Amazing.
Analog computers unable to take square root of 2 with more than 5 decimals precision (better than G = (17/12 + 24/17)/2 = 577/408 = 1.414215686), becouse the accuracy is 10^{-4}. They don't have more than 1000-100000 "digital" levels.
When they for example solving diferential equation then coeficients arn't bigger than 0.236, 0.589.
http://www.cisl.columbia.edu/grads/gcowan/vlsianalog.pdf "Accuracy: Analog computers are not very accurate. Solutions are usually
accurate to no better than 0.1 % - 1 %. These errors stem from the following:
˛ Thermal, flicker and shot noise.
˛ Nonlinear transfer characteristics of blocks that should be linear.
˛ Other deviations from the intended input-to-output transfer characteristics of
memoryless blocks. For example: imperfect multiplication of multipliers, finite
output resistance of current mode circuits, or o®sets of variable gain amplifiers.
˛ Finite DC gain of integrators.
˛ Input offsets of integrators.
˛ Finite bandwidth of memoryless blocks and non-dominant poles in the integra-
tors.
˛ Granularity in the setting of coe±cients such as integrator time constants or
gains of variable gain amplifiers.
˛ Coupling between signal wires."
Only most 5 decimal digits can manage analog computer:
"6.2.2 First Order Nonlinear SDEs: Particle in a Potential
Well
Mathematical Description: The diferential equation investigated here is an ex-
ample of an Ornstein-Uhlenbeck process and is given by:
x' = -gradU(x) + n(t) (6.26)
grad denotes the gradient operation. For this example, x is a scalar and
U(x) =
0.185x^4 + 0.0139x^3 - 0.251x^2 giving
gradU(x) = 0.740x^3 + 0.0416x^2 - 0.502x. The
function n(t) is a random variable with zero mean and a Gaussian distribution.
Physical Interpretation and Qualitative Behaviour: U(x) is referred to
as a potential function. This system describes a particle moving in a double potential
well. x is the horizontal displacement of the particle, while gradU(x) is the steepness of
the well. For this example, the cubic gradient has roots at:
-0.852, 0.796 and 0. The
first and second root give rise to stable equilibria. This means that in the absence of
large noise, if the particle is near one of these roots, it will stay near one of them, and
if the noise were reduced to zero, the particle would converge to one of these values of
x. The root at zero is an unstable equilibrium, and hence, infinitesimally small noise
will perturb the particle away from it. This is analogous to an inverted pendulum.
Mathematically, the inverted vertical state is a solution, but the smallest noise will
disturb the pendulum from its balance."
5 digits is 100000 levels, so thats why analog computer can't manage with more than 5 digits, becouse nature don't leting it, becouse nature itself is noisy...
Quantum computer qubit indeed is analog, becouse qubit can be rotatated with arbitrary degree, but qubit rotation precision is limited with 0.99999, becouse qubit rotation is analog. So after 100000 qubit rotations, qubit will loose all information and will obey like random bit (if will be measured). With more qubits rotation precision also decreasing, becouse qubits then rotates also each over and so 1000 qubits quantum computer will break down after 100 rotations of each qubit.
Qubits are rotated from one binary state to another binary state, but this rotations still is analog and not binary/digital.
If entanglement between atoms or electrons realy was estimated, then possible that 2 or more qubits quantum computer isn't faster and don't giving speedup, becouse number of qubits rotations is far more than 10000.
But what if qubits also rotating another atoms or electrons - the noise?..
