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Find $f(x)$, the unknown function satisfying $$f(x) = \sum\limits_{n=0}^{+\infty}\frac{x^{4n}}{(4n)!}$$

I'm looking for a direct solution which is different from mine, if possible.

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@AndréNicolas Would you please explain further? I don't understand. Do you mean to plug $x=1,-1,i,i$ into the series? Or to plug in $x,-x,ix,-ix$? –  FrenzY DT. Oct 26 '12 at 14:30
    
@AndréNicolas Plugging in $x$, $-x$, $ix$, $-ix$ into $f$ yields $f(x)$, $f(x)$, $f(x)$, $f(x)$. –  Did Oct 26 '12 at 14:40
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Wrong $f$! Let $g(x)=e^x$. Write down the series for $g(x)$. Evaluate at $x$, $-x$, $ix$, $-ix$. Add up. The coefficients of $x^{4k+i}$ are $0$ for $i=1,2,3$ and $\frac{4}{(4k)!}$ for $i=0$. –  André Nicolas Oct 26 '12 at 14:46
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5 Answers 5

up vote 1 down vote accepted

By discrete Fourier transform.

Let $e(x)=\mathrm e^x=\displaystyle\sum\limits_{n\geqslant0}\frac{x^n}{n!}$. Note that, for every integer $n$, $\displaystyle\sum\limits_\zeta\zeta^n$ is $4$ when $n=0\pmod{4}$ and is $0$ otherwise, where the sum is over the set of roots $\{\zeta\in\mathbb C\mid \zeta^4=1\}=\{1,-1,\mathrm i,-\mathrm i\}$. Hence, $$ \sum\limits_\zeta e(\zeta x)=\sum\limits_{n\geqslant0}\frac{x^n}{n!}\cdot\sum\limits_\zeta\zeta^n=4\cdot\sum\limits_{n\geqslant0}\frac{x^{4n}}{(4n)!}. $$ In other words, $$ \sum\limits_{n\geqslant0}\frac{x^{4n}}{(4n)!}=\frac{e(x)+e(-x)+e(\mathrm ix)+e(-\mathrm ix)}4=\frac{\cosh(x)+\cos(x)}2. $$


Likewise, for every $k$ in $\{1,2,3\}$, $\displaystyle\sum\limits_\zeta\zeta^{n-k}$ is $4$ when $n=k\pmod{4}$ and is $0$ otherwise, hence $$ \sum\limits_\zeta\zeta^{-k}\cdot e(\zeta x)=\sum\limits_{n\geqslant0}\frac{x^n}{n!}\cdot\sum\limits_\zeta\zeta^{n-k}=4\cdot\sum\limits_{n\geqslant0}\frac{x^{4n+k}}{(4n+k)!}. $$ In other words, $$ \sum\limits_{n\geqslant0}\frac{x^{4n+k}}{(4n+k)!}=\frac{e(x)+(-1)^ke(-x)+(-\mathrm i)^{k}e(\mathrm ix)+\mathrm i^{k}e(-\mathrm ix)}4. $$ For example, $$ \sum\limits_{n\geqslant0}\frac{x^{4n+3}}{(4n+3)!}=\frac{\sinh(x)-\sin(x)}2. $$

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I'm not sure what this means: $\sum\limits_zz^n$. Is it $1^n+i^n+(-1)^n+(-i)^n ?$ –  FrenzY DT. Oct 26 '12 at 13:59
    
It is. $ $ $ $ $ $ –  Did Oct 26 '12 at 14:00
    
Original: Why $\sum_{n\ge 0}\frac{x^n}{n!}\cdot\sum_\zeta\zeta^n=4\cdot\sum\limits_{n\ge 0}\frac{x^{4n}}{(4n)!}$? Edit: Silly me! I forgot about the modulo argument! –  FrenzY DT. Oct 26 '12 at 14:14
    
Why is this answer related to DFT? –  FrenzY DT. Oct 26 '12 at 14:16
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Because the discrete Fourier transform of $(x_n)_{0\leqslant n\leqslant3}$ is $(y_k)_{0\leqslant k\leqslant3}$ defined by $y_k=\sum\limits_{n=0}^3x_n\zeta^{-kn}$. The beginning of the post is based on the fact that $(4,0,0,0)$ is the Fourier transform of $(1,1,1,1)$. –  Did Oct 26 '12 at 14:36
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This series converges on $\mathbb R$ because $\displaystyle \lim_{n\to+\infty}\frac{1}{(4n)!}^{\frac{1}{4n}} = \lim_{n\to+\infty}\frac{1}{\sqrt[n]{n!}}=0 $. By observation:

The sum of $\mathrm e^x$ and $\mathrm e^{-x}$ has no odd-powered terms: $\displaystyle \mathrm e^{x}+\mathrm e^{-x} = 2\cosh x = 2\sum\limits_{k=0}^{+\infty}\frac{x^{2n}}{(2n)!}$.

Also that we can cancel terms with power $4n-2$ with the help of: $\displaystyle \cos x = \sum\limits_{k=0}^{+\infty}(-1)^{k}\frac{x^{2n}}{(2n)!}$.

Thus, $\displaystyle \frac{1}{4}(\mathrm e^x+\mathrm e^{-x}+2\cos{x}) = \sum\limits_{k=0}^{+\infty}\frac{x^{4n}}{(4n)!} = f(x)$.

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I tested the convergence in the wrong way. I think a root test should be used instead. –  FrenzY DT. Oct 26 '12 at 13:27
    
The ratio test should suffice. –  Pedro Tamaroff Oct 26 '12 at 13:40
    
@PeterTamaroff Is it ok to ratio test even if there are missing terms? –  FrenzY DT. Oct 26 '12 at 13:41
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The fourth derivative of the function is the function itself. Then apply the well-known theory for solution of a linear ODE with constant coefficients. Of course you get the same answer as the other methods. But this method (find a DE it satisfies) can be applied in many cases where you want to sum a power series.

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Recalling the Laplace transform of a function $f$,

$$ F(s) = \int_{0}^{\infty} f(x) e^{-sx} dx \,, $$

and using the fact that the Laplace transform of $x^m$ is given by $\frac{\Gamma(m+1)}{x^{m+1}}$, we can compute the Laplace transform of $f(x)=\sum\limits_{n=0}^{+\infty}\frac{x^{4n}}{(4n)!}$ as

$$ F(s)=\sum_{k=0}^{\infty} \frac{1}{s^{4k+1}}=\frac{s^3}{s^4-1}$$

$$= \frac{1}{4(s-1)}+\frac{1}{4(s+1)}+\frac{1}{4(s+i)}+\frac{1}{4(s-i)}\rightarrow (1)\,. $$

Taking the inverse Laplace transform of $(1)$ yields,

$$ \frac{1}{2}\,\cos \left( x \right) +\frac{1}{2}\,\cosh \left( x \right) \,. $$

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I'm not familiar with Laplace transform. Would you please tell me how I could understand it intuitively? –  FrenzY DT. Oct 26 '12 at 14:50
    
@FrenzYDT: Laplace transform is a very useful technique and it has a lot of applications such as solving differential equations. You should practice it to learn it. –  Mhenni Benghorbal Oct 26 '12 at 15:41
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Mhenni: Seriously, do you consider that your comment answers @FrenzyDT's query? –  Did Oct 26 '12 at 18:01
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We know, $e^x= \sum\limits_{n=0}^{\infty}\frac{x^n}{n!}$

Let's consider, the equation, $y^r-1=0--->(1)$

So, the roots are $e^{\frac{2ki\pi}r}=y_k$(say) where $1\le k\le r$,

then $\sum y_k=0$

also, $\sum y_k^s=r$ if $r\mid s$, else $\sum y_k^s=0$ (find here)

As, $ e^{y_kx}= \sum\limits_{n=0}^{+\infty}\frac{(y_kx)^n}{n!}$

summing for $k=1$ to $r, \sum_{1\le k\le r} e^{y_kx}= \sum\limits_{n=0}^{+\infty}\frac{x^n}{n!}(\sum y_k^n)=n\sum\limits_{n=0}^{+\infty}\frac{x^{rn}}{(rn)!}$

$y_{r-k}=e^{\frac{2(r-k)\pi i}r}=e^{\frac{-2k\pi i}r}=y_{-k}$

If $r$ is even, $$\sum_{1\le k\le r} e^{y_kx} =e^x+e^{-x}+\sum_{1\le k\le \frac r2 -1}(e^{\frac{2kx\pi i}r}+e^{\frac{2(r-k)x\pi i}r}) =2\cosh x+\sum_{1\le k\le \frac r 2 - 1}(e^{\frac{2kx\pi i}r}+e^{\frac{-2kx\pi i}r}) =2\cosh x+2\sum_{1\le k\le \frac r 2 - 1}\cos \frac{2kx\pi }r$$

If $r$ is odd, $$\sum_{1\le k\le r} e^{y_kx} =e^x+\sum_{1\le k\le \frac {r-1} 2}(e^{\frac{2kx\pi i}r}+e^{\frac{2(r-k)x\pi i}r}) =e^x+\sum_{1\le k\le \frac {r-1} 2}(e^{\frac{2kx\pi i}r}+e^{\frac{-2kx\pi i}r}) =e^x+2\sum_{1\le k\le \frac {r-1} 2}\cos \frac{2kx\pi }r$$

$r=4\implies 4\sum\limits_{n=0}^{+\infty}\frac{x^{4n}}{(4n)!}$ $=2\cosh x+2\cos x$

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