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Let $\{z_k^0\}_{k=1}^n$ be $n$ points in the complex plane and consider the following iteration: $$z_k^{m+1}=\frac 1 2 (z_k^m +z_{k+1}^m)$$ for $k=1,2,3,\ldots,n-1$.

$$z_k^{m+1}=\frac 1 2 (z_n^m+z_1^m)$$ for $k=n$.

Show that $\lim_{m \rightarrow \infty} z_k^m = \frac 1 n \sum_{j=1}^n z_j^0$.

I tried to iterate the iteration but I can't see the magic in there; besides, when I iterated, I got the coefficients like $\frac 1 {2^n}$, rather than $\frac 1 n$.

Thanks.

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    Are the exponents powers or indices? – 2017-01-11
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    Observe that in the recurrence, the sum of the coefficients does not change. Therefore, you can't get coefficients like $\frac{1}{2^n}$ because those don't sum to $1$. – 2017-01-11
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    When you write \Sigma_{j=1}^n you see this: $$\Sigma_{j=1}^n$$ But when you write \sum_{j=1}^n you see this: $$\sum_{j=1}^n$$ In an inline rather than displayed context with the latter you see this: $\sum_{j=1}^n.$ Standard usage for this purpose is \sum, and I edited accrodingly. – 2017-01-11
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    @MichaelBurM They are indices. – 2017-01-12
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    @MichaelHardM Thank you! Always forget that. – 2017-01-12

2 Answers 2

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Sketch:

  1. Consider $z_k^m$ as a sum of $z_i^0$'s. In particular, $$ z_k^m=\sum a_i z_i^0. $$ The $a_i$'s are nonnegative and sum to $1$ since every expansion preserves the sum of the coefficients.

  2. Consider $z_{k+1}^m$ as a sum of $z_i^0$'s (or $z_1^m$ if $k=n$). In particular, $$ z_{k+1}^m=\sum b_i z_i^0. $$ Then, $a_i=b_{i+1}$ for $i

  3. Claim: Let $z_k^{m+1}$ be written as a sum of $z_i^0$'s as $$ z_k^{m+1}=\sum c_iz_i^0. $$ If the $a_i$'s are not all zero, then either $\max\{a_i\}>\max\{c_i\}$ or the number of $a_i$'s which attain the maximum is greater than the number of $c_i$'s which attain the maximum.

  4. I'm going to get the indices wrong, but $a_i$ is something like $$ \frac{1}{2^m}\sum_{\substack{j=0\\j\equiv i+k\pmod n}}^m\binom{m}{j}. $$ You can use this to add coefficients together (along with binomial rules).

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The denominator in question is indeed $n$. The iteration can be rewritten as $\mathbf z^{m+1}=C\mathbf z^m$, where $$ C=\pmatrix{\frac12&\frac12\\ &\ddots&\ddots\\ &&\frac12&\frac12\\ \frac12&&&\frac12}. $$ Since $C$ is primitive and doubly stochastic, $\lim_{m\to\infty}C^m$ exists and the limit is a doubly stochastic matrix with identical rows and identical columns. Hence $\lim_{m\to\infty}C^m=\frac1n\mathbf1\mathbf1^T$ (where $\mathbf1$ is all-one vector) and $\lim_{m\to\infty}\mathbf z^m=\lim_{m\to\infty}C^m\mathbf z^0=\frac1n\mathbf1\mathbf1^T\mathbf z^0=\frac{\sum_k z_k^0}{n}\mathbf 1$.

Alternatively, you may explicitly calculate $C^m$ using the properties of circulant matries.