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I want to show that the power series

$$\sum_{n = 0}^{\infty} a_{n}x^{n} = \frac{1}{3 - \cos(x)}$$

has a radius $R > 1$? I don't think the ratio test will work here. From Googling, I found the Cauchy-Hadamard Theorem, and I was wondering if I could somehow apply it to this. Can someone please help me? I'm not so familiar with $\limsup$.

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This can be seen as a heuristic way which may not be necessarily correct.

There is a slightly easier way using the expansion for $\dfrac{1}{1-x}$.

$\dfrac{1}{3 - \cos(x)}=\dfrac{1}{3}\left(\dfrac{1}{1-\dfrac{\cos(x)}{3}}\right)=\dfrac{1}{3}\displaystyle\sum_{n=0}^\infty\left(\dfrac{\cos(x)}{3}\right)^n$ exists if $\left|\dfrac{\cos(x)}{3}\right|<1 \implies |\cos(x)|<3$ which is true for all $x\in\mathbb{R}$.

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To find the power series we calculate using Taylor series

$f(0)=\dfrac{1}{2}, \: f'(x)=\dfrac{-\sin x}{(3-\cos x)^2}\implies f'(0)=0, \: \\f''(x)=\dfrac{(3-\cos x)(-\cos x)+2\sin^2 x(3-\cos x)}{(3-\cos x)^4}\implies f''(0)=\dfrac{1}{4}$.

Continuing this way we get $\quad$ $\dfrac{1}{3 - \cos(x)}=\dfrac{1}{2}-\dfrac{x^2}{8}+\dfrac{x^4}{24}+\cdots$

$\cos(z)=\dfrac{e^{iz}+e^{-iz}}{2}=3\implies e^{2iz}-6e^{iz}+1=0 \implies e^{iz}=\dfrac{6\pm\sqrt{36-4}}{2}=3\pm2\sqrt{2} $ $\implies z=2\pi n -i\ln(3\pm2\sqrt{2}),\: n\in\mathbb{Z}$.

Now since radius of curvature is defined such that the function is analytic in $\{z: |z - z_0| < r\}$, so the radius of convergence is at least $r$. We infer that if $f$ has a pole at $c$ then it is analytic for $|z-z_o|<c$, so radius of curvature is exactly the distance to the closest pole.

Thus $|z|=|\ln(3+2\sqrt{2})|$. Thus we see that $R=|\ln(3+2\sqrt{2})|>1$.

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  • $\begingroup$ that is very clever. $\endgroup$ – joseph Nov 16 '18 at 13:21
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    $\begingroup$ I don't think this works. The power series of $1/(3- \cos x)$ is not $\frac{1}{3} \sum_n (\frac{\cos x}{3})^n$, because it is not a power series. If you expand it, it does not converge uniformly on the whole real line. Actually the power series has a finite radius of convergence. That's because otherwise it would converge uniformly on the whole $\Bbb C$: but there is the problem that on complex numbers there is a solution for $\cos z=3$. $\endgroup$ – Crostul Nov 16 '18 at 13:26
  • $\begingroup$ @Crostul: Isnt this true for $x\in\mathbb{R}$ ? How does the power series look like in the case $x\in\mathbb{C}$? $\endgroup$ – Yadati Kiran Nov 16 '18 at 13:31
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    $\begingroup$ The radius of convergence of that power series is bounded by $\ln (3+2\sqrt 2) = 1.76...$. I think that is exactly the radius of convergence, though I'm not sure. You can see it graphically at wolframalpha.com/input/?i=power+series+of+(3-cos+z)%5E(-1) . $\endgroup$ – Crostul Nov 16 '18 at 13:33
  • $\begingroup$ @Crostul: I shall check in the case for $x\in\mathbb{C}$. But what about when $x\in\mathbb{R}$ $\endgroup$ – Yadati Kiran Nov 16 '18 at 13:38
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The poles of the function occur at

$$\cos z=3$$ or $$z=\pm i\text{ arcosh}(3)+2k\pi=\pm i\log(3+\sqrt8)+2k\pi.$$

As the radius of convergence is the distance to the closest pole,

$$R=\log(3+\sqrt8)\approx1.76274717.$$

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You may also invoke Cauchy's integral theorem. $f(z)=\frac{1}{3-\cos(z)}$ is holomorphic over $|z|\leq\frac{3}{2}$, since

$$ \left|3-\cos(z)\right|\geq 2-\sum_{n\geq 1}\frac{|z|^{2n}}{(2n)!}=3-\cosh(|z|) $$ and $K=3-\cosh\left(\tfrac{3}{2}\right)>0$. In particular

$$ |a_n|=\left|\frac{1}{2\pi i}\oint_{|z|=\frac{3}{2}}\frac{dz}{z^{n+1}(3-\cos z)}\right|\leq \frac{1}{K\left(\frac{3}{2}\right)^n} $$ by the triangle inequality, and the radius of convergence of $\sum a_n z^n$ is at least $\frac{3}{2}$.
The same argument works also by replacing $\frac{3}{2}$ with $\frac{7}{4}$, since $\cosh\frac{7}{4}$ still is less than three.

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