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I just found the following Taylor series expansions around $z=0$ for the following functions:

  • $\displaystyle \frac{1}{z^{2}-5z+6} = \frac{1}{(z-2)(z-3)} = \frac{-1}{(z-2)} + \frac{1}{(z-3)} = \sum_{n=0}^{\infty}\frac{z^{n}}{2^{n+1}} - \sum_{n=0}^{\infty}\frac{z^{n}}{3^{n+1}} = \sum_{n=0}^{\infty}\left( \frac{1}{2^{n+1}}-\frac{1}{3^{n+1}}\right)z^{n}$

  • $\displaystyle \frac{1}{1-z-z^{2}} = \frac{1}{\left[z-\left(\frac{1-\sqrt{5}}{-2} \right) \right]\left[z-\left(\frac{1+\sqrt{5}}{-2} \right)\right]} = \frac{2\sqrt{5}}{5(1-\sqrt{5})}\frac{1}{1-\left(\frac{-2}{1-\sqrt{5}} \right)z} - \frac{2\sqrt{5}}{5(1+\sqrt{5})}\frac{1}{1-\left(\frac{-2}{1+\sqrt{5}} \right)z} = \frac{\sqrt{5}}{5} \sum_{n=0}^{\infty}\left(\frac{(-1)^{n}2^{n+1}}{(1-\sqrt{5})^{n+1}} - \frac{(-1)^{n}2^{n+1}}{(1+\sqrt{5})^{n+1}} \right)z^{n}$

I wanted to confirm that the radius of convergence for the first series is $|z|<2$ and that the radius of convergence for the second series is $|z|<\frac{-1+\sqrt{5}}{2}$.

I know that for a series $\sum_{n=1}^{\infty}a_{n}z^{n}$ with radius of convergence $R_{1}$ and $\sum_{n=1}^{\infty}b_{n}z^{n}$ with radius of convergence $R_{2}$, the radius of convergence of the series $\sum_{n=1}^{\infty}(a_{n}+b_{n})z^{n}$ is some $R$ where $R \geq \min(R_{1}, R_{2})$. And it can even be the case that if both of the series have finite $R_{1}$ and $R_{2}$, that $R$ can be infinite!

So, I wanted to make sure this was not the case here, and that I have the correct radii of convergence for both. If not, how do I go about finding them (preferably in the least icky way possible)?

Thank you.

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You can use Hadamard's rule: $$\frac 1R=\limsup_n a_n^{1/n}=\limsup_n \biggl(\frac1{2^{n+1}}-\frac1{3^{n+1}}\biggr)^{\!1/n}\!.$$ Now rewrite this expression as $$\frac1{2^{(n+1)/n}}\biggl(1-\frac{2^{n+1}}{3^{n+1}}\biggr)^{\!1/n}=\frac1{2^{1+1/n}}\biggl(1-\Bigl(\frac23\Bigr)^{n+1}\biggr)^{1/n}\!.$$ The first factor tends to $\frac12$. As to the second factor, $$\frac23<1-\Bigl(\frac23\Bigr)^{n+1}<1,\enspace\text{whence}\quad\Bigl(\frac23\Bigr)^{1/n}<\biggl(1-\Bigl(\frac23\Bigr)^{n+1}\biggr)^{\!1/n}<1, $$ and by the squeeze principle, it tends to $1$. Thus $\limsup_n a_n^{1/n}=\frac12$, and $R=2$.

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  • $\begingroup$ that's very helpful for that one. But what about the ugly-looking one? $\endgroup$ – ALannister Apr 15 '16 at 0:42
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From the representation \begin{align*} \frac{1}{z^2-5z+6}&=\frac{1}{(z-2)(z-3)}\tag{1}\\ \frac{1}{1-z-z^2}&=\frac{1}{\left(z-\left(\frac{1-\sqrt{5}}{-2}\right)\right)\left(z-\left(\frac{1+\sqrt{5}}{-2}\right)\right)}\tag{2}\\ \end{align*} we can already deduce the radius of convergence.

Both expressions have two simple poles. Recall the radius of convergence is the distance from the center of the series expansion to the nearest singularity.

In case of (1) we have two simple poles at $z=2$ and $z=3$. So, the radius of convergence $R$ is $$R=\min\{2,3\}=2$$.

In case of (2) we have two simple poles at $z=\frac{1\pm\sqrt{5}}{2}$. So, the radius of convergence $R$ is

$$R=\min\left\{\left|\frac{1\pm\sqrt{5}}{2}\right|\right\}=-\frac{1+\sqrt{5}}{2}$$

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