The Incomplete Gamma and Confluent Hypergeometric Functions (Part 7)

We are finally in position to directly prove 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.

Step 1. Our surprising “starting” point is the combinatorical identity

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

which we proved in the previous post.

Step 2. We now begin to manipulate this identity. Replacing $a$ with $n-a$ , we find

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

for $0 \le n-a \le n-1$ , or $1 \le a \le n$ . Therefore,

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

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

If we replace $n$ by $a+n$ , we obtain

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

for $a \ge 1$ (since we are guaranteed that $a \le a+n$ if $a \ge 1$ ).

Step 3. We now turn to the direct integration of the incomplete gamma function:

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

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

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

From the work in Step 2, we may substitute:

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

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

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

We now use the formula for the product of two power series:

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

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

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

I’ll venture to say that there are several steps with this deductive derivation that look positively miraculous. However, when we were working backwards in the previous posts of this series, these miraculous steps were actually logical next steps.

And that’s the end of the direct computation of the incomplete gamma function.

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.

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

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.

We have shown that the motion of a planet around the Sun, expressed in polar coordinates $(r,\theta)$ with the Sun at the origin, under general relativity follows the initial-value problem

$u''(\theta) + u(\theta) = \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2} + \frac{2\delta \epsilon \cos \theta}{\alpha^2} + \frac{\delta \epsilon^2 \cos 2\theta}{2\alpha^2}$ ,

$u(0) = \displaystyle \frac{1}{P}$ ,

$u'(0) = 0$ ,

where $u = \displaystyle \frac{1}{r}$ , $\displaystyle \frac{1}{\alpha} = \frac{GMm^2}{\ell^2}$ , $\delta = \displaystyle \frac{3GM}{c^2}$ , $G$ is the gravitational constant of the universe, $m$ is the mass of the planet, $M$ is the mass of the Sun, $\ell$ is the constant angular momentum of the planet, $c$ is the speed of light, and $P$ is the smallest distance of the planet from the Sun during its orbit (i.e., at perihelion).

In this post, we will use the guesses

$u(\theta) = f(\theta) \cos \theta \qquad \hbox{or} u(\theta) = f(\theta) \sin \theta$

that arose from the technique/trick of reduction of order, where $f(\theta)$ is some unknown function, to find the general solution of the differential equation

$u^{(4)} + 2u'' + u = 0$ .

To do this, we will need to use the Product Rule for higher-order derivatives that was derived in the previous post:

$(fg)'' = f'' g + 2 f' g' + f g''$

and

$(fg)^{(4)} = f^{(4)} g + 4 f''' g' + 6 f'' g'' + 4f' g''' + f g^{(4)}$ .

In these formulas, Pascal’s triangle makes a somewhat surprising appearance; indeed, this pattern can be proven with mathematical induction.

We begin with $u(\theta) = f(\theta) \cos \theta$ . If $g(\theta) = \cos \theta$ , then

$g'(\theta) = - \sin \theta$ ,

$g''(\theta) = -\cos \theta$ ,

$g'''(\theta) = \sin \theta$ ,

$g^{(4)}(\theta) = \cos \theta$ .

Substituting into the fourth-order differential equation, we find the differential equation becomes

$(f \cos \theta)^{(4)} + 2 (f \cos \theta)'' + f \cos \theta = 0$

$f^{(4)} \cos \theta - 4 f''' \sin \theta - 6 f'' \cos \theta + 4 f' \sin \theta + f \cos \theta + 2 f'' \cos \theta - 4 f' \sin \theta - 2 f \cos \theta + f \cos \theta = 0$

$f^{(4)} \cos \theta - 4 f''' \sin \theta - 6 f'' \cos \theta + 2 f'' \cos \theta = 0$

$f^{(4)} \cos \theta - 4 f''' \sin \theta - 4 f'' \cos \theta = 0$

The important observation is that the terms containing $f$ and $f'$ cancelled each other. This new differential equation doesn’t look like much of an improvement over the original fourth-order differential equation, but we can make a key observation: if $f'' = 0$ , then differentiating twice more trivially yields $f''' = 0$ and $f^{(4)} = 0$ . Said another way: if $f'' = 0$ , then $u(\theta) = f(\theta) \cos \theta$ will be a solution of the original differential equation.

Integrating twice, we can find $f$ :

$f''(\theta) = 0$

$f'(\theta) = c_1$

$f(\theta) = c_1 \theta + c_2$ .

Therefore, a solution of the original differential equation will be

$u(\theta) = c_1 \theta \cos \theta + c_2 \cos \theta$ .

We now repeat the logic for $u(\theta) = f(\theta) \sin \theta$ :

$(f \sin \theta)^{(4)} + 2 (f \sin \theta)'' + f \sin \theta = 0$

$f^{(4)} \sin \theta + 4 f''' \cos \theta - 6 f'' \sin \theta - 4 f' \cos\theta + f \sin \theta + 2 f'' \sin \theta + 4 f' \cos \theta - 2 f \sin \theta + f \sin \theta = 0$

$f^{(4)} \sin\theta + 4 f''' \cos \theta - 6 f'' \sin \theta + 2 f'' \sin \theta = 0$

$f^{(4)} \sin\theta - 4 f''' \cos\theta - 4 f'' \sin\theta = 0$ .

Once again, a solution of this new differential equation will be $f(\theta) = c_3 \theta + c_4$ , so that $f'' = f''' = f^{(4)} = 0$ . Therefore, another solution of the original differential equation will be

$u(\theta) = c_3 \theta \sin \theta + c_4 \sin \theta$ .

Adding these provides the general solution of the differential equation:

$u(\theta) = c_1 \theta \cos \theta + c_2 \cos \theta + c_3 \theta \sin \theta + c_4 \sin \theta$ .

Except for the order of the constants, this matches the solution that was presented earlier by using techniques taught in a proper course in differential equations.

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

We have shown that the motion of a planet around the Sun, expressed in polar coordinates $(r,\theta)$ with the Sun at the origin, under general relativity follows the initial-value problem

$u(0) = \displaystyle \frac{1}{P}$ ,

$u'(0) = 0$ ,

In the previous post, I used a standard technique from differential equations to find the general solution of

$u^{(4)} + 2u'' + u = 0$ .

to be

$u(theta) = c_1 \cos \theta + c_2 \sin \theta + c_3 \theta \cos \theta + c_4 \theta \sin \theta$ .

However, as much as possible in this series, I want to take the perspective of a talented calculus student who has not yet taken differential equations — so that the conclusion above is far from obvious. How could this be reasonable coaxed out of such a student?

To begin, we observe that the characteristic equation is

$r^4 + 2r^2 + 1 = 0$ ,

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

Clearly this has the same roots as the simpler equation $r^2 + 1 = 0$ , which corresponds to the second-order differential equation $u'' + u = 0$ . We’ve already seen that $u_1(\theta) = \cos \theta$ and $u_2(\theta) = \sin \theta$ are solutions of this differential equation; perhaps they might also be solutions of the more complicated differential equation also? The answer, of course, is yes:

$u_1^{(4)} + 2 u_1'' + u_1 = \cos \theta - 2 \cos \theta + \cos \theta = 0$

and

$u_2^{(4)} + 2u_2'' + u_2 = \sin \theta - 2 \sin \theta + \sin \theta = 0$ .

The far trickier part is finding the two additional solutions. To find these, we use a standard trick/technique called reduction of order. In this technique, we guess that any additional solutions much have the form of either

$u(\theta) = f(\theta) \cos \theta \qquad \hbox{or} \qquad u(\theta) = f(\theta) \sin \theta$ ,

where $f(\theta)$ is some unknown function that we’re multiplying by the solutions we already have. We then substitute this into the differential equation $u^{(4)} + 2u'' + u = 0$ to form a new differential equation for the unknown $f$ , which we can (hopefully) solve.

Doing this will require multiple applications of the Product Rule for differentiation. We already know that

$(fg)' = f' g + f g'$ .

We now differentiate again, using the Product Rule, to find $(fg)''$ :

$(fg)'' = ( [fg]')' = (f'g)' + (fg')'$

$= f''g + f' g' + f' g' + f g''$

$= f'' g + 2 f' g' + f g''$ .

We now differential twice more to find $(fg)^{(4)}$ :

$(fg)''' = ( [fg]'')' = (f''g)' + 2(f'g')' + (fg'')'$

$= f'''g + f'' g' + 2f'' g' + 2f' g'' + f' g'' + f g'''$

$= f''' g + 3 f'' g' + 3 f' g'' + f g'''$ .

A good student may be able to guess the pattern for the next derivative:

$(fg)^{(4)} = ( [fg]''')' = (f'''g)' + 3(f''g')' +3(f'g'')' + (fg''')'$

$= f^{(4)}g + f''' g' + 3f''' g' + 3f'' g'' + 3f'' g'' + 3f'g''' + f' g''' + f g^{(4)}$

$= f^{(4)} g + 4 f''' g' + 6 f'' g'' + 4f' g''' + f g^{(4)}$ .

In this way, Pascal’s triangle makes a somewhat surprising appearance; indeed, this pattern can be proven with mathematical induction.

In the next post, we’ll apply this to the solution of the fourth-order differential equation.

Engaging students: Using Pascal’s triangle

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 Jaeda Ransom. Her topic, from Precalculus: using Pascal’s triangle.

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

A great activity that involves Pascal’s Triangle would be the sticky note triangle activity. For this activity students will be recreating an enlarged version of Pascal’s Triangle. To complete this activity students will need a poster of Pascal’s Triangle, poster board, markers, sticky notes, classroom wall (optional), and tape (optional). The teacher’s role is to show students Pascal’s Triangle, along with an explanation of how it was made. Students will be working in pairs and grabbing the necessary materials needed to complete this activity.On the poster board the students will recreate Pascal’s Triangle. Students will write a number 1 on a sticky note and place it at the top of the posterboard, they will then write 2 number 1’s on a sticky note and place it directly under. The students will continue recreating the triangle on their poster board until they run out of space. You can also consider having students use smaller sticky notes so that students are engaged with creating more rows.

What interesting things can you say about the people who contributed to the discovery and/or the development of this topic?

Pascal’s Triangle was named after French mathematician Blaise Pascal. At just the age of 16 years old Pascal wrote a significant treatise on the subject of projective geometry marking him as a child prodigy. Amongst that, Pascal also corresponded with other mathematicians on probability theory, which vastly encouraged the development of modern economics and social science. Pascal was also one of the first two inventors of the mechanical calculator when he started pioneering work on calculating machines, these were called Pascal’s calculators and later Pascalines. Pascal impressively created and invented all of this as a teenager. Though the Pascal Triangle was named after Blaise Pascal, this theory was established well before Pascal in India, Persia, China, Germany, and Italy. As a matter of fact, in China they still call it the Yang Hui’s triangle, named after Chinese mathematician Yang Hui who presented the triangle in the 13th century, though the triangle was known in China since the early 11th century.

How can this topic be used in your students’ future courses in mathematics or science?

This topic can be used in my students future mathematics course to introduce binomial expansions, where it is known that Pascal’s Triangle determines the coefficients that arise in binomial expansion. The coefficients aᵢ in a binomial expansion represents the number of row n in the Pascal’s Triangle. Thus, $a_i = \displaystyle {n \choose i}$ .

Another useful application of this topic is in the calculations of combinations. The equation to find the combination is also the formula to find a cell for Pascal’s Triangle. So, instead of performing the calculations using the equation a student can simply use Pascal’s Triangle. In doing this you can continue a lesson over probability or even do an activity using Pascal’s Triangle while implicating probability questions.

Resources:

https://en.wikipedia.org/wiki/Pascal%27s_triangle#Formula

https://study.com/academy/lesson/pascals-triangle-activities-games.html

Slightly Incorrect Ugly Mathematical Christmas T-Shirts: Index

I’m doing something that I should have done a long time ago: collecting a series of posts into one single post. The following links comprised my series on slightly incorrect ugly mathematical Christmas T-shirts.

Part 1: Missing digits in the expansion of $\pi$ .

Part 2: Incorrect computation of Pascal’s triangle.

Part 3: Incorrect name of Pascal’s triangle.

Slightly Incorrect Ugly Mathematical Christmas T-Shirts: Part 2

This was another T-shirt that I found in my search for the perfect ugly mathematical Christmas sweater: https://www.amazon.com/Pascals-Triangle-Math-Christmas-shirt/dp/B07KJS5SM2/I love the artistry of this shirt; the “ornaments” at the corners of the hexagons and the presents under the tree are nice touches.

There’s only one small problem:

$\displaystyle {8 \choose 3} = \displaystyle {8 \choose 5} = \displaystyle \frac{8!}{3! \times 5!} = 56$ .

Oops.

Engaging students: Using Pascal’s triangle

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 Rachel Delflache. Her topic, from Precalculus: using Pascal’s triangle.

How does this topic expand what your students would have learned in previous courses?

In previous courses students have learned how to expand binomials, however after $(x+y)^3$ the process of expanding the binomial by hand can become tedious. Pascal’s triangle allows for a simpler way to expand binomials. When counting the rows, the top row is row 0, and is equal to one. This correlates to $(x+y)^0 =1$ . Similarly, row 2 is 1 2 1, correlating to $(x+y)^2 = 1x^2 + 2xy + 1y^2$ . The pattern can be used to find any binomial expansion, as long as the correct row is found. The powers in each term also follow a pattern, for example look at $(x+y)^4$ :

$1x^4y^0 + 4x^3y^1 + 6x^2y^2 + 4x^1y^3 + 1x^0y^4$

In this expansion it can be seen that in the first term of the expansion the first monomial is raised to the original power, and in each term the power of the first monomial decreases by one. Conversely, the second monomial is raised to the power of 0 in the first term of the expansion, and increases by a power of 1 for each subsequent term in the expansion until it is equal to the original power of the binomial.

Sierpinski’s Triangle is triangle that was characterized by Wacław Sieriński in 1915. Sierpinski’s triangle is a fractal of an equilateral triangle which is subdivided recursively. A fractal is a design that is geometrically constructed so that it is similar to itself at different angles. In this particular construction, the original shape is an equilateral triangle which is subdivided into four smaller triangles. Then the middle triangle is whited out. Each black triangle is then subdivided again, and the patter continues as illustrated below.

Sierpinski’s triangle can be created using Pascal’s triangle by shading in the odd numbers and leaving the even numbers white. The following video shows this creation in practice.

What are the contributions of various cultures to this topic?

The pattern of Pascal’s triangle can be seen as far back as the 11th century. In the 11th century Pascal’s triangle was studied in both Persia and China by Oman Khayyam and Jia Xian, respectively. While Xian did not study Pascal’s triangle exactly, he did study a triangular representation of coefficients. Xian’s triangle was further studied in 13th century China by Yang Hui, who made it more widely known, which is why Pascal’s triangle is commonly called the Yanghui triangle in China. Pascal’s triangle was later studies in the 17th century by Blaise Pascal, for whom it was named for. While Pascal did not discover the number patter, he did discover many new uses for the pattern which were published in his book Traité du Triangle Arithméthique. It is due to the discovery of these uses that the triangle was named for Pascal.

Reference:
• https://en.wikipedia.org/wiki/Pascal%27s_triangle
• http://mathforum.org/workshops/usi/pascal/images/fill.comb.gif
• https://www.britannica.com/biography/Blaise-Pascal#toc445406main
• https://en.wikipedia.org/wiki/Sierpinski_triangle