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Starting with $f(x) = sin x$ centered at $a = \frac{\pi}{6}$, I got the following Taylor Series expansion: $$\sin x = \frac{1}{2} + \frac{\sqrt{3}}{2} \Big(x - \frac{\pi}{6}\Big) - \frac{1}{4} \Big(x - \frac{\pi}{6}\Big)^2 - \frac{\sqrt{3}}{12} \Big(x - \frac{\pi}{6}\Big)^3 + \ldots$$

My goal here is to write the power-series representation and then use the ratio test to find the radius of convergence. It seems that each term has a factor of $\frac{1}{2}$, $\frac{1}{n!}$ starting at $n = 0$, and $\Big(x - \frac{\pi}{6}\Big)^n$ starting at $n = 0$. From here, I haven't been able to figure out how to deal with the alternating signs for every third term or the alternating $\sqrt{3}$ (since we swap back and forth from $\sin$ to $\cos$ if we continuously take derivatives.)

I also tried writing $\sin x$ in the general form of a power series, $\frac{1}{1-r}$ (using $r$ in place of $x$, though it's usually writing with $x$), but then I ran into the issue of not knowing what to do with $\sin x$.

I'd appreciate any help anyone could offer.

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  • $\begingroup$ Are you looking for an explicit expansion of $\sin x$ near $x=\frac \pi6$ or are you chiefly interested to prove that the sought expansion has an infinite radius of convergence? $\endgroup$ Jan 2, 2018 at 8:01
  • $\begingroup$ I actually didn't know what the radius of convergence was and was trying to find this. I computed the Taylor expansion of this, as above, as the first step of the problem, but this didn't seem to lend itself to any clear conclusions on convergence. $\endgroup$
    – user465188
    Jan 2, 2018 at 8:03

1 Answer 1

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Hint: The series will be $$\frac{1}{2}\sum_{n=0}^\infty\frac{(-1)^n(x-\pi/6)^{2n+1}}{(2n+1)!}+\frac{\sqrt 3}{2}\sum_{n=0}^\infty\frac{(-1)^n(x-\pi/6)^{2n}}{(2n)!}.$$

This can be obtained by the Maclaurin series of $\sin x$ and $\cos x$. If we let $t=x-\pi/6$, then $\sin x=\sin (t+\pi/6)=\frac{1}{2}\sin t+\frac{\sqrt 3}{2}\cos t$. Since the Maclaurin series of $\sin t$ and $\cos t$ are known we can find it and then put $t=x-\pi/6$. It is also known that the radius of convergence of both series are infinity.

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