Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Prove that $f(z)=\sum_{n=1}^\infty\frac{\sin(nz)}{2^n}$ is analytic on $A=\{z\in\mathbb{C}:|\text{Im}(z)|<\log(2)\}$

I tried expanding $\sin(nz)$ in terms of $e^{inz}$ but that did not help me unless I am doing something wrong. I know Wierstrass's M-Test comes in to play.

share|improve this question
    
Using $\frac{1}{2i}(e^{inz}-e^{-inz})$ works. The constant is irrelevant. Do the two exponentials separately. Let $z=x+iy$. Then $|e^{inz}|=e^{-ny}$. Only bad if $y$ negative, large. OK if $e^{-y}/2 \lt 1$, becuse of $M$-test. For the $e^{-iz}$ part, looking at $e^{y}/2$. –  André Nicolas Jun 29 '12 at 20:59
    
could you give me more hints? –  john Jun 29 '12 at 21:10
    
The norm of $(n+1)$-th term over $n$-th term is $e^{-y}/2$. This is $\lt 1$ if $y \ge 0$, or if $y \lt 0$ but $e^{-y}<2$, meaning that $y \gt -\log 2$. For the other term, we want $e^y \lt 2$, meaning $y \lt \log 2$. For both to be OK, want $-\log 2\lt y\lt \log 2$. –  André Nicolas Jun 29 '12 at 21:15

3 Answers 3

up vote 2 down vote accepted

Using $\sin nz=\frac{1}{2i}(e^{inz}-e^{-inz})$ works. The constant is irrelevant for the convergence.

Deal with the two exponentials separately. Let $z=x+iy$. Then $|e^{inz}|=e^{-ny}$, and $|e^{i((n+1)z}|=e^{-(n+1)y}$.

Thus, remembering about the $2^n$ in the denominator, we see that the norm of the ratio of two consecutive terms is $\frac{e^{-y}}{2}$. This norm is $\lt 1$ precisely if $e^{-y} \lt 2$, that is, if $y\gt -\log 2$.

In the same way, for the term in $e^{-iz}$, the norm of the ratio of two consecutive terms is $\lt 1$ precisely if $y \lt \log 2$. Thus, by the Weierstrass $M$-test, we have analyticity if $-\log 2\lt y\lt \log 2$.

share|improve this answer
    
Since the OP referred to Weierstrass $M$-test, I assumed the OP knew that in our region we have uniform convergence of a series of analytic functions. –  André Nicolas Jun 29 '12 at 21:44

We have $$ f(z)=\frac{1}{2i}\sum_{n=0}^\infty\frac{e^{inz}-e^{-inz}}{2^n} =\frac{1}{2i}\sum_{n=0}^\infty\left[\left(\frac{e^{iz}}{2}\right)^n-\left(\frac{e^{-iz}}{2}\right)^n\right] $$ If $|e^{\pm iz}/2|=\frac{1}{2}e^{\mp\text{Im}(z)}<1$, i.e. $\mp\text{Im}(z)<\log2$ or $|\text{Im}(z)|<\log2$ then the series converges and \begin{eqnarray} f(z)&=&\frac{1}{2i}\left[\frac{e^{iz}}{2}\frac{1}{1-e^{iz}/2}-\frac{e^{-iz}}{2}\frac{1}{1-e^{-iz}/2}\right]\cr &=&\frac{1}{2i}\left(\frac{e^{iz}}{2-e^{iz}}-\frac{e^{-iz}}{2-e^{-iz}}\right)\cr &=&\frac{2e^{iz}-1-2e^{-iz}+1}{2i(4-2e^{iz}-2e^{-iz}+1)}\cr &=&\frac{4i\sin z}{2i(5-4\cos z)}=\frac{2\sin z}{5-4\cos z}. \end{eqnarray} This function is obviously analytic!

share|improve this answer

Fix $\epsilon>0$.

If $|\Im(z)|<\log(2-\epsilon)$, then $$|\sin(nz)|=\left |\frac{e^{inz}-e^{-inz}}{2i}\right |\leq\frac{1}{2}\left (|e^{inz}|+|e^{-inz}|\right )=\frac{1}{2}\left (e^{-n\Im(z)}+e^{n\Im(z)}\right )<(2-\epsilon)^n,$$

so that the series converges uniformly on $A_\epsilon=\{z\in\mathbb{C}:|\text{Im}(z)|<\log(2-\epsilon)\}$ by the M-test.

Since the result of differentiating the series in the definition of $f$ term-by-term is $\sum_{n=1}^\infty\frac{n\sin(nz)}{2^n}$ which converges absolutely on $A_\epsilon$, it follows that differentiating term-by-term is legitimate and that $f$ is complex differentiable on $A_\epsilon$.

Since $\epsilon$ was arbitrary, we conclude that $f$ is complex differentiable on $A$. In particular, $f$ is holomorphic and hence analytic on $A$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.