Solving Problems Submitted to MAA Journals (Part 5e)

The following problem appeared in Volume 96, Issue 3 (2023) of Mathematics Magazine.

Evaluate the following sums in closed form:

$f(x) = \displaystyle \sum_{n=0}^\infty \left( \cos x - 1 + \frac{x^2}{2!} - \frac{x^4}{4!} \dots + (-1)^{n-1} \frac{x^{2n}}{(2n)!} \right)$

and

$g(x) = \displaystyle \sum_{n=0}^\infty \left( \sin x - x + \frac{x^3}{3!} - \frac{x^5}{5!} \dots + (-1)^{n-1} \frac{x^{2n+1}}{(2n+1)!} \right)$ .

By using the Taylor series expansions of $\sin x$ and $\cos x$ and flipping the order of a double sum, I was able to show that

$f(x) = -\displaystyle \frac{x \sin x}{2} \qquad \hbox{and} \qquad g(x) = \frac{x\cos x - \sin x}{2}$ .

I immediately got to thinking: there’s nothing particularly special about $\sin x$ and $\cos x$ for this analysis. Is there a way of generalizing this result to all functions with a Taylor series expansion?

Suppose

$h(x) = \displaystyle \sum_{k=0}^\infty a_k x^k$ ,

and let’s use the same technique to evaluate

$\displaystyle \sum_{n=0}^\infty \left( h(x) - \sum_{k=0}^n a_k x^k \right) = \sum_{n=0}^\infty \sum_{k=n+1}^\infty a_k x^k$

$= \displaystyle \sum_{k=1}^\infty \sum_{n=0}^{k-1} a_k x^k$

$= \displaystyle \sum_{k=1}^\infty k a_k x^k$

$= x \displaystyle \sum_{k=1}^\infty k a_k x^{k-1}$

$= x \displaystyle \sum_{k=1}^\infty \left(a_k x^k \right)'$

$= x \displaystyle \left[ (a_0)' + \sum_{k=1}^\infty \left(a_k x^k \right)' \right]$

$= x \displaystyle \sum_{k=0}^\infty \left(a_k x^k \right)'$

$= x \displaystyle \left( \sum_{k=0}^\infty a_k x^k \right)'$

$= x h'(x)$ .

To see why this matches our above results, let’s start with $h(x) = \cos x$ and write out the full Taylor series expansion, including zero coefficients:

$\cos x = 1 + 0x - \displaystyle \frac{x^2}{2!} + 0x^3 + \frac{x^4}{4!} + 0x^5 - \frac{x^6}{6!} \dots$ ,

so that

$x (\cos x)' = \displaystyle \sum_{n=0}^\infty \left( \cos x - \sum_{k=0}^n a_k x^k \right)$

$-x \sin x= \displaystyle \left(\cos x - 1 \right) + \left(\cos x - 1 + 0x \right) + \left( \cos x -1 + 0x + \frac{x^2}{2!} \right) + \left( \cos x -1 + 0x + \frac{x^2}{2!} + 0x^3 \right)$

$\displaystyle + \left( \cos x -1 + 0x + \frac{x^2}{2!} + 0x^3 - \frac{x^4}{4!} \right) + \left( \cos x -1 + 0x + \frac{x^2}{2!} + 0x^3 - \frac{x^4}{4!} + 0x^5 \right) \dots$

After dropping the zero terms and collecting, we obtain

$-x \sin x= \displaystyle 2 \left(\cos x - 1 \right) + 2 \left( \cos x -1 + \frac{x^2}{2!} \right) + 2 \left( \cos x -1 + \frac{x^2}{2!} - \frac{x^4}{4!} \right) \dots$

$-x \sin x = 2 f(x)$

$\displaystyle -\frac{x \sin x}{2} = f(x)$ .

A similar calculation would apply to any even function $h(x)$ .

We repeat for

$h(x) = \sin x = 0 + x + 0x^2 - \displaystyle \frac{x^3}{3!} + 0x^4 + \frac{x^5}{5!} + 0x^6 - \frac{x^7}{7!} \dots$ ,

so that

$x (\sin x)' = (\sin x - 0) + (\sin x - 0 - x) + (\sin x - 0 - x + 0x^2)$

$+ \displaystyle \left( \sin x - 0 - x + 0x^2 + \frac{x^3}{3!} \right) + \left( \sin x - 0 - x + 0x^2 + \frac{x^3}{3!} + 0x^4 \right)$

$+ \displaystyle \left( \sin x - 0 - x + 0x^2 + \frac{x^3}{3!} + 0x^4 - \frac{x^5}{5!} \right) + \left( \sin x - 0 - x + 0x^2 + \frac{x^3}{3!} + 0x^4 - \frac{x^5}{5!} + 0 x^6 \right) \dots$ ,

$x\cos x - \sin x = 2(\sin x - x) + \displaystyle 2\left(\sin x - x + \frac{x^3}{3!} \right) + 2 \left( \sin x - x + \frac{x^3}{3!} - \frac{x^5}{5!} \right) \dots$

$x \cos x - \sin x = 2 g(x)$

$\displaystyle \frac{x \cos x - \sin x}{2} = g(x)$ .

A similar argument applies for any odd function $h(x)$ .

Solving Problems Submitted to MAA Journals (Part 5e)

Published by John Quintanilla

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