Solving Problems Submitted to MAA Journals (Part 5d)

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

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

In the previous two posts, I showed that

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

the technique that I used was using the Taylor series expansions of $\sin x$ and $\cos x$ to write $f(x)$ and $g(x)$ as double sums and then interchanging the order of summation.

In the post, I share an alternate way of solving for $f(x)$ and $g(x)$ . I wish I could take credit for this, but I first learned the idea from my daughter. If we differentiate $g(x)$ , we obtain

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

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

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

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

$= f(x)$ .

Something similar happens when differentiating the series for $f(x)$ ; however, it’s not quite so simple because of the $-1$ term. I begin by separating the $n=0$ term from the sum, so that a sum from $n =1$ to $\infty$ remains:

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

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

I then differentiate as before:

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

$= -\sin x + \displaystyle \sum_{n=1}^\infty \left( -\sin x + \frac{2x}{2!} - \frac{4x^3}{4!} \dots + (-1)^{n-1} \frac{(2n) x^{2n-1}}{(2n)!} \right)$

$= -\sin x + \displaystyle \sum_{n=1}^\infty \left( -\sin x + \frac{2x}{2 \cdot 1!} - \frac{4x^3}{4 \cdot 3!} \dots + (-1)^{n-1} \frac{(2n) x^{2n-1}}{(2n)(2n-1)!} \right)$

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

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

At this point, we reindex the sum. We make the replacement $k = n - 1$ , so that $n = k+1$ and $k$ varies from $k=0$ to $\infty$ . After the replacement, we then change the dummy index from $k$ back to $n$ .

$f'(x) = -\sin x - \displaystyle \sum_{k=0}^\infty \left( \sin x - x + \frac{x^3}{3!} + \dots - (-1)^{(k+1)-1} \frac{x^{2(k+1)-1}}{(2(k+1)-1)!} \right)$

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

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

With a slight alteration to the $(-1)^n$ term, this sum is exactly the definition of $g(x)$ :

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

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

$= -\sin x - g(x)$ .

Summarizing, we have shown that $g'(x) = f(x)$ and $f'(x) = -\sin x - g(x)$ . Differentiating $f'(x)$ a second time, we obtain

$f''(x) = -\cos x - g'(x) = -\cos x - f(x)$

$f''(x) + f(x) = -\cos x$ .

This last equation is a second-order nonhomogeneous linear differential equation with constant coefficients. A particular solution, using the method of undetermined coefficients, must have the form $F(x) = Ax\cos x + Bx \sin x$ . Substituting, we see that

$[Ax \cos x + B x \sin x]'' + A x \cos x + Bx \sin x = -\cos x$

$-2A \sin x - Ax \cos x + 2B \cos x - B x \sin x + Ax \cos x + B x \sin x = -\cos x$

$-2A \sin x + 2B \cos x = -\cos x$

We see that $A = 0$ and $B = -1/2,$ which then lead to the particular solution

$F(x) = -\displaystyle \frac{1}{2} x \sin x$

Since $\cos x$ and $\sin x$ are solutions of the associated homogeneous equation $f''(x) + f(x) = 0$ , we conclude that

$f(x) = c_1 \cos x + c_2 \sin x - \displaystyle \frac{1}{2} x \sin x$ ,

where the values of $c_1$ and $c_2$ depend on the initial conditions on $f$ . As it turns out, it is straightforward to compute $f(0)$ and $f'(0)$ , so we will choose $x=0$ for the initial conditions. We observe that $f(0)$ and $g(0)$ are both clearly equal to 0, so that $f'(0) = -\sin 0 - g(0) = 0$ as well.