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

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Theorem

\(\ds e^{\cos x}\) \(=\) \(\ds e \paren {e^{\cos x - 1} }\)
\(\ds \) \(=\) \(\ds e \paren {\sum_{n \mathop = 0}^\infty \frac {\paren {-1}^m \map {P_2} {2 m} } {2 m!} x^{2 m} }\) where $\map {P_2} {2 m}$ is the partition of the set of size $2 m$ into even blocks
\(\ds \) \(=\) \(\ds e \paren {1 - \dfrac {x^2} 2 + \dfrac {x^4} 6 - \dfrac {31 x^6} {720} + \cdots}\)

for all $x \in \R$.


Proof

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

$\ds \map f x = \sum_{n \mathop = 1}^\infty \frac {c_n} {n!} x^n $

then:

$\ds e^{\map f x} = \sum_{k \mathop = 0}^\infty \frac {\map {B_k} {c_1, c_2, \ldots, c_k} } {k!} x^k$

where $\map {B_k} {c_1, c_2, \ldots, c_k}$ is the $k$th complete Bell Polynomial.

From the Power Series Expansion for Cosine Function, we get:

$\ds \cos x - 1 = \sum_{p \mathop = 1}^\infty \frac {\paren {-1}^p} {2 p!} x^{2 p}$

Using this we may plug the sequence $\tuple {0, -1, 0, 1, 0, -1}$ into the first $6$ complete Bell Polynomials.

An arbitrary even Complete Bell Polynomial, will take the form:

$\ds \map {B_n} {x_1, x_2, x_3, \ldots, x_n} = \sum \frac {n!} {k_1! k_2! \dotsm k_n!} \paren {\frac {x_1} {1!} }^{c_1} \paren {\frac {x_2} {2!} }^{c_2} \dotsm \paren {\frac {x_n} {n!} }^{c_n}$

where the sum is taken over all n-tuples $\tuple {c_1, \ldots, c_n}$ such that $c_1 + 2 c_2 + \dotsb + n c_n = n$

In other words, it is taken over every integer partition of $n$.

Here we have it that

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

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:

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

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 $n \equiv 2 \pmod 4$, then an odd number of integers congruent to $2 \pmod 4$ must be chosen for every partition.

Thus all summands will be a negative.

A similar argument holds for the case $n \equiv 0 \pmod 4$.

Thus for our given values of $x_k$, we are only summing over the partitions of $2 m$ such that all summands in the partition are even, and because all the summands in the polynomial are the same sign we have:

$\map {B_{2 m} } {0, -1, 0, \ldots, \pm 1} = \paren {-1}^m \map {B_{2 m} } {0, 1, 0, 1, \ldots, 1}$

Thus from the definition of Bell Polynomials as a sum over all incomplete Bell polynomials we have:

$\map {B_{2 m} } {0, -1, 0, \ldots , \pm x_{2 m} } = \paren {-1}^m \map {P_2} {2 m}$

where $\map {P_2} {2 m}$ is all partitions of a set of size $2 m$ into even blocks.

Thus we have:

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

where $P_2 \paren {2 m}$ is the partition of the set of size $2 m$ into even blocks.

$\blacksquare$


Also see


Sources

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