Solving Problems Submitted to MAA Journals (Part 2a)

The following problem appeared in Volume 53, Issue 4 (2022) of The College Mathematics Journal. This was the first problem that was I able to solve in over 30 years of subscribing to MAA journals.

Suppose that $X$ and $Y$ are independent, uniform random variables over $[0,1]$ . Now define the random variable $Z$ by

$Z = (Y-X) {\bf 1}(Y \ge X) + (1-X+Y) {\bf 1}(Y<X)$ .

Prove that $Z$ is uniform over $[0,1]$ . Here, ${\bf 1}[S]$ is the indicator function that is equal to 1 if $S$ is true and 0 otherwise.

The first thing that went through my mind was something like, “This looks odd. But it’s a probability problem using concepts from a senior-level but undergraduate probability course. This was once my field of specialization. I had better be able to get this.”

My second thought was that one way of proving that $Z$ is uniform on $[0,1]$ is showing that $P(Z \le t) = t$ if $0 \le t \le 1$ .

My third thought was that $Z$ really had a two-part definition:

$Z = \bigg\{ \begin{array}{ll} Y-X, & Y \ge X \\ 1-X+Y, & Y <X \end{array}$

So I got started by dividing this probability into the two cases:

$P(Z \le t) = P(Z \le t, Y \ge X ~\hbox{or}~ Y < X)$

$= P(Z \le t, Y \ge X ~\hbox{or}~ Z \le t, Y < X)$

$= P(Z \le t, Y \ge X) + P(Z \le t, Y < X)$

$= P(Y-X \le t, Y \ge X) + P(1 - X + Y \le t, Y < X)$

$= P(Y \le X + t, Y \ge X) + P( Y \le X + t - 1, Y < X)$

$= P(X \le Y \le X + t) + P(Y \le X + t - 1)$ .

In the last step, since $0 \le t \le 1$ , the events $Y \le X + t - 1$ and $Y < X$ are redundant: if $Y \le X + t - 1$ , then $Y$ will automatically be less than $X$ . Therefore, it’s safe to remove $Y < X$ from the last probability.

Ordinarily, such probabilities are computed by double integrals over the joint probability density function of $X$ and $Y$ , which usually isn’t easy. However, in this case, since $X$ and $Y$ are independent and uniform over $[0,1]$ , the ordered pair $(X,Y)$ is uniform on the unit square $[0,1] \times [0,1]$ . Therefore, probabilities can be found by simply computing areas.

In this case, since the area of the unit square is 1, $P(Z \le t)$ is equal to the sum of the areas of

$\{(x,y) \in [0,1] \times [0,1] : x \le y \le x + t \}$ ,

which is depicted in green below, and

$\{(x,y) \in [0,1] \times [0,1] : y \le x + t - 1 \}$ ,

which is depicted in purple.

First, the area in green is a trapezoid. The $y-$ intercept of the line $y = x+t$ is $(0,t)$ , and the two lengths of $t$ and $1-t$ on the upper left of the square are found from this $y-$ intercept. The area of the green trapezoid is easiest found by subtracting the areas of two isosceles right triangles:

$latex \displaystyle \frac{(1)^2}{2} – \frac{(1-t)^2}{2} = \frac{1-1+2t-t^2}{2} = \frac{2t-t^2}{2}.

Second, the area in purple is an isosceles right triangle. The $y-$ intercept of the line $y=x+t-1$ is $(0,t-1)$ , so that the distance from the $y-$ intercept to the origin is $1-t$ . From this, the two lengths of $1-t$ and $t$ are found. Therefore, the area of the purple right triangle is $\displaystyle \frac{t^2}{2}$ .

Adding, we conclude that

$P(Z \le t) = \displaystyle \frac{2t-t^2}{2} + \frac{t^2}{2} = \frac{2t}{2} = t$ .

Therefore, $Z$ is uniform over $[0,1]$ .

A closing note: after going 0-for-4000 in my previous 30+ years of attempting problems submitted to MAA journals, I was unbelievably excited to finally get one. As I recall, it took me less than an hour to get the above solution, although writing up the solution cleanly took longer.

However, the above was only Part 1 of a two-part problem, so I knew I just had to get the second part before submitting. That’ll be the subject of the next post.

Solving Problems Submitted to MAA Journals (Part 1)

I first became a member of the Mathematical Association of America in 1988. My mentor in high school gave me a one-year membership as a high school graduation present, and I’ve maintained my membership ever since. Most years, I’ve been a subscriber to three journals: The American Mathematical Monthly, Mathematics Magazine, and College Mathematics Journal.

A feature for each of these journals is the Problems/Solutions section. In a nutshell, readers devise and submit original problems in mathematics for other readers to solve; the editors usually allow readers to submit solutions for a few months after the problems are first published. Between the three journals, something like 120 problems are submitted annually by readers.

And, historically, I had absolutely no success in solving these problems. Said another way: over my first 30+ years as an MAA member, I went something like 0-for-4000 at solving these submitted problems. This gnawed at me for years, especially when I read the solutions offered by other readers, maybe a year after the problem originally appeared, and thought to myself, “Why didn’t I think of that?”

Well, to be perfectly honest, that’s still my usual. However, in the past couple of years, I actually managed to solve a handful of problems that appeared in Mathematics Magazine and College Mathematics Journal, to my great surprise and delight. I don’t know what happened. Maybe I’ve just got better at problem solving. Maybe solving the first one or two boosted my confidence. Maybe success breeds success. Maybe all the hard problems have already been printed and the journals’ editors have nothing left to publish except relatively easier problems.

In this short series, I’ll try to reconstruct my thought processes and flashes of inspiration that led to these solutions.

500,000 page views

I’m taking a break from my usual posts on mathematics and mathematics education to note a symbolic milestone: meangreenmath.com has had more than 500,000 total page views since its inception in June 2013. Many thanks to the followers of this blog, and I hope that you’ll continue to find this blog to be a useful resource to you.

Thirty most viewed individual posts:

Thirty most viewed series:

2048 and algebra
Another poorly written word problem
Area of a triangle and volume of common shapes
Arithmetic and geometric series
Calculators and complex numbers
Common Core, subtraction, and the open number line
Computing e to any power
Confirming Einstein’s theory of general relativity with calculus
Day One of my Calculus I class
Different definitions of e
Exponential growth and decay
Facebook birthday problem
Fun lecture on geometric series
How I impressed my wife: $\displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$
Inverse Functions
Langley’s Adventitious Angles
Lessons from teaching gifted elementary students
My favorite one-liners
My mathematical magic show
Parabolas from string art
Predicate logic and popular culture
Proving theorems and special cases
Reminding students about Taylor series
Slightly incorrect ugly mathematical Christmas T-shirts
Square roots and logarithms without a calculator
The antiderivative of $\displaystyle \frac{1}{x^4+1}$
Thoughts on 1/7 and other rational numbers
Thoughts on numerical integration
Wason selection task
Why does $x^0 = 1$ and $x^{-n} = 1/x^n$ ?

Thirty most viewed posts (guest presenters):

If I’m still here at that time, I’ll make a summary post like this again when this blog has over 1,000,000 page views.

My Favorite One-Liners: Index

I’m doing something that I should have done a long time ago: collecting a series of posts into one single post. The links below show my series on my favorite one-liners.

Mathematical Wisecracks for Almost Any Occasion: Part 2, Part 7, Part 8, Part 12, Part 21, Part 28, Part 29, Part 41, Part 46, Part 53, Part 60, Part 63, Part 65, Part 71, Part 79, Part 84, Part 85, Part 100, Part 101, Part 108, Part 109, Part 114, Part 118, Part 121

All-Purpose Anecdotes: Part 38, Part 50, Part 64, Part 70, Part 92, Part 94

Addressing Misconceptions: Part 3, Part 4, Part 11, Part 14, Part 15, Part 18, Part 30, Part 32, Part 33, Part 37, Part 45, Part 59

Tricky Steps in a Calculation: Part 5, Part 6

Greek alphabet and choice of variables: Part 40, Part 43, Part 56

Homework and exams: Part 39, Part 47, Part 55, Part 57, Part 58, Part 66, Part 77, Part 78, Part 91, Part 96, Part 97, Part 107

Inequalities: Part 99

Simplification: Part 10, Part 102, Part 103

Polynomials: Part 19, Part 48, Part 49, Part 81, Part 90

Inverses: Part 16

Exponential and Logarithmic Functions: Part 1, Part 42, Part 68, Part 80, Part 110

Trigonometry: Part 9, Part 69, Part 76, Part 106, Part 120

Complex numbers: Part 54, Part 67, Part 86, Part 112, Part 113

Sequences and Series: Part 20, Part 35, Part 111

Combinatorics: Part 27

Statistics: Part 22, Part 23, Part 36, Part 51, Part 52, Part 61, Part 95, Part 116

Probability: Part 26, Part 31, Part 62, Part 93, Part 122

Calculus: Part 24, Part 25, Part 72, Part 73, Part 74, Part 75, Part 83, Part 87, Part 88, Part 104, Part 115, Part 117

Logic and Proofs: Part 13, Part 17, Part 34, Part 44, Part 89, Part 98, Part 119

Differential Equations: Part 82, Part 105

Confirming Einstein’s General Theory of Relativity with Calculus: Index

I’m doing something that I should have done a long time ago: collecting a series of posts into one single post. The links below show my series on general relativity and the precession of Mercury’s orbit.

Part 1: Introduction

Part 1a: Introduction
Part 1b: Observed precession of Mercury
Part 1c: Outline of argument

Part 2: Precession, polar coordinates, and conic sections

Part 2a: Graphically exploring precession
Part 2b: Polar coordinates and ellipses
Part 2c: Polar coordinates, circles, and parabolas
Part 2d: Polar coordinates and hyperbolas

Part 3: Method of successive approximations

Part 4: Principles from physics

Part 4a: Angular momentum
Part 4b: Acceleration in polar coordinates
Part 4c: Newton’s Second Law and Newton’s Law of Gravitation

Part 5: Orbits under Newtonian mechanics

Part 5a: Confirmation of solution
Part 5b: Derivation with calculus
Part 5c: Derivation with differential equations and the method of undetermined coefficients
Part 5d: Derivation with differential equations and variation of parameters

Part 6: Orbits under general relativity

Part 6a: New differential equation under general relativity
Part 6b: Confirmation of solution
Part 6c: Derivation with variation of parameters
Parts 6d, 6e, 6f, 6g, 6h, 6i, 6j: Rationale for the method of undetermined coefficients
Part 6k: Derivation with undetermined coefficients

Part 7: Computing precession

Parts 7a, 7b, 7c, 7d: Predicting precession
Part 7e: Computing precession

Part 8: Second- and third-order solutions with the method of successive approximations

Part 9: Pedagogical thoughts

Earlier this year, I presented these ideas for the UNT Math Department’s Undergraduate Mathematics Colloquium Series. The video of my lecture is below.

My Mathematical Magic Show: Index

I’m doing something that I should have done a long time ago: collecting a series of posts into one single post. The links below show the mathematical magic show that I’ll perform from time to time.

Part 1: Introduction.

Part 2a, Part 2b, and Part 2c: The 1089 trick.

Part 3a, Part 3b, and Part 3c: A geometric magic trick.

Part 4a: Part 4b, Part 4c, and Part 4d: A trick using binary numbers.

Part 5a, Part 5b, Part 5c, and Part 5d: A trick using the rule for checking if a number is a multiple of 9.

Part 7: The Fitch-Cheney card trick, which is perhaps the slickest mathematical card trick ever devised.

Part 8a, Part 8b, and Part 8c: A trick using Pascal’s triangle.

Part 9: Mentally computing $n$ given $n^5$ if $10 \le n \le 99$ .

Part 10: A mathematical optical illusion.

Part 11: The 27-card trick, which requires representing numbers in base 3.

Part 6: The Grand Finale.

And, for the sake of completeness, here’s a picture of me just before I performed an abbreviated version of this show for UNT’s Preview Day for high school students thinking about enrolling at my university.

Lagrange Points and Polynomial Equations: Part 5

This series was motivated by a terrific article that I read in the American Mathematical Monthly about Lagrange points, which are (from Wikipedia) “points of equilibrium for small-mass objects under the gravitational influence of two massive orbiting bodies.” There are five such points in the Sun-Earth system, called $L_1$ , $L_2$ , $L_3$ , $L_4$ , and $L_5$ .

The article points out a delicious historical factoid: Lagrange had a slight careless mistake in his derivation!

From the article:

Equation (d) would be just the tool to use to determine where to locate the JWST [James Webb Space Telescope, which is now in orbit about $L_2$ ], except for one thing: Lagrange got it wrong!… Do you see it? His algebra in converting $1 - \displaystyle \frac{1}{(m-1)^3}$ to common denominator form is incorrect… Fortunately, at some point in the two-and-a-half centuries between Lagrange’s work and the launch of JWST, this error has been recognized and corrected.

This little historical anecdote illustrates that, despite our best efforts, even the best of us are susceptible to careless mistakes. The simplification should have been

$q' = \displaystyle \left[ 1 - \frac{1}{(m-1)^3} \right] \cdot \frac{1}{r^3}$

$= \displaystyle \frac{(m-1)^3 - 1}{(m-1)^3} \cdot \frac{1}{r^3}$

$= \displaystyle \frac{m^3 - 3m^2 + 3m - 1 - 1}{(m-1)^3} \cdot \frac{1}{r^3}$

$= \displaystyle \frac{m^3 - 3m^2 + 3m - 2}{(m-1)^3} \cdot \frac{1}{r^3}$ .

(Parenthetically, The article also notes a clear but unintended typesetting error, as the correct but smudged exponent of 3 in the first equation became an incorrect exponent of 2 in the second.)

Lagrange Points and Polynomial Equations: Part 4

From Wikipedia, Lagrange points are points of equilibrium for small-mass objects under the gravitational influence of two massive orbiting bodies. There are five such points in the Sun-Earth system, called $L_1$ , $L_2$ , $L_3$ , $L_4$ , and $L_5$ .

The stable equilibrium points $L_4$ and $L_5$ are easiest to explain: they are the corners of equilateral triangles in the plane of Earth’s orbit. The points $L_1$ and $L_2$ are also equilibrium points, but they are unstable. Nevertheless, they have practical applications for spaceflight.

As we’ve seen, the positions of $L_1$ and $L_2$ can be found by numerically solving the fifth-order polynomial equations

$t^5 - (3-\mu) t^4 + (3-2\mu)t^3 - \mu t^2 + 2\mu t - \mu = 0$

and

$t^5 + (3-\mu) t^4 + (3-2\mu)t^3 - \mu t^2 - 2\mu t - \mu = 0$ ,

respectively. In these equations, $\mu = \displaystyle \frac{m_2}{m_1+m_2}$ where $m_1$ is the mass of the Sun and $m_2$ is the mass of Earth. Also, $t$ is the distance from the Earth to $L_1$ or $L_2$ measured as a proportion of the distance from the Sun to Earth.

We’ve also seen that, for the Sun and Earth, $mu \approx 3.00346 \times 10^{-6}$ , and numerically solving the above quintics yields $t \approx 0.000997$ for $L_1$ and $t \approx 0.01004$ for $L_2$ . In other words, $L_1$ and $L_2$ are approximately the same distance from Earth but in opposite directions.

There’s a good reason why the positive real roots of these two similar quintics are almost equal. We know that $t$ will be a lot closer to 0 than 1 because, for gravity to balance, the Lagrange points have to be a lot closer to Earth than the Sun. For this reason, the terms $\mu t^2$ and $2\mu t$ will be a lot smaller than $\mu$ , and so those two terms can be safely ignored in a first-order approximation. Also, the terms $t^5$ and $(3-\mu)t^4$ will be a lot smaller than $(3-2\mu)t^3$ , and so those two terms can also be safely ignored in a first-order approximation. Furthermore, since $\mu$ is also close to 0, the coefficient $(3-2\mu)$ can be safely replaced by just $3$ .

Consequently, the solution of both quintic equations should be close to the solution of the cubic equation

$3t^3 - \mu = 0$ ,

which is straightforward to solve:

$3t^3 = \mu$

$t^3 = \displaystyle \frac{\mu}{3}$

$t = \displaystyle \sqrt[3]{ \frac{\mu}{3} }$ .

If $\mu = 3.00346 \times 10^{-6}$ , we obtain $t \approx 0.010004$ , which is indeed reasonably close to the actual solutions for $L_1$ and $L_2$ . Indeed, this may be used as the first approximation in Newton’s method to quickly numerically evaluate the actual solutions of the two quintic polynomials.

Happy Pythagoras Day!

Let me take a break from my current series of posts to wish everyone an early Happy Pythagoras Day! Tomorrow will be 10/26/24 (or 26/10/24 in other parts of the world), and $26^2 = 10^2 + 24^2$ .

Bonus points if you can figure out (without Googling) when the next Pythagoras Day will be.

Lagrange Points and Polynomial Equations: Part 3

The position of $L_2$ can be found by numerically solving the fifth-order polynomial equation

$(m_1+m_2)t^5+(3m_1+2m_2)t^4+(3m_1+m_2)t^3$

$-(m_2+3m_3)t^2-(2m_2+3m_3)t-(m_2+m_3)=0$ .

In this equation, $m_1$ is the mass of the Sun, $m_2$ is the mass of Earth, $m_3$ is the mass of the spacecraft, and $t$ is the distance from the Earth to $L_2$ measured as a proportion of the distance from the Sun to Earth. In other words, if the distance from the Sun to Earth is 1 unit, then the distance from the Earth to $L_2$ is $t$ units. The above equation is derived using principles from physics which are not elaborated upon here.

We notice that the coefficients of $t^5$ , $t^4$ , and $t^3$ are all positive, while the coefficients of $t^2$ , $t$ , and the constant term are all negative. Therefore, since there is only one change in sign, this equation has only one positive real root by Descartes’ Rule of Signs.

Since $m_3$ is orders of magnitude smaller than both $m_1$ and $m_2$ , this may safely approximated by

$(m_1+m_2)t^5+(3m_1+2m_2)t^4+(3m_1+m_2)t^3 - m_2 t^2- 2m_2 t-m_2=0$ .

This new equation can be rewritten as

$t^5 + \displaystyle \frac{3m_1 + 2m_2}{m_1 + m_2} t^4 + \frac{3m_1+m_2}{m_1+m_2} t^3 - \frac{m_2}{m_1+m_2} t^2 - \frac{2m_2}{m_1+m_2} t- \frac{m_2}{m_1+m_2} = 0$ .

If we define

$\mu = \displaystyle \frac{m_2}{m_1+m_2}$ ,

we see that

$\displaystyle \frac{3m_1 + 2m_2}{m_1 + m_2} = \frac{3m_1 + 3m_2}{m_1 + m_2} - \frac{m_2}{m_1 + m_2} = 3 - \mu$

and

$\displaystyle \frac{3m_1 + m_2}{m_1 + m_2} = \frac{3m_1 + 3m_2}{m_1 + m_2} - \frac{2m_2}{m_1 + m_2} = 3 - 2\mu$ ,

so that the equation may be written as

$t^5 + (3-\mu) t^4 + (3-2\mu) - \mu t^2 - 2\mu t - \mu = 0$ ,

matching the equation found at Wikipedia.

For the Sun and Earth, $m_1 \approx 1.9885 \times 10^{30} ~ \hbox{kg}$ and $m_2 \approx 5.9724 \times 10^{24} ~ \hbox{kg}$ , so that

$mu = \displaystyle \frac{5.9724 \times 10^{24}}{1.9885 \times 10^{30} + 5.9724 \times 10^{24}} \approx 3.00346 \times 10^{-6}$ .

This yields a quintic equation that is hopeless to solve using standard techniques from Precalculus, but the root can be found graphically by seeing where the function crosses the $x-$ axis (or, in this case, the $t-$ axis):

As it turns out, the root is $t \approx 0.01004$ , so that $L_2$ is located $1.004\%$ of the distance from the Earth to the Sun in the direction away from the Sun.