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For a complex sequence $\{z_n\} $ defined by the recurrence relation $$z_{n+1}=\frac12 \Bigg(z_n +\frac{1}{z_n} \Bigg) $$ where $\arg z_n \in (-\frac{\pi}{2},\frac{\pi}{2})$, prove that it converges to $1$.

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    $\begingroup$ What's $z{{}}$? $\endgroup$ – Git Gud Jun 29 '13 at 13:30
  • $\begingroup$ Should that just be $\lim z_n = 1$? It's the same thing, but clearer. $\endgroup$ – Thomas Andrews Jun 29 '13 at 13:32
  • $\begingroup$ Presumably, instead of $z$ you meant $z_0$ $\endgroup$ – Thomas Andrews Jun 29 '13 at 13:34
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    $\begingroup$ @shimee Please don't use the question's body and the title's body as one. Write the whole question in question's body. $\endgroup$ – Git Gud Jun 29 '13 at 13:35
  • $\begingroup$ This is a particular example of Newton Raphson method to calculate square root of $A > 0$ by iteration $$x_{n + 1} = \frac{1}{2}\left(x_{n} + \frac{A}{x_{n}}\right)$$ where initial value $x_{0}$ can be chosen as any positive number. In this question $A = 1$ so that the sequence converges to $\sqrt{1} = 1$. $\endgroup$ – Paramanand Singh Jul 7 '13 at 9:40
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From the recursion relation we get, $$\left(\frac{z_{n+1}+1}{z_{n+1}-1}\right)=\left(\frac{z_{n}+1}{z_{n}-1}\right)^2, \forall n\geq1 $$ So iteratively, we get, $$\left(\frac{z_{n+1}+1}{z_{n+1}-1}\right)=\left(\frac{z_{0}+1}{z_{0}-1}\right)^{2^{n+1}}$$ Rearranging, we get, $$z_{n+1}=\frac{1+w^{2^{n+1}}}{1-w^{2^{n+1}}}$$ where $$w=\frac{z_0-1}{z_0+1}$$ Clearly, $|w|<1$ and $\arg{w}\in[-\pi/2, \pi/2]$ (I'm omitting the proofs but they are not difficult). Hence $$\lim_{n\rightarrow \infty}w^{2^{n+1}}=0 e^{j0}=0 $$ Thus $$\lim_{n\rightarrow \infty}z_n=1 \hspace{0.6cm}\Box$$

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  • $\begingroup$ Very beautiful solution. I think when $\arg{z_0}\in[-\pi/2, \pi/2]$ then $|w|<1$ is evident since $z_0$ is closer to $1$ than $-1$ and $|z_0-1|<|z_0+1|$. Here $\arg{w}\in[-\pi/2, \pi/2]$ is not necessary. $|w|<1$ is enough. $\endgroup$ – Arashium Feb 3 '15 at 6:52
  • $\begingroup$ Yes @Arashium, $\arg w$ is not necessary here, I just mentioned that for completeness. $\endgroup$ – Samrat Mukhopadhyay Feb 3 '15 at 9:07

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