# Estimation with complex exponent

Context (not necessary to understand the task). I'm working through so called Eigenvalue Estimation Problem from the field of Quantum Computing. More precisely, I try to measure the inaccuracy of the quantum circuit when an eigenvalue measured is irrational (the circuit is designed for rational phases).

The task can be reduced to the estimation with the lower bound of an (modulus of) expression:

$$\frac 1 {2^m} \frac {1 - e^{2\pi i \delta 2^m}} {1 - e^{2\pi i \delta }}$$

Here $m$ is a natural number, $\delta$ is a real from $[0,1)$, $i$ is the imaginary unit.

It is known that the desired lower bound is $2 / \pi$. It is suggested to get upper bound for denominator (which is clearly $2\pi\delta$) and lower bound for numerator (which is problematic for me) and then divide them. It is known that numerator $N$ can be estimated like this: $N > 2\pi\delta2^m(2/\pi) > 4\delta 2^m$. Can anyone explain, how to obtain the bound in the left part of the inequality?

• I think that desired estimation is wrong. See code: cpp.sh/8qiog Dec 19, 2016 at 19:12
• Key role plays 2^(-m) factor, result lower bound cannot be without it Dec 19, 2016 at 19:19
• I suppose, that your estimation will be right only in the case, where we have fixed number m (it plays role as constant), and b converges to +0. Dec 19, 2016 at 19:23

It is another approach which finds more exact lower bound (you can see it in my program for example)

It gives lower bound 1 (as limit)

Note, it will works only if b converges to +0 (for very small b and fixed m)

The solution turns to be direct consequence of Jordan's inequality:

$$\sin x \geqslant x \frac 2 {\pi}$$

We would like to obtain lower bound for $|1 - e^{2\pi i \delta 2^m}|$, that is $|AC|$ on the following graphics with unit circle:

$OB$ is altitude and, at the same time, bisector and median, so

$$|AC| = 2 |AB| = 2 \sin {\frac a 2} \geqslant 2 \frac a 2 \frac 2 {\pi} = \frac {2a} \pi .$$

Given that $a = 2\pi \delta 2^m$ we have desired $|AC| = |1 - e^{2\pi i \delta 2^m}| \geqslant 4 \delta 2^m$. The whole fraction from the post turns to be estimated as follows:

$$\frac 1 {2^m} \frac {1 - e^{2\pi i \delta 2^m}} {1 - e^{2\pi i \delta }} \geqslant \frac 1 {2^m} \frac {4 \delta 2^m} {2 \pi \delta} = \frac 2 {\pi}.$$