The Incomplete Gamma and Confluent Hypergeometric Functions (Part 6)

In the previous posts, I showed that the curious integral

$\displaystyle \int_0^z t^{a-1} e^{-t} \, dt = \displaystyle e^{-z} \sum_{s=0}^\infty \frac{(a-1)!}{(a+s)!} z^{a+s}$ ,

where the right-hand side is a special case of the confluent hypergeometric function when $a$ is a positive integer, can be confirmed if I can show that

$\displaystyle \sum_{s=0}^a (-1)^s \binom{n}{s} = (-1)^a \binom{n-1}{a}$ ,

where $0 \le a \le n-1$ . We now prove this combinatorical identity.

Case 1. If $a=0$ , then

$\displaystyle \sum_{s=0}^0 (-1)^s \binom{n}{s} = (-1)^0 \binom{n}{0} = 1 = (-1)^0 \binom{n-1}{0}$ .

Case 2. If $1 \le a \le n-1$ , then

$\displaystyle \sum_{s=0}^a (-1)^s \binom{n}{s} = (-1)^0 \binom{n}{0} + \sum_{s=1}^{a} (-1)^s \binom{n}{s}$

$\displaystyle = 1 + \sum_{s=1}^{a} (-1)^s \binom{n}{s}$

$\displaystyle = 1 + \sum_{s=1}^a (-1)^s \left[\binom{n-1}{s-1} + \binom{n-1}{s}\right]$

$\displaystyle = 1 + \sum_{s=1}^a (-1)^s \binom{n-1}{s-1} + \sum_{s=1}^{a-1} (-1)^s \binom{n-1}{s}$

$\displaystyle = 1 + \sum_{s=0}^{a-1} (-1)^{s+1} \binom{n-1}{s} + \sum_{s=1}^a (-1)^s \binom{n-1}{s}$

$\displaystyle = 1 + (-1)^1 \binom{n-1}{0} + \sum_{s=1}^{a-1} (-1)^{s+1} \binom{n-1}{s} + \sum_{s=1}^{a-1} (-1)^s \binom{n-1}{s} + (-1)^a \binom{n-1}{a}$

$\displaystyle = 1 -1 + (-1)^a \binom{n-1}{a} + \sum_{s=1}^{a-1} (-1)^{s+1} \binom{n-1}{s} + \sum_{s=1}^{a-1} (-1)^s \binom{n-1}{s}$

$\displaystyle = (-1)^a \binom{n-1}{a} +\sum_{s=1}^{a-1} \left[(-1)^{s+1} \binom{n-1}{s} + (-1)^s \binom{n-1}{s} \right]$

$\displaystyle = (-1)^a \binom{n-1}{a} + \sum_{s=1}^{a-1} (-1)^s \binom{n-1}{s} \left[ -1 + 1 \right]$

$\displaystyle = (-1)^a \binom{n-1}{a} + \sum_{s=1}^{a-1} 0$

$\displaystyle = (-1)^a \binom{n-1}{a}$

In retrospect, I’m really surprised that I don’t remember seeing this identity before. The proof uses many of the techniques that we teach in discrete mathematics, including peeling off a term from either the start or end of a series, reindexing a sum, and the identity for constructing the terms in Pascal’s triangle in terms of the binomial coefficients. So I would’ve thought that I would’ve come across this, either in my own studies as a student or else in a textbook while preparing to teach discrete mathematics.

The observant reader may have noted that the “proof” over the past few posts isn’t really a proof since I started with the equation that I wanted to prove and then ended up with a true statement. While working backwards helped me see what was going on, this logic is ultimately flawed since it’s possible for false statements to imply true statements (another principle from discrete mathematics). I’ll clean this up in the next post.

The Incomplete Gamma and Confluent Hypergeometric Functions (Part 5)

In the previous two posts, I show that the curious integral

$\displaystyle \int_0^z t^{a-1} e^{-t} \, dt = \displaystyle e^{-z} \sum_{s=0}^\infty \frac{(a-1)!}{(a+s)!} z^{a+s}$ ,

where the right-hand side is a special case of the confluent hypergeometric function when $a$ is a positive integer, can be confirmed if I can show that

$\displaystyle \sum_{s=0}^a (-1)^s \binom{n}{s} =^? (-1)^a \binom{n-1}{a}$ .

When I first arrived at this equation, I noticed that this was really a property about the numbers in Pascal’s triangle: if we alternate adding and subtracting the binomial coefficients $\displaystyle \binom{n}{s}$ on the $n$ th row of Pascal’s triangle, the result is supposed to be an entry in the previous row (perhaps multiplied by $-1$ ).

For example, using the 10th row of Pascal’s triangle:

$1 - 10 = -9$ , which is negative the number to the “northeast” of 10.
$1 - 10 + 45 = 36$ , which is the number northeast of 45.
$1 - 10 + 45 - 120 = -84$ , which is negative the number northeast of 120.
$1 - 10 + 45 -120 + 210 = 126$ , which is the number northeast of 210.

And so on.

These computations suggest a proof of this identity. Any term in the interior of Pascal’s triangle can be found by adding the two terms immediately above it. Indeed, we can see this working out in the sequence of sums listed above.

I’ll write up the formal proof of the identity in the next post.

The Incomplete Gamma and Confluent Hypergeometric Functions (Part 4)

In the previous post, I show that the curious integral

$\displaystyle \int_0^z t^{a-1} e^{-t} \, dt = \displaystyle e^{-z} \sum_{s=0}^\infty \frac{(a-1)!}{(a+s)!} z^{a+s}$ ,

where the right-hand side is a special case of the confluent hypergeometric function when $a$ is a positive integer, can be confirmed if I can show that

$\displaystyle \sum_{s=0}^n \frac{(-1)^s (a-1)!}{s! (a+n-s)!} =^? \frac{(-1)^n}{n! (a+n)}$ .

I’m using the symbol $=^?$ to emphasize that I haven’t proven that this equality is true. To be honest, I didn’t immediately believe that this worked; however, I was psychologically convinced after using Mathematica to compute this sum for about a dozen values of $n$ and $a$ .

To attempt a proof, we first note that if $n=0$ , then

$\displaystyle \sum_{s=0}^0 \frac{(-1)^s (a-1)!}{s! (a+0-s)!} = (-1)^0 \frac{(a-1)!}{0!a!} = \frac{(-1)^0}{0! (a+0)}$ ,

and so the equality works if $n=0$ . So, for the following, we will assume that $n \ge 1$ . I tried replacing $n$ with $n-a$ in the above equation to hopefully simplify the summation a little bit:

$\displaystyle \sum_{s=0}^{n-a} \frac{(-1)^s (a-1)!}{s! (a+n-a-s)!} =^? \frac{(-1)^{n-a}}{(n-a)! (a+n-a)}$

$\displaystyle \sum_{s=0}^{n-a} \frac{(-1)^s (a-1)!}{s! (n-s)!} =^? \frac{(-1)^{n-a}}{n \cdot (n-a)!}$

$\displaystyle \sum_{s=0}^{n-a} \frac{(-1)^s}{s! (n-s)!} =^? \frac{(-1)^{n-a}}{n \cdot (a-1)!(n-a)!}$

The left-hand side looks like a binomial coefficient, which suggest multiplying both sides by $n!$ :

$\displaystyle \sum_{s=0}^{n-a} (-1)^s \frac{n!}{s! (n-s)!} =^? \frac{(-1)^{n-a} n!}{n \cdot (a-1)!(n-a)!}$

$\displaystyle \sum_{s=0}^{n-a} (-1)^s \binom{n}{s} =^? \frac{(-1)^{n-a} (n-1)!}{(a-1)!(n-a)!}$

Surprise, surprise: the right-hand side is also a binomial coefficient since $(a-1)+(n-a) = n-1$ :

$\displaystyle \sum_{s=0}^{n-a} (-1)^s \binom{n}{s} =^? (-1)^{n-a} \binom{n-1}{n-a}$ .

Now we’re getting somewhere. To again make a sum a little simpler, let’s replace $a$ with $n-a$ :

$\displaystyle \sum_{s=0}^{n-(n-a)} (-1)^s \binom{n}{s} =^? (-1)^{n-(n-a)} \binom{n-1}{n-(n-a)}$

$\displaystyle \sum_{s=0}^a (-1)^s \binom{n}{s} =^? (-1)^a \binom{n-1}{a}$

Therefore, it appears that the confirming the complicated integral at the top of this post reduces to this equality involving binomial coefficients. In the next post, I’ll directly confirm this equality.

The Incomplete Gamma and Confluent Hypergeometric Functions (Part 3)

In the previous post, I confirmed the curious integral

$\displaystyle \int_0^z t^{a-1} e^{-t} \, dt = \displaystyle e^{-z} \sum_{s=0}^\infty \frac{(a-1)!}{(a+s)!} z^{a+s}$ ,

where the right-hand side is a special case of the confluent hypergeometric function when $a$ is a positive integer, by differentiating the right-hand side. However, the confirmation psychologically felt very unsatisfactory — we basically guessed the answer and then confirmed that it worked.

A seemingly better way to approach the integral is to use the Taylor series representation of $e^{-t}$ to integrate the left-hand side term-by-term:

$\displaystyle \int_0^z t^{a-1} e^{-t} \, dt = \int_0^z t^{a-1} \sum_{n=0}^\infty \frac{(-t)^n}{n!} \, dt$

$= \displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{n!} \int_0^z t^{a+n-1} \, dt$

$= \displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{n!} \frac{z^{a+n}}{a+n}$ .

Well, that doesn’t look like the right-hand side of the top equation. However, the right-hand side of the top equation also has a $e^{-z}$ in it. Let’s also convert that to its Taylor series expansion and then use the formula for multiplying two infinite series:

$\displaystyle e^{-z} \sum_{s=0}^\infty \frac{(a-1)!}{(a+s)!} z^{a+s} = \left( \sum_{s=0}^\infty \frac{(-z)^s}{s!} \right) \left( \sum_{s=0}^\infty \frac{(a-1)!}{(a+s)!} z^{a+s} \right)$

$= \displaystyle z^a \left( \sum_{s=0}^\infty \frac{(-1)^s z^s}{s!} \right) \left( \sum_{s=0}^\infty \frac{(a-1)!}{(a+s)!} z^s \right)$

$= \displaystyle z^a \sum_{n=0}^\infty \sum_{s=0}^n \frac{(-1)^s z^s}{s!} \frac{(a-1)!}{(a+n-s)!} z^{n-s}$

$= \displaystyle \sum_{n=0}^\infty \sum_{s=0}^n \frac{(-1)^s}{s!} \frac{(a-1)!}{(a+n-s)!} z^{a+n}$

Summarizing, apparently the following two infinite series are supposed to be equal:

$\displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{n!} \frac{z^{a+n}}{a+n} = \sum_{n=0}^\infty \sum_{s=0}^n \frac{(-1)^s}{s!} \frac{(a-1)!}{(a+n-s)!} z^{a+n}$ ,

or, matching coefficients of $z^{a+n}$ ,

$\displaystyle \frac{(-1)^n}{n! (a+n)} = \sum_{s=0}^n \frac{(-1)^s (a-1)!}{s! (a+n-s)!}$ .

When I first came to this equality, my immediate reaction was to throw up my hands and assume I made a calculation error someplace — I had a hard time believing that this sum from $s=0$ to $s=n$ was true. However, after using Mathematica to evaluate this sum for about a dozen different values of $n$ and $a$ , I was able to psychologically assure myself that this identity was somehow true.

But why does this awkward summation work? This is no longer a question about integration: it’s a question about a finite sum with factorials. I continue this exploration in the next post.

The Incomplete Gamma and Confluent Hypergeometric Functions (Part 2)

In this series of posts, I confirm this curious integral:

$\displaystyle \int_0^z t^{a-1} e^{-t} \, dt = \displaystyle \frac{z^a e^{-z}}{a} M(1, 1+a, z)$ ,

where the confluent hypergeometric function $M(a,b,z)$ is

$M(a,b,z) = \displaystyle 1+\sum_{s=1}^\infty \frac{a(a+1)\dots(a+s-1)}{b(b+1)\dots (b+s-1)} \frac{z^s}{s!}$ .

This integral can be confirmed — unsatisfactorily confirmed, but confirmed — by differentiating the right-hand side. For the sake of simplicity, I restrict my attention to the case when $a$ is a positive integer. To begin, the right-hand side is

$\displaystyle \frac{z^a e^{-z}}{a} M(1, 1+a, z) = \displaystyle \frac{z^a e^{-z}}{a} \left[1 + \sum_{s=1}^\infty \frac{1 \cdot 2 \cdot \dots \cdot s}{(a+1)(a+2)\dots (a+s)} \frac{z^s}{s!} \right]$

$= \displaystyle \frac{z^a e^{-z}}{a} \left[1 + \sum_{s=1}^\infty \frac{1}{(a+1)(a+2)\dots (a+s)} z^s \right]$

$= \displaystyle \frac{z^a e^{-z}}{a} \left[1 + \sum_{s=1}^\infty \frac{a!}{(a+s)!} z^s \right]$

$= \displaystyle \frac{z^a e^{-z}}{a} \sum_{s=0}^\infty \frac{a!}{(a+s)!} z^s$

$= \displaystyle e^{-z} \sum_{s=0}^\infty \frac{(a-1)!}{(a+s)!} z^{a+s}$ .

We now differentiate, first by using the Product Rule and then differentiating the series term-by-term (blatantly ignoring the need to confirm that term-by-term differentiation applies to this series):

$\displaystyle \frac{d}{dz} \left[\frac{z^a e^{-z}}{a} M(1, 1+a, z) \right] = \displaystyle \frac{d}{dz} \left[ e^{-z} \sum_{s=0}^\infty \frac{(a-1)!}{(a+s)!} z^{a+s} \right]$

$= -e^{-z} \displaystyle \sum_{s=0}^\infty \frac{(a-1)!}{(a+s)!} z^{a+s} + e^{-z} \frac{d}{dz} \left[ \sum_{s=0}^\infty \frac{(a-1)!}{(a+s)!} z^{a+s} \right]$

$= -e^{-z} \displaystyle \sum_{s=0}^\infty \frac{(a-1)!}{(a+s)!} z^{a+s} + e^{-z} \sum_{s=0}^\infty \frac{(a-1)!}{(a+s)!} \frac{d}{dz} z^{a+s}$

$= -e^{-z} \displaystyle \sum_{s=0}^\infty \frac{(a-1)!}{(a+s)!} z^{a+s} + e^{-z} \sum_{s=0}^\infty \frac{(a-1)!}{(a+s)!} (a+s) z^{a+s-1}$

$= -e^{-z} \displaystyle \sum_{s=0}^\infty \frac{(a-1)!}{(a+s)!} z^{a+s} + e^{-z} \sum_{s=0}^\infty \frac{(a-1)!}{(a+s-1)!} z^{a+s-1}$ .

We now shift the index of the first series:

$\displaystyle \frac{d}{dz} \left[\frac{z^a e^{-z}}{a} M(1, 1+a, z) \right] =-e^{-z} \sum_{s=1}^\infty \frac{(a-1)!}{(a+s-1)!} z^{a+s-1} + e^{-z} \sum_{s=0}^\infty \frac{(a-1)!}{(a+s-1)!} z^{a+s-1}$ .

By separating the $s=0$ term of the second series, the right-hand side becomes:

$\displaystyle -e^{-z} \sum_{s=1}^\infty \frac{(a-1)!}{(a+s-1)!} z^{a+s-1} + e^{-z} \frac{(a-1)!}{(a-1)!} z^{a-1} + e^{-z} \sum_{s=1}^\infty \frac{(a-1)!}{(a+s-1)!} z^{a+s-1}$ = e^{-z} z^{a-1}$

since the two infinite series cancel. We have thus shown that

$\displaystyle \frac{d}{dz} \left[\frac{z^a e^{-z}}{a} M(1, 1+a, z) \right] = \frac{e^{-z} z^{a-1}}{a}$ .

Therefore, we may integrate the right-hand side:

$\displaystyle \int_0^z t^{a-1} e^{-t} \, dt = \left[\frac{t^a e^{-t}}{a} M(1, 1+a, t) \right]_0^z$

$\displaystyle = \frac{z^a e^{-z}}{a} M(1, 1+a, z) - \frac{0^a e^{0}}{a} M(1, 1+a, 0)$

$\displaystyle = \frac{z^a e^{-z}}{a} M(1, 1+a, z)$ .

While this confirms the equality, this derivation still feels very unsatisfactory — we basically guessed the answer and then confirmed that it worked. In the next few posts, I’ll consider the direct verification of this series.

The Incomplete Gamma and Confluent Hypergeometric Functions (Part 1)

Yes, the title of this post is a mouthful.

While working on a research project, a trail of citations led me to this curious equality in the Digital Library of Mathematical Functions:

$\gamma(a,z) = \displaystyle \frac{z^a e^{-z}}{a} M(1, 1+a, z)$ ,

where the incomplete gamma function $\gamma(a,z)$ is

$\gamma(a,z) = \displaystyle \int_0^z t^{a-1} e^{-t} \, dt$

and the confluent hypergeometric function $M(a,b,z)$ is

$M(a,b,z) = \displaystyle 1+\sum_{s=1}^\infty \frac{a(a+1)\dots(a+s-1)}{b(b+1)\dots (b+s-1)} \frac{z^s}{s!}$ .

While I didn’t doubt that this was true — I don’t doubt this has been long established — I had an annoying problem: I didn’t really believe it. The gamma function

$\Gamma(a) = \displaystyle \int_0^\infty t^{a-1} e^{-t} \, dt$

is a well-known function with the famous property that

$\Gamma(n+1) = n!$

for non-negative integers $n$ ; this is often seen in calculus textbooks as an advanced challenge using integration by parts. The incomplete gamma function $\gamma(a,z)$ has the same look as $\Gamma(a)$ , except that the range of integration is from $0$ to $z$ (and not $\infty$ ). The gamma function appears all over the place in mathematics courses.

The confluent hypergeometric function, on the other hand, typically arises in mathematical physics as the solution of the differential equation

$z f''(z) + (b-z) f'(z) - af(z) = 0$ .

As I’m not a mathematical physicist, I won’t presume to state why this particular differential equation is important — except that it appears to be a niche equation that arises in very specialized applications.

So I had a hard time psychologically accepting that these two functions were in any way related.

While ultimately unimportant for advancing mathematics, this series will be about the journey I took to directly confirm the above equality.

Confirming Einstein’s Theory of General Relativity With Calculus, Part 9: Pedagogical Thoughts

At long last, we have reached the end of this series of posts.

The derivation is elementary; I’m confident that I could have understood this derivation had I seen it when I was in high school. That said, the word “elementary” in mathematics can be a bit loaded — this means that it is based on simple ideas that are perhaps used in a profound and surprising way. Perhaps my favorite quote along these lines was this understated gem from the book Three Pearls of Number Theory after the conclusion of a very complicated proof in Chapter 1:

You see how complicated an entirely elementary construction can sometimes be. And yet this is not an extreme case; in the next chapter you will encounter just as elementary a construction which is considerably more complicated.

Here are the elementary ideas from calculus, precalculus, and high school physics that were used in this series:

Physics
- Conservation of angular momentum
- Newton’s Second Law
- Newton’s Law of Gravitation
Precalculus
- Completing the square
- Quadratic formula
- Factoring polynomials
- Complex roots of polynomials
- Bounds on $\cos \theta$ and $\sin \theta$
- Period of $\cos \theta$ and $\sin \theta$
- Zeroes of $\cos \theta$ and $\sin \theta$
- Trigonometric identities (Pythagorean, sum and difference, double-angle)
- Conic sections
- Graphing in polar coordinates
- Two-dimensional vectors
- Dot products of two-dimensional vectors (especially perpendicular vectors)
- Euler’s equation
Calculus
- The Chain Rule
- Derivatives of $\cos \theta$ and $\sin \theta$
- Linearizations of $\cos x$ , $\sin x$ , and $1/(1-x)$ near $x \approx 0$ (or, more generally, their Taylor series approximations)
- Derivative of $e^x$
- Solving initial-value problems
- Integration by $u-$ substitution

While these ideas from calculus are elementary, they were certainly used in clever and unusual ways throughout the derivation.

I should add that although the derivation was elementary, certain parts of the derivation could be made easier by appealing to standard concepts from differential equations.

One more thought. While this series of post was inspired by a calculation that appeared in an undergraduate physics textbook, I had thought that this series might be worthy of publication in a mathematical journal as an historical example of an important problem that can be solved by elementary tools. Unfortunately for me, Hieu D. Nguyen’s terrific article Rearing Its Ugly Head: The Cosmological Constant and Newton’s Greatest Blunder in The American Mathematical Monthly is already in the record.

Confirming Einstein’s Theory of General Relativity With Calculus, Part 6i: Rationale for Method of Undetermined Coefficients VI

In this series, I’m discussing how ideas from calculus and precalculus (with a touch of differential equations) can predict the precession in Mercury’s orbit and thus confirm Einstein’s theory of general relativity. The origins of this series came from a class project that I assigned to my Differential Equations students maybe 20 years ago.

In the last few posts, I’ve used a standard technique from differential equations: to solve the $n$ th order homogeneous differential equation with constant coefficients

$a_n y^{(n)} + \dots + a_3 y''' + a_2 y'' + a_1 y' + a_0 y = 0$ ,

we first solve the characteristic equation

$a_n r^n + \dots + a_3 r^3 + a_2 r^2 + a_1 r + a_0 = 0$

using techniques from Precalculus. The form of the roots $r$ determines the solutions of the differential equation.

While this is a standard technique from differential equations, the perspective I’m taking in this series is scaffolding the techniques used to predict the precession in a planet’s orbit using only techniques from Calculus and Precalculus. So let me discuss why the above technique works, assuming that the characteristic equation does not have repeated roots. (The repeated roots case is a little more complicated but is not needed for the present series of posts.)

We begin by guessing that the above differential equation has a solution of the form $y = e^{rt}$ . Differentiating, we find $y' = re^{rt}$ , $y'' = r^2 e^{rt}$ , etc. Therefore, the differential equation becomes

$a_n r^n e^{rt} + \dots + a_3 r^3 e^{rt} + a_2 r^2 e^{rt} + a_1 r e^{rt} + a_0 e^{rt} = 0$

$e^{rt} \left(a_n r^n + \dots + a_3 r^3 + a_2 r^2 + a_1 r + a_0 \right) = 0$

$a_n r^n + \dots + a_3 r^3 + a_2 r^2 + a_1 r + a_0 = 0$

The last step does not “lose” any possible solutions for $r$ since $e^{rt}$ can never be equal to $0$ . Therefore, solving the differential equation reduces to finding the roots of this polynomial, which can be done using standard techniques from Precalculus.

For example, one of the differential equations that we’ve encountered is $y''+y=0$ . The characteristic equation is $r^2+1=0$ , which has roots $r=\pm i$ . Therefore, two solutions to the differential equation are $e^{it}$ and $e^{-it}$ , so that the general solution is

$y = c_1 e^{it} + c_2 e^{-it}$ .

To write this in a more conventional way, we use Euler’s formula $e^{ix} = \cos x + i \sin x$ , so that

$y = c_1 (\cos t + i \sin t) + c_2 (\cos (-t) + i \sin (-t))$

$= c_1 \cos t + i c_1 \sin t + c_2 \cos t - i c_2 \sin t$

$= (c_1 + c_2) \cos t + (ic_1 - ic_2) \sin t$

$= C_1 \cos t + C_2 \sin t$ .

Likewise, in the previous post, we encountered the fourth-order differential equation $y^{(4)}+5y''+4y = 0$ . To find the roots of the characteristic equation, we factor:

$r^4 + 5r^2 + 4r = 0$

$(r^2+1)(r^2+4) = 0$

$r^2 +1 = 0 \qquad \hbox{or} \qquad \hbox{or} r^2 + 4 = 0$

$r = \pm i \qquad \hbox{or} \qquad r = \pm 2i$ .

Therefore, four solutions of this differential equation are $e^{it}$ , $e^{-it}$ , $e^{2it}$ , and $e^{-2it}$ , so that the general solution is

$y = c_1 e^{it} + c_2 e^{-it} + c_3 e^{2it} + c_4 e{-2it}$ .

Using Euler’s formula as before, this can be rewritten as

$y = C_1 \cos t + C_2 \sin t + C_3 \cos 2t + C_4 \sin 2t$ .

Engaging students: Exponential Growth and Decay

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Ashlyn Farley. Her topic, from Precalculus: exponential growth and decay.

The most current example of exponential growth and decay is with the global pandemic, Covid 19. One example is that The Washington Post wrote an article stating that “The spread of coronavirus boils down to a simple math lesson.” The article goes on to explain what exponential growth is and how that applies to Covid 19. Another website, ourworldindata.org, has a graph of the daily new cases of Covid 19. This graph allows one to see the information for multiple countries, and starts on January 28th 2020 until Today, whatever day that you may be viewing it. Many other news sources also have graphs and information on the growth, and decay in some cases, of the pandemic situation. Teachers can use this information to easily make a connection from math class to the real world.

One idea of teaching graphing exponential functions so that it is engaging is to use a project over the zombie apocalypse. The spread of a disease is a common and great example of exponential functions, so although this disease is pretend, the idea can be applied in the real world, like with a global pandemic. Three examples of projects are:

News Reporters
- This project has the students analyzing data they received to best report to the people who are dealing with the outbreak. It allows students to learn how to read the graphs of exponential functions, understand the functions, integrate technology into the class by creating news reports, and practice an actual career.
Government Officials
- This project has the students running a simulation of their city. They are to use the statistics of a city to see what the impact of a zombie outbreak would be. After finding the best and worst case scenario, they are to write a letter to the mayor of the city that explains the scenarios so that government can implement plans to keep the outbreak to a minimum. This allows the students a chance to practice analyzing exponential functions, modifying exponential functions, and informing others of the meaning of the functions and modifications.
Scientists
- This project has the students predicting the outcome of a zombie outbreak, finding a cure, and determining at what point is the zombie population controlled. The students will get practice with the exponential functions, making changes to the functions, finding the point of “control”, as well as creating an action plan.

Each of these projects can be used separately or can be combined to create one major project to learn about exponential functions and their graphs. The goal is to get students excited about learning math instead of dreading it. Math is used daily, even if the students don’t realize it, so the understanding of real-life implications is very important for a teacher to bring into the classroom.

Of the many websites, one key website for educators trying to make lessons engaging is YouTube. YouTube has songs, such as the Exponential Function Music Video, explanatory videos, such as from Kahn Academy, and allows students to create their own videos about the topics. Explanatory videos may help students get a specific idea they didn’t quite understand in class, music is very catchy allowing quick memorization of information, and creating videos shows that the students truly have an understanding of the material. By giving the students multiple types of representation of the material, allows all types of learners a chance to understand the material. Multiple representations is very important in keeping students engaged in the class and having them truly learn the material.

Resources:

https://www.teacherspayteachers.com/Product/Zombie-Apocalypse-Exponential-Function-Pandemics-21st-Century-Math-Project-767712?epik=dj0yJnU9UnRuNHVLLUxrV0JkTVJQc1ZFY0szb3JJNXRyenQwb2omcD0wJm49aEQ2UjFHVUcyYm5FakE1ZXhSXzhpQSZ0PUFBQUFBR0ZnTWRB

https://medium.com/innovative-instruction/math-mini-project-idea-the-zombie-apocalypse-5ddd0e6af389#.sph1x08k8

https://www.teacherspayteachers.com/Product/Exponential-Growth-and-Decay-Activity-Exponential-Functions-Zombie-Apocalypse-2609226

https://ourworldindata.org/coronavirus/country/united-states

https://www.washingtonpost.com/weather/2020/03/27/what-does-exponential-growth-mean-context-covid-19/

https://www.youtube.com/watch?v=txMFwOLjZjQ

Engaging students: Solving exponential equations

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Austin Stone. His topic, from Precalculus: solving exponential equations.

What interesting (i.e., uncontrived) word problems using this topic can your students do now?

Exponential equations can be used in lots of different kinds of word problems. One that is pretty common but is very useful for students involves interest rate. “Megan has $20,000 to invest for 5 years and she found an interest rate of 5%. How much money will she have at the end of 5 years if the interest rate compounds monthly?” I would give them the formula A=P(1+r/n)^rt. It is pretty easy to convince students that this is a real-world problem and would get the students engaged about exponential equations. You can also reword the problem to ask for how much Megan started with, what the rate is, or how much time the money was in there. That way students get used to solving equations when the variable is in the exponent and when it is not. This also can lead into or us prior knowledge of natural log to solve for the variable in the exponent.

How could you as a teacher create an activity or project that involves your topic?

Using the basis of the problem I mentioned above, a teacher could create a Project Based Instruction lesson using this idea. The teacher can set up a scenario where, over the course of a week or two, the students would have to decide which bank to make an investment in by calculating how much money they would profit at each bank. The students would have to research different banks and their interest rate. The teacher could also give each group different scenarios where some groups have more money to invest. Students would have to figure out how long they would like to invest. The teacher would give Do It Yourselves and Workshops that deal with solving exponential equations and also getting used to natural log. They would then make a presentation explaining what bank they have chosen and why. They would also have to explain the math that they would have used.

How has this topic appeared in the news?

To say that exponential equations have been in the news lately would be an understatement. It has virtually been the news this year. COVID-19 is a virus and viruses spread exponentially. This would get students engaged immediately because the topic would be relatable to their own lives. Doctors and scientists try to figure out different ways to “flatten the curve”, which essentially means to make the spread of the virus not exponential anymore. We have all heard people on the news telling the public how to stop the virus from spreading and how not make people around you at risk of contracting it (contributing the exponential spread). We all have most likely seen a doctor or scientist show a graph of the virus’s spread and their predictions on how it will look in the upcoming weeks. This would give students a chance to see that what they are learning can be applied to very crucial things going on in the world around them.

References

Exponential Functions