't asked to do real work on such things, so you can claim to be reading whatever you like.
Besides, when I do test you with a question which is laughably easy, you fail it! Says a lot, doesn't it?
Besides, when I do test you with a question which is laughably easy, you fail it! Says a lot, doesn't it?
QUOTE (Ivars+Nov 19 2007, 10:07 AM)
I am trying to show that limits are limits-to the knowledge.
Your knowledge, yes.
QUOTE (Ivars+Nov 19 2007, 10:07 AM)
It is not rigorous that matters, but results. Self consistency can be brought in later- If results are true, (as in Euler case) , self consistency will be found,.
Without justification, how do you know your results are valid? Euler did use rigour, it just wasn't put in the form we now call 'rigour'. He knew how to integrate but he didn't know about 'Rieman integrability'. So all his results from integration (Gamma, Beta functions etc) were valid provided integration was.
You offer no rigour AT ALL.
You offer no rigour AT ALL.
QUOTE (xaos+Nov 19 2007, 06:31 AM)
the geometric progression S=1+z+z^2+...=1/(1-z) is only defined on the open unit disc centered at 0 as there is an unremoveable singularity at z=1. so z=-1 doesn't work as an example.
whatever ivars is saying, if it involves 1=0, he can show whatever he wants.
Hej xaos
It works perfectly well in all cases, just not as a normal summation.
I never said 1=0 . n/1=0/0 does not mean 1=0, but that those 2 zeroes are different if compared as ratio, insted of arithmetic comparison where 0+0=0 always.
I was just quoting few things from Euler about ratios of infinitesimals (0-oes of different orders) and infinities (of different order) that he used systematically and never (almost?) failed to produce correct results.
whatever ivars is saying, if it involves 1=0, he can show whatever he wants.
Hej xaos
It works perfectly well in all cases, just not as a normal summation.
I never said 1=0 . n/1=0/0 does not mean 1=0, but that those 2 zeroes are different if compared as ratio, insted of arithmetic comparison where 0+0=0 always.
I was just quoting few things from Euler about ratios of infinitesimals (0-oes of different orders) and infinities (of different order) that he used systematically and never (almost?) failed to produce correct results.
QUOTE (Ivars+Nov 19 2007, 10:19 AM)
It works perfectly well in all cases, just not as a normal summation.
No, it depends on your approach to things.
1-1+1-1+1-1.... is undefined, on it's own.
However, if you define the function in the complex plane f(z) = 1/(1-z) then you find it's valid everywhere but at z=1, which is a singularity. This function has the Taylor expansion of 1+z+z^2+z^3+.... and so you have an analytic continuation of the summation of Real numbers 1+x+x^2+x^3+... = 1/(1-x) to be true everywhere by z=x=1.
Analytic continuation is a common thing in complex analysis and it's done for the Gamma function (it analytically continues the notion of factorials into non-integers and complex numbers), but then you don't know this, do you?
Just as 0/0 is meaningless but you can construct limits which have both numerator and denominator going to 0 and get a meaningful result, 1+x+x^2+x^3+.... is not valie outside |x|<1, but via analytic continuation you can make it valid everywhere but a few singularities.
This is all stuff taught on the lecture course Euler once asked you to answer some questions from. See what you miss out on because you avoid learning?
No, it depends on your approach to things.
1-1+1-1+1-1.... is undefined, on it's own.
However, if you define the function in the complex plane f(z) = 1/(1-z) then you find it's valid everywhere but at z=1, which is a singularity. This function has the Taylor expansion of 1+z+z^2+z^3+.... and so you have an analytic continuation of the summation of Real numbers 1+x+x^2+x^3+... = 1/(1-x) to be true everywhere by z=x=1.
Analytic continuation is a common thing in complex analysis and it's done for the Gamma function (it analytically continues the notion of factorials into non-integers and complex numbers), but then you don't know this, do you?
Just as 0/0 is meaningless but you can construct limits which have both numerator and denominator going to 0 and get a meaningful result, 1+x+x^2+x^3+.... is not valie outside |x|<1, but via analytic continuation you can make it valid everywhere but a few singularities.
This is all stuff taught on the lecture course Euler once asked you to answer some questions from. See what you miss out on because you avoid learning?
my complex analysis is a bit rusty.
for what i remember, z=(cosx,siny) is a point on the unit circle, and z^n=(cosnx,sinny) never zero, are points cyclic about the unit circle, then every so often 1+z+z^2+...+z^(m-1) adds up to zero., this means for all a multiples n*m of m similarly, z^(n*m)+z^(n*m+1)+...+z^((n*m+(m-1))=0.
nonconvergent cyclic infinite sums cannot be used to define a single point number: the z (not 1) mapped onetoone with 1/(1-z). one has a multivalued limit set (cyclic), the other is simply single valued.
there are several conditions we need so that we get 1+z+z^2+...z^k, as k goes to infinity, to be single valued rather than cyclic (and so we can identify it with 1/(1-z)). one, the modulus of of the summands z^n must contract to zero (true if abs(z)<1) sufficiently fast (true if abs(z)<1), and each partial sum must be bounded (true if abs(z)<1).
for what i remember, z=(cosx,siny) is a point on the unit circle, and z^n=(cosnx,sinny) never zero, are points cyclic about the unit circle, then every so often 1+z+z^2+...+z^(m-1) adds up to zero., this means for all a multiples n*m of m similarly, z^(n*m)+z^(n*m+1)+...+z^((n*m+(m-1))=0.
nonconvergent cyclic infinite sums cannot be used to define a single point number: the z (not 1) mapped onetoone with 1/(1-z). one has a multivalued limit set (cyclic), the other is simply single valued.
there are several conditions we need so that we get 1+z+z^2+...z^k, as k goes to infinity, to be single valued rather than cyclic (and so we can identify it with 1/(1-z)). one, the modulus of of the summands z^n must contract to zero (true if abs(z)<1) sufficiently fast (true if abs(z)<1), and each partial sum must be bounded (true if abs(z)<1).
hej xaos,
see how comlicated it is?
Meanwhile, a little sidestep, but still about tetration-very interesting topic:
h is the infinite tetration function, so that h(1/4) = (1/4)^(((1/4)^((1/4)^............ coming from the right so e.g 3^3^3 when tetrated, is 3^27, not 27^3. W is Lambert function which I struck upon 1 year ago thinking it must be related to scaling of something..
W(2ln2)=ln2
W(3ln3)=ln3
W(4ln4)=ln4
W(nln(n))=ln n , n>=1
So h(1/4) = h((1/2)^2) = W( -ln (1/2)^2)/-ln(1/2)^2=W(2ln2)/2ln2= 1/2
h(1/27) = h((1/3)^3) = 1/3
h(1/256) = h(1/4)^4) = h((1/2)^2^2^2) = 1/4
h(1/3125)=h(1/5)^5=1/5
h(1/n^n)=1/n
Which is the same as taking n-th root of n^n.
so that :
h(1) +h(1/4) + h(1/27) + h(1/256) ...+h(1/n^n)... = harmonic series
h(1)^2+h((1/4))^2+h((1/27))^2 +...+ h(1/n^n))^2...= pi^2/6 and so on for higher powers in series.
We also have W(-ln2/2) = -ln2
So h (2^1/2) = W(-ln2/2)/-1/2ln2 = -ln2/(-1/2ln2)= 2 and simmetry ends at e^1/e
Multiplying h(1/x^x) * h(x^1/x) = 1 at each x if -1/e<x<e.
so h(x^1/x) = 1/h(1/x^x). h(1/x^x) converges for all -<1/e x<1/e so all positive integers can be produced as
n=1/(h(1/n^n))), n>=1 , fractions
1/n=h(1/n^n))
In that sense h(1/n^n) works opposite integers -when n is in denominator, h(1/n^n) is in numerator.
also if x>=2, or may be x>=e^1/e, ln x = Integral (1/x) = integral h((1/x^x))
we can also make:
m/n = h(1/n^n) / h(1/m^m)
practically all integer numbers and rational fractions can be substituted by h of something, also functions like ln .
Especially, h(e^pi/2) = +- i .......... but i^4=1 so in the end
1= (+h(e^pi/2))^4, -1 = (+h(e^pi/2))^2
2= (+h(e^pi/2))^4+ (h(e^pi/2))^4
So even i can be replaced by h(e^pi/2) so that
i = e^+h(e^pi/2)*pi/2=+h(e^pi/2)
My feeling is that the whole number axis except transcendental numbers like e,pi, 2 can be constructed from infinite tetration-may be that is how nature works? To get rid of real number axis.
Take any series with ANY integers, or rational fractions, or series of x where x>=2 , replace x with h(1/x^x) and see what happens.
for 1, use h(e^pi/2)^4 or h(1/1^1) , for -1- h(/-1^-1)?
Just an idea
Best regards,
Ivars
see how comlicated it is?
Meanwhile, a little sidestep, but still about tetration-very interesting topic:
h is the infinite tetration function, so that h(1/4) = (1/4)^(((1/4)^((1/4)^............ coming from the right so e.g 3^3^3 when tetrated, is 3^27, not 27^3. W is Lambert function which I struck upon 1 year ago thinking it must be related to scaling of something..
W(2ln2)=ln2
W(3ln3)=ln3
W(4ln4)=ln4
W(nln(n))=ln n , n>=1
So h(1/4) = h((1/2)^2) = W( -ln (1/2)^2)/-ln(1/2)^2=W(2ln2)/2ln2= 1/2
h(1/27) = h((1/3)^3) = 1/3
h(1/256) = h(1/4)^4) = h((1/2)^2^2^2) = 1/4
h(1/3125)=h(1/5)^5=1/5
h(1/n^n)=1/n
Which is the same as taking n-th root of n^n.
so that :
h(1) +h(1/4) + h(1/27) + h(1/256) ...+h(1/n^n)... = harmonic series
h(1)^2+h((1/4))^2+h((1/27))^2 +...+ h(1/n^n))^2...= pi^2/6 and so on for higher powers in series.
We also have W(-ln2/2) = -ln2
So h (2^1/2) = W(-ln2/2)/-1/2ln2 = -ln2/(-1/2ln2)= 2 and simmetry ends at e^1/e
Multiplying h(1/x^x) * h(x^1/x) = 1 at each x if -1/e<x<e.
so h(x^1/x) = 1/h(1/x^x). h(1/x^x) converges for all -<1/e x<1/e so all positive integers can be produced as
n=1/(h(1/n^n))), n>=1 , fractions
1/n=h(1/n^n))
In that sense h(1/n^n) works opposite integers -when n is in denominator, h(1/n^n) is in numerator.
also if x>=2, or may be x>=e^1/e, ln x = Integral (1/x) = integral h((1/x^x))
we can also make:
m/n = h(1/n^n) / h(1/m^m)
practically all integer numbers and rational fractions can be substituted by h of something, also functions like ln .
Especially, h(e^pi/2) = +- i .......... but i^4=1 so in the end
1= (+h(e^pi/2))^4, -1 = (+h(e^pi/2))^2
2= (+h(e^pi/2))^4+ (h(e^pi/2))^4
So even i can be replaced by h(e^pi/2) so that
i = e^+h(e^pi/2)*pi/2=+h(e^pi/2)
My feeling is that the whole number axis except transcendental numbers like e,pi, 2 can be constructed from infinite tetration-may be that is how nature works? To get rid of real number axis.
Take any series with ANY integers, or rational fractions, or series of x where x>=2 , replace x with h(1/x^x) and see what happens.
for 1, use h(e^pi/2)^4 or h(1/1^1) , for -1- h(/-1^-1)?
Just an idea
Best regards,
Ivars
Proceeding still a little further with a question in the end:
h(z) = - W(-ln(z))/lnz:
h(1/n^n)=1/n
h(1) +h(1/4) + h(1/27) + h(1/256) ...+h(1/n^n)... = harmonic series
h(1)^2+h((1/4))^2+h((1/27))^2 +...+ h(1/n^n))^2...= pi^2/6 and so on for higher powers in series.
Multplying h(1/x^x) * h(x^1/x) = 1 at each x if -1/e<x<e.
To make a more general case, we can consider:
x= z^y so that 1/x = z^-y; then from h(1/x^x)=1/x ->
h(z^-y)^(z^y) = z^-y
but we know that h(e^pi/2)^h(e^pi/2) = i^i = -i^-i = e^-pi/2
on other hand, i^-i =i^(1/i) = e^( ipi/2*-i)= e^pi/2
so:
h(e^pi/2)^1/h(e^pi/2) = e^pi/2
then we can generalize to z^y:
h(z^y)^h(z^y) = z^-y
h(z^y)^1/h(z^y) = z^y
so that:
h(z^y)^h(z^y)*h(z^y)^1/h(z^y) =1 or
h(z^y)^ ((h(z^y))^2+1/h(z^y)) =1 which is possible always if
(h(z^y))^2+1=0 -> h(z^y) = +- i; +- 2pik. We know that the first solution is:
z^y = e^pi/2 or z=e, y=pi/2 and h(e^pi/2) = +- i .
But we also know that for z^y = e^pi/2 other branches of Lambert function leads to values h(e^pi/2) different from +-2pik.
Question:
Do there exist any roots z and y for equations:
h(z^y) = +-I*2pi*k? for all k, or just for some single special k?
h(z) = - W(-ln(z))/lnz:
h(1/n^n)=1/n
h(1) +h(1/4) + h(1/27) + h(1/256) ...+h(1/n^n)... = harmonic series
h(1)^2+h((1/4))^2+h((1/27))^2 +...+ h(1/n^n))^2...= pi^2/6 and so on for higher powers in series.
Multplying h(1/x^x) * h(x^1/x) = 1 at each x if -1/e<x<e.
To make a more general case, we can consider:
x= z^y so that 1/x = z^-y; then from h(1/x^x)=1/x ->
h(z^-y)^(z^y) = z^-y
but we know that h(e^pi/2)^h(e^pi/2) = i^i = -i^-i = e^-pi/2
on other hand, i^-i =i^(1/i) = e^( ipi/2*-i)= e^pi/2
so:
h(e^pi/2)^1/h(e^pi/2) = e^pi/2
then we can generalize to z^y:
h(z^y)^h(z^y) = z^-y
h(z^y)^1/h(z^y) = z^y
so that:
h(z^y)^h(z^y)*h(z^y)^1/h(z^y) =1 or
h(z^y)^ ((h(z^y))^2+1/h(z^y)) =1 which is possible always if
(h(z^y))^2+1=0 -> h(z^y) = +- i; +- 2pik. We know that the first solution is:
z^y = e^pi/2 or z=e, y=pi/2 and h(e^pi/2) = +- i .
But we also know that for z^y = e^pi/2 other branches of Lambert function leads to values h(e^pi/2) different from +-2pik.
Question:
Do there exist any roots z and y for equations:
h(z^y) = +-I*2pi*k? for all k, or just for some single special k?
Ivars Wrote:
Do there exist any roots z and y for equations:
h(z^y) = e^(i*pi/2+-I*2pi*k) ? for all k, or just for some single special k?
Solution to this is:
y*lnz = (ln (I +-2pi*k*I))/( I+-2pi*k*I)
when k=+1: y*lnz = (ln (I +2pi*I)/(I+2pi*I)
if z=e, y=ln (I+- 2*k*pi*I)/(I +-2pi*k*I) which gives y=pi/2 for k=0.
However, if z= (I+-2*k*pi*I) , y = 1/(I+-2pikI)
So:
h((I+-2k*pi*I)^(1/(I+-2pikI )))= e^((pi/2+2kpi)* I)
So these should be values of argument of h (y^z) which have purely imaginary values =+-I +- 2pikI; if k=0 we have
h((I^1/I) = h(I^-I) = h(e^pi/2) = +- i which is true.
for k=1 z^y = (I+2piI)^1/(I+2piI) = 1.194877 - 0.334069*i
k=2 z^y = (I+4piI)^1/(I+4piI)=1.102079 - 0.214479*i
k=100 z^y=1.004829 - 0.018345*i
k=1000000 z^y=1.000005 - 0.000040*i
so at k= +infinity z^y = 1 and h(z)=1 meaning:
e^(pi/2+2pik)*I = 1 when k-> +infinity
Probably I am making a mistake somewhere...
Do there exist any roots z and y for equations:
h(z^y) = e^(i*pi/2+-I*2pi*k) ? for all k, or just for some single special k?
Solution to this is:
y*lnz = (ln (I +-2pi*k*I))/( I+-2pi*k*I)
when k=+1: y*lnz = (ln (I +2pi*I)/(I+2pi*I)
if z=e, y=ln (I+- 2*k*pi*I)/(I +-2pi*k*I) which gives y=pi/2 for k=0.
However, if z= (I+-2*k*pi*I) , y = 1/(I+-2pikI)
So:
h((I+-2k*pi*I)^(1/(I+-2pikI )))= e^((pi/2+2kpi)* I)
So these should be values of argument of h (y^z) which have purely imaginary values =+-I +- 2pikI; if k=0 we have
h((I^1/I) = h(I^-I) = h(e^pi/2) = +- i which is true.
for k=1 z^y = (I+2piI)^1/(I+2piI) = 1.194877 - 0.334069*i
k=2 z^y = (I+4piI)^1/(I+4piI)=1.102079 - 0.214479*i
k=100 z^y=1.004829 - 0.018345*i
k=1000000 z^y=1.000005 - 0.000040*i
so at k= +infinity z^y = 1 and h(z)=1 meaning:
e^(pi/2+2pik)*I = 1 when k-> +infinity
Probably I am making a mistake somewhere...
QUOTE (Ivars+Nov 18 2007, 09:18 AM)
The definition of h(z) by - W(-ln(0)/ln(0) does not seem very helpful here, but perhaps to ln(0) can be formally attributed some value as well?
As you will see I am unable to get that far.
We can also invert it and end up with -1/2, of course.
-1-1-1-1-1- = - infinity
and -1-1-1-1-1- = - infinity
Would that make the sign "irrelevant"? The "value" you are giving it could be off by 1!!!!!!
However, I am still fuzzy on the whole thing... you define a "process" that uses addition, then give it a value of 1/2 (or -1/2).
Then you define a completely different process and try to give it a value, and that is what is not working: is that what the question is, Ivars? Do I understand the problem?
We can also invert it and end up with -1/2, of course.
-1-1-1-1-1- = - infinity
and -1-1-1-1-1- = - infinity
Would that make the sign "irrelevant"? The "value" you are giving it could be off by 1!!!!!!
However, I am still fuzzy on the whole thing... you define a "process" that uses addition, then give it a value of 1/2 (or -1/2).
Then you define a completely different process and try to give it a value, and that is what is not working: is that what the question is, Ivars? Do I understand the problem?
8) the conjecture is, that there could exist a value atributable to the h(0) when 0^0=1 as a value of 0,1,0,1,0,1.............and that this value also gives the correct answer to 0^0 when 0/0 = 1.
We can~t simultaneously treat "h(0)" a value and as a process because we will get lost in uncertainties.
If it is process, the value's sign will change as that of the "1/2" above, from negative to positive, in a predetermined manner, yes, no, yes, no... but I am able to further that thought very little.
(because the sign is an infinitesimal of an equation when it is a sign; when it is a **numeric value** it must follow the behaviour predetermined by the hyerarchical superior, a number... and the *number** of the value, previously, was 1, 0, 1, 0, 1...)
Once it is determined that the sign must change following hierarchical rules, you then do like my daughter and determine the indeterminate: "some1 is at the door", "some2 is at the door", "some3 is at the door" --she does it in Portuguese too: 1guem chegou, 2guem, 3guem!!!
But the hierarchical rules of numbers you started with, even in the definition of a process, started with little numbers and went on ascending.
So that the "value" you are looking for is defined by a series of plus and minus signs, and a series of ascending numbers, and a stable series of multiplications and divisions. I have no idea what that means, needless to say.
Sorry, Ivars, it may almost look like science but that is the best I can do
I think I didn~t really understand the problem, but then that never kept me from searching for answers before 
As you will see I am unable to get that far.
QUOTE
5) If we look at sum 1-1+1-1+1-1+1..............we see that partial sums develop like 0,1,0,1 .....similar to case h(0) when 0^0=1.
(...)6) but there is a difference:
we can look at 1-1+1-1+1 ....... as sum of
1+1+1+1+1.........= + infinity
and -1-1-1-1-1- = - infinity
So value 1/2 is actually attributed to sum -infinity+infinity= 1/2.
(...)6) but there is a difference:
we can look at 1-1+1-1+1 ....... as sum of
1+1+1+1+1.........= + infinity
and -1-1-1-1-1- = - infinity
So value 1/2 is actually attributed to sum -infinity+infinity= 1/2.
We can also invert it and end up with -1/2, of course.
-1-1-1-1-1- = - infinity
and -1-1-1-1-1- = - infinity
Would that make the sign "irrelevant"? The "value" you are giving it could be off by 1!!!!!!
However, I am still fuzzy on the whole thing... you define a "process" that uses addition, then give it a value of 1/2 (or -1/2).
Then you define a completely different process and try to give it a value, and that is what is not working: is that what the question is, Ivars? Do I understand the problem?
QUOTE (->
| QUOTE |
| 5) If we look at sum 1-1+1-1+1-1+1..............we see that partial sums develop like 0,1,0,1 .....similar to case h(0) when 0^0=1. (...)6) but there is a difference: we can look at 1-1+1-1+1 ....... as sum of 1+1+1+1+1.........= + infinity and -1-1-1-1-1- = - infinity So value 1/2 is actually attributed to sum -infinity+infinity= 1/2. |
We can also invert it and end up with -1/2, of course.
-1-1-1-1-1- = - infinity
and -1-1-1-1-1- = - infinity
Would that make the sign "irrelevant"? The "value" you are giving it could be off by 1!!!!!!
However, I am still fuzzy on the whole thing... you define a "process" that uses addition, then give it a value of 1/2 (or -1/2).
Then you define a completely different process and try to give it a value, and that is what is not working: is that what the question is, Ivars? Do I understand the problem?
8) the conjecture is, that there could exist a value atributable to the h(0) when 0^0=1 as a value of 0,1,0,1,0,1.............and that this value also gives the correct answer to 0^0 when 0/0 = 1.
We can~t simultaneously treat "h(0)" a value and as a process because we will get lost in uncertainties.
If it is process, the value's sign will change as that of the "1/2" above, from negative to positive, in a predetermined manner, yes, no, yes, no... but I am able to further that thought very little.
(because the sign is an infinitesimal of an equation when it is a sign; when it is a **numeric value** it must follow the behaviour predetermined by the hyerarchical superior, a number... and the *number** of the value, previously, was 1, 0, 1, 0, 1...)
QUOTE
9) This value seems not to be 1/2. What could it be? pi/2? pi/4? 1/e? i/2? sqrt(i)? something else?
Once it is determined that the sign must change following hierarchical rules, you then do like my daughter and determine the indeterminate: "some1 is at the door", "some2 is at the door", "some3 is at the door" --she does it in Portuguese too: 1guem chegou, 2guem, 3guem!!!
But the hierarchical rules of numbers you started with, even in the definition of a process, started with little numbers and went on ascending.
So that the "value" you are looking for is defined by a series of plus and minus signs, and a series of ascending numbers, and a stable series of multiplications and divisions. I have no idea what that means, needless to say.
Sorry, Ivars, it may almost look like science but that is the best I can do
I think I didn~t really understand the problem, but then that never kept me from searching for answers before
Dear Ivars,
Here's are a few thoughts for you (/edit> related to 0^0^0^0 in purely philosophical vein, obviously not related to the specifics of the h(z^y) = e^(i*pi/2+-I*2pi*k) equation...
(but first let's establish the "rule" first in order that we might then break it.)
===================================
Exponentiation: x^0 = 1, except that the case x = 0 may be left undefined in some contexts. For all positive real x, 0^x = 0
http://en.wikipedia.org/wiki/0_(number)
===================================
Now if...
0^x = 0
AND
x^0 = 1
... would it not make a certain amount of sense ala GRANDI'S SERIES (thank you to Guest00 for teaching me the name of that series...)...
===============================
The infinite series 1 - 1 + 1 - 1 +
"after Italian mathematician, philosopher, and priest Guido Grandi, who gave a memorable treatment of the series in 1703" which "is a divergent series, meaning that it lacks a sum in the usual sense" but "its Cesāro sum is .5."
http://en.wikipedia.org/wiki/Grandi's_series
===============================
... to treat
0^0 in similar manner ala the general concept of the Cesaro sum?
Thus...
0^0
= BOTH 1 and 0...
with an AVERAGE of .5
[1 + 0]/2
= .5
Any thoughts?
I might add, by the logic above does it change anything if instead of expressing the 0^0^0^0 equation in that specific manner, we break the rules again saying that the additive principal of the law of exponents does not apply to zero by expressing it thusly?
0^(0*0*0) = 0^0 = .5 (=sin (pi/6))
versus
.5^(0*0) = .5^(0*0) = .5^(0) = 1
Or how about expressing it thusly?
((0^0)^(0^0)) = .5^.5 = 0.707106781 = sin 45 degrees? (=sin (pi/4))
========
OR
========
0^0^0^0
= ((1^0)^0) + ((0^0)^0)/ 2
= ((1)^0) + (((0)^0) + ((1)^0)/2)/2
= ((1) + (((0 + 1)/2) + ((1))/2))/2
= ((1) + (.5) + (.5))/2
= 2/2
= 1 (=sin (pi/2))
Thoughts?
Cheers,
Raphie
Here's are a few thoughts for you (/edit> related to 0^0^0^0 in purely philosophical vein, obviously not related to the specifics of the h(z^y) = e^(i*pi/2+-I*2pi*k) equation...
(but first let's establish the "rule" first in order that we might then break it.)
===================================
Exponentiation: x^0 = 1, except that the case x = 0 may be left undefined in some contexts. For all positive real x, 0^x = 0
http://en.wikipedia.org/wiki/0_(number)
===================================
Now if...
0^x = 0
AND
x^0 = 1
... would it not make a certain amount of sense ala GRANDI'S SERIES (thank you to Guest00 for teaching me the name of that series...)...
===============================
The infinite series 1 - 1 + 1 - 1 +
"after Italian mathematician, philosopher, and priest Guido Grandi, who gave a memorable treatment of the series in 1703" which "is a divergent series, meaning that it lacks a sum in the usual sense" but "its Cesāro sum is .5."
http://en.wikipedia.org/wiki/Grandi's_series
===============================
... to treat
0^0 in similar manner ala the general concept of the Cesaro sum?
Thus...
0^0
= BOTH 1 and 0...
with an AVERAGE of .5
[1 + 0]/2
= .5
Any thoughts?
I might add, by the logic above does it change anything if instead of expressing the 0^0^0^0 equation in that specific manner, we break the rules again saying that the additive principal of the law of exponents does not apply to zero by expressing it thusly?
0^(0*0*0) = 0^0 = .5 (=sin (pi/6))
versus
.5^(0*0) = .5^(0*0) = .5^(0) = 1
Or how about expressing it thusly?
((0^0)^(0^0)) = .5^.5 = 0.707106781 = sin 45 degrees? (=sin (pi/4))
========
OR
========
0^0^0^0
= ((1^0)^0) + ((0^0)^0)/ 2
= ((1)^0) + (((0)^0) + ((1)^0)/2)/2
= ((1) + (((0 + 1)/2) + ((1))/2))/2
= ((1) + (.5) + (.5))/2
= 2/2
= 1 (=sin (pi/2))
Thoughts?
Cheers,
Raphie
Raphie Frank,
0^0 by itself is undefined in the way that we have currently defined our mathematics. x^y where x and y both go to zero can arise as a limit of a function, in which case it will have a specific value (or may still be undefined). Before you start redefining things, you need to have a thorough understanding of the current system.
This post is not meant to insult you, just that you have mentioned before that you have not made any extensive study of mathematics, yet you still persist in trying to redefine it.
0^0 by itself is undefined in the way that we have currently defined our mathematics. x^y where x and y both go to zero can arise as a limit of a function, in which case it will have a specific value (or may still be undefined). Before you start redefining things, you need to have a thorough understanding of the current system.
This post is not meant to insult you, just that you have mentioned before that you have not made any extensive study of mathematics, yet you still persist in trying to redefine it.
I absolutely understand that NeoDevin and by the way took a brief look at that link you posted earlier (your work perchance?). It seems very well put together and I will download it almost certainly. In any case, I am simply asking questions because, as you know, I am working on some ideas and am, in a sense, my own guinea pig. No insult whatsoever taken, but one point: When you mention "thoroughly understanding the current system," I indeed hope to one day, but there is a time sensitive pressure to what I am doing as well as some methodological considerations I won't here go into. This said, I am not looking, by and large, to "redefine," (although I've a notion or two...) but to supplement.
One question I have, for instance, is WHY can we not treat the numbers in the manner I just did? Particularly with respect to 0^0, if not the 0^0^0^0?
Cheers,
Raphie
P.S. I ask the "Whys" now in part because these are the types of questions, for instance, that other relational thinkers ask and I did not ask when I was younger. There is a reason, for instance, most artists I know avoid mathematics like the Plague. They ask "Why can't I do it that way, instead?" and in return are told "Because that's the way it is." Pedagogically speaking, that approach just does not fly with them. :-)
One question I have, for instance, is WHY can we not treat the numbers in the manner I just did? Particularly with respect to 0^0, if not the 0^0^0^0?
Cheers,
Raphie
P.S. I ask the "Whys" now in part because these are the types of questions, for instance, that other relational thinkers ask and I did not ask when I was younger. There is a reason, for instance, most artists I know avoid mathematics like the Plague. They ask "Why can't I do it that way, instead?" and in return are told "Because that's the way it is." Pedagogically speaking, that approach just does not fly with them. :-)
And, Neodevin, here is another WHY for you:
0/0 is also UNDEFINED under the current rules. In my imaginary playground of the mind...
0/0
= 0/x = 0
AND
x/0 = infinity
BOTH
Thus...
0/0 == [0 + infinity]/2
= infinity/2
And why? Because, I see it as providing a useful reference and/or equilibrium point (i.e. a "relative zero") as well serving as a convenient metaphor for the Big Bang. Dividing by zero is a bit like "inflation" while 0/x "tastes" a lot like entropy to me.
And (0^0)/(0^0)?
== .5/.5 == 1
That one might express this, for instance, also as .1F_5/.1F_5 (F_n= nth Fib number) or (.1*((phi) + (phi^-1))^2)/(.1*((phi) + (phi^-1))^2), it so happens, also allows us to set "zeroes" from which to "branch out" along other numerical "routes."
Best,
Raphie
0/0 is also UNDEFINED under the current rules. In my imaginary playground of the mind...
0/0
= 0/x = 0
AND
x/0 = infinity
BOTH
Thus...
0/0 == [0 + infinity]/2
= infinity/2
And why? Because, I see it as providing a useful reference and/or equilibrium point (i.e. a "relative zero") as well serving as a convenient metaphor for the Big Bang. Dividing by zero is a bit like "inflation" while 0/x "tastes" a lot like entropy to me.
And (0^0)/(0^0)?
== .5/.5 == 1
That one might express this, for instance, also as .1F_5/.1F_5 (F_n= nth Fib number) or (.1*((phi) + (phi^-1))^2)/(.1*((phi) + (phi^-1))^2), it so happens, also allows us to set "zeroes" from which to "branch out" along other numerical "routes."
Best,
Raphie
QUOTE (Raphie Frank+Dec 4 2007, 03:51 AM)
0/0 is also UNDEFINED under the current rules. In my imaginary playground of the mind...
0/0
= 0/x = 0
AND
x/0 = infinity
BOTH
Thus...
0/0 == [0 + infinity]/2
= infinity/2
And why? Because, I see it as providing a useful reference and/or equilibrium point (i.e. a "relative zero") as well serving as a convenient metaphor for the Big Bang. Dividing by zero is a bit like "inflation" while 0/x "tastes" a lot like entropy to me.
Hi, Ralphie and Ivars:
Ralphie, if you try to knock your mīs sideways with me I will hit you, man!
Youīd be wasting your time.
However, procedurally, dividing a number has a "value" of an infinitesimal of that number. If we check prime codes, for instance, we end up with:
2 times 5:
10101010101010101010... times
11110111101111011110... equals
10100010101010001010... 10, 20, 40, 80, 100, etc.
and that would apply to any prime times 2. so that 2 times 19 would be
10101010101010101010101010101010101010...
11111111111111111101111111111111111110...
10101010101010101000101010101010101010...
Procedurally, when it was time for division, the only thing that would make this "number" be worth a multiple of 19 --and not at all a plain 2-- would be the extra 0 in the middle. It's its placement that gives it its value, which is infinitesimal if we consider that the number has no end. So its position has acquired infinitesimal numberiness that is representative of any sequence (prime times 2).
But the "code", the sequence of 1's and 0's that make up a prime number, is irretrievable by this procedure of division. If you start with
1010101010101010100... you wouldn~t end up with
1111111111111111110... in a million years! The procedure only goes forward in time.
Now we have a contradiction, in that both the position gives away the numberiness, and the code of 1's and 0's hides it from the method. Then the position, of course, is the equivalent of the division, hence my statement that dividing a number has, intrinsically, a "value" of an infinitesimal of that number.
So that 0/0 is not quite undefined as it is... any number! It~s the difference between "none of the above" and "all of the above".
Now, if "all of the above", then if
Not that I think infinity is impossible to divide by two, of course! But procedurally the statement does~t sustain itself as an "equilibrium point", that is, as a neutron. Whereas Ivars, as I understood it, was attempting to re-create an atom and got the three parts right.
(important note: you might conceivably have put your glasses on and turned on your electron microscope and noticed, after careful scientific observation and plenty of charts, that we are filled with private languages and bullshit in this forum: I am no exception --worse, I may change my mind about every word I said if I decide later that I am wrong, or if I strongly suspect, as I do in this case, that the thoughts are underdeveloped. I just wanted to say my piece in my own language... You let it go and I will forgive you for knocking your m's around )
0/0
= 0/x = 0
AND
x/0 = infinity
BOTH
Thus...
0/0 == [0 + infinity]/2
= infinity/2
And why? Because, I see it as providing a useful reference and/or equilibrium point (i.e. a "relative zero") as well serving as a convenient metaphor for the Big Bang. Dividing by zero is a bit like "inflation" while 0/x "tastes" a lot like entropy to me.
Hi, Ralphie and Ivars:
Ralphie, if you try to knock your mīs sideways with me I will hit you, man!
Youīd be wasting your time.However, procedurally, dividing a number has a "value" of an infinitesimal of that number. If we check prime codes, for instance, we end up with:
2 times 5:
10101010101010101010... times
11110111101111011110... equals
10100010101010001010... 10, 20, 40, 80, 100, etc.
and that would apply to any prime times 2. so that 2 times 19 would be
10101010101010101010101010101010101010...
11111111111111111101111111111111111110...
10101010101010101000101010101010101010...
Procedurally, when it was time for division, the only thing that would make this "number" be worth a multiple of 19 --and not at all a plain 2-- would be the extra 0 in the middle. It's its placement that gives it its value, which is infinitesimal if we consider that the number has no end. So its position has acquired infinitesimal numberiness that is representative of any sequence (prime times 2).
But the "code", the sequence of 1's and 0's that make up a prime number, is irretrievable by this procedure of division. If you start with
1010101010101010100... you wouldn~t end up with
1111111111111111110... in a million years! The procedure only goes forward in time.
Now we have a contradiction, in that both the position gives away the numberiness, and the code of 1's and 0's hides it from the method. Then the position, of course, is the equivalent of the division, hence my statement that dividing a number has, intrinsically, a "value" of an infinitesimal of that number.
So that 0/0 is not quite undefined as it is... any number! It~s the difference between "none of the above" and "all of the above".
Now, if "all of the above", then if
QUOTE
0/0 == [0 + infinity]/2
then the second part after the == would be "meaningless" in the sense of infinity plus 1 versus 1 plus infinity, in which case division by two would fail procedurally. Not that I think infinity is impossible to divide by two, of course! But procedurally the statement does~t sustain itself as an "equilibrium point", that is, as a neutron. Whereas Ivars, as I understood it, was attempting to re-create an atom and got the three parts right.
(important note: you might conceivably have put your glasses on and turned on your electron microscope and noticed, after careful scientific observation and plenty of charts, that we are filled with private languages and bullshit in this forum: I am no exception --worse, I may change my mind about every word I said if I decide later that I am wrong, or if I strongly suspect, as I do in this case, that the thoughts are underdeveloped. I just wanted to say my piece in my own language... You let it go and I will forgive you for knocking your m's around
)
QUOTE (Raphie Frank+Dec 4 2007, 07:23 AM)
I absolutely understand that NeoDevin and by the way took a brief look at that link you posted earlier (your work perchance?). It seems very well put together and I will download it almost certainly.
I'll respond to the rest later, gotta run to catch a bus right away.
No, not my work, the work of my first year calculus prof (I'm assuming you're referring to the notes I linked to in the 0.9r thread?). He is one of the best prof's I have ever had.
I'll respond to the rest later, gotta run to catch a bus right away.
No, not my work, the work of my first year calculus prof (I'm assuming you're referring to the notes I linked to in the 0.9r thread?). He is one of the best prof's I have ever had.
QUOTE (Raphie Frank+Dec 4 2007, 07:23 AM)
One question I have, for instance, is WHY can we not treat the numbers in the manner I just did? Particularly with respect to 0^0, if not the 0^0^0^0?
You can make any definitions you wish to make, as long as they're consistant with the rest of mathematics. The two points you bring up are not defined with the current system of mathematics simply because, depending on the situation, they have different values.
You can make any definitions you wish to make, as long as they're consistant with the rest of mathematics. The two points you bring up are not defined with the current system of mathematics simply because, depending on the situation, they have different values.
QUOTE (Raphie Frank+Dec 4 2007, 06:34 AM)
... to treat
0^0 in similar manner ala the general concept of the Cesaro sum?
Thus...
0^0
= BOTH 1 and 0...
with an AVERAGE of .5
[1 + 0]/2
= .5
There are many (infinite) other cases where something can go towards 0^0 without going to either 1 or 0. For example:
x^(1/ln(x)) goes to e, as x goes to zero. (try and work it out, this one can be tricky if you don't know what you're doing)
By changing the natural logarithm there to any other base, we get that this converges to that base instead (at least for any positive real base, I won't promise anything for negative or complex).
Yet clearly x goes to 0, and 1/ln(x) goes to zero, so this expression goes to 0^0, but does not approach either 0 or 1. I have just given you infinite examples where this does not limit to 0 or 1.
On the other hand, we could still simply define it to be 1, or 0, or 0.5, but then we would have to comment every time such a limit comes up in a problem as a special case, where it does not go to whatever we define it to be. So it is undefined, because it creates far more problems to define it to be a certain value, than to say it's undefined, but various functions can limit to it, and have different values.
0^0 in similar manner ala the general concept of the Cesaro sum?
Thus...
0^0
= BOTH 1 and 0...
with an AVERAGE of .5
[1 + 0]/2
= .5
There are many (infinite) other cases where something can go towards 0^0 without going to either 1 or 0. For example:
x^(1/ln(x)) goes to e, as x goes to zero. (try and work it out, this one can be tricky if you don't know what you're doing)
By changing the natural logarithm there to any other base, we get that this converges to that base instead (at least for any positive real base, I won't promise anything for negative or complex).
Yet clearly x goes to 0, and 1/ln(x) goes to zero, so this expression goes to 0^0, but does not approach either 0 or 1. I have just given you infinite examples where this does not limit to 0 or 1.
On the other hand, we could still simply define it to be 1, or 0, or 0.5, but then we would have to comment every time such a limit comes up in a problem as a special case, where it does not go to whatever we define it to be. So it is undefined, because it creates far more problems to define it to be a certain value, than to say it's undefined, but various functions can limit to it, and have different values.
QUOTE (Raphie Frank+Dec 4 2007, 07:51 AM)
0/0
= 0/x = 0
AND
x/0 = infinity
BOTH
Thus...
0/0 == [0 + infinity]/2
= infinity/2
We will discuss countable infinities for a moment, to show you why your method of manipulating infinity leads to problems.
If we take the natural numbers (that is, the counting numbers, positive integers), there are infinite elements to it. Therefore the cardinality of this set is infinite.
We then take only the even natural numbers, clearly for every even number there are two corresponding natural numbers, so there are ininity/2 even natural numbers.
and yet, we can clearly map one onto the other
1, 2, 3, 4, ...
2, 4, 6, 8, ...
Because we can match them up, and have none left over, there must therefore be the same number of even numbers as natural numbers.
Therefore, infinity/2 = infinity.
We can further take the natural numbers missing the first n numbers, (n+1, n+2, ...), we will call the number of elements in this set infinity_n (that's a subscript)
Clearly there we can match these up as follows
1 -> n+1
2 -> n+2
3 -> n+3
...
So there are the same number of elements in this set as in the natural numbers (and therefore also as in the even natural numbers). At the same time there are clearly n numbers in the natural numbers which are not in this set.
So
infinity_n = infinity/2
infinity - infinity/2 = infinity/2
infinity - infinity_n = n
therefore infinity = infinity/2 = n.
Trying to manipulate infinities like that leads to all sorts of contradictions which are not so easily defined away. For example, by trying to manipulate infinities as you have, I have just proven that every (since the above logic holds for any choice of n) natural number is equal to infinity.
Other than the contradictions leading from treating infinities like that, as in the previous example, many functions can go to 0/0 and still have a well defined value.
For a further discussion of how to treat 0/0, see L'Hopitals Rule on p.98 of the notes which I linked to in the other thread. Again, this is material which is covered in almost every first year university calculus class.
In certain cases, functions which limit to 0^0 (or even 0^(0^(0^(0...))) can be defined, and can be defined to have any value you want. The same is true for 0/0 (or you could say also for 0/(0/(0/(0...)))... Oh great, now we're going to get a thread on that one too). Anyways, the point is, that it is far more useful to have 0/0 or 0^0 undefined than it is to define them to have any particular value.
I hope this all makes sense, if any of it is unclear, feel free to ask and I will try to clarify/correct.
= 0/x = 0
AND
x/0 = infinity
BOTH
Thus...
0/0 == [0 + infinity]/2
= infinity/2
We will discuss countable infinities for a moment, to show you why your method of manipulating infinity leads to problems.
If we take the natural numbers (that is, the counting numbers, positive integers), there are infinite elements to it. Therefore the cardinality of this set is infinite.
We then take only the even natural numbers, clearly for every even number there are two corresponding natural numbers, so there are ininity/2 even natural numbers.
and yet, we can clearly map one onto the other
1, 2, 3, 4, ...
2, 4, 6, 8, ...
Because we can match them up, and have none left over, there must therefore be the same number of even numbers as natural numbers.
Therefore, infinity/2 = infinity.
We can further take the natural numbers missing the first n numbers, (n+1, n+2, ...), we will call the number of elements in this set infinity_n (that's a subscript)
Clearly there we can match these up as follows
1 -> n+1
2 -> n+2
3 -> n+3
...
So there are the same number of elements in this set as in the natural numbers (and therefore also as in the even natural numbers). At the same time there are clearly n numbers in the natural numbers which are not in this set.
So
infinity_n = infinity/2
infinity - infinity/2 = infinity/2
infinity - infinity_n = n
therefore infinity = infinity/2 = n.
Trying to manipulate infinities like that leads to all sorts of contradictions which are not so easily defined away. For example, by trying to manipulate infinities as you have, I have just proven that every (since the above logic holds for any choice of n) natural number is equal to infinity.
Other than the contradictions leading from treating infinities like that, as in the previous example, many functions can go to 0/0 and still have a well defined value.
For a further discussion of how to treat 0/0, see L'Hopitals Rule on p.98 of the notes which I linked to in the other thread. Again, this is material which is covered in almost every first year university calculus class.
In certain cases, functions which limit to 0^0 (or even 0^(0^(0^(0...))) can be defined, and can be defined to have any value you want. The same is true for 0/0 (or you could say also for 0/(0/(0/(0...)))... Oh great, now we're going to get a thread on that one too). Anyways, the point is, that it is far more useful to have 0/0 or 0^0 undefined than it is to define them to have any particular value.
I hope this all makes sense, if any of it is unclear, feel free to ask and I will try to clarify/correct.
few more interestingf simple relations:
roots of
h(z^y) = e^(i*pi/2+-I*2pi*k) ?[/b] for all k, or just for some single special k?
We have in fact 3 pairs of z and y for which solutions exist for k=0:
z=e, y=pi/2 , z^y= e^pi/2 h(e^pi/2)= +-i
z=i, y=-i, z^y = i^-i= e^pi/2 h (i^-i)=+-i
z=-i, y=i, z^y = -i^i = e^pi/2 h(-i^i) = +-i
it also follows from formula h(1/(i^i)) = 1/i = -i
and h(1/(-i^-i)) = 1/-i = i.
so this formula h(1/z^z))=1/z must be valid not only for integer and real x, but also for imaginary and maybe complex z.
I still find this interesting, especially as i=e^(h(1/i^i) *pi/2) is a 90 degree rotation operator but involves infinite number of exponating h(1/i^i) = 1/(i^i)^(1/i^i)) ^...etc = (e^pi/2)^(((e^pi/2)^((e^pi/2)^(e^pi/2)........
There seems to be no other function in math that is able to return purely imaginary values from purely real (like h(e^pi/2)) . The question is, are then transcendental and irrational numbers really purely real?
roots of
h(z^y) = e^(i*pi/2+-I*2pi*k) ?[/b] for all k, or just for some single special k?
We have in fact 3 pairs of z and y for which solutions exist for k=0:
z=e, y=pi/2 , z^y= e^pi/2 h(e^pi/2)= +-i
z=i, y=-i, z^y = i^-i= e^pi/2 h (i^-i)=+-i
z=-i, y=i, z^y = -i^i = e^pi/2 h(-i^i) = +-i
it also follows from formula h(1/(i^i)) = 1/i = -i
and h(1/(-i^-i)) = 1/-i = i.
so this formula h(1/z^z))=1/z must be valid not only for integer and real x, but also for imaginary and maybe complex z.
I still find this interesting, especially as i=e^(h(1/i^i) *pi/2) is a 90 degree rotation operator but involves infinite number of exponating h(1/i^i) = 1/(i^i)^(1/i^i)) ^...etc = (e^pi/2)^(((e^pi/2)^((e^pi/2)^(e^pi/2)........
There seems to be no other function in math that is able to return purely imaginary values from purely real (like h(e^pi/2)) . The question is, are then transcendental and irrational numbers really purely real?
QUOTE (Ivars+Dec 9 2007, 11:06 AM)
h(z^y) = e^(i*pi/2+-I*2pi*k) ?[/b] for all k, or just for some single special k?
I am sorry, Ivars, I am out of the loop again!!!!!
Does this question mean that k could somehow possibly be anything other than an infinitesimal multiple of h(z^y)/?!?!?!
(It~s just that I don~t jump into chaos without understanding what I am missing ...
I am sorry, Ivars, I am out of the loop again!!!!!
Does this question mean that k could somehow possibly be anything other than an infinitesimal multiple of h(z^y)/?!?!?!
(It~s just that I don~t jump into chaos without understanding what I am missing
...
QUOTE (IAMoraes+Dec 10 2007, 02:09 AM)
I am sorry, Ivars, I am out of the loop again!!!!!
Does this question mean that k could somehow possibly be anything other than an infinitesimal multiple of h(z^y)/?!?!?!
(It~s just that I don~t jump into chaos without understanding what I am missing ...
hej IAM,
I do not know. I just have found 3 solutions ( if we consider z=i and z=-i and z=e different solutions, although they all give e^pi/2 as argument for h) for k=0.
Since +- I= e^(+-I*(pi/2+-2k*pi)). I just wondered if there are values of z and y that would lead to e.g.
h(z^y)=e^+-I*5pi/2
because other branches of Lambert function and logarithm leads to values of h(e^pi/2) which does not have the periodicity of original I (+-2 pi) - as h is defined by:
h = -W(-ln(z^y)/ln(z^y)
In fact, it gets complex from purely imaginary. At least that is what I was told in tetration forum:
Code:
LambertW(-3,-Pi/2)=-2.198342630-13.98120831*I
LambertW(-2,-Pi/2)=-1.604290913-7.647192276*I
LambertW(-1,-Pi/2)=-1.570796327*I
LambertW(0,-Pi/2)=1.570796327*I
LambertW(1,-Pi/2)=-1.604290913+7.647192276*I
LambertW(2,-Pi/2)=-2.198342630+13.98120831*I
where the first argument denotes the branch.
The numbering of the branches is of course convention.
So if there are other root z^y giving h(z^y)= e^+-I*5pi/2 , it is not e^pi/2.
Infinitesimal multiple? What do you mean by that? That each rotation of I adds and infinitesimal in 3rd dimension while we do not see it on 2D complex plane? Like an infinitesimal step of a screw? Or it adds a new dimension each time, perhaps?
Where can I read about it?
Does this question mean that k could somehow possibly be anything other than an infinitesimal multiple of h(z^y)/?!?!?!
(It~s just that I don~t jump into chaos without understanding what I am missing
... hej IAM,
I do not know. I just have found 3 solutions ( if we consider z=i and z=-i and z=e different solutions, although they all give e^pi/2 as argument for h) for k=0.
Since +- I= e^(+-I*(pi/2+-2k*pi)). I just wondered if there are values of z and y that would lead to e.g.
h(z^y)=e^+-I*5pi/2
because other branches of Lambert function and logarithm leads to values of h(e^pi/2) which does not have the periodicity of original I (+-2 pi) - as h is defined by:
h = -W(-ln(z^y)/ln(z^y)
In fact, it gets complex from purely imaginary. At least that is what I was told in tetration forum:
QUOTE
Code:
LambertW(-3,-Pi/2)=-2.198342630-13.98120831*I
LambertW(-2,-Pi/2)=-1.604290913-7.647192276*I
LambertW(-1,-Pi/2)=-1.570796327*I
LambertW(0,-Pi/2)=1.570796327*I
LambertW(1,-Pi/2)=-1.604290913+7.647192276*I
LambertW(2,-Pi/2)=-2.198342630+13.98120831*I
where the first argument denotes the branch.
The numbering of the branches is of course convention.
So if there are other root z^y giving h(z^y)= e^+-I*5pi/2 , it is not e^pi/2.
Infinitesimal multiple? What do you mean by that? That each rotation of I adds and infinitesimal in 3rd dimension while we do not see it on 2D complex plane? Like an infinitesimal step of a screw? Or it adds a new dimension each time, perhaps?
Where can I read about it?
k is pany integer you idiots. The particular choice is dependent upon your branch cut.
This comes directly in when you're defining log(e^(iθ)) ~ iθ ~ iθ+2i.pi ~ θ-2i.pi ~ θi+2i.k.pi
You have to make a branch cut within the complex plane so that the function is single values. If it's not single values, it isn't a function! Then you get into the whole very interesting (not that either of you would know) area of complex analysis, which involves see how functions (particularly integrals of functions) vary within the complex plane. Cauchy's deformation and residue theorems being two of the most important and most insightful results in maths and physics.
I read a 'few' things in a book on this stuff once. I heard a few things in some rooms called 'lecture theatres' a few times too. All of those things point out how much you are stumbling about in the dark.
How long you been at this now? At least 12 months. In that time a student could have learnt all this material and advanced way beyond it, on only 2 or 3 hours of work a week. You obviously put in more time than that but you also so obviously waste it. Well done. You must feel so proud.
This comes directly in when you're defining log(e^(iθ)) ~ iθ ~ iθ+2i.pi ~ θ-2i.pi ~ θi+2i.k.pi
You have to make a branch cut within the complex plane so that the function is single values. If it's not single values, it isn't a function! Then you get into the whole very interesting (not that either of you would know) area of complex analysis, which involves see how functions (particularly integrals of functions) vary within the complex plane. Cauchy's deformation and residue theorems being two of the most important and most insightful results in maths and physics.
QUOTE
Where can I read about it?
I'm remided of a scene from Family Guy. A situation occurs and Peter says "Don't worry, I read something like this in a book once....". Peter, the dog, replies "Are you sure it was a book? Are you sure it wasn't ... nothing?". Peter's face drops and he says "Oh yeah". I read a 'few' things in a book on this stuff once. I heard a few things in some rooms called 'lecture theatres' a few times too. All of those things point out how much you are stumbling about in the dark.
How long you been at this now? At least 12 months. In that time a student could have learnt all this material and advanced way beyond it, on only 2 or 3 hours of work a week. You obviously put in more time than that but you also so obviously waste it. Well done. You must feel so proud.
So what was the message, Alphanumeric?
Which z^y(k) are roots of h(z^y)= e^I*((pi/2)+2pik) for any k?
Do You have a dog
? Why is he unhappy
Which z^y(k) are roots of h(z^y)= e^I*((pi/2)+2pik) for any k?
Do You have a dog
? Why is he unhappy
i'm still struggling to understand what point you're trying to make. here we are dozen or so messages later, and i'm still struggling. you're alternating between two function types, throwing in complex analysis as if it were childs play, and sides stepping mine and others' attempts at reconciling some kind of understanding.
z^z in complex space is not a trivial Riemann surface. ignoring this and jumping ahead to z^(z^z) is like skipping the first 200 pages of the book you were supposed to be studying for the exam. it isn't that easy.
example. z^z=exp(z*lnz). often i am seeing this identity ignored. you need to consolidate what you're trying to say into a coherent formalism.
z^z in complex space is not a trivial Riemann surface. ignoring this and jumping ahead to z^(z^z) is like skipping the first 200 pages of the book you were supposed to be studying for the exam. it isn't that easy.
example. z^z=exp(z*lnz). often i am seeing this identity ignored. you need to consolidate what you're trying to say into a coherent formalism.
QUOTE (Ivars+Dec 10 2007, 11:49 AM)
Tetration Forum:
That is a salad as far as I understand it, Ivars. Which is fine since I don't understand it
We can't simultaneously 1-have a number become a direction as it approaches the infinitesimal AND 2-have that direction be branch-numberable, WITHOUT 3-coming to the conclusion that the positions of time and matter have been reversed and that therefore the "purely conventional" numbering has valid, solid, physical meaning.
As far as I am concerned, I can't follow your "narrative", therefore I switch it into another field, where I come to the conclusion that you are successful at twisting the atom inside out, but unfortunately you are also expecting to end up with matter and... it is not going to happen. Now, you have 3 results that are valid: I have no problem with them only as long as one of them is a neutron, the other is a proton inside-out, and the other is an electron inside-out. Since I am not a physicist, I am up against a wall. I can't further the subject unless someone thinks along those lines and gives me a bone to chew
But if number became direction suddenly, then pi would jump from geometry into the number field, and would become something else, like e... and since I am not a mathematician either I am even more chewless!
Even though I don't understand the question, I just know those switches are there, right here, right in front of us. Those "switches" are aleph transition points, but that aleph talk is still another philosophical re-telling of the same old story, one that I am introducing because I am unable to solve or even understand anything you are saying by regular means! In short, I am cheating by switching narratives.
(I will now drop out of sight and off the thread since I am neither helping it nor making progress.)
QUOTE
Code:
LambertW(-3,-Pi/2)=-2.198342630-13.98120831*I
LambertW(-2,-Pi/2)=-1.604290913-7.647192276*I
LambertW(-1,-Pi/2)=-1.570796327*I
LambertW(0,-Pi/2)=1.570796327*I
LambertW(1,-Pi/2)=-1.604290913+7.647192276*I
LambertW(2,-Pi/2)=-2.198342630+13.98120831*I
where the first argument denotes the branch.
The numbering of the branches is of course convention.
LambertW(-3,-Pi/2)=-2.198342630-13.98120831*I
LambertW(-2,-Pi/2)=-1.604290913-7.647192276*I
LambertW(-1,-Pi/2)=-1.570796327*I
LambertW(0,-Pi/2)=1.570796327*I
LambertW(1,-Pi/2)=-1.604290913+7.647192276*I
LambertW(2,-Pi/2)=-2.198342630+13.98120831*I
where the first argument denotes the branch.
The numbering of the branches is of course convention.
That is a salad as far as I understand it, Ivars. Which is fine since I don't understand it
We can't simultaneously 1-have a number become a direction as it approaches the infinitesimal AND 2-have that direction be branch-numberable, WITHOUT 3-coming to the conclusion that the positions of time and matter have been reversed and that therefore the "purely conventional" numbering has valid, solid, physical meaning.
As far as I am concerned, I can't follow your "narrative", therefore I switch it into another field, where I come to the conclusion that you are successful at twisting the atom inside out, but unfortunately you are also expecting to end up with matter and... it is not going to happen. Now, you have 3 results that are valid: I have no problem with them only as long as one of them is a neutron, the other is a proton inside-out, and the other is an electron inside-out. Since I am not a physicist, I am up against a wall. I can't further the subject unless someone thinks along those lines and gives me a bone to chew
But if number became direction suddenly, then pi would jump from geometry into the number field, and would become something else, like e... and since I am not a mathematician either I am even more chewless!
Even though I don't understand the question, I just know those switches are there, right here, right in front of us. Those "switches" are aleph transition points, but that aleph talk is still another philosophical re-telling of the same old story, one that I am introducing because I am unable to solve or even understand anything you are saying by regular means! In short, I am cheating by switching narratives.
(I will now drop out of sight and off the thread since I am neither helping it nor making progress.)
QUOTE (xaos+Dec 12 2007, 03:04 AM)
i'm still struggling to understand what point you're trying to make. here we are dozen or so messages later, and i'm still struggling. you're alternating between two function types, throwing in complex analysis as if it were childs play, and sides stepping mine and others' attempts at reconciling some kind of understanding.
z^z in complex space is not a trivial Riemann surface. ignoring this and jumping ahead to z^(z^z) is like skipping the first 200 pages of the book you were supposed to be studying for the exam. it isn't that easy.
example. z^z=exp(z*lnz). often i am seeing this identity ignored. you need to consolidate what you're trying to say into a coherent formalism.
hej xaos
i might be missing more than first 200 pages.. So i do not know if the analytic continuation of h(z) is still contested.
I know that in the region of convergence z=real < e^1/e = 1,444667861
h(z) = - W(-lnz)/lnz gives the same result as infinite teatration z^(z^(z^(z^......
I know that complex z returns complex h what ever it may mean.
Now, i also have noticed in tetration pages that analytic continuation of h(z) when z= real and > e^1/e exists.
This is very interesting as logic tells that such tetrations should diverge, but function returns complex value from real z!
Especially, which was already noticed by Euler, h(z) = +- i when z=e^(pi/2)
I think it is very nice formula: h(e^pi/2) = +- i = +- sqrt(-1). So is h(e^pi/2) a root of x^2+1=0? Or not? Is it equal to i, or just coincides in 2 branches of W, 0 and -1 ? I could not get clear answer but it seems h(e^pi/2) is not identical to i.
So I am asking the next question:
Are there real (or complex) values of z>e^1/e ( or complex) expressable as x^y ( like e^pi/2) which makes h(z) = e^+-I*5pi/2, e^+-I*9pi/2, e^+-I*13pi/2 ..........etc. I could not find any, nor get an answer about all branches of h(z) for all k- though there is some series expansion , but that has again problems with convergence as it diverges, naturally, at z>e^1/e.
if we look at e^pi/2 as a logarithm of e^(e^pi/2) than it will have also imaginary values which might be the right inputs in h( e^pi/2+ imaginary values)= i?
Another thing, i was trying to find some properties of h(z=x^y) which would not require knowledge of branches etc- so I can skip the first 200 pages and work directly with h(z) for any z. For example:
I found that h(1/(n^n)) = 1/n for ALL n > = 1. This is true. so
1/ h(1/(n^n)) = n and
but also h(n^(1/n))) = n if n^1/n <= e^(1/e) which is true for all n>=1 so
h(1/(n^n))*h(n^(1/n))=1 and
(1/h(1/(n^n)) )^ h (n^(1/n)) = n^n
Now I tried to replace one n with x, another with y as in z=e^pi/2:
h(x^-y)) * h(x^(1/y))=1
and other formulas which would allow to find h(x^-y) if h(x^y) and h( x^1/y) would be known without numerical calculations, analytically.
z^z in complex space is not a trivial Riemann surface. ignoring this and jumping ahead to z^(z^z) is like skipping the first 200 pages of the book you were supposed to be studying for the exam. it isn't that easy.
example. z^z=exp(z*lnz). often i am seeing this identity ignored. you need to consolidate what you're trying to say into a coherent formalism.
hej xaos
i might be missing more than first 200 pages.. So i do not know if the analytic continuation of h(z) is still contested.
I know that in the region of convergence z=real < e^1/e = 1,444667861
h(z) = - W(-lnz)/lnz gives the same result as infinite teatration z^(z^(z^(z^......
I know that complex z returns complex h what ever it may mean.
Now, i also have noticed in tetration pages that analytic continuation of h(z) when z= real and > e^1/e exists.
This is very interesting as logic tells that such tetrations should diverge, but function returns complex value from real z!
Especially, which was already noticed by Euler, h(z) = +- i when z=e^(pi/2)
I think it is very nice formula: h(e^pi/2) = +- i = +- sqrt(-1). So is h(e^pi/2) a root of x^2+1=0? Or not? Is it equal to i, or just coincides in 2 branches of W, 0 and -1 ? I could not get clear answer but it seems h(e^pi/2) is not identical to i.
So I am asking the next question:
Are there real (or complex) values of z>e^1/e ( or complex) expressable as x^y ( like e^pi/2) which makes h(z) = e^+-I*5pi/2, e^+-I*9pi/2, e^+-I*13pi/2 ..........etc. I could not find any, nor get an answer about all branches of h(z) for all k- though there is some series expansion , but that has again problems with convergence as it diverges, naturally, at z>e^1/e.
if we look at e^pi/2 as a logarithm of e^(e^pi/2) than it will have also imaginary values which might be the right inputs in h( e^pi/2+ imaginary values)= i?
Another thing, i was trying to find some properties of h(z=x^y) which would not require knowledge of branches etc- so I can skip the first 200 pages and work directly with h(z) for any z. For example:
I found that h(1/(n^n)) = 1/n for ALL n > = 1. This is true. so
1/ h(1/(n^n)) = n and
but also h(n^(1/n))) = n if n^1/n <= e^(1/e) which is true for all n>=1 so
h(1/(n^n))*h(n^(1/n))=1 and
(1/h(1/(n^n)) )^ h (n^(1/n)) = n^n
Now I tried to replace one n with x, another with y as in z=e^pi/2:
h(x^-y)) * h(x^(1/y))=1
and other formulas which would allow to find h(x^-y) if h(x^y) and h( x^1/y) would be known without numerical calculations, analytically.
Few more obvius, but interesting relations involving also infinite tetration
e^i = i^(2/pi)
e=i^(2/i*pi) = i^(1/lni)
1/e = i^(2i/pi)
h(1/e) = h(i^(2i/pi) = h(1/(i^(1/lni))) = h(1/(x^(1/lnx)))= Omega constant.
x=i^(lnx/lni) = i^li(x) li (x) = logarithm with base i
e^i = i^(2/pi)
e=i^(2/i*pi) = i^(1/lni)
1/e = i^(2i/pi)
h(1/e) = h(i^(2i/pi) = h(1/(i^(1/lni))) = h(1/(x^(1/lnx)))= Omega constant.
x=i^(lnx/lni) = i^li(x) li (x) = logarithm with base i
Is this Chaitin's Omega or something else I don't know about?!?!
If it is Chaitin's, then Omega should be arrived at by a "mindless automaton" --without any computation whatsoever-- because of the way your system would evolve and things would shift around and cease to be themselves so they could remain being themselves (shoot, this is harder to explain than I thought!).
I think Omega's binary form is some sort of prime-code wrap process, like Ulam's unmitigated mess, I mean, spiral. I was trying to do this the day before yesterday but I get bored... I get bored pretty quickly...
(Edit: I just meant that the maximum entropy state of ANY system, even Chaitin's, MUST collapse.)
If it is Chaitin's, then Omega should be arrived at by a "mindless automaton" --without any computation whatsoever-- because of the way your system would evolve and things would shift around and cease to be themselves so they could remain being themselves (shoot, this is harder to explain than I thought!).
I think Omega's binary form is some sort of prime-code wrap process, like Ulam's unmitigated mess, I mean, spiral. I was trying to do this the day before yesterday but I get bored... I get bored pretty quickly...
(Edit: I just meant that the maximum entropy state of ANY system, even Chaitin's, MUST collapse.)
QUOTE (IAMoraes+Dec 27 2007, 09:20 PM)
Is this Chaitin's Omega or something else I don't know about?!?!
I think Omega's binary form is some sort of prime-code wrap process, like Ulam's unmitigated mess, I mean, spiral. I was trying to do this the day before yesterday but I get bored... I get bored pretty quickly...
(Edit: I just meant that the maximum entropy state of ANY system, even Chaitin's, MUST collapse.)
hej IAM,
Well it may be a spiral in number space when we find it out, but in usual decimal form it is just a transcendental number to which all infinite power towers
h(1/(x^lnx))= h(1/e) converge.
Power Tower
Omega constant
The value of Ω is approximately 0.5671432904097838729999686622 (sequence A030178 in OEIS). It has properties that are akin to those of the golden ratio, in that
e^(-Omega) = Omega
or equivalently,
ln(1 / Ω) = Ω.
Power tower was or is the missing operation as it links all (many) scales while normal real and complex algebra works usually in any one scale.
I think Omega's binary form is some sort of prime-code wrap process, like Ulam's unmitigated mess, I mean, spiral. I was trying to do this the day before yesterday but I get bored... I get bored pretty quickly...
(Edit: I just meant that the maximum entropy state of ANY system, even Chaitin's, MUST collapse.)
hej IAM,
Well it may be a spiral in number space when we find it out, but in usual decimal form it is just a transcendental number to which all infinite power towers
h(1/(x^lnx))= h(1/e) converge.
Power Tower
Omega constant
The value of Ω is approximately 0.5671432904097838729999686622 (sequence A030178 in OEIS). It has properties that are akin to those of the golden ratio, in that
e^(-Omega) = Omega
or equivalently,
ln(1 / Ω) = Ω.
Power tower was or is the missing operation as it links all (many) scales while normal real and complex algebra works usually in any one scale.
I have been looking at relation for infinitesimal exponentiation of base a:
by definition, if we accept I as infinitesimal which is not 0, I^I = (1+J) where J is infinitesimal as well (since I^0=1). Then log base I ( 1+J) = I log base I (I) = I.
Euler made a proposition which he proved that J=kI, where k = 1 for base e so that ln(1+I)=I where I is just infinitesimal.
If we take imaginary Unit I as being the infinitesimal in base e , what is k?
From Euler's derivations, k as function of base a is :
k= 2( (a-1)/(a+1) + (a-1)^3/3*(a+1)^3+ (a-1)^5/5*(a+1)^5 ....
As we work with base I, where I = imaginary Unit, we have to evaluate term
(I-1)/(I+1) = (sqrt(2)* e^I*((3pi/4)+2pik)/ sqrt(2)* e^I* ((pi/4)+2pin) = e^I* (3pi/4-pi/4+2pik-2pin) = e^I*pi/2= i as a single value, not periodic, since all 2pik and 2pin cancel out.
So k= 2 ( I+I^3 /3+ I^5/5+I^7/7 +...) = 2*I ( 1-1/3+1/5-1/7.......) = 2*I* pi where I is JUST e^I*pi/2, single value.
Then J as infinitesimal result J = (I*pi/2 ) * I so J is also infinitesimal but turned by 90 degrees via multiplication of I by I plus it has module pi/2 and of I^I =(1+J) = (1+ (I*pi/2)*I) and
But we know that I^I = e^-pi/2 so that ln(I^I) = -pi/2= (I*pi/2)*I:
But we also know that I^I = (1+I*pi/2*I) so that ln(I^I) = ln(1+I^2*pi/2) = k*I= J= I*pi/2*I= -pi/2 if I is imaginary infinitesimal such that I^2 = -1.
So assumption that I may be imaginary infinitesimal works with logarithms, but the unique note here is that we are not dealing with ALL periodic values of I, but just I, such that is linked to 1 in this scale so that I=(1-I)/(1+I); In a similar vein, unique value of -I = (1+I)/(1-I) can be obtained.
The conclusion is ( or proposition was) : Mathematics works in SCALES which we do not notice for a reason that they are imaginary ( meaning in our minds),so we assume the same scale exists for all operations, which is not true, and in each scale the relation ship between Unit lentgh 1 and imaginary infinitesimal is defined by I=(1-I)/(1+I). This imaginary infinitesimal is Unique for that scale and by rotating it by +2pi we move into next scale, perhaps via multiplying by J= I*pi/2 where 1 is also different, but we do not notice it, as in math, we mix all scales in our heads - which is not possible in physics, and therefore the discrepancy.
On other hand, I^I= e^-pi/2 means that taken over all scales, imaginary unit is linked to e. However, if we stay in 1 scale, we do not need e for the base of natural logarithms, we can have k=I*pi/2 as defining natural k in that scale where I is infinitesimal.
by definition, if we accept I as infinitesimal which is not 0, I^I = (1+J) where J is infinitesimal as well (since I^0=1). Then log base I ( 1+J) = I log base I (I) = I.
Euler made a proposition which he proved that J=kI, where k = 1 for base e so that ln(1+I)=I where I is just infinitesimal.
If we take imaginary Unit I as being the infinitesimal in base e , what is k?
From Euler's derivations, k as function of base a is :
k= 2( (a-1)/(a+1) + (a-1)^3/3*(a+1)^3+ (a-1)^5/5*(a+1)^5 ....
As we work with base I, where I = imaginary Unit, we have to evaluate term
(I-1)/(I+1) = (sqrt(2)* e^I*((3pi/4)+2pik)/ sqrt(2)* e^I* ((pi/4)+2pin) = e^I* (3pi/4-pi/4+2pik-2pin) = e^I*pi/2= i as a single value, not periodic, since all 2pik and 2pin cancel out.
So k= 2 ( I+I^3 /3+ I^5/5+I^7/7 +...) = 2*I ( 1-1/3+1/5-1/7.......) = 2*I* pi where I is JUST e^I*pi/2, single value.
Then J as infinitesimal result J = (I*pi/2 ) * I so J is also infinitesimal but turned by 90 degrees via multiplication of I by I plus it has module pi/2 and of I^I =(1+J) = (1+ (I*pi/2)*I) and
But we know that I^I = e^-pi/2 so that ln(I^I) = -pi/2= (I*pi/2)*I:
But we also know that I^I = (1+I*pi/2*I) so that ln(I^I) = ln(1+I^2*pi/2) = k*I= J= I*pi/2*I= -pi/2 if I is imaginary infinitesimal such that I^2 = -1.
So assumption that I may be imaginary infinitesimal works with logarithms, but the unique note here is that we are not dealing with ALL periodic values of I, but just I, such that is linked to 1 in this scale so that I=(1-I)/(1+I); In a similar vein, unique value of -I = (1+I)/(1-I) can be obtained.
The conclusion is ( or proposition was) : Mathematics works in SCALES which we do not notice for a reason that they are imaginary ( meaning in our minds),so we assume the same scale exists for all operations, which is not true, and in each scale the relation ship between Unit lentgh 1 and imaginary infinitesimal is defined by I=(1-I)/(1+I). This imaginary infinitesimal is Unique for that scale and by rotating it by +2pi we move into next scale, perhaps via multiplying by J= I*pi/2 where 1 is also different, but we do not notice it, as in math, we mix all scales in our heads - which is not possible in physics, and therefore the discrepancy.
On other hand, I^I= e^-pi/2 means that taken over all scales, imaginary unit is linked to e. However, if we stay in 1 scale, we do not need e for the base of natural logarithms, we can have k=I*pi/2 as defining natural k in that scale where I is infinitesimal.
Maybe I am spoiled. Maybe I am used to having answers for questions I haven't asked. Maybe I am used to finding all answers nicely and neatly stacked, in order of "creation". Maybe I am too stunningly good looking for my own good though I have bad breath... and so on...
I simply can not believe that whether or not I understand it, a mere question on sub-infinitesimals would stump moi, Chaitin's evil twin -and to change things a little bit and be creative, this time I don't understand it. As I have told you, I can only manipulate symbols: they mean nothing to me. I can show you everything you just said on paper, and try to check if it makes sense or not to you, not to me --since I don't have analytical skill. As for knowledge to handle it, it means nothing to me, I have none. But I won't break the prime code.
A **compressible** theory of infinitesimal exponentiation of base a. I don't have that, I only have ones and zeros. A number to any power of its divisors is equal to itself, that is, the prime code is the same, just the number of repeats is different. I.e., 2, 4 and 8 would be 10, 1010, 10101010, OR all would be 10 repeated forever. If a number is taken to a power and it approaches infinitesimal instead of approaching infinity, its prime code would be the same. So I would fully expect all infinitesimals to be lying nicely and neatly, in order, waiting for me to discover them.
So "I" can NOT be zero but it can't simply be a decimal point followed by (...)000000001 because that is still understandable and therefore not too good(!!!!). So I know what it can not be in decimal.
A **compressible** theory of infinitesimal exponentiation of base a. I don't have that, I only have ones and zeros. A number to any power of its divisors is equal to itself, that is, the prime code is the same, just the number of repeats is different. I.e., 2, 4 and 8 would be 10, 1010, 10101010, OR all would be 10 repeated forever. If a number is taken to a power and it approaches infinitesimal instead of approaching infinity, its prime code would be the same. So I would fully expect all infinitesimals to be lying nicely and neatly, in order, waiting for me to discover them.
So "I" can NOT be zero but it can't simply be a decimal point followed by (...)000000001 because that is still understandable and therefore not too good(!!!!). So I know what it can not be in decimal.
by definition, if we accept I as infinitesimal which is not 0
Time out! What is the code, Kenneth? We know it can not be the code for regular zero, which is at a 90 degree angle from 1, and also is at -1 and shift left:
...
0111101111011110... this is 5/infinity
0101010101010101... this is 4/infinity
0110110110110110... this is 3/infinity
0101010101010101... this is 2/infinity
0111111111111111... this is 1/infinity
1000000000000000... this is zero/infinity
0111111111111111... this is 1/infinity
0101010101010101... this is 2/infinity
0110110110110110... this is 3/infinity
0101010101010101... this is 4/infinity
0111101111011110... this is 5/infinity
...
xxxxxxxThe diagonal that crosses the capital 't' at the start of this sentence is **11** followed by an infinite string of zeros (and preceded by them, but I didn't bother to write them down). That "number" is greater than zero (whose representation on a non-doubled prime graph is 100000000000...).
If we are going to simplify that code to an imaginary value that is greater than zero and smaller than 0.000000(...)0001 then we can do two things:
--that is a number 3. We know that because the code for 3 is 110. (This is a short version of the "explanation".)
--that is a value x that has never been and can never have been multiplied OR multipliable by 2 or any of its multiples. This is why:
1.2.3.4.5.6.7.8 etc. in prime base is equal to 100000000000... which is the same code for zero. Of course, if you never ever multiply it by 2, the code would be 110000000000... because the "particle" is not addressable within this code. So we are agreed that the number 2 **as a divisor or multiple** does NOT appear within this number --whatever the number is in decimal mode, there will be plenty of digits that scream (2) at you: that is not the what I am saying, I am talking about the internal structure of the number.
We can now Explain Every Single Repeating Pattern In All Of The History Of Mankind... as long as the missing infinitesimal is 2!
We can also generalize this Oh So Great Finding to all bases beyond 3: as long as numeric base x is used, there will be no way to find out at a glance if any number below x is prime. Duh!
Back to you:
If we accept I as an infinitesimal of the FORM .....0001100000... then this code, to the power of itself, is equal to 1 plus an infinitesimal, that is, in any base it would be 1.000000(...)0001. However, the negation of the code for 1, which is 111111... would be 0, which is 10000000... from which we had subtracted/negated the *second* "particle" -the number 2- much before.
If we accept I as an infinitesimal of the FORM .....0001100000... then this code, to the power of itself, is equal to 1 plus an infinitesimal, that is, in any base it would be 1.000000(...)0001. However, the negation of the code for 1, which is 111111... would be 0, which is 10000000... from which we had subtracted/negated the *second* "particle" -the number 2- much before.
Euler made a proposition which he proved that J=kI, where k = 1 for base e so that ln(1+I)=I where I is just infinitesimal.
I can't tell whether or not what I said agrees with this since I still don't know where e is coming from.
As far as I can tell, the code for I is ...00011000... k should be a **code** in two identical parts, one which runs to the left and one to the right. (I have no idea what that means, but that is my result!)
As far as I can tell, the code for I is ...00011000... k should be a **code** in two identical parts, one which runs to the left and one to the right. (I have no idea what that means, but that is my result!)
From Euler's derivations, k as function of base a is :
k= 2( (a-1)/(a+1) + (a-1)^3/3*(a+1)^3+ (a-1)^5/5*(a+1)^5 ....
Then it is explained. There was no 2 in the definition of I before, we negated it, suddenly it's back here... being definitionally negated again, though it is being called "2" still... I am lost.
(And the square root of 2 is not and can not have a different code other than 10101010101010...)
??????
??????
So k= 2 ( I+I^3 /3+ I^5/5+I^7/7 +...) = 2*I ( 1-1/3+1/5-1/7.......) = 2*I* pi where I is JUST e^I*pi/2, single value.
Does I negate 2 or does it not? I can't tell.
If J equals ((...0000110000... times the prime line) divided by 2 (!!!!!) then J is also an infinitesimal turned 90 degrees. But we had agreed that 2 was negated before, now it is back, stopped dead on its track by a ...0000110000.... In terms of "code" we are repeating the same thing over and over again.
As far as I am concerned you are talking electrons, Ivars. I don't understand much more because the meanings of the math collapse into what I already know.
If J equals ((...0000110000... times the prime line) divided by 2 (!!!!!) then J is also an infinitesimal turned 90 degrees. But we had agreed that 2 was negated before, now it is back, stopped dead on its track by a ...0000110000.... In terms of "code" we are repeating the same thing over and over again.
As far as I am concerned you are talking electrons, Ivars. I don't understand much more because the meanings of the math collapse into what I already know.
But we know that I^I = e^-pi/2 so that ln(I^I) = -pi/2= (I*pi/2)*I:
????
???? I can't translate it since as I said
???? I can't translate it since as I said
So assumption that I may be imaginary infinitesimal works with logarithms, but the unique note here is that we are not dealing with ALL periodic values of I, but just I, such that is linked to 1 in this scale so that I=(1-I)/(1+I); In a similar vein, unique value of -I = (1+I)/(1-I) can be obtained.
The meanings DO collapse. The bis and pieces of what I could check sound right, as far as I can tell, because of that.
The meanings DO collapse. The bis and pieces of what I could check sound right, as far as I can tell, because of that.
On other hand, I^I= e^-pi/2 means that taken over all scales, imaginary unit is linked to e. However, if we stay in 1 scale, we do not need e for the base of natural logarithms, we can have k=I*pi/2 as defining natural k in that scale where I is infinitesimal.
Great! Can we throw e out the window already?
However, just one more note: I fully expect all answers to be sitting down waiting to be asked to come forward, nicely and neatly. That includes all irrational numbers and all physical constants. Nice and neat, in order. I can't get it off my head that a prime spiral would *not* have ALL answers.
Consider the wrap:
1:
1110101000101... --1,2,3,4,5,6,7,8,9,10,11,12,13...
2:
1111011011010... --1,3,5,7,9,11,13,15,17,19,21,23,25...
1000000000000... --2,4,6,8,10,12,14,16,18,20,22,24...
3:
101010100010... --1,4,7,10,13,16,19,22,25...
110101010100... --2,5,8,11,14,17,20,23,26...
100000000000... --3,6,9,12,15,18,21,24,27...
etc...
Since all integer arrangements of the prime line *logically* ORed (computational smallest comparison) to one another would give us the complete graph, and then, since the procedure is *logically* ("analytically", this time) equivalent to wrapping around a central pole the prime numbers, then...
This should make perfect irrational sense if reversed, as in wrapping the prime codes one by one around a center point, 1, 10, 110, 1010, 11110, 100010, etc... Everything, all sorts of irrational numbers, should be here in those wraps around a single point, whether or not it starts with the code for 1.
All irrationals must collapse.
They should be there as complete coordinates, in order, ready for the taking. It should be there because I expect them to be there, because I expect all answers to be written already, because I expect even the irrational to be rationalizable.
And because I am Chaitin's evil twin!
and so on...I simply can not believe that whether or not I understand it, a mere question on sub-infinitesimals would stump moi, Chaitin's evil twin -and to change things a little bit and be creative, this time I don't understand it. As I have told you, I can only manipulate symbols: they mean nothing to me. I can show you everything you just said on paper, and try to check if it makes sense or not to you, not to me --since I don't have analytical skill. As for knowledge to handle it, it means nothing to me, I have none. But I won't break the prime code.
QUOTE
I have been looking at relation for infinitesimal exponentiation of base a:
A **compressible** theory of infinitesimal exponentiation of base a. I don't have that, I only have ones and zeros. A number to any power of its divisors is equal to itself, that is, the prime code is the same, just the number of repeats is different. I.e., 2, 4 and 8 would be 10, 1010, 10101010, OR all would be 10 repeated forever. If a number is taken to a power and it approaches infinitesimal instead of approaching infinity, its prime code would be the same. So I would fully expect all infinitesimals to be lying nicely and neatly, in order, waiting for me to discover them.
So "I" can NOT be zero but it can't simply be a decimal point followed by (...)000000001 because that is still understandable and therefore not too good(!!!!). So I know what it can not be in decimal.
QUOTE (->
| QUOTE |
| I have been looking at relation for infinitesimal exponentiation of base a: |
A **compressible** theory of infinitesimal exponentiation of base a. I don't have that, I only have ones and zeros. A number to any power of its divisors is equal to itself, that is, the prime code is the same, just the number of repeats is different. I.e., 2, 4 and 8 would be 10, 1010, 10101010, OR all would be 10 repeated forever. If a number is taken to a power and it approaches infinitesimal instead of approaching infinity, its prime code would be the same. So I would fully expect all infinitesimals to be lying nicely and neatly, in order, waiting for me to discover them.
So "I" can NOT be zero but it can't simply be a decimal point followed by (...)000000001 because that is still understandable and therefore not too good(!!!!). So I know what it can not be in decimal.
by definition, if we accept I as infinitesimal which is not 0
Time out! What is the code, Kenneth? We know it can not be the code for regular zero, which is at a 90 degree angle from 1, and also is at -1 and shift left:
...
0111101111011110... this is 5/infinity
0101010101010101... this is 4/infinity
0110110110110110... this is 3/infinity
0101010101010101... this is 2/infinity
0111111111111111... this is 1/infinity
1000000000000000... this is zero/infinity
0111111111111111... this is 1/infinity
0101010101010101... this is 2/infinity
0110110110110110... this is 3/infinity
0101010101010101... this is 4/infinity
0111101111011110... this is 5/infinity
...
xxxxxxxThe diagonal that crosses the capital 't' at the start of this sentence is **11** followed by an infinite string of zeros (and preceded by them, but I didn't bother to write them down). That "number" is greater than zero (whose representation on a non-doubled prime graph is 100000000000...).
If we are going to simplify that code to an imaginary value that is greater than zero and smaller than 0.000000(...)0001 then we can do two things:
--that is a number 3. We know that because the code for 3 is 110. (This is a short version of the "explanation".)
--that is a value x that has never been and can never have been multiplied OR multipliable by 2 or any of its multiples. This is why:
1.2.3.4.5.6.7.8 etc. in prime base is equal to 100000000000... which is the same code for zero. Of course, if you never ever multiply it by 2, the code would be 110000000000... because the "particle" is not addressable within this code. So we are agreed that the number 2 **as a divisor or multiple** does NOT appear within this number --whatever the number is in decimal mode, there will be plenty of digits that scream (2) at you: that is not the what I am saying, I am talking about the internal structure of the number.
We can now Explain Every Single Repeating Pattern In All Of The History Of Mankind... as long as the missing infinitesimal is 2!
We can also generalize this Oh So Great Finding to all bases beyond 3: as long as numeric base x is used, there will be no way to find out at a glance if any number below x is prime. Duh!
Back to you:
QUOTE
by definition, if we accept I as infinitesimal which is not 0, I^I = (1+J) where J is infinitesimal as well (since I^0=1). Then log base I ( 1+J) = I log base I (I) = I.
If we accept I as an infinitesimal of the FORM .....0001100000... then this code, to the power of itself, is equal to 1 plus an infinitesimal, that is, in any base it would be 1.000000(...)0001. However, the negation of the code for 1, which is 111111... would be 0, which is 10000000... from which we had subtracted/negated the *second* "particle" -the number 2- much before.
QUOTE (->
| QUOTE |
| by definition, if we accept I as infinitesimal which is not 0, I^I = (1+J) where J is infinitesimal as well (since I^0=1). Then log base I ( 1+J) = I log base I (I) = I. |
If we accept I as an infinitesimal of the FORM .....0001100000... then this code, to the power of itself, is equal to 1 plus an infinitesimal, that is, in any base it would be 1.000000(...)0001. However, the negation of the code for 1, which is 111111... would be 0, which is 10000000... from which we had subtracted/negated the *second* "particle" -the number 2- much before.
Euler made a proposition which he proved that J=kI, where k = 1 for base e so that ln(1+I)=I where I is just infinitesimal.
I can't tell whether or not what I said agrees with this since I still don't know where e is coming from.
QUOTE
If we take imaginary Unit I as being the infinitesimal in base e , what is k?
As far as I can tell, the code for I is ...00011000... k should be a **code** in two identical parts, one which runs to the left and one to the right. (I have no idea what that means, but that is my result!)
QUOTE (->
| QUOTE |
| If we take imaginary Unit I as being the infinitesimal in base e , what is k? |
As far as I can tell, the code for I is ...00011000... k should be a **code** in two identical parts, one which runs to the left and one to the right. (I have no idea what that means, but that is my result!)
From Euler's derivations, k as function of base a is :
k= 2( (a-1)/(a+1) + (a-1)^3/3*(a+1)^3+ (a-1)^5/5*(a+1)^5 ....
Then it is explained. There was no 2 in the definition of I before, we negated it, suddenly it's back here... being definitionally negated again, though it is being called "2" still... I am lost.
(And the square root of 2 is not and can not have a different code other than 10101010101010...)
QUOTE
As we work with base I, where I = imaginary Unit, we have to evaluate term
(I-1)/(I+1) = (sqrt(2)* e^I*((3pi/4)+2pik)/ sqrt(2)* e^I* ((pi/4)+2pin) = e^I* (3pi/4-pi/4+2pik-2pin) = e^I*pi/2= i as a single value, not periodic, since all 2pik and 2pin cancel out.
(I-1)/(I+1) = (sqrt(2)* e^I*((3pi/4)+2pik)/ sqrt(2)* e^I* ((pi/4)+2pin) = e^I* (3pi/4-pi/4+2pik-2pin) = e^I*pi/2= i as a single value, not periodic, since all 2pik and 2pin cancel out.
??????
QUOTE (->
| QUOTE |
| As we work with base I, where I = imaginary Unit, we have to evaluate term (I-1)/(I+1) = (sqrt(2)* e^I*((3pi/4)+2pik)/ sqrt(2)* e^I* ((pi/4)+2pin) = e^I* (3pi/4-pi/4+2pik-2pin) = e^I*pi/2= i as a single value, not periodic, since all 2pik and 2pin cancel out. |
??????
So k= 2 ( I+I^3 /3+ I^5/5+I^7/7 +...) = 2*I ( 1-1/3+1/5-1/7.......) = 2*I* pi where I is JUST e^I*pi/2, single value.
Does I negate 2 or does it not? I can't tell.
QUOTE
Then J as infinitesimal result J = (I*pi/2 ) * I so J is also infinitesimal but turned by 90 degrees via multiplication of I by I plus it has module pi/2 and of I^I =(1+J) = (1+ (I*pi/2)*I)
If J equals ((...0000110000... times the prime line) divided by 2 (!!!!!) then J is also an infinitesimal turned 90 degrees. But we had agreed that 2 was negated before, now it is back, stopped dead on its track by a ...0000110000.... In terms of "code" we are repeating the same thing over and over again.
As far as I am concerned you are talking electrons, Ivars. I don't understand much more because the meanings of the math collapse into what I already know.
QUOTE (->
| QUOTE |
| Then J as infinitesimal result J = (I*pi/2 ) * I so J is also infinitesimal but turned by 90 degrees via multiplication of I by I plus it has module pi/2 and of I^I =(1+J) = (1+ (I*pi/2)*I) |
If J equals ((...0000110000... times the prime line) divided by 2 (!!!!!) then J is also an infinitesimal turned 90 degrees. But we had agreed that 2 was negated before, now it is back, stopped dead on its track by a ...0000110000.... In terms of "code" we are repeating the same thing over and over again.
As far as I am concerned you are talking electrons, Ivars. I don't understand much more because the meanings of the math collapse into what I already know.
But we know that I^I = e^-pi/2 so that ln(I^I) = -pi/2= (I*pi/2)*I:
????
QUOTE
But we also know that I^I = (1+I*pi/2*I) so that ln(I^I) = ln(1+I^2*pi/2) = k*I= J= I*pi/2*I= -pi/2 if I is imaginary infinitesimal such that I^2 = -1.
???? I can't translate it since as I said
QUOTE (->
| QUOTE |
| But we also know that I^I = (1+I*pi/2*I) so that ln(I^I) = ln(1+I^2*pi/2) = k*I= J= I*pi/2*I= -pi/2 if I is imaginary infinitesimal such that I^2 = -1. |
???? I can't translate it since as I said
So assumption that I may be imaginary infinitesimal works with logarithms, but the unique note here is that we are not dealing with ALL periodic values of I, but just I, such that is linked to 1 in this scale so that I=(1-I)/(1+I); In a similar vein, unique value of -I = (1+I)/(1-I) can be obtained.
QUOTE
The conclusion is ( or proposition was) : Mathematics works in SCALES which we do not notice for a reason that they are imaginary ( meaning in our minds),so we assume the same scale exists for all operations, which is not true, and in each scale the relation ship between Unit lentgh 1 and imaginary infinitesimal is defined by I=(1-I)/(1+I). This imaginary infinitesimal is Unique for that scale and by rotating it by +2pi we move into next scale, perhaps via multiplying by J= I*pi/2 where 1 is also different, but we do not notice it, as in math, we mix all scales in our heads - which is not possible in physics, and therefore the discrepancy.
The meanings DO collapse. The bis and pieces of what I could check sound right, as far as I can tell, because of that.
QUOTE (->
| QUOTE |
| The conclusion is ( or proposition was) : Mathematics works in SCALES which we do not notice for a reason that they are imaginary ( meaning in our minds),so we assume the same scale exists for all operations, which is not true, and in each scale the relation ship between Unit lentgh 1 and imaginary infinitesimal is defined by I=(1-I)/(1+I). This imaginary infinitesimal is Unique for that scale and by rotating it by +2pi we move into next scale, perhaps via multiplying by J= I*pi/2 where 1 is also different, but we do not notice it, as in math, we mix all scales in our heads - which is not possible in physics, and therefore the discrepancy. |
The meanings DO collapse. The bis and pieces of what I could check sound right, as far as I can tell, because of that.
On other hand, I^I= e^-pi/2 means that taken over all scales, imaginary unit is linked to e. However, if we stay in 1 scale, we do not need e for the base of natural logarithms, we can have k=I*pi/2 as defining natural k in that scale where I is infinitesimal.
Great! Can we throw e out the window already?
However, just one more note: I fully expect all answers to be sitting down waiting to be asked to come forward, nicely and neatly. That includes all irrational numbers and all physical constants. Nice and neat, in order. I can't get it off my head that a prime spiral would *not* have ALL answers.
Consider the wrap:
1:
1110101000101... --1,2,3,4,5,6,7,8,9,10,11,12,13...
2:
1111011011010... --1,3,5,7,9,11,13,15,17,19,21,23,25...
1000000000000... --2,4,6,8,10,12,14,16,18,20,22,24...
3:
101010100010... --1,4,7,10,13,16,19,22,25...
110101010100... --2,5,8,11,14,17,20,23,26...
100000000000... --3,6,9,12,15,18,21,24,27...
etc...
Since all integer arrangements of the prime line *logically* ORed (computational smallest comparison) to one another would give us the complete graph, and then, since the procedure is *logically* ("analytically", this time) equivalent to wrapping around a central pole the prime numbers, then...
This should make perfect irrational sense if reversed, as in wrapping the prime codes one by one around a center point, 1, 10, 110, 1010, 11110, 100010, etc... Everything, all sorts of irrational numbers, should be here in those wraps around a single point, whether or not it starts with the code for 1.
All irrationals must collapse.
They should be there as complete coordinates, in order, ready for the taking. It should be there because I expect them to be there, because I expect all answers to be written already, because I expect even the irrational to be rationalizable.
And because I am Chaitin's evil twin!
hej IAM,
I still can not read You, sorry. What am I missing is how You make those 10101 number lines-can You explain SLOWLY? Or is it like art, vision, feeling?
In prime base meaning each digit represents a prime? 1,2,3,5,7,11,13? So that
2=0100000000000000
4=101000000000000
5=00010000000000
8=1000100000000000000
it looks different from what You produce...
I still can not read You, sorry. What am I missing is how You make those 10101 number lines-can You explain SLOWLY? Or is it like art, vision, feeling?
In prime base meaning each digit represents a prime? 1,2,3,5,7,11,13? So that
2=0100000000000000
4=101000000000000
5=00010000000000
8=1000100000000000000
it looks different from what You produce...
1 is not prime.
QUOTE (Ivars+Dec 29 2007, 02:45 PM)
I still can not read You, sorry. What am I missing is how You make those 10101 number lines-can You explain SLOWLY? Or is it like art, vision, feeling?
In prime base meaning each digit represents a prime?
A range of possibilities.
In prime base (that should be "base" because it is not quite informational, and it is more than informational) the "numbers" agree with their coordinate. This is the number line:
1,2,3,4,5,6,7...
This is the prime line:
1,1,1,0,1,0,1...
Thus
1110101000101... is exactly equal to 1,2,3,4,5,6,7,8,9,10,11,12,13...
The ones and zeros are there just to denote the presence of a prime. Now you divide it by two and give it "mass":
1111011011010... --1,3,5,7,9,11,13,15,17,19,21,23,25...
1000000000000... --2,4,6,8,10,12,14,16,18,20,22,24...
and 3
101010100010... --1,4,7,10,13,16,19,22,25...
110101010100... --2,5,8,11,14,17,20,23,26...
100000000000... --3,6,9,12,15,18,21,24,27...
and 4
11011001011... --1,5,9,13,17,21,25,29,33,37,41...
10000000000... --2,6,10,14,18,22,26,30,34,38,42...
11101101001... --3,7,11,15,19,23,27,31,35,39,43...
00000000000... --4,8,12,16,20,24,28,32,36,40,44...
etc...
Then you put then on top of one another and end up with a prime matrix. That is what I call the prime graph, derived from the prime line.
Then, since all points are exactly equal (because it's a fractal where every single 1 is equal to the next except in angle) what is left for you to do is to play with the first 1 or play with coordinates.
When you play with the first 1, all of the meanings of that "mass" collapse on top of e=mc squared (it's a long story that doesn't make much sense, with ones and zeros, but I have checked it over and over again: the meanings are collapsed) because to take the first one out you split the graph into two equal parts, and to add a point you double its "mass" and end up with a copy going down adn another going up. You could perfectly switch readings too, and say that when something bumps into that first point the graph turns inside out.
It's because of the meaning collapse that you can read any theory in it and check if it collapses into another theoretical structure --where "meaning collapse" means that if I were building the universe a very small group of elementals would have to **mean** everything that can be expressed from within that mathematical system. Meaning collapse is not system collapse, by the way, since a system collapse would be when thishere type of mathematics doesn't know that .9r is equal to 1, and thatthere system of mathematics doesn't know that they are different.
It is because of meaning collapse that I am "checking" your mathematics (oh, yeah, that will fly! ) without needing to understand it any of it except the most basics of the theoretic structure. I just read ones and zeros, and YOU collapse the meanings into what you are saying and check if you are right or wrong --the prime graph is a mathematical construct and can't be wrong, and "what I am saying" is irrelevant because I am not saying anything, I am reporting a matrix comparison. I stand by that only --I can't and I don't have any background to check your meaning collapse into infinite tetration.
When you switch to whole coordinate symbolism you end up with "number"equals "atom". For multiple reasons I already expected it much before I ever found the prime graph. Except I have no idea how to prove it even after all this time. I could make a case for ionization to be prime based but only if 2 or less numbers are involved, and I would make a fool of myself because I don't know what ionization means, but in any event the moment they become a composite number of 3 or more divisors all hell breaks lose and there is no way I can get around it.
Look what I, OH POOR ME, have to put up with:
"
The code for any prime is a periodic series of 1's: 1111111111111111...
You can't simultaneously say that 1 is NOT prime and that 0.9r equals 1, Alphanumeric. Either one or the other. If .xxxxxxxx... magically becomes 1 in base x+1, then if a code for a prime can NOT ever be broken, then if the code for 1 is 1111111111... then if a code for a number to any power is equal to itself repeated forever, and then if you absolutely know that logically there is a zero "at the end" which can never be reached, the same way that there is an infinitesimal missing "at the end" of .9r... THEN
... the "limit" for 1 is primality. Two can play this little game.
YOU SEE WHAT I HAVE TO PUT UP WITH, IVARS? AI, COMO SUFRO! COMO SUFRO!
In prime base meaning each digit represents a prime?
A range of possibilities.
In prime base (that should be "base" because it is not quite informational, and it is more than informational) the "numbers" agree with their coordinate. This is the number line:
1,2,3,4,5,6,7...
This is the prime line:
1,1,1,0,1,0,1...
Thus
1110101000101... is exactly equal to 1,2,3,4,5,6,7,8,9,10,11,12,13...
The ones and zeros are there just to denote the presence of a prime. Now you divide it by two and give it "mass":
1111011011010... --1,3,5,7,9,11,13,15,17,19,21,23,25...
1000000000000... --2,4,6,8,10,12,14,16,18,20,22,24...
and 3
101010100010... --1,4,7,10,13,16,19,22,25...
110101010100... --2,5,8,11,14,17,20,23,26...
100000000000... --3,6,9,12,15,18,21,24,27...
and 4
11011001011... --1,5,9,13,17,21,25,29,33,37,41...
10000000000... --2,6,10,14,18,22,26,30,34,38,42...
11101101001... --3,7,11,15,19,23,27,31,35,39,43...
00000000000... --4,8,12,16,20,24,28,32,36,40,44...
etc...
Then you put then on top of one another and end up with a prime matrix. That is what I call the prime graph, derived from the prime line.
Then, since all points are exactly equal (because it's a fractal where every single 1 is equal to the next except in angle) what is left for you to do is to play with the first 1 or play with coordinates.
When you play with the first 1, all of the meanings of that "mass" collapse on top of e=mc squared (it's a long story that doesn't make much sense, with ones and zeros, but I have checked it over and over again: the meanings are collapsed) because to take the first one out you split the graph into two equal parts, and to add a point you double its "mass" and end up with a copy going down adn another going up. You could perfectly switch readings too, and say that when something bumps into that first point the graph turns inside out.
It's because of the meaning collapse that you can read any theory in it and check if it collapses into another theoretical structure --where "meaning collapse" means that if I were building the universe a very small group of elementals would have to **mean** everything that can be expressed from within that mathematical system. Meaning collapse is not system collapse, by the way, since a system collapse would be when thishere type of mathematics doesn't know that .9r is equal to 1, and thatthere system of mathematics doesn't know that they are different.
It is because of meaning collapse that I am "checking" your mathematics (oh, yeah, that will fly!
) without needing to understand it any of it except the most basics of the theoretic structure. I just read ones and zeros, and YOU collapse the meanings into what you are saying and check if you are right or wrong --the prime graph is a mathematical construct and can't be wrong, and "what I am saying" is irrelevant because I am not saying anything, I am reporting a matrix comparison. I stand by that only --I can't and I don't have any background to check your meaning collapse into infinite tetration.When you switch to whole coordinate symbolism you end up with "number"equals "atom". For multiple reasons I already expected it much before I ever found the prime graph. Except I have no idea how to prove it even after all this time. I could make a case for ionization to be prime based but only if 2 or less numbers are involved, and I would make a fool of myself because I don't know what ionization means, but in any event the moment they become a composite number of 3 or more divisors all hell breaks lose and there is no way I can get around it.
Look what I, OH POOR ME, have to put up with:
"
QUOTE
1 is not prime."
The code for any prime is a periodic series of 1's: 1111111111111111...
You can't simultaneously say that 1 is NOT prime and that 0.9r equals 1, Alphanumeric. Either one or the other. If .xxxxxxxx... magically becomes 1 in base x+1, then if a code for a prime can NOT ever be broken, then if the code for 1 is 1111111111... then if a code for a number to any power is equal to itself repeated forever, and then if you absolutely know that logically there is a zero "at the end" which can never be reached, the same way that there is an infinitesimal missing "at the end" of .9r... THEN
... the "limit" for 1 is primality. Two can play this little game.
YOU SEE WHAT I HAVE TO PUT UP WITH, IVARS? AI, COMO SUFRO! COMO SUFRO!
QUOTE (IAMoraes+Dec 29 2007, 08:55 PM)
You can't simultaneously say that 1 is NOT prime and that 0.9r equals 1, Alphanumeric. Either one or the other. If .xxxxxxxx... magically becomes 1 in base x+1, then if a code for a prime can NOT ever be broken, then if the code for 1 is 1111111111... then if a code for a number to any power is equal to itself repeated forever, and then if you absolutely know that logically there is a zero "at the end" which can never be reached, the same way that there is an infinitesimal missing "at the end" of .9r... THEN
... the "limit" for 1 is primality. Two can play this little game.
The proof that 0.9r=1 or that 1.9r=2 or that 2.9r=3 or that 3.9r=4 etc has nothing to do with the primality of 1, 2, 3, 4 etc
3.9r=4 is true, 0.9r=1 is true. Neither 1 or 4 is prime.
1.9r=2 is true, 2.9r=3 is true. Both 2 and 3 are prime.
Your incoherent ramblings do nothing to support your case.
In a nod to slightly deeper mathematics, 1 is known as a unit. When considering factorisations, any number has a unique factorisation in terms of prime numbers. Therefore
n =(p1^n1)(p2^n2)....(pk^nk)
where p1, p2, .... , pk are the prime factors of n and n1, ... , nk are their multiplicities.
Prime factoriatisation is unique. Therefore if
n =(p1^n1)(p2^n2)....(pk^nk) = n =(q1^m1)(q2^m2)....(qj^mj)
then the following are true :
k=j
pc = qd for one and only one c and d pair and then nc=md
For instance
12 = (2^2)(3^1) = (a^n)(b^m)
Either a=2 (so n=2) and b=3 (so m=1) or vice versa.
If 1 is prime, this is no longer true, since 12 = 12*1, so there's no unique factoratisation.
Prime factorisation is unique because 1 is not prime.
... the "limit" for 1 is primality. Two can play this little game.
The proof that 0.9r=1 or that 1.9r=2 or that 2.9r=3 or that 3.9r=4 etc has nothing to do with the primality of 1, 2, 3, 4 etc
3.9r=4 is true, 0.9r=1 is true. Neither 1 or 4 is prime.
1.9r=2 is true, 2.9r=3 is true. Both 2 and 3 are prime.
Your incoherent ramblings do nothing to support your case.
In a nod to slightly deeper mathematics, 1 is known as a unit. When considering factorisations, any number has a unique factorisation in terms of prime numbers. Therefore
n =(p1^n1)(p2^n2)....(pk^nk)
where p1, p2, .... , pk are the prime factors of n and n1, ... , nk are their multiplicities.
Prime factoriatisation is unique. Therefore if
n =(p1^n1)(p2^n2)....(pk^nk) = n =(q1^m1)(q2^m2)....(qj^mj)
then the following are true :
k=j
pc = qd for one and only one c and d pair and then nc=md
For instance
12 = (2^2)(3^1) = (a^n)(b^m)
Either a=2 (so n=2) and b=3 (so m=1) or vice versa.
If 1 is prime, this is no longer true, since 12 = 12*1, so there's no unique factoratisation.
Prime factorisation is unique because 1 is not prime.
QUOTE (AlphaNumeric+Dec 29 2007, 04:37 PM)
The proof that 0.9r=1 or that 1.9r=2 or that 2.9r=3 or that 3.9r=4 etc has nothing to do with the primality of 1, 2, 3, 4 etc
Ok.
Ok.
Thanks to both of You.
What AlphaNumeric wrote, is clear. 1 is not prime as unique factorization existance in prime factors is used to define what prime is and it equals the condition of prime being divisible only by itself AND 1 at the same time.
What IAM is saying is less clear but at least I understand now how the lines of 1 and 0 are built to represent sequencies of numbers resulting from certain operations ( like taking evey third number) and denoting in a resulting series every prime with 1, every non prime with 0. Probably if You make a lot of pictures out of this You can see some patterns and get some gut feeling about it.
Would be interesting to know what complex value patterns we can get out of prime infinite tetration since all primes > e^(1/e)=1,444667861
so all values of h(prime) will be complex numbers.
What AlphaNumeric wrote, is clear. 1 is not prime as unique factorization existance in prime factors is used to define what prime is and it equals the condition of prime being divisible only by itself AND 1 at the same time.
What IAM is saying is less clear but at least I understand now how the lines of 1 and 0 are built to represent sequencies of numbers resulting from certain operations ( like taking evey third number) and denoting in a resulting series every prime with 1, every non prime with 0. Probably if You make a lot of pictures out of this You can see some patterns and get some gut feeling about it.
Would be interesting to know what complex value patterns we can get out of prime infinite tetration since all primes > e^(1/e)=1,444667861
so all values of h(prime) will be complex numbers.
QUOTE (Ivars+Dec 29 2007, 04:55 PM)
What AlphaNumeric wrote, is clear. 1 is not prime as unique factorization existance in prime factors is used to define what prime is and it equals the condition of prime being divisible only by itself AND 1 at the same time.
I have no gut feelings, I report what I am looking at!
Here we go backward:
3 is prime = 1
2 is prime = 1
1 is not prime = 0
One, one, zero. Where did we last see this? Oh, yeah, now I remember...
The singularity of ...00000011000000... has shifted around and instead of being at I where it was supposed to be, it is now at 1.
This is a logical trap, Ivars, caused directly by our inability to express the totality of our universe with the mathematics within it. I can't do infinite regression.
(Edit: if you have to switch alephs (mathematics) in order to prove that 1 is NOT prime, then you didn't keep to Your Eternal Promisses to the real integer set. In short, you went overboard AND over the limit you set for other people... but not for yourself! AAAAAHHHHHHHAAAAAA!
... NOT switching logical system was the whole *cause** of the unending conversation about "limits" and .9r-equals-1-no-it-doesn't, wasn't it?)
QUOTE
What IAM is saying is less clear but at least I understand now how the lines of 1 and 0 are built to represent sequencies of numbers resulting from certain operations ( like taking evey third number) and denoting in a resulting series every prime with 1, every non prime with 0. Probably if You make a lot of pictures out of this You can see some patterns and get some gut feeling about it.
I have no gut feelings, I report what I am looking at!
Here we go backward:
3 is prime = 1
2 is prime = 1
1 is not prime = 0
One, one, zero. Where did we last see this? Oh, yeah, now I remember...
The singularity of ...00000011000000... has shifted around and instead of being at I where it was supposed to be, it is now at 1.
This is a logical trap, Ivars, caused directly by our inability to express the totality of our universe with the mathematics within it. I can't do infinite regression.
(Edit: if you have to switch alephs (mathematics) in order to prove that 1 is NOT prime, then you didn't keep to Your Eternal Promisses to the real integer set. In short, you went overboard AND over the limit you set for other people... but not for yourself! AAAAAHHHHHHHAAAAAA!
... NOT switching logical system was the whole *cause** of the unending conversation about "limits" and .9r-equals-1-no-it-doesn't, wasn't it?)
hej IAM
Speaking about things turning itself inside out at number 3, the places of number 1 and 2.
Continuos variable transmission with flexible shaft
Waiting for Your reaction:)
Speaking about things turning itself inside out at number 3, the places of number 1 and 2.
Continuos variable transmission with flexible shaft
Waiting for Your reaction:)
QUOTE (Ivars+Dec 30 2007, 04:32 AM)
Waiting for Your reaction:)
Obviously, since a moebius strip cut in half has a ring twice as big as the other, I would want to see the physics of that!
Fabulous link!
Obviously, since a moebius strip cut in half has a ring twice as big as the other, I would want to see the physics of that!
Fabulous link!
hej IAM
The idea how I got there lies via differentiation as a property of differential gear ( and so also planetary gear) and Klein bottle as a cell of Universe - Klein bottle is 2 Mobius strips glued together, and You can see how this device in the link is starting to form the handle of Klein bottle.
And now, Imagine tha planetary gear (the one with radius 1/2 of the big gear with internal teeth) is and infinitesimal thickness ORB-consisting of infinitely many discs inside it , may be sized ala FORD cycles, but aranged inside concentric stripes (e.g. maybe 5 of them).
Then, think of these circles as vortexes of infintesimals of various orders- what do You think?
Appolonian gaskets
Ford circles
These links does not give the exact picture of what I imagine, but I can not find a closer one.
The idea how I got there lies via differentiation as a property of differential gear ( and so also planetary gear) and Klein bottle as a cell of Universe - Klein bottle is 2 Mobius strips glued together, and You can see how this device in the link is starting to form the handle of Klein bottle.
And now, Imagine tha planetary gear (the one with radius 1/2 of the big gear with internal teeth) is and infinitesimal thickness ORB-consisting of infinitely many discs inside it , may be sized ala FORD cycles, but aranged inside concentric stripes (e.g. maybe 5 of them).
Then, think of these circles as vortexes of infintesimals of various orders- what do You think?
Appolonian gaskets
Ford circles
These links does not give the exact picture of what I imagine, but I can not find a closer one.
QUOTE (Ivars+Dec 30 2007, 05:08 AM)
And now, Imagine tha planetary gear (the one with radius 1/2 of the big gear with internal teeth) is and infinitesimal thickness ORB-consisting of infinitely many discs inside it , may be sized ala FORD cycles, but aranged inside concentric stripes (e.g. maybe 5 of them).
Then, think of these circles as vortexes of infintesimals of various orders- what do You think
It sounds true in a Mandelbrot kind of way... everyone of my cats is trying to come up with an image just like that!
I have to switch theories to say something understandable: the five circles are people.
The infinitely many discs are already where I can use them, they are what the people are doing, their actions.
What is the "verb" of those five people inside the ring? What are they saying? Are they passing words or an object around? To choose from my extremely limited repertoire, you have "acquire"/"give" (final and permanent) or you have "attract"/"repel".
What is their network? Token ring or parallel? In token ring, one passes the "token" to the other for an action to be performed, but the token is not spent, it just passes it around to the next guy, and the final weight of the whole system never changes. On "acquire", or parallel network, you can input from any one person or any number of persons, and the system eventually ends unbalanced and comes to a full stop.
On "attract" the system is designed to innocently NOT do anything. It just waits for an input while it is doing its own structure, and when the "token" gets in, there already is something waiting for it, and the token is spent, but weight is acquired because of the prep-work --the "innocently not doing anything"-- and the output is whatever the person wishes to output, from nothing to the whole thing. If the output is the token plus the structure acquired from the person, then the weight of the whole system increases.
Then there is the "expel" the token too, when it doesn't fit the person. This expel is permanent, the token doesn't get acquired by anyone else, it is expelled from the system lossless-ly.
This would be the same thing in the infinitesimal world as in ours because the "people", that is, the circles, must be necessarily working in a network, as a group, and the whole system MUST have an identifying verb.
Without it, you have logical degeneration until everything starts again from the very beginning.
I need more info. I suspect my thoughts are structured in the wrong manner to be accurate: I am getting old.
(Tangent of the day: and that is why I loathe what advertising did to the world)
Then, think of these circles as vortexes of infintesimals of various orders- what do You think
It sounds true in a Mandelbrot kind of way... everyone of my cats is trying to come up with an image just like that!
I have to switch theories to say something understandable: the five circles are people.
The infinitely many discs are already where I can use them, they are what the people are doing, their actions.
What is the "verb" of those five people inside the ring? What are they saying? Are they passing words or an object around? To choose from my extremely limited repertoire, you have "acquire"/"give" (final and permanent) or you have "attract"/"repel".
What is their network? Token ring or parallel? In token ring, one passes the "token" to the other for an action to be performed, but the token is not spent, it just passes it around to the next guy, and the final weight of the whole system never changes. On "acquire", or parallel network, you can input from any one person or any number of persons, and the system eventually ends unbalanced and comes to a full stop.
On "attract" the system is designed to innocently NOT do anything. It just waits for an input while it is doing its own structure, and when the "token" gets in, there already is something waiting for it, and the token is spent, but weight is acquired because of the prep-work --the "innocently not doing anything"-- and the output is whatever the person wishes to output, from nothing to the whole thing. If the output is the token plus the structure acquired from the person, then the weight of the whole system increases.
Then there is the "expel" the token too, when it doesn't fit the person. This expel is permanent, the token doesn't get acquired by anyone else, it is expelled from the system lossless-ly.
This would be the same thing in the infinitesimal world as in ours because the "people", that is, the circles, must be necessarily working in a network, as a group, and the whole system MUST have an identifying verb.
Without it, you have logical degeneration until everything starts again from the very beginning.
I need more info. I suspect my thoughts are structured in the wrong manner to be accurate: I am getting old.
(Tangent of the day: and that is why I loathe what advertising did to the world)
QUOTE (IAMoraes+Dec 30 2007, 10:13 AM)
On "attract" the system is designed to innocently NOT do anything. It just waits for an input while it is doing its own structure, and when the "token" gets in, there already is something waiting for it, and the token is spent, but weight is acquired because of the prep-work --the "innocently not doing anything"-- and the output is whatever the person wishes to output, from nothing to the whole thing. If the output is the token plus the structure acquired from the person, then the weight of the whole system increases.
Hej IAM
This is definitely one operation it has to perform- be sensitive by doing nothing ( e.g spinning internally) to what might fit it, and use it properly to expand itself a bit.
Hej IAM
This is definitely one operation it has to perform- be sensitive by doing nothing ( e.g spinning internally) to what might fit it, and use it properly to expand itself a bit.
This is an infinitesimal line segment produced from infinitesimal epicycles inside circle of a circle whose diameter is 1/2 of the ring, and drawing pen is fixed on the outside of inner ring, You get a finite piece of straight line = diameter of bigger circle which is the result of motion of iside ring.
Spirograph
If not, fix sliders at tool at: 44, -22, 22.
To get a point, You have to increase the size of inner circle to be equal to the outer circle. 44,-44, any-in this case offset does not change anything since inner and outer circles are the same, and does not move relative to each other. Also, You may get a point if you freeze the picture at the moment when pen drawing line is changing direction.
Spherical triangle at 90, -30, 30.
Interesting device.
Spirograph
If not, fix sliders at tool at: 44, -22, 22.
To get a point, You have to increase the size of inner circle to be equal to the outer circle. 44,-44, any-in this case offset does not change anything since inner and outer circles are the same, and does not move relative to each other. Also, You may get a point if you freeze the picture at the moment when pen drawing line is changing direction.
Spherical triangle at 90, -30, 30.
Interesting device.
QUOTE (Ivars+Dec 30 2007, 07:16 AM)
Spirograph
The spirograph was one of my favorite toys ever. I think it reached the market when I was about 10 or 11 years old. It's more than interesting, it is fabulous.
(I am assuming that you are familiar with the spirograph from here on)
Let's forget teeth for a second.
if (x times y) mod z = 1 then plot z, x (or z, y, it doesn't make a difference)
So that when z is 1, every number will end in 1. So that line will plot a solid 11111111...
When z is 2, every other point will be plotted on the whole line:
10101010101...
When z is 3, the output is
110110110110...
So that is how we build the prime graph. We know that every prime IS IN ITS COORDINATE. And we know that every prime n is made up of n plots (1) and a single no plot (0). That is the length of the truths needed for each and every prime.
Going back to teeth, though. If we change the number of tee
The spirograph was one of my favorite toys ever. I think it reached the market when I was about 10 or 11 years old. It's more than interesting, it is fabulous.
(I am assuming that you are familiar with the spirograph from here on)
Let's forget teeth for a second.
if (x times y) mod z = 1 then plot z, x (or z, y, it doesn't make a difference)
So that when z is 1, every number will end in 1. So that line will plot a solid 11111111...
When z is 2, every other point will be plotted on the whole line:
10101010101...
When z is 3, the output is
110110110110...
So that is how we build the prime graph. We know that every prime IS IN ITS COORDINATE. And we know that every prime n is made up of n plots (1) and a single no plot (0). That is the length of the truths needed for each and every prime.
Going back to teeth, though. If we change the number of tee