How to find the series $\sum_{n=0}^{\infty} \frac{(-1)^n x^{4n}}{(4n!)}$

Question

It is known that, $$\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}=\cos(x)$$ and $$\sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}=\cosh(x)$$

From this follows that $$\sum_{n=0}^{\infty} \frac{x^{4n}}{(4n)!}=\frac{\cos(x)+\cosh(x)}{2}$$

How to find $$\sum_{n=0}^{\infty} \frac{(-1)^n x^{4n}}{(4n)!}$$

Answer 1

To expand a little on @Angina Seng's point, essentially the thing to notice is that if we can find the fourth root of ${-1}$, and replace $x$ with $x$ multiplied by this fourth root you get

$${\frac{((-1)^{\frac{1}{4}}x)^{4n}}{(4n)!}=\frac{((-1)^{\frac{1}{4}})^{4n}x^{4n}}{(4n)!}=\frac{(-1)^nx^{4n}}{(4n)!}}$$

There are actually four different fourth roots of ${-1}$:

$${e^{\frac{i\pi}{4}},e^{\frac{-i\pi}{4}},e^{\frac{3i\pi}{4}},e^{\frac{-3i\pi}{4}}}$$

So any one of these will suffice. If we take the principal root (${e^{\frac{i\pi}{4}}}$) though, this would make your answer

$${\sum_{n=0}^{\infty}(-1)^n\frac{x^{4n}}{(4n)!}=\frac{\cos\left(e^{\frac{i\pi}{4}}x\right)+\cosh\left(e^{\frac{i\pi}{4}}x\right)}{2}}$$

Just as others have said. But we don't have to pick that root,

$${\sum_{n=0}^{\infty}(-1)^n\frac{x^{4n}}{(4n)!}=\frac{\cos\left(e^{\frac{3i\pi}{4}}x\right)+\cosh\left(e^{\frac{3i\pi}{4}}x\right)}{2}}$$

is also just as valid

Edit: I won't write the full working (I will leave it as an exercise to you) - but we can actually further then simplify the result to

$${\sum_{n=0}^{\infty}(-1)^n\frac{x^{4n}}{(4n)!}=\cos\left(\frac{x}{\sqrt{2}}\right)\cosh\left(\frac{x}{\sqrt{2}}\right)}$$

Answer 2

For $0\le j\le 3$ define $S_j(z):=\sum_{n\ge0}\frac{z^{4n+j}}{(4n+j)!}$ so, for $0\le k\le3$,$$\sum_ji^{jk}S_j=\sum_{jn}\frac{(zi^k)^{4n+j}}{(4n+j)!}=e^{i^kz}.$$In other words,$$\begin{align}S_0+S_1+S_2+S_3&=e^z,\\S_0+iS_1-S_2-iS_3&=e^{iz},\\S_0-S_1+S_2-S_3&=e^{-z},\\S_0-iS_1-S_2+iS_3&=e^{-iz}.\end{align}$$Solving simultaneous equations gives each $S_j(z)$; you want $S_0(e^{\pi i/4}x)$. In particular $S_0+S_2=\cosh z,\,S_0-S_2=i\sin z$ so $S_0=\tfrac12(\cosh z+i\sin z)$, making your series $\tfrac12(\cosh\tfrac{1+i}{\sqrt{2}}x+i\sin\tfrac{1+i}{\sqrt{2}}x)$, which you're welcome to rewrite as you wish.

Answer 3

Use $(-1)^{\frac{1}{4}}$ to be able to use the series, so you get $$\frac{\cos (\frac{(1+i)x}{\sqrt 2})+\cosh (\frac{(1+i)x}{\sqrt 2})}{2}=\sum_{n=0}^\infty\frac{(-1)^nx^{4n}}{(4n)!} $$ Then use $$ \cos(a+ib)=\cos a\cosh b-i\sin a\sinh b $$ and $$ \cosh(a+ib)=\cos b\cosh a+i\sin b\sinh a. $$ In our case $a=b$ so $\cos(a+ia)+\cosh(a+ib)=2\cos a\cosh a$. Set $a=\frac{x}{\sqrt 2}$ and we get the solution $$ \sum_{n=0}^\infty\frac{(-1)^nx^{4n}}{(4n)!}=\cos \frac{x}{\sqrt 2}\cosh \frac{x}{\sqrt 2}. $$