My Favorite One-Liners: Part 35

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

Every once in a while, I’ll discuss something in class which is at least tangentially related to an unsolved problems in mathematics. For example, when discussing infinite series, I’ll ask my students to debate whether or not this series converges:

1 + \frac{1}{10} + \frac{1}{100} + \frac{1}{1000} + \dots

Of course, this one converges since it’s an infinite geometric series. Then we’ll move on to an infinite series that is not geometric:

1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \dots,

where the denominators are all perfect squares. I’ll have my students guess about whether or not this one converges. It turns out that it does, and the answer is exactly what my students should expect the answer to be, \pi^2/6.

Then I tell my students, that was a joke (usually to relieved laughter).

Next, I’ll put up the series

1 + \frac{1}{8} + \frac{1}{27} + \frac{1}{64} + \dots,

where the denominators are all perfect cubes. I’ll have my students guess about whether or not this one converges. Usually someone will see that this one has to converge since the previous one converged and the terms of this one are pairwise smaller than the previous series — an intuitive use of the Dominated Convergence Test. Then, I’ll ask, what does this converge to?

The answer is, nobody knows. It can be calculated to very high precision with modern computers, of course, but it’s unknown whether there’s a simple expression for this sum.

So, concluding the story whenever I present an unsolved problem, I’ll tell my students,

If you figure out the answer, call me, and call me collect.

My Favorite One-Liners: Part 20

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

Perhaps the world’s most famous infinite series is

S = \displaystyle \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots

as this is the subject of Zeno’s paradox. When I teach infinite series in class, I often engage the students by reminding students about Zeno’s paradox and then show them this clip from the 1994 movie I.Q.

This clip is almost always a big hit with my students. After showing this clip, I’ll conclude, “When I was single, this was part of my repertoire of pick-up lines… but it never worked.”

Even after showing this clip, some students resist the idea that an infinite series can have a finite answer. For such students, I use a physical demonstration: I walk half-way across the classroom, then a quarter, and so on… until I walk head-first into a walk at full walking speed. The resulting loud thud usually confirms for students that an infinite sum can indeed have a finite answer.

For further reading, see my series on arithmetic and geometric series.

 

 

What I Learned from Reading “Gamma: Exploring Euler’s Constant” by Julian Havil: Part 3

At the time of this writing, it is unknown if there are infinitely many twin primes, which are prime numbers that differ by 2 (like 3 and 5, 5 and 7, 11 and 13, 17 and 19, etc.) However, significant progress has been made in recent years. However, it is known (Gamma, page 30) the sum of the reciprocals of the twin primes converges:

\displaystyle \left( \frac{1}{3} + \frac{1}{5} \right) + \left( \frac{1}{5} + \frac{1}{7} \right) + \left( \frac{1}{11} + \frac{1}{13} \right) + \left( \frac{1}{17} + \frac{1}{19} \right) = 1.9021605824\dots.

This constant is known as Brun’s constant (see also Mathworld). In the process of computing this number, the infamous 1994 Pentium bug was found.

Although this sum is finite, it’s still unknown if there are infinitely many twin primes since it’s possible for an infinite sum to converge (like a geometric series).

green line

When I researching for my series of posts on conditional convergence, especially examples related to the constant \gamma, the reference Gamma: Exploring Euler’s Constant by Julian Havil kept popping up. Finally, I decided to splurge for the book, expecting a decent popular account of this number. After all, I’m a professional mathematician, and I took a graduate level class in analytic number theory. In short, I don’t expect to learn a whole lot when reading a popular science book other than perhaps some new pedagogical insights.

Boy, was I wrong. As I turned every page, it seemed I hit a new factoid that I had not known before.

In this series, I’d like to compile some of my favorites — while giving the book a very high recommendation.

Another poorly written word problem (Part 9)

Textbooks have included the occasional awful problem ever since Pebbles Flintstone and Bamm-Bamm Rubble chiseled their homework on slate tablets while attending Bedrock Elementary. But even with the understanding that there have been children have been doing awful homework problems since the dawn of time (and long before the advent of the Common Core), this one is a doozy.

While the ball is on the 20-yard line, a defensive end is suddenly cursed so that he commits a penalty every down that causes the following:

a. The ball is moved half the distance to the goal line, and
b. The down is replayed.

Show that the ball will eventually travel the entire 20-yard distance to the goal.

Sigh. The textbook expects students to use the formula for an infinite geometric series

\displaystyle \sum_{n=0}^\infty ar^n = \displaystyle \frac{a}{1-r}

with a = 10 and r = 0.5. However, this series only works if there are an infinite number of terms, so that any finite partial sum will be less than 20. Therefore, saying that the ball “will eventually travel” all 20 yards is misleading, as this implies that this happens after a finite amount of time.

 

 

Useless Numerology for 2016: Index

I’m doing something that I should have done a long time ago: collecting a series of posts into one single post.

Part 1: Introduction.

Part 2: 2016 = 2^{10}+2^9+2^8+2^7+2^6+2^5 = 2^{11} - 2^5.

Part 3: 2016 = 1 + 2 + \dots + 63.

Part 4: 2016 = (1 + 2 + \dots + 9)^2 - (1+2)^2 = 3^3 + 4^3 + 5^3 + 6^3 + 7^3 + 8^3 + 9^3

Part 5: 2016 = \displaystyle \sum_{n=0}^{63} (-1)^{n+1} n^2

 

 

 

Zeno’s Hairodox

Source: http://joegp.com/halfcut/

How I Impressed My Wife: Part 4f

Previously in this series, I have used two different techniques to show that

Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x} = \displaystyle \frac{2\pi}{|b|}.

Originally, my wife had asked me to compute this integral by hand because Mathematica 4 and Mathematica 8 gave different answers. At the time, I eventually obtained the solution by multiplying the top and bottom of the integrand by \sec^2 x and then employing the substitution u = \tan x (after using trig identities to adjust the limits of integration).
But this wasn’t the only method I tried. Indeed, I tried two or three different methods before deciding they were too messy and trying something different. So, for the rest of this series, I’d like to explore different ways that the above integral can be computed.
green linePreviously in this series, I have used two different techniques to show that

Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x} = \displaystyle \frac{2\pi}{|b|}.

Originally, my wife had asked me to compute this integral by hand because Mathematica 4 and Mathematica 8 gave different answers. At the time, I eventually obtained the solution by multiplying the top and bottom of the integrand by \sec^2 x and then employing the substitution u = \tan x (after using trig identities to adjust the limits of integration).
But this wasn’t the only method I tried. Indeed, I tried two or three different methods before deciding they were too messy and trying something different. So, for the rest of this series, I’d like to explore different ways that the above integral can be computed.
green lineHere’s my progress so far:

Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}

= \displaystyle \int_0^{2\pi} \frac{2 \, dx}{1+\cos 2x + 2 a \sin 2x + (a^2 + b^2)(1-\cos 2x)}

= 2 \displaystyle \int_0^{2\pi} \frac{d\theta}{(1+a^2+b^2) + 2 a \sin \theta + (1 - a^2 - b^2) \cos \theta}

= 2 \displaystyle \int_{0}^{2\pi} \frac{d\theta}{S + R \cos (\theta - \alpha)}

= 2 \displaystyle \int_{0}^{2\pi} \frac{d\phi}{S + R \cos \phi}

= \displaystyle -\frac{4i}{R} \oint_C \frac{dz}{z^2 + 2\frac{S}{R}z + 1},

where this last integral is taken over the complex plane on the unit circle, a closed contour oriented counterclockwise. In these formulas, R = \sqrt{(2a)^2 + (1-a^2-b^2)^2} and S = 1 + a^2 + b^2. (Also, \alpha is a certain angle that is now irrelevant at this point in the calculation).

This contour integral looks complicated; however, it’s an amazing fact that integrals over closed contours can be easily evaluated by only looking at the poles of the integrand. In recent posts, I established that there was only one pole inside the contour, and the residue at this pole was equal to \displaystyle \frac{R}{ 2 \sqrt{S^2-R^2} }.

This residue can be used to evaluate the contour integral. Ordinarily, integrals are computed by subtracting the values of the antiderivative at the endpoints. However, there is an alternate way of computing a contour integral using residues. It turns out that the value of the contour integral is 2\pi i times the sum of the residues within the contour; see Wikipedia and Mathworld for more information.

Therefore,

Q = \displaystyle -\frac{4i}{R} \oint_C \frac{dz}{(z - r_1)(z- r_2)}

= \displaystyle -\frac{4i}{R} \cdot 2\pi i \cdot \frac{R}{ 2 \sqrt{S^2-R^2} }

= \displaystyle \frac{4\pi}{\sqrt{S^2-R^2}}

Next, I use some algebra to simplify the denominator:

S^2 - R^2 = (1+a^2+b^2)^2 - (1-a^2-b^2)^2 - (2a)^2

S^2 - R^2 = [(1 + a^2 + b^2) + (1-a^2-b^2)][(1 + a^2 + b^2) - (1 - a^2 -b^2)] - 4a^2

S^2 - R^2 = 2[2 a^2 + 2b^2] - 4a^2

S^2 - R^2 = 4b^2

Therefore,

Q = \displaystyle \frac{4\pi}{\sqrt{4b^2}} = \displaystyle \frac{4\pi}{2|b|} = \frac{2\pi}{|b|}

Once again, this matches the solution found with the previous methods… and I was careful to avoid a common algebraic mistake.

green lineIn tomorrow’s post, I’ll discuss an alternative way of computing the residue.

How I Impressed My Wife: Part 4e

Previously in this series, I have used two different techniques to show that

Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x} = \displaystyle \frac{2\pi}{|b|}.

Originally, my wife had asked me to compute this integral by hand because Mathematica 4 and Mathematica 8 gave different answers. At the time, I eventually obtained the solution by multiplying the top and bottom of the integrand by \sec^2 x and then employing the substitution u = \tan x (after using trig identities to adjust the limits of integration).
But this wasn’t the only method I tried. Indeed, I tried two or three different methods before deciding they were too messy and trying something different. So, for the rest of this series, I’d like to explore different ways that the above integral can be computed.
green lineHere’s my progress so far:

Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}

= \displaystyle \int_0^{2\pi} \frac{2 \, dx}{1+\cos 2x + 2 a \sin 2x + (a^2 + b^2)(1-\cos 2x)}

= 2 \displaystyle \int_0^{2\pi} \frac{d\theta}{(1+a^2+b^2) + 2 a \sin \theta + (1 - a^2 - b^2) \cos \theta}

= 2 \displaystyle \int_{0}^{2\pi} \frac{d\theta}{S + R \cos (\theta - \alpha)}

= 2 \displaystyle \int_{0}^{2\pi} \frac{d\phi}{S + R \cos \phi}

= \displaystyle -\frac{4i}{R} \oint_C \frac{dz}{z^2 + 2\frac{S}{R}z + 1}

= \displaystyle -\frac{4i}{R} \oint_C \frac{dz}{(z - r_1)(z- r_2)}

where this last integral is taken over the complex plane on the unit circle, a closed contour oriented counterclockwise. Also,

r_1 = \displaystyle \frac{-S + \sqrt{S^2 -R^2}}{R}

and

r_2 = \displaystyle \frac{-S - \sqrt{S^2 -R^2}}{R},

are the two distinct roots of the denominator (as long as b \ne 0). In these formulas,R = \sqrt{(2a)^2 + (1-a^2-b^2)^2} and S = 1 + a^2 + b^2. (Also, \alpha is a certain angle that is now irrelevant at this point in the calculation).

This contour integral looks complicated; however, it’s an amazing fact that integrals over closed contours can be easily evaluated by only looking at the poles of the integrand. In yesterday’s post, I established that r_1 lies inside the contour, but r_2 lies outside of the contour.

The next step of the calculation is finding the residue at r_1; see Wikipedia and Mathworld for more information. This means rewriting the rational function

\displaystyle \frac{1}{(z - r_1)(z - r_2)}

as a power series (technically, a Laurent series) about the point z = r_1. This can be done by using the formula for an infinite geometric series (see here, here, and here):

\displaystyle \frac{1}{(z - r_1)(z - r_2)} = \displaystyle \frac{1}{z-r_1} \times \frac{1}{z-r_2}

= \displaystyle \frac{-1}{z-r_1} \times \frac{1}{r_2-z}

= \displaystyle \frac{-1}{z-r_1} \times \frac{1}{(r_2-r_1) - (z-r_1)}

= \displaystyle \frac{-1}{z-r_1} \times \frac{1}{r_2-r_1} \times \frac{ 1}{ 1 - \displaystyle \frac{z-r_1}{r_2-r_1} }

= \displaystyle \frac{-1}{z-r_1} \times \frac{1}{r_2-r_1} \left[ 1 + \left( \displaystyle \frac{z-r_1}{r_2-r_1} \right) + \left( \displaystyle \frac{z-r_1}{r_2-r_1} \right)^2 + \left( \displaystyle \frac{z-r_1}{r_2-r_1} \right)^3 + \dots \right]

= \displaystyle \frac{-1}{z-r_1} \times \frac{1}{r_2-r_1} - \frac{1}{(r_2-r_1)^2} - \frac{z-r_1}{(r_2-r_1)^3} - \frac{(z-r_1)^2}{(r_2-r_1)^4} \dots

The residue of the function at z = r_1 is defined to be the constant multiplying the \displaystyle \frac{1}{z-r_1} term in the above series. Therefore,

The residue at x = r_1 is \displaystyle \frac{-1}{r_2-r_1} = \displaystyle \frac{1}{r_1-r_2}

From the definitions of r_1 and r_2 above,

\displaystyle \frac{1}{r_1-r_2} = \displaystyle \frac{1}{\displaystyle \frac{-S + \sqrt{S^2 -R^2}}{R} - \frac{-S - \sqrt{S^2 -R^2}}{R}}

= \displaystyle \frac{1}{ ~ 2 \displaystyle \frac{\sqrt{S^2-R^2}}{R} ~ }

= \displaystyle \frac{R}{ 2 \sqrt{S^2-R^2} }

green lineNow that I’ve identified the residue of the only root that lies inside of the contour, we are in position to evaluate the contour integral above. I’ll discuss this in tomorrow’s post.

Fun lecture on geometric series: Index

I’m using the Twelve Days of Christmas (and perhaps a few extra days besides) to do something that I should have done a long time ago: collect past series of posts into a single, easy-to-reference post. The following posts formed my series concerning one of my favorite lectures concerning various applications of geometric series.

Part 1: Introduction to generating functions.

Part 2: Enumeration problems; or counting how many ways $2.00 can be formed using pennies, nickels, dimes, and quarters. (The answer is 1463.)

Part 3: The generating function for the Fibonacci sequence.

Part 4: Using a generating function to find a closed-form expression for the (ahem) Quintanilla sequence, a close but somewhat less famous relative of the Fibonacci sequence.

Part 5: Reproving the formula for the Quintanilla sequence using mathematical induction.

 

 

 

Arithmetic and Geometric Series: Index

I’m using the Twelve Days of Christmas (and perhaps a few extra days besides) to do something that I should have done a long time ago: collect past series of posts into a single, easy-to-reference post. The following posts formed my series on how I remind students about Taylor series. I often use this series in a class like Differential Equations, when Taylor series are needed but my class has simply forgotten about what a Taylor series is and why it’s important.

Part 1: Deriving the formulas for the nth term of arithmetic and geometric sequences.

Part 2: Pedagogical thoughts on conceptual barriers that students often face when encountering sequences and series.

Part 3: The story of how young Carl Frederich Gauss, at age 10, figured out how to add the integers from 1 to 100 in his head.

Part 4: Deriving the formula for an arithmetic series.

Part 5: Deriving the formula for an arithmetic series, using mathematical induction. Also, extensions to other series.

Part 6: Deriving the formula for an arithmetic series, using telescoping series. Also, extensions to other series.

Part 7: Pedagogical thoughts on assessing students’ depth of understanding the formula for an arithmetic series.

Part 8: Deriving the formula for a finite geometric series.

Part 9: Infinite geometric series and Xeno’s paradox.

Part 10: Deriving the formula for an infinite geometric series.

Part 11: Applications of infinite geometric series in future mathematics courses.

Part 12: Other commonly-arising infinite series.