Upper bound of $\sum_{n=1}^N |1-z^n|$ where $|z| \leq 1$

Question

How to derive an upper bound of

$$\sum_{n=1}^N |1-z^n|$$ where $z\in\mathbb{C}$ and $|z| \leq 1$?

A trivial upper bound would be $2N$ since each $|1-z^n| \leq 2$. But I am hoping for tighter bounds. I ran some numerical experiments and believe the bound should be $\frac{3}{2}N$ but don't know how to prove it.

Answer 1

Here is an upper bound:

According to the link given by @Gerd in comment (Maximum of sum of finite modulus of analytic function.), the maximum of $\sum_{n=1}^N |1 - z^n|$ is attained on $|z| = 1$.

Let $z = \mathrm{e}^{\mathrm{i}\theta}$ with $\theta \in [0, 2\pi]$. Using AM-QM, we have \begin{align*} \sum_{n=1}^N |1 - z^n| &\le \sqrt{N\sum_{n=1}^N |1 - z^n|^2}\\ &= \sqrt{N\sum_{n=1}^N (2 - 2\cos n\theta)}\\ &= \sqrt{2N^2 + N \frac{\sin \frac{\theta}{2} - \sin\frac{(2N + 1)\theta}{2}}{\sin \frac{\theta}{2}}}. \tag{1} \end{align*}

Fact 1: Let $x\in [0, \pi/2]$ and $N \ge 20$. Then $$\frac{\sin x - \sin (2N + 1)x}{\sin x} \le \frac{N}{2}.$$ (The proof is not difficult.)

By Fact 1 and (1), we have, for all $N \ge 20$, $$\sum_{n=1}^N |1 - z^n| \le \sqrt{5/2}\, N.$$

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Some thoughts:

We have \begin{align} \sum_{n=1}^N |1 - z^n| &= \sum_{n=1}^N \sqrt{2 - 2\cos n\theta }\\ &= \sum_{n=1}^N 2\left|\sin \frac{n\theta}{2}\right|\\ &= \frac{4}{\pi}N - \sum_{n=1}^N \sum_{k=1}^\infty \frac{8}{\pi(4k^2 - 1)}\cos kn\theta\\ &= \frac{4}{\pi}N - \sum_{k=1}^\infty \sum_{n=1}^N \frac{8}{\pi(4k^2 - 1)}\cos kn\theta\\ &= \frac{4}{\pi}N + \sum_{k=1}^\infty \frac{8}{\pi(4k^2 - 1)} \frac{\sin\frac{k\theta}{2} - \sin \frac{(2N + 1)k \theta}{2}}{2\sin \frac{k\theta}{2}}\\ &= \frac{4}{\pi}N + \sum_{k=1}^\infty \frac{4}{\pi(4k^2 - 1)} + \sum_{k=1}^\infty \frac{4}{\pi(4k^2 - 1)} \frac{ - \sin \frac{(2N + 1)k \theta}{2}}{\sin \frac{k\theta}{2}}\\ &= \frac{4}{\pi}N + \frac{2}{\pi} + \sum_{k=1}^\infty \frac{4}{\pi(4k^2 - 1)} \frac{ - \sin \frac{(2N + 1)k \theta}{2}}{\sin \frac{k\theta}{2}} \end{align} where we have used the identity $$\left|\sin y\right| = \frac{2}{\pi}-\sum_{k = 1}^\infty \frac{4}{\pi(4k^2-1)}\cos(2k y).$$ (Note: The RHS is the Fourier expansion of the LHS.)

We need to find bounds for $$\sum_{k=1}^\infty \frac{4}{\pi(4k^2 - 1)} \frac{ - \sin \frac{(2N + 1)k \theta}{2}}{\sin \frac{k\theta}{2}}.$$

Answer 2

The behavior in the limit (as $N\to\infty$) looks interesting: the value $$L:=\lim_{N\to\infty}\frac1N\sup_{|z|\leqslant 1}\sum_{n=1}^N|1-z^n|\approx1.4492227\ldots\color{red}{<1.5}$$ equals $2\sin\lambda$, where $\lambda$ is the smallest positive root of $\cos\lambda+\lambda\sin\lambda=1$.

Here is a sketch towards a proof. As already noted, the $\sup_{|z|\leqslant 1}$ is attained at $|z|=1$: $$\sup_{|z|\leqslant 1}\sum_{n=1}^N|1-z^n|\underset{\big[z=e^{2it}\big]}{=}2\sup_{0\leqslant t\leqslant\pi/2}\sum_{n=1}^N|\sin nt|.$$ Suppose the maximum is attained at $t=t_N$, and let $\lambda_N=Nt_N$. Then it turns out that $\lambda_N<\pi$ for all $N$, and that the limit $\lambda=\lim_{N\to\infty}\lambda_N$ exists. Assuming this is proven, $t=t_N$ is the smallest positive root of $\sum_{n=1}^N n\cos nt=0$, that is of $(N+1)\cos Nt-N\cos(N+1)t=1$, so that $$\left(N+1-N\cos\frac{\lambda_N}N\right)\cos\lambda_N+N\sin\frac{\lambda_N}N\sin\lambda_N=1,$$ which gives $\cos\lambda+\lambda\sin\lambda=1$ after $N\to\infty$. And then $$L=\lim_{N\to\infty}\frac{\cos\frac{t_N}2-\cos\left(N+\frac12\right)t_N}{N\sin\frac{t_N}2}=\frac{1-\cos\lambda}{\lambda/2}=2\sin\lambda.$$

Answer 3

Some thoughts:

As already noted, the maximum of $\sum_{n=1}^N |1 - z^n|$ is attained on $|z| = 1$.

Let $z = \mathrm{e}^{\mathrm{i}\theta}$ with $\theta \in [0, 2\pi]$. We have $$\sum_{n=1}^N |1 - z^n| = \sum_{n=1}^N 2\left|\sin \frac{n\theta}{2}\right|.$$

Conjecture 1: The maximum of $\sum_{n=1}^N 2\left|\sin \frac{n\theta}{2}\right|$ on $[0, 2\pi]$ is attained at some $x_0 \in [0, 2\pi/N]$. (Note: Numerical experiment supports the claim. )

If Conjecture 1 is true, when $\theta \in [0, 2\pi/N]$, we have $$ \sum_{n=1}^N 2\left|\sin \frac{n\theta}{2}\right| = \sum_{n=1}^N 2\sin \frac{n\theta}{2} = \frac{2\sin^2 \frac{N\theta }{4} \cos \frac{\theta}{4}}{\sin \frac{\theta}{4}} + \sin \frac{N\theta}{2} = \frac{2\sin^2 y \cos \frac{y}{N}}{\sin \frac{y}{N}} + \sin 2y $$ where $y = N\theta/4 \in [0, \pi/2]$. Then we have $$\frac{2\sin^2 y \cos \frac{y}{N}}{\sin \frac{y}{N}} + \sin 2y \le \frac{2\sin^2 y}{\frac{2y}{\pi}\sin \frac{\pi}{2N}} + 1 = \frac{\pi}{\sin \frac{\pi}{2N}}\frac{\sin^2 y}{y} + 1 \le \frac{\pi}{\sin \frac{\pi}{2N}}\, \frac{29}{40} + 1$$ where we have used $\sin \frac{y}{N} \ge \frac{2y}{\pi}\sin \frac{\pi}{2N} $ and $\frac{\sin^2 y}{y} \le \frac{29}{40}$ for all $y\in [0, \pi/2]$. It is not difficult to prove that, for all $N > 21$, $$\frac{\pi}{\sin \frac{\pi}{2N}}\, \frac{29}{40} + 1 < \frac{3}{2}N.$$

Answer 4

This is not an an exact answer: I just want to present a physical-geometrical approach to the problem which can help to "visualize" and guide to a rigorous mathematical solution.

Please allow me to slightly change the notation and use a more practical small $n$ instead of $N$.
So let's consider $$ \left\{ \matrix{ S_n = \sum\limits_{1 \le k \le n} {\left| {1 - z^k } \right|} = \sum\limits_{1 \le k \le n} {\left| {z^k - 1} \right|} = \sum\limits_{0 \le k \le n} {\left| {z^k - 1} \right|} \hfill \cr z = re^{i\alpha } \quad \left| {\;r \le 1} \right. \hfill \cr} \right. $$

Now, we can look at $\left| {w - 1} \right|$ as being the distance from $U=(1,0) = 1+i0$, and thus the first moment of a unit mass placed at $w$ with respect to $U$.
Then $$ \sum\limits_{0 \le k \le n} {\left| {e^{ik\alpha } - 1} \right|} $$ is the sum of the moments of $n+1$ unit masses, placed on the unit circle at constant angular separation $\alpha$ with the first being at $U$.
If $r < 1$ the $n+1$ unit masses will be placed instead on the log-spiral an starting from $U$. $$ \rho = \left| {r^k e^{ik\alpha } } \right| = \left| {r^{\left( {k\alpha } \right)/\alpha } e^{i\left( {k\alpha } \right)} } \right| = r^{\theta /\alpha } = e^{{{\ln r} \over \alpha }\theta } = e^{\ln r\;k} $$ again with constant angular separation, $\alpha$ or better $k$ if we eliminate $\alpha$ from the constant at the exponent, and again starting from $U$ and proceeding towards the origin.

Therefore $S_n$ is the moment of a mass $n+1$ placed at the barycenter of the configuration of the single masses.

It becomes then geometrically clear that

• For large $n$ and $r=1$ the largest distance of the baricenter from $U$ is attained when most of the points are lying in the 2nd and 3rd quadrant; this means to correspondingly decrease $\alpha$ so that the almost continuous circular arc is about $270^{circ}$.
• In the limit, for $n \to \infty$, $S_n /(n+1)$ will tend to the barycenter of an arc of unitary mass of which to maximize the barycenter distance.
• For a spiral configuration, also the barycenter distance will be maximized when most of the points are in the 2nd and 3rd quadrant
• It is also geometrically "visible" that given a maximized spiral configuration, we can radially expand it to the circle, thereby increasing the distance from $U$ except for the points close to that.