Solving Problems Submitted to MAA Journals (Part 5c)

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 post, we showed that $f(x) = - \frac{1}{2} x \sin x$ by writing the series as a double sum and then reversing the order of summation. We proceed with very similar logic to evaluate $g(x)$ . Since

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

is the Taylor series expansion of $\sin x$ , we may write $g(x)$ as

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

$= \displaystyle \sum_{n=0}^\infty \sum_{k=n+1}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!}$

As before, we employ one of my favorite techniques from the bag of tricks: reversing the order of summation. Also as before, the inner sum is inner sum is independent of $n$ , and so the inner sum is simply equal to the summand times the number of terms. We see that

$g(x) = \displaystyle \sum_{k=1}^\infty \sum_{n=0}^{k-1} (-1)^k \frac{x^{2k+1}}{(2k+1)!}$

$= \displaystyle \sum_{k=1}^\infty (-1)^k \cdot k \frac{x^{2k+1}}{(2k+1)!}$

$= \displaystyle \frac{1}{2} \sum_{k=1}^\infty (-1)^k \cdot 2k \frac{x^{2k+1}}{(2k+1)!}$ .

At this point, the solution for $g(x)$ diverges from the previous solution for $f(x)$ . I want to cancel the factor of $2k$ in the summand; however, the denominator is

$(2k+1)! = (2k+1)(2k)!$ ,

and $2k$ doesn’t cancel cleanly with $(2k+1)$ . Hypothetically, I could cancel as follows:

$\displaystyle \frac{2k}{(2k+1)!} = \frac{2k}{(2k+1)(2k)(2k-1)!} = \frac{1}{(2k+1)(2k-1)!}$ ,

but that introduces an extra $(2k+1)$ in the denominator that I’d rather avoid.

So, instead, I’ll write $2k$ as $(2k+1)-1$ and then distribute and split into two different sums:

$g(x) = \displaystyle \frac{1}{2} \sum_{k=1}^\infty (-1)^k \cdot 2k \frac{x^{2k+1}}{(2k+1)!}$

$= \displaystyle \frac{1}{2} \sum_{k=1}^\infty (-1)^k (2k+1-1) \frac{x^{2k+1}}{(2k+1)!}$

$= \displaystyle \frac{1}{2} \sum_{k=1}^\infty \left[ (-1)^k (2k+1) \frac{x^{2k+1}}{(2k+1)!} - (-1)^k \cdot 1 \frac{x^{2k+1}}{(2k+1)!} \right]$

$= \displaystyle \frac{1}{2} \sum_{k=1}^\infty (-1)^k (2k+1) \frac{x^{2k+1}}{(2k+1)!} - \frac{1}{2} \sum_{k=1}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!}$

$= \displaystyle \frac{1}{2} \sum_{k=1}^\infty (-1)^k (2k+1) \frac{x^{2k+1}}{(2k+1)(2k)!} - \frac{1}{2} \sum_{k=1}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!}$

$= \displaystyle \frac{1}{2} \sum_{k=1}^\infty (-1)^k \frac{x^{2k+1}}{(2k)!} - \frac{1}{2} \sum_{k=1}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!}$ .

At this point, I factored out a power of $x$ from the first sum. In this way, the two sums are the Taylor series expansions of $\cos x$ and $\sin x$ :

$g(x) = \displaystyle \frac{x}{2} \sum_{k=1}^\infty (-1)^k \cdot \frac{x^{2k}}{(2k)!} - \frac{1}{2} \sum_{k=1}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!}$

$= \displaystyle \frac{x}{2} \cos x - \frac{1}{2} \sin x$

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

This was sufficiently complicated that I was unable to guess this solution by experimenting with Mathematica; nevertheless, Mathematica can give graphical confirmation of the solution since the graphs of the two expressions overlap perfectly.

Solving Problems Submitted to MAA Journals (Part 5b)

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

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

We start with $f(x)$ and the Taylor series

$\cos x = \displaystyle \sum_{k=0}^\infty (-1)^k \frac{x^{2k}}{(2k)!}$ .

With this, $f(x)$ can be written as

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

$= \displaystyle \sum_{n=0}^\infty \sum_{k=n+1}^\infty (-1)^k \frac{x^{2k}}{(2k)!}$ .

At this point, my immediate thought was one of my favorite techniques from the bag of tricks: reversing the order of summation. (Two or three chapters of my Ph.D. theses derived from knowing when to apply this technique.) We see that

$f(x) = \displaystyle \sum_{k=1}^\infty \sum_{n=0}^{k-1} (-1)^k \frac{x^{2k}}{(2k)!}$ .

At this point, the inner sum is independent of $n$ , and so the inner sum is simply equal to the summand times the number of terms. Since there are $k$ terms for the inner sum ( $n = 0, 1, \dots, k-1$ ), we see

$f(x) = \displaystyle \sum_{k=1}^\infty (-1)^k \cdot k \frac{x^{2k}}{(2k)!}$ .

To simplify, we multiply top and bottom by 2 so that the first term of $(2k)!$ cancels:

$f(x) = \displaystyle \frac{1}{2} \sum_{k=1}^\infty (-1)^k \cdot 2k \frac{x^{2k}}{(2k)(2k-1)!}$

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

At this point, I factored out a $(-1)$ and a power of $x$ to make the sum match the Taylor series for $\sin x$ :

$f(x) = \displaystyle -\frac{x}{2} \sum_{k=1}^\infty (-1)^{k-1} \frac{x^{2k-1}}{(2k-1)!} = -\frac{x \sin x}{2}$ .

I was unsurprised but comforted that this matched the guess I had made by experimenting with Mathematica.

Solving Problems Submitted to MAA Journals (Part 5a)

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

Evaluate the following sums in closed form:

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

and

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

When I first read this problem, I immediately noticed that

$\displaystyle 1 - \frac{x^2}{2!} + \frac{x^4}{4!} \dots - (-1)^{n-1} \frac{x^{2n}}{(2n)!} \right)$

is a Taylor polynomial of $\cos x$ and

$\displaystyle x - \frac{x^3}{3!} + \frac{x^5}{5!} \dots - (-1)^{n-1} \frac{x^{2n+1}}{(2n+1)!} \right)$

is a Taylor polynomial of $\sin x$ . In other words, the given expressions are the sums of the tail-sums of the Taylor series for $\cos x$ and $\sin x$ .

As usual when stumped, I used technology to guide me. Here’s the graph of the first sum, adding the first 50 terms.

I immediately notice that the function oscillates, which makes me suspect that the answer involves either $\cos x$ or $\sin x$ . I also notice that the sizes of oscillations increase as $|x|$ increases, so that the answer should have the form $g(x) \cos x$ or $g(x) \sin x$ , where $g$ is an increasing function. I also notice that the graph is symmetric about the origin, so that the function is even. I also notice that the graph passes through the origin.

So, taking all of that in, one of my first guesses was $y = x \sin x$ , which is satisfies all of the above criteria.

That’s not it, but it’s not far off. The oscillations of my guess in orange are too big and they’re inverted from the actual graph in blue. After some guessing, I eventually landed on $y = -\frac{1}{2} x \sin x$ .

That was a very good sign… the two graphs were pretty much on top of each other. That’s not a proof that $-\frac{1}{2} x \sin x$ is the answer, of course, but it’s certainly a good indicator.

I didn’t have the same luck with the other sum; I could graph it but wasn’t able to just guess what the curve could be.

Solving Problems Submitted to MAA Journals (Part 4)

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

Let $A_1, \dots, A_n$ be arbitrary events in a probability field. Denote by $B_k$ the event that at least $k$ of $A_1, \dots A_n$ occur. Prove that $\displaystyle \sum_{k=1}^n P(B_k) = \sum_{k=1}^n P(A_k)$ .

I’ll admit when I first read this problem, I didn’t believe it. I had to draw a couple of Venn diagrams to convince myself that it actually worked:

Of course, pictures are not proofs, so I started giving the problem more thought.

I wish I could say where I got the inspiration from, but I got the idea to define a new random variable $N$ to be the number of events from $A_1, \dots A_n$ that occur. With this definition, $B_k$ becomes the event that $N \ge k$ , so that

$\displaystyle \sum_{k=1}^n P(B_k) = \sum_{k=1}^n P(N \ge k)$

At this point, my Spidey Sense went off: that’s the tail-sum formula for expectation! Since $N$ is a non-negative integer-valued random variable, the mean of $N$ can be computed by

$E(N) = \displaystyle \sum_{k=1}^n P(N \ge k)$ .

Said another way, $E(N) = \displaystyle \sum_{k=1}^n P(B_k)$ .

Therefore, to solve the problem, it remains to show that $\displaystyle \sum_{k=1}^n P(A_k)$ is also equal to $E(N)$ . To do this, I employed the standard technique from the bag of tricks of writing $N$ as the sum of indicator random variables. Define

$I_k = \displaystyle \bigg\{ \begin{array}{ll} 1, & A_k \hbox{~occurs} \\ 0, & A_k \hbox{~does not occur} \end{array}$

Then $N = I_1 + \dots + I_n$ , so that

$E(N) = \displaystyle \sum_{k=1}^n E(I_k) =\sum_{k=1}^n [1 \cdot P(A_k) + 0 \cdot P(A_k^c)] =\sum_{k=1}^n P(A_k)$ .

Equating the two expressions for $E(N)$ , we conclude that $\displaystyle \sum_{k=1}^n P(B_k) = \sum_{k=1}^n P(A_k)$ , as claimed.

Solving Problems Submitted to MAA Journals (Part 3)

The following problem appeared in Volume 53, Issue 4 (2022) of The College Mathematics Journal.

Define, for every non-negative integer $n$ , the $n$ th Catalan number by

$C_n := \displaystyle \frac{1}{n+1} {2n \choose n}$ .

Consider the sequence of complex polynomials in $z$ defined by $z_k := z_{k-1}^2 + z$ for every non-negative integer $k$ , where $z_0 := z$ . It is clear that $z_k$ has degree $2^k$ and thus has the representation

$z_k =\displaystyle \sum_{n=1}^{2^k} M_{n,k} z^n$ ,

where each $M_{n,k}$ is a positive integer. Prove that $M_{n,k} = C_{n-1}$ for $1 \le n \le k+1$ .

This problem appeared in the same issue as the probability problem considered in the previous two posts. Looking back, I think that the confidence that I gained by solving that problem gave me the persistence to solve this problem as well.

My first thought when reading this problem was something like “This involves sums, polynomials, and binomial coefficients. And since the sequence is recursively defined, it’s probably going to involve a proof by mathematical induction. I can do this.”

My second thought was to use Mathematica to develop my own intuition and to confirm that the claimed pattern actually worked for the first few values of $z_k$ .

As claimed in the statement of the problem, each $z_k$ is a polynomial of degree $2^k$ without a nontrivial constant term. Also, for each $z_k$ , the term of degree $n$ , for $1 \le n \le k+1$ , has a coefficient that is independent of $k$ which equal to $C_{n-1}$ . For example, for $z_4$ , the coefficient of $z^5$ (in orange above) is equal to

$C_4 = \displaystyle \frac{1}{5} {8 \choose 4} = \frac{8!}{4! 4! \cdot 5} = \frac{40320}{2880} = 14$ ,

and the problem claims that the coefficient of $z^5$ will remain 14 for $z_5, z_6, z_7, \dots$

Confident that the pattern actually worked, all that remained was pushing through the proof by induction.

We proceed by induction on $k$ . The statement clearly holds for $k=1$ :

$z_1 = z_0^2 + z = z + z^2 = C_0 z + C_1 z^2$ .

Although not necessary, I’ll add for good measure that

$z_2 = z_1^2 + z = (z^2+z)^2 + z = z + z^2 + 2z^3 + z^4 = C_0 z + C_1 z^2 + C_2 z^3 + z^4$

and

$z_3 = z^2 + z = (z^4+2z^3+z^2+z)^2 + z$

$= z + z^2 + 2z^3 + 5z^4 + 6z^5 + 6z^6 + 4z^7 + z^8$

$= C_0 z + C_1 z^2 + C_2 z^3 + C_3 z^4 + 6z^5 + 6z^6 + 4z^7 + z^8.$

This next calculation illustrates what’s coming later. In the previous calculation, the coefficient of $z^4$ is found by multiplying out

$(z^4+2z^3+z^2+z)(z^4+2z^3+z^2+z)$ .

This is accomplished by examining all pairs, one from the left product and one from the right product, so that the exponent works out to be $z^4$ . In this case, it’s

$(2z^3)(z) + (z^2)(z^2) + (z)(2z^3) = 5z^4$ .

For the inductive step, we assume that, for some $k \ge 1$ , $M_{n,k} = C_{n-1}$ for all $1 \le n \le k+1$ , and we define

$z_{k+1} = z + \left( M_{1,k} z + M_{2,k} z^2 + M_{3,k} z^3 + \dots + M_{2^k,k} z^{2^k} \right)^2$

Our goal is to show that $M_{n,k+1} = C_{n-1}$ for $n = 1, 2, \dots, k+2$ .

For $n=1$ , the coefficient $M_{1,k+1}$ of $z$ in $z_{k+1}$ is clearly 1, or $C_0$ .

For $2 \le n \le k+2$ , the coefficient $M_{n,k+1}$ of $z^n$ in $z_{k+1}$ can be found by expanding the above square. Every product of the form $M_{j,k} z^j \cdot M_{n-j,k} z^{n-j}$ will contribute to the term $M_{n,k+1} z^n$ . Since $n \le k+2 \le 2^k+1$ (since $k \ge 1$ ), the values of $j$ that will contribute to this term will be $j = 1, 2, \dots, n-1$ . (Ordinarily, the $z^0$ and $z^n$ terms would also contribute; however, there is no $z^0$ term in the expression being squared). Therefore, after using the induction hypothesis and reindexing, we find