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First, we have $f(x) = \sin x$ and want to write the taylor series centered at $a = \frac{\pi}{6}$. I found the first several terms of this to be \begin{align*} \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 \end{align*} I then wanted to write a power-series representation and determine the radius of convergence. After struggling with this for quite some bit, I ended up writing this as the sum of two power series. One dealt with every odd value of $n$ and one with every even value of $n$. For each even value, I got \begin{align*} \sum\limits_{n=0}^{\infty} \frac{(-1)^n \Big(x - \frac{\pi}{6}\Big)^{2n}}{2 \cdot (n+1)!} \end{align*} This will generate $\frac{\sqrt{3}}{2} \Big(x - \frac{\pi}{6}\Big), - \frac{\sqrt{3}}{12} \Big(x - \frac{\pi}{6}\Big)^3, \ldots$ and so on. For each odd value, I get \begin{align*} \sum\limits_{n=0}^{\infty} \frac{(-1)\sqrt{3} \Big(x - \frac{\pi}{6}\Big)^{2n+1}}{2\cdot (2n+1)!} \end{align*} This generates $\frac{1}{2}, - \frac{1}{4} \Big(x - \frac{\pi}{6}\Big)^2, \ldots$ and so on.

I tried combining these and running the ratio test, but the expression just doesn't seem workable. Another idea I've considered is running the ratio test on both of these series separately -- both of which should have a radius of convergence of $\infty$ -- and then concluding that the the entire Taylor series has a radius of convergence of $\infty$.

My main question is whether this above step of logic -- check both series, and then make a conclusion as to the sum -- is valid. But if there is any other way to do this I would greatly appreciate it.

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  • $\begingroup$ You could add the terms of each sum together in parentheses, and put just one summation symbol in front of it. And that would converge because each term converges independently. So yes, this is a valid approach. $\endgroup$ – Myridium Jan 5 '18 at 4:35
  • $\begingroup$ If I understand correctly, what you're saying, it's that convergence of $\sum a_n$ and $\sum b_n$ implies that $\sum (a_n + b_n)$ converges, correct? But could I go a step further and say that because $\sum a_n$ and $\sum b_n$ have the same radius of convergence, $R = \infty$, that $\sum (a_n + b_n)$ has a radius of convergence of $R = \infty$? Now that I write it this way, it seems like it should follow just logically, but I'd be very curious if there are any formal theorems for this, just to be sure I'm not making a mistake. $\endgroup$ – Matt.P Jan 5 '18 at 4:49
  • $\begingroup$ Yes, that is absolutely correct. It is true that $\sum a_n + \sum b_n = \sum(a_n + b_n)$ when each sum converges. It is possible that the latter converges while the former diverge, but not the converse. If each individual sum is defined and converges, then the combined sum converges also, and to the same value. You know that the former always converge (i.e. radius of $\infty$), and so then does the latter. I'm sorry I don't know how to put any more rigour behind that. $\endgroup$ – Myridium Jan 5 '18 at 4:53
  • $\begingroup$ Excellent. Thank you very much. $\endgroup$ – Matt.P Jan 5 '18 at 4:54
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    $\begingroup$ No problem. To generalise, if the radius of convergence of $\sum a_n$ is $r_a$ and likewise for $r_b$, the the radius of convergence of the combined sum is at least $\mathrm{min}(r_a,r_b)$, though in general it may be larger. $\endgroup$ – Myridium Jan 5 '18 at 4:56

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