# Power Series Expansion for Exponential of Cosine of x/Proof 2

## Contents

## Theorem

For all $x \in \R$:

- $\displaystyle e^{\cos \left({x}\right)} = e \left({e^{cos \left({x}\right)- 1 } }\right) = e \left({\sum_{n \mathop = 0}^\infty \frac{ \left({-1}\right)^m P_2 \left({2 m}\right)} {2 m!} x^{2 m} }\right)$

where $P_2 \left({2 m}\right)$ is the partition of the set of size 2m into even blocks.

## Proof

A result in combinatorics known as the Exponential formula states that if

- $f(x) = \sum_{n=1}^{\infty} \frac{c_n}{n!} x^n $

then

- $e^{f(x)} = \sum_{k=0}^{\infty} \frac{B_k(c_1,c_2,...c_k)}{k!}x^k $

Where $B_k(c_1,c_2,...c_k)$ is the k-th complete Bell Polynomial. From the Power Series Expansion for Cosine Function, we get

- $ \cos(x)-1 = \sum_{p=1}^{\infty} \frac{(-1)^p}{2p!} x^{2p} $

Using this we may plug the sequence 0,-1,0,1,0,-1 into the first 6 complete Bell Polynomials an arbitrary even Complete Bell Polynomial, will take the from

- $ B_n(x_1,x_2,x_3,....,x_n)=\sum \frac{n!}{k_1!k_2!...k_n!}(\frac{x_1}{1!})^{c_1}(\frac{x_2}{2!})^{c_2}....(\frac{x_n}{n!})^{c_n}$

where the sum is taken over all n-tuples $(c_1,....,c_n)$ such that $c_1+2c_2....+nc_n=n$ in other words, it is taken over every integer partion of n. Here we have it that

- $x_{4n+0}=1$
- $x_{4n+1}=0$
- $x_{4n+2}=-1$
- $x_{4n+3}=0$

Because there is no way to form a partition of an odd number without using an odd number, all the summands in the odd Complete Bell polynomials contain a 0, thus they equal 0. Explicitly calculating the first 6 even complete Bell Polynomials we get

- $ B_0(0)=1$
- $ B_2(0,-1)=0^2-1=-1$
- $ B_4(0,-1,0,1)=0^4+6(0)^2(-1)+4(0)(0)+3(-1)^2 + 1=4$
- $ B_6(0,-1,0,1,0,-1)=0^6+6(-1)(0)+15(-1)(1)+10(0)^2+15(0)^2(1)+15(-1)^3+60(0)(-1)(0)+20(0)^3(0)+45(0)^2(-1)^2+15(0)^4(-1)+(-1)=-31$

Note that in all the summands for the above polynomials, they are either 0 or of the same sign. This is provable to always be the case, as if $n\equiv 2 (mod 4)$, then an odd number of integers congruent to 2 mod 4 must be chosen for every partition, thus all summands will be a negative. A similar argument holds for the case $n\equiv 0 (mod 4)$ Thus for our given values of x_k, we are only summing over the partitions of 2m such that all summands in the partition are even, and because all the summands in the polynomial are the same sign we have $B_2m(0,-1,0....,\pm 1)=(-1)^m B_2m(0,1,0,1,....,1).$ Thus from the definition of Bell Polynomials as a sum over all incomplete Bell polynomials we have :$B_2m(0,-1,0....,\pm x_2m)=(-1)^m P_{2}(2m)$, where $P_{2}(2m)$ is all partitions of a set of size 2m into even blocks. Thus we have

- $ e^{\cos(x)}=e(e^{cos(x)-1})= e(\sum_{n=0}^{\infty} \frac{(-1)^m P_2(2m)}{2m!}x^{2m})$

where $P_2(2m)$ is the partition of the set of size 2m into even blocks.

$\blacksquare$

## Also see

## Sources

- Bell polynomial. E.K. Lloyd (originator), Encyclopedia of Mathematics. URL: http://www.encyclopediaofmath.org/index.php?title=Bell_polynomial&oldid=17635