1

Sumfunction of $\sum_{n=0}^{\infty} \frac{z^{4n}}{4n!}$ and $\sum_{n=1}^{\infty} \frac{z^{-n}}{4n}$.

The first one looks like the cosine hyperbolicus but the $4n!$ anoys me. I tried using $\cosh(z^2)$ or something in that form but that doesn't work, for the second problem i tried to find something in the form of log(1+z) but that did give me problems with the $(-1)^n$. usually if i try to find the sumfunction i go through this phase:

1) try to find sumfunctions which look alike 2) if one does not work, try to differentiate to find your solution!

Any hints for the first sum function, the second function will eventually work i think.

Kees Til
  • 1,958

5 Answers5

2

Since the second one is $\frac{1}{4} \sum_{k \geq 1} \frac{s^k}{k}$ for $s =\frac{1}{z}$, it's just Maclaurin series expansion for $\frac{1}{4} \log \frac{1}{1-s}$ for $|s|<1$.

Alex
  • 19,262
2

The first one is $\frac12 (\cos z + \cosh z)$. (Every second term will cancel when you expand.)

Hans Lundmark
  • 53,395
  • how did you come up with this? Did you just write out the first terms of the taylor series and tried to manipulate those, or did you write out the series where you started with in this form? – Kees Til Mar 17 '15 at 10:33
  • Suppose you were given the series for $\cosh z$. You recognize that it agrees with the series for $e^z$, except that every second term is missing, but since you can find the partner series $e^{-z}$ which is like $e^z$ but with the opposite sign on every second term, you can just average them to get what you want: $\cosh z= \frac12 (e^z + e^{-z})$. And it's the same idea here: you recognize that your series is like $\cosh z$ with every second term missing, and you know that $\cos z$ is a partner series with alternating signs. – Hans Lundmark Mar 17 '15 at 11:08
  • For a more general method (say if you have every third term of a known series), see this question: http://math.stackexchange.com/questions/11739/ – Hans Lundmark Mar 17 '15 at 11:09
0

\begin{align} f(z)&=\sum_{n=0}^{\infty} \frac{z^{4n}}{(4n)!}\\ f'(z)&=\sum_{n=1}^{\infty} \frac{z^{4n-1}}{(4n-1)!}\\ f^{(2)}(z)&=\sum_{n=1}^{\infty} \frac{z^{4n-2}}{(4n-2)!}\\ f^{(3)}(z)&=\sum_{n=1}^{\infty} \frac{z^{4n-3}}{(4n-3)!}\\ f^{(4)}(z)&=\sum_{n=1}^{\infty} \frac{z^{4n-4}}{(4n-4)!}\\ &=\sum_{n=0}^{\infty} \frac{z^{4n}}{(4n)!}=f(x)\\ \end{align} Now note that \begin{align} f(0)&=1\\ f'(0)&=0\\ f^{(2)}(0)&=0\\ f^{(3)}(0)&=0\\ \end{align} Therefore we proceed to solve the following boundary condition differential equation $$f^{(4)}-f(x)=0$$ Taking a Laplace transform we obtain $$s^4F(s)-s^3f(0)-s^2f'(0)-sf^{(2)}(0)-f^{(3)}(0)-F(s)=0$$ or $$F(s)=\frac{s^3}{s^4-1}=\frac14\frac{1}{s-1}+\frac14\frac{1}{s+1}+\frac12\frac{s}{s^2+1}$$ thus taking an inverse Laplace transform we obtain $$f(z)=\frac{e^{-z}}{4}+\frac{e^{z}}{4}+\frac12 \cos z =\frac12(\cos z + \cosh z)$$

Note: Stefan also mentioned this above.

The second was treated already.

Math-fun
  • 9,507
0

The first series can be approached with the discrete Fourier transform. Since $$ f(n) = \frac{1}{4}\left(1^n+(-1)^n+i^n+(-i)^n\right)$$ is the indicator function of the integers being multiples of four, we have: $$\sum_{n\geq 0}\frac{x^{4n}}{(4n)!}=\sum_{n\equiv 0\pmod{4}}\frac{x^n}{n!}=\frac{1}{4}\left(e^x+e^{-x}+e^{ix}+e^{-ix}\right) = \frac{1}{2}\left(\cos x+\cosh x\right).$$ The second series is just: $$\frac{1}{4}\sum_{n\geq 1}\frac{(1/z)^n}{n}=-\frac{1}{4}\log\left(1-\frac{1}{z}\right).$$

Jack D'Aurizio
  • 353,855
0

For the second sum use this this general result: $$ \sum_{n=1}^{+\infty}\frac{1}{nz^n},\;\;\;\; z\in\mathbb C\;. $$ We immediately see that $|z|>1$, in order to have absolute convergence.

We recall first two results:

$\bullet\;\;$First: $$ \log(1+z)=\sum_{n=1}^{+\infty}(-1)^{n+1}\frac{z^n}{n},\;\;\;\forall |z|<1 $$ $\bullet\;\;$Second: $$ \prod_{n=0}^{+\infty}\left(1+z^{2^{n}}\right)= \sum_{n=0}^{+\infty}z^{n}=\frac{1}{1-z},\;\;\;\forall |z|<1 $$ The last one can be proved, showing by induction that $\prod_{k=0}^{N}\left(1+z^{2^{k}}\right)=\sum_{k=0}^{2^{N+1}-1}z^{k}$.

Ok: \begin{align*} \sum_{n=1}^{+\infty}\frac{1}{nz^n}=& \sum_{n=1}^{+\infty}\frac{1}{n}\left(\frac{1}{z}\right)^n\\ =&\underbrace{\sum_{k=0}^{+\infty}\frac{1}{2k+1}\left(\frac{1}{z}\right)^{2k+1}- \sum_{k=1}^{+\infty}\frac{1}{2k}\left(\frac{1}{z}\right)^{2k}}_{\log\left(1+\frac{1}{z}\right)}+ 2\sum_{k=1}^{+\infty}\frac{1}{2k}\left(\frac{1}{z}\right)^{2k}\\ =&\log\left(1+\frac{1}{z}\right)+ \sum_{k=1}^{+\infty}\frac{1}{k}\left(\frac{1}{z^2}\right)^{k}\\ =&\log\left(1+\frac{1}{z}\right)+ \log\left(1+\frac{1}{z^2}\right)+\cdots\\ =&\sum_{n=0}^{+\infty}\log\left(1+\frac{1}{z^{2^n}}\right)\\ =&\log\left(\prod_{n=0}^{+\infty}\left(1+\left(\frac{1}{z}\right)^{2^n}\right)\right)\\ =&\log\left(\frac{z}{z-1}\right) \end{align*}

Joe
  • 11,745