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I am trying to find the radius of convergence for the following function

$$ f(x)=\sin(\pi x/4)$$

I already found the Maclaurin series of the function and applied the ratio test but seems I cant get the radius of convergence right. I find the radius

$R = 4/\pi$

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If you want somebody to find the mistake, you need to show the work that might contain the mistake. Often it is hard to guess just from the result. –  Ross Millikan Apr 23 '12 at 2:48
2  
Perhaps you forgot the $1/n!$ factor in the coefficients of the Maclaurin series? –  anon Apr 23 '12 at 2:51
    
excuse me if I am stupid, do you mean $f(x)$ as a complex function? a real function $\sin[x]$ always converges.. –  Kerry Apr 23 '12 at 2:56
    
I found the Maclaurin series /$sum (-1)^n * pi^(2n+1)/(2n+1)! * x^(2n+1)/4^(2n+1) How do i find the Radius of convergence now? –  Kleon Kita Apr 23 '12 at 2:58
1  
The sine function is analytic. –  Neal Apr 23 '12 at 3:07
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2 Answers

You have that the derivatives around $x=0$ of $y=\sin x$ are

$$\{ y^{(n)}(0)\}=\{0,1,0,-1 ,\dots \}$$

Since you have a multiplicative factor of $\dfrac{\pi }{4}$ this changes to

$$\{ y^{(n)}(0)\}=\left\{0,\dfrac{\pi }{4},0,-\dfrac{\pi^3 }{4^3} ,\dots \right\}$$

As a consequence you have that the coefficients are

$$c_n=\frac{(-1)^n}{(2n+1)!} \left(\frac{\pi}{4}\right)^{2n+1}$$

You can readily check that

$$\lim \frac{c_{n+1}}{c_{n}}=0$$

from which the radius is $\infty$, i.e., the whole extended real line.

As a general result, the convergence radius of the series for

$$y=\sin(ax+b)$$

is the whole real line.

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Thank man i just found it. I was calculating it the wrong way. –  Kleon Kita Apr 23 '12 at 3:12
    
@David Sorry, It was a typo. I'll fix it. –  Pedro Tamaroff Apr 23 '12 at 16:17
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The radius of convergence for a complex function is defined by $R=a_{n}^{-1/n}$ in the Taylor series. In your case $a_{n}$ is $(-1)^{k}*[\frac{\pi}{2}]^{2k+1}*\frac{1}{(2k+1)!}$. It is obvious that $R$ should be infinite.

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