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Find the sum of $\sum_{n=1}^{\infty} \dfrac{x^{2n}}{(2n)!}$ on its interval of convergence.

We can see that the domain of convergence is $D=R$. Then let:

$$f(x)=\sum_{n=1}^{\infty} \dfrac{x^{2n}}{(2n)!}$$

$$f'(x)=\sum_{n=1}^{\infty} \dfrac{x^{2n-1}}{(2n-1)!}$$

$$f''(x)=\sum_{n=1}^{\infty} \dfrac{x^{2n-2}}{(2n-2)!}$$ Thus $f''(x)=f(x)$, solve this differential equation, we'll get the solution.

Is my solution right? I just begin to study the power series. Thank you so much.

Martin Sleziak
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anvo
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    It's right, and you can directly say that $f=\cosh-1$ – Aforest Oct 23 '16 at 02:25
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    That solution will work, but it is incomplete. You have found a differential equation that $f$ satisfies, but the differential equation has other solutions too. You need to be able to solve the equation and find a way to isolate which solution is $f$. – Aaron Oct 23 '16 at 02:25
  • Oh I see, thank you so much :D – anvo Oct 23 '16 at 02:25
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    Search relevant posts here: https://approach0.xyz/search/?q=%24%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%20%5Cdfrac%7Bx%5E%7B2n%7D%7D%7B(2n)!%7D%24&p=1 – Wei Zhong Oct 23 '16 at 03:46
  • Related: See the comments here: http://math.stackexchange.com/questions/12340/what-is-limit-of-sum-limits-n-0-infty-frac12n/12341#12341 – Hans Lundmark Oct 23 '16 at 09:13

5 Answers5

3

Hint:

$$\cosh x=\sum _{n=0}^{\infty }{\frac {x^{2n}}{(2n)!}}=1+\sum _{n=1}^{\infty }{\frac {x^{2n}}{(2n)!}} $$

Community
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E.H.E
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hint: Consider $f(x) = \dfrac{e^x+ e^{-x}}{2}$

Martin Sleziak
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DeepSea
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Note that your equation should be $f''(x)=f(x)+1$.

Using the series for $e^x$, we get $$ \begin{align} e^x&=\sum_{k=0}^\infty\frac{x^k}{k!}\tag{1}\\ e^{-x}&=\sum_{k=0}^\infty(-1)^k\frac{x^k}{k!}\tag{2} \end{align} $$ Average $(1)$ and $(2)$ $$ \frac{e^x+e^{-x}}2=\sum_{k=0}^\infty\frac{x^{2k}}{(2k)!}\tag{3} $$ Subtract $1$ $$ \frac{e^x-2+e^{-x}}2=\sum_{k=1}^\infty\frac{x^{2k}}{(2k)!}\tag{4} $$ $(4)$ can be written as $\cosh(x)-1$.

robjohn
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $$\bbox[10px,#ffe,border:1px dotted navy]{\ds{% \mbox{Besides your 'differential equation technique', you can perform a direct evaluation as follows:}}} $$

\begin{align} \sum_{n = 1}^{\infty}{x^{2n} \over \pars{2n}!} & = \sum_{n = 2}^{\infty}{x^{n} \over n!}\,{1 + \pars{-1}^{n} \over 2} = {1 \over 2}\pars{\sum_{n = 0}^{\infty}{x^{n} \over n!} - 1 - x} + {1 \over 2}\pars{\sum_{n = 0}^{\infty}{\pars{-x}^{n} \over n!} - 1 + x} \\[5mm] & = -1 + {1 \over 2}\,\expo{x} + {1 \over 2}\,\expo{-x} =\ \bbox[#ffe,10px,border:1px dotted navy]{\ds{\cosh\pars{x} - 1}} \end{align}

Felix Marin
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Let $\omega=\exp\left(\pi i\right) $. Then $$e^{\omega x}=\sum_{n\geq0}\frac{x^{n}\omega^{n}}{n!}=\sum_{n\geq0}\frac{x^{2n}}{\left(2n\right)!}-\sum_{n\geq0}\frac{x^{2n+1}}{\left(2n+1\right)!} $$ and $$e^{-\omega x}=\sum_{n\geq0}\frac{x^{n}\left(-\omega\right)^{n}}{n!}=\sum_{n\geq0}\frac{x^{2n}}{\left(2n\right)!}+\sum_{n\geq0}\frac{x^{2n+1}}{\left(2n+1\right)!} $$ hence $$\sum_{n\geq0}\frac{x^{2n}}{\left(2n\right)!}=\frac{e^{\omega x}+e^{-\omega x}}{2}=\frac{e^{-x}+e^{x}}{2}=\color{red}{\cosh\left(x\right)}.$$

Marco Cantarini
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