# Power Series, Taylor/Maclaurin and n-th derivative

a) Find the convergence radius of the power series $$\sum_\limits{n=1}^∞\frac{(n+1)^{n^2}}{3^nn^{n^2}}x^n$$

b)Find all the $$x\in \mathbb{R}$$ for which the power series $$\sum_\limits{n=1}^∞\frac{(x+1)^n}{\sqrt{4^nn}}$$ converge.

c)Using the Maclaurin expansion of $$e^x$$, find the power series of $$g(x)=e^{5x^2}$$ with center $$x_0=0$$. Also for every $$n\in \mathbb{Z}$$ find a formula for $$g^{(n)}(0)$$.

My work so far:

a) I found $$R=\frac{3}{e}$$ with $$R={(\limsup_{n\rightarrow∞}\sqrt[n]{|a_n|})}^{-1}$$.

b) From the ratio test and $$n\rightarrow ∞$$ we get $$|(x+1)\frac{1}{2}|$$. So the power series converge for $$|(x+1)\frac{1}{2}|<1$$. Thus $$-3 is the interval of convergence of the power series.

c) The power series of $$g(x)$$ using the Maclaurin expansion of $$e^x$$ is $${\sum_\limits{n=0}^ ∞}\frac{5^n}{n!}x^{2n}$$

Is my work correct so far? I don't know what to do for the n-th derivative of $$g(x)$$. Any help would be appreciated.

The Maclaurin series for $f$ is: $$f(x)=\sum\limits_{n \in \mathbb{N}} \left.\frac{\mathrm{d}^nf(x)}{\mathrm{d}x^n}\right|_{x=0}\frac{x^n}{n!}$$ I think this will be enough for you to finish the excercise c.

So, the first few terms of the Maclaurin series are: $$g^{(0)}(0)\frac{x^0}{0!}+g^{(1)}(0)\frac{x^1}{1!}+g^{(2)}(0)\frac{x^2}{2!}+\dots$$ $$g^{(0)}(0)+g^{(1)}(0)x+g^{(2)}(0)\frac{x^2}{2}+\dots$$ While your function is: $$5^0\frac{x^{2*0}}{0!}+5^1\frac{x^{2*1}}{1!}+5^2\frac{x^{2*2}}{2!}+\dots$$ $$1+5x^2+\frac{25}{2}x^4+\dots$$

• If i get this right ${\sum_{n=0}^ ∞}\frac{5^n}{n!}x^{2n}={\sum_{n=0}^∞}\frac{\text{d}^nf(x)}{\text{d}x^n}|_{x=0}\frac{x^n}{n}$, thus $f^{(n)}(0)=5^nx^n$ ??? Mar 7, 2018 at 20:45
• Almost! Why don't you try to write out the first few terms of both sums? Mar 7, 2018 at 20:46
• I edited the convergence radius i forgot to write $(x+1)\frac{1}{2}$. Mar 7, 2018 at 20:48
• I'm gonna remove that from my answer then. And also, the left side ($f^{(n)}(0)$) is not a function of $x$, but the right side ($5^n x^n$) is, and that's a problem. I think you will notice the pattern by writing out the terms, maybe up to $n=2$. Mar 7, 2018 at 20:50
• I get the same result $f^{(n)}(x)=5^nx^n$. i'm sorry if it's obvious and i don't get it. Mar 7, 2018 at 20:59

Finding the n-th derivative of $g$ is easy, since you have already found the Maclaurin expansion:

• Since $g$ is an even function, $g^{(n)}(0)=0$, if $n$ is odd.
• From the Maclaurin expansion, if $n$ is even:
• \begin{align*} g(x) =\sum_{n=0}^{\infty}\frac{g^{(n)}(0)}{n!}x^n =\sum_{v=0}^\infty \frac{5^v}{v!}x^{2v}\Rightarrow x^n=x^{2v}\Rightarrow n=2v \end{align*}
• \begin{align*} g(x) =\sum_{n=0}^{\infty}\frac{g^{(n)}(0)}{n!}x^n =\sum_{v=0}^\infty\frac{5^v}{v!}x^{2v}\Rightarrow \frac{g^{(n)}(0)}{n!}=\frac{5^v}{v!}\Rightarrow g^{(n)}(0) =\frac{5^vn!}{v!} =\frac{\sqrt{5^n}n!}{\left(\frac{n}{2}\right)!} \end{align*}

A piecewise definition of the n-th derivative of $g$ would be: $$g^{(n)}(0)=\begin{cases}\frac{\sqrt{5^n}n!}{\left(\frac{n}{2}\right)!}&n \to even\\ 0&n \to odd\end{cases}$$