Intuition for the Dual in Linear Programming

One of the most important theoretical results in linear programming is that every LP has a corresponding dual program. Where, exactly, this dual comes from can often seem mysterious. Several years ago I answered a question on a couple of Stack Exchange sites giving an intuitive explanation for where the dual comes from. Those posts seem to have been appreciated, so I thought I would reproduce my answer here.

Suppose we have a primal problem as follows.

$Primal = \begin{Bmatrix} \max &5x_1 - 6x_2 \\ s.t. &2x_1 - x_2 = 1\\ &x_1 + 3x_2 \leq 9 \\ &x_1 \geq 0 \end{Bmatrix}$

Now, suppose we want to use the primal’s constraints as a way to find an upper bound on the optimal value of the primal. If we multiply the first constraint by 9, the second constraint by 1, and add them together, we get $9(2x_1 - x_2) + 1(x_1 +3 x_2)$ for the left-hand side and $9(1) + 1(9)$ for the right-hand side. Since the first constraint is an equality and the second is an inequality, this implies

$19x_1 - 6x_2 \leq 18.$

But since $x_1 \geq 0$, it’s also true that $5x_1 \leq 19x_1$, and so

$\displaystyle 5x_1 - 6x_2 \leq 19x_1 - 6x_2 \leq 18.$

Therefore, 18 is an upper-bound on the optimal value of the primal problem.

Surely we can do better than that, though. Instead of just guessing 9 and 1 as the multipliers, let’s let them be variables. Thus we’re looking for multipliers $y_1$ and $y_2$ to force

$\displaystyle 5x_1 - 6x_2 \leq y_1(2x_1-x_2) + y_2(x_1 + 3x_2) \leq y_1(1) + y_2(9).$

Now, in order for this pair of inequalities to hold, what has to be true about $y_1$ and $y_2$? Let’s take the two inequalities one at a time.

The first inequality: $5x_1 - 6x_2 \leq y_1(2x_1-x_2) + y_2(x_1 + 3x_2)$

We have to track the coefficients of the $x_1$ and $x_2$ variables separately. First, we need the total $x_1$ coefficient on the right-hand side to be at least 5. Getting exactly 5 would be great, but since $x_1 \geq 0$, anything larger than 5 would also satisfy the inequality for $x_1$. Mathematically speaking, this means that we need $2y_1 + y_2 \geq 5$.

On the other hand, to ensure the inequality for the $x_2$ variable we need the total $x_2$ coefficient on the right-hand side to be exactly $-6$. Since $x_2$ could be positive, we can’t go lower than $-6$, and since $x_2$ could be negative, we can’t go higher than $-6$ (as the negative value for $x_2$ would flip the direction of the inequality). So for the first inequality to work for the $x_2$ variable, we’ve got to have $-y_1 + 3y_2 = -6$.

The second inequality: $y_1(2x_1-x_2) + y_2(x_1 + 3x_2) \leq y_1(1) + y_2(9)$

Here we have to track the $y_1$ and $y_2$ variables separately. The $y_1$ variable comes from the first constraint, which is an equality constraint. It doesn’t matter if $y_1$ is positive or negative, the equality constraint still holds. Thus $y_1$ is unrestricted in sign. However, the $y_2$ variable comes from the second constraint, which is a less-than-or-equal to constraint. If we were to multiply the second constraint by a negative number that would flip its direction and change it to a greater-than-or-equal constraint. To keep with our goal of upper-bounding the primal objective, we can’t let that happen. So the $y_2$ variable can’t be negative. Thus we must have $y_2 \geq 0$.

Finally, we want to make the right-hand side of the second inequality as small as possible, as we want the tightest upper-bound possible on the primal objective. So we want to minimize $y_1 + 9y_2$.

Putting all of these restrictions on $y_1$ and $y_2$ together we find that the problem of using the primal’s constraints to find the best upper bound on the optimal primal objective entails solving the following linear program:

$\begin{matrix} \text{Minimize } &y_1 + 9y_2 \\ \text{ subject to } &2y_1 + y_2 \geq 5 \\ &-y_1 + 3y_2 = -6\\ &y_2 \geq 0 \end{matrix}$

And that’s the dual.

It’s probably worth summarizing the implications of this argument for all possible forms of the primal and dual. The following table is taken from p. 214 of Introduction to Operations Research, 8th edition, by Hillier and Lieberman. They refer to this as the SOB method, where SOB stands for Sensible, Odd, or Bizarre, depending on how likely one would find that particular constraint or variable restriction in a maximization or minimization problem.

             Primal Problem                           Dual Problem
(or Dual Problem)                        (or Primal Problem)

Maximization                             Minimization

Sensible     <= constraint            paired with     nonnegative variable
Odd          =  constraint            paired with     unconstrained variable
Bizarre      >= constraint            paired with     nonpositive variable

Sensible     nonnegative variable     paired with     >= constraint
Odd          unconstrained variable   paired with     = constraint
Bizarre      nonpositive variable     paired with     <= constraint


The Sum of Cubes is the Square of the Sum

It’s fairly well-known, to those who know it, that

$\displaystyle \left(\sum_{k=1}^n k \right)^2 = \frac{n^2(n+1)^2}{4} = \sum_{k=1}^n k^3$.

In other words, the square of the sum of the first n positive integers equals the sum of the cubes of the first n positive integers.

It’s probably less well-known that a similar relationship holds for $\tau$, the function that counts the number of divisors of an integer:

$\displaystyle \left(\sum_{d|n} \tau(d) \right)^2 = \sum_{d|n} \tau(d)^3$.

In this post we’re going to prove this formula for $\tau$.

First, it’s pretty easy to see what the value of $\tau$ is for prime powers; i.e., integers of the form $p^e$, where p is prime. Since the only divisors of $p^e$ are $1, p, p^2, \ldots, p^e$, the number of divisors of $p^e$ is given by $\tau(p^e) = e+1$.

Let’s check the identity we’re trying to prove when $n = p^e$. We have

$\displaystyle \left(\sum_{d|p^e} \tau(d) \right)^2 = \left(\tau(1) + \tau(p) + \tau(p^2) + \cdots + \tau(p^e)\right)^2 \\ = \left(1 + 2 + 3 + \cdots + (e+1)\right)^2 = \left( \frac{(e+1)(e+2)}{2}\right)^2.$

We also have

$\displaystyle \sum_{d|p^e} \tau(d)^3 = \tau(1)^3 + \tau(p)^3 +\tau(p^2)^3 + \cdots + \tau(p^e)^3 \\ = 1^3 + 2^3 + 3^3 + \cdots + (e+1)^3 = \left( \frac{(e+1)(e+2)}{2}\right)^2.$

Clearly, the two expressions are equal, so the identity we’re trying to prove holds in the prime-power case. (And, in fact, the derivation uses the identity about the sum of the first several positive integers mentioned at the beginning of the post!)

Let $f(n) = \sum_{d|n} \tau(d)$ and $F(n) = \sum_{d|n} \tau(d)^3$. What we’ve shown thus far is that $f(p^e) = F(p^e)$.

Now, we’re going to pull some elementary number theory. One fact about $\tau$ is that it is multiplicative; i.e., $\tau(mn) = \tau(m) \tau(n)$ when m and n are relatively prime. This is one of the first properties you learn about $\tau$ once you learn its definition, and we’re not going to prove it here.

It turns out that both f and F are multiplicative as well! First, the product of two multiplicative functions is also multiplicative. (This is a one-line proof using the definition of multiplicative.) So $\tau(d)^3$ is multiplicative.

Another property of multiplicative functions is that if g is multiplicative, then $\sum_{d|n} g(d)$ is also multiplicative. This is a special case of the more general result that the Dirichlet convolution of two multiplicative functions is multiplicative. (In the Dirichlet convolution $g \star h$, take h to be the identity function; i.e., $h(n) = 1$.) This means that our functions f and F defined in the previous paragraph are both multiplicative.

Therefore, with the prime-power factorization of n given by $n = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k}$, we have

$\displaystyle \left(\sum_{d|n} \tau(d) \right)^2 = f(n) = f(p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k}) = f(p_1^{e_1}) f(p_2^{e_2}) \cdots f(p_k^{e_k}) \\= F(p_1^{e_1}) F(p_2^{e_2}) \cdots F(p_k^{e_k}) = F(p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k}) = F(n) = \sum_{d|n} \tau(d)^3.$

For an even further generalization, see Barbeau and Seraj [1]. (The proof we give in this post follows the same basic argument as the proof of Proposition 1 in Barbeau and Seraj’s paper.)

1. Barbeau, Edward, and Samer Seraj, “Sum of cubes is square of sum,” Notes on Number Theory and Discrete Mathematics 19(1), 2013, pages 1–13.

Happy Birthday, Benoit Mandelbrot

Today’s Google doodle honors mathematician Benoit Mandelbrot. He would have been 96 today. If you’re interested in learning more about his life and work, the Google doodle link contains a short summary. If you want to go deeper, you can read the Wikipedia article on him.

I only have one small thing to add. A few years ago I wrote an interactive text-based game, A Beauty Cold and Austere. You can read some reviews of it and play it here, and you can read some of my thoughts about it here. The connection to Mandelbrot is that I used part of the Mandelbrot set for the cover art for the game, as you can see below.

No Integer Solutions to a Mordell Equation

Equations of the form $x^3 = y^2 + k$ are called Mordell equations.  In this post we’re going to prove that the equation $x^3 = y^2 -7$ has no integer solutions, using (with one exception) nothing more complicated than congruences.

Theorem: There are no integer solutions to the equation $x^3 = y^2 -7$.

Proof.

Case 1.  First, suppose that there is a solution in which x is even.  Since $2|x$, $8|x^3$.  Looking at the equation modulo 8, then, we have

$\displaystyle 0 \equiv y^2 - 7 \pmod{8} \implies y^2 \equiv 7 \pmod{8}$.

However, if we square the residues $\{0, 1, 2, 3, 4, 5, 6, 7\}$ and reduce them modulo 8 we have $\{0, 1, 4, 1, 0, 1, 4, 1\}$.  So there is no integer y that when squared is congruent to 7 modulo 8.  Therefore, there is no solution to $x^3 = y^2 -7$ when is even.

Case 2.  Now, suppose there is a solution in which x is odd.  If so, then we have $x \equiv 3 \pmod{4}$ or $x \equiv 1 \pmod{4}$.

If $x \equiv 3 \pmod{4}$ then we have $y^2 \equiv 27 + 7 \equiv 34 \equiv 2 \pmod{4}$.  However, if we square the residues $\{0, 1, 2, 3\}$ and reduce them modulo 4 we get $\{0, 1, 0, 1\}$.  So there is no integer y that when squared is congruent to 2 modulo 4.  Thus there is no solution to $x^3 = y^2 - 7$ when $x \equiv 3 \pmod{4}$.

Now, suppose $x \equiv 1 \pmod{4}$.  Applying some algebra to the original equation, we have

$\displaystyle y^2 = x^3 + 7 \implies y^2 + 1 = x^3 + 8 = (x+2)(x^2 - 2x + 4)$.

By assumption, $x + 2 \equiv 3 \pmod{4}$.  If $x + 2$ has prime factors that are all equivalent to 1 modulo 4, then their product (i.e., $x+2$) would be equivalent to 1 modulo 4.  Thus $x + 2$ has at least one prime factor q that is congruent to 3 modulo 4.

However, if $q|x+2$, then $q|y^2+1$.   This means that $y^2 \equiv -1 \pmod{q}$.  However, thanks to Euler’s criterion, only odd primes q such that $q \equiv 1 \pmod{4}$ can have a solution to $y^2 \equiv -1 \pmod{q}$.  Since $q \equiv 3 \pmod{4}$, $y^2 \equiv -1 \pmod{q}$ has no solution.  Thus there is no solution to $x^3 = y^2 - 7$ when $x \equiv 1 \pmod{4}$, either.

Since we’ve covered all cases, there can be no solution in integers to the Mordell equation $x^3 = y^2 + k$.

Some Video Learning Suggestions

During this COVID-tide many of us have been seeking out online learning resources.  I’ve done so quite a bit in the past few months, and I thought I would do a post to recommend some of these.  They are all sites that feature videos.  I’ll start with the math ones and then go on to other subjects.

Math

• Khan Academy.  This one should be obvious to anyone who knows anything about learning math online.  The videos on this site are quite good – among the best online math videos I’ve seen.  The site also features regular self-assessments ranging from quick quizzes to end-of-course tests so that you can gauge your learning.  From what I can tell, Khan Academy is the gold standard for online learning, and I have no problems recommending it to my children (ages 9, 9, and 12) or to my college students.  (While Khan Academy got their start with math, they’ve branched out to many other subjects as well.)
• Numberphile.  This is a YouTube channel sponsored by the Mathematical Sciences Research Institute.  It features a wide variety of mathematical topics generally aimed at those who know something about math and are curious to learn more.  I’ve only watched two Numberphile videos – both on Möbius strips – but they were both quite good, and I even learned something from them!  (I also showed them to my children.)  I wouldn’t normally recommend a YouTube channel based on just two videos, but their quality plus the fact that my college students have often spoken highly of Numberphile is sufficient to make me comfortable recommending this channel.
• Prodigy.  This is an online role-playing game in which players have to solve math problems in order to defeat monsters in battle.  While I have some pedagogical criticisms of Prodigy, the game has good production values for its target audience (mostly elementary school children), it successfully ramps up the difficulty level as players demonstrate greater math knowledge, and all three of my kids have had fun with it.

Geography

• Geography Now.  Paul Barbato deserves an award for this YouTube channel, in which he covers all the countries in the world in alphabetical order.  He’s funny and informative, giving an overview of the physical geography, the people, the culture, and the political relations of each country he discusses.  The production values are much better than any of the other general-interest geography channels I’ve seen, too.  He’s only up to the Seychelles at this point, so if you’re interested in countries like Spain or the UK you’ll have to wait a bit, but this has been the go-to site for geography homeschooling with my kids the past several months.  (“Hast du gluten-free?”  “Nein!” still brings a chuckle in my household.)
• Touropia.  This is the best YouTube travel channel I’ve found.  I generally pair a country’s Touropia “Top Ten Places to Visit” video with the country’s corresponding Geography Now video for two different perspectives on the same country.

Science

• National Geographic.  As you can imagine, this venerable institution has some high-quality videos out there.  I’m recommending them under “science” because we’ve only watched their science videos.
• Mystery Science.  This is a series of hands-on video lessons for K-5.  My children’s teachers used them in the classroom pre-COVID, and I’ve used several as well for homeschooling.  The science is solid and at the appropriate level, and the hands-on activities help engage the children.  When I was using them more regularly back in the spring, some of the video lessons were available for free and others you had to purchase.  As of right now they are offering a limited number of free memberships that presumably allow you to access all the videos.

History

• Crash Course in World History.  This YouTube series features an irreverent overview of world history that should appeal to people of all ages.  I haven’t watched any of their videos all the way through, but I’ve seen segments, my wife has used them when she’s homeschooling, and my kids have enjoyed them.  This link is just to the first video in the world history series, and they have other series as well (including U.S. history).  “Except for the Mongols!” has become a catchphrase in my house.

Philosophy

• Philosophy of the Humanities.  These YouTube videos simply feature Leiden University professor Victor Gijsbers giving a series of lectures on the philosophy of science, history, knowledge, and the humanities in general.  There’s nothing fancy here, video-wise, but Victor is an amazingly clear lecturer, elucidating intricate philosophical concepts with clarity and easy-to-understand examples.  This is the only recommendation in this post that I haven’t used with my kids; this is advanced high school level material at the very least.  Favorite quote: “Hegel is evidently a comic historian.  Which, by the way, doesn’t mean we will laugh a lot when we read Hegel.  Because, believe me, we don’t.”  (I actually know Victor in another context: He and I have both written interactive fiction!)

English

• Schoolhouse Rock.  If you were an American child in the late 70s or early 80s you will know exactly what this video series is, as it was ubiquitous during prime Saturday-morning cartoon-watching.  My kids and I viewed their videos on the eight parts of speech, but they have some history videos and even a solid video on how a bill becomes law as well.  Enjoy the 70s-style animation and music, and expect to emerge with a few extra earwigs.  (“Conjunction junction, what’s your function?” and half of the interjections video continue to be repeated in my house.)  The link is to the YouTube channel, which doesn’t have all the videos.  However, they are easy to find online.

French

• Get Started with French Like a Boss!.  I am not anywhere near fluent in French, but I did study it in high school and college, and I decided homeschooling would be a great opportunity to introduce my kids to a foreign language.  Lya, the host of this video that introduces basic French words and phrases, is funny and a little goofy in an endearing way.  She’s really good about using the vocabulary in different contexts, too, to the point that my 12-year-old was able to start picking up basic French sentence structure just from Lya’s examples.  This video is part of a larger series, but it’s the only one from that series I’ve seen.
• Rock ‘N Learn.  This brightly-animated series aimed at young children pairs French vocabulary with corresponding English vocabulary.  My kids are older than the target audience for the series, but the vocabulary is right at their level, and they’re old enough that they can laugh about how goofy it is.  Rock ‘N Learn actually covers a wide variety of subjects, but we’ve only watched it for French.  (Also, they appear to have a series of French language videos for teens and adults, but I haven’t watched those yet.)

A Lesson on Converting Between Different Bases

We’re in the time of COVID-19, and that has meant taking far more direct responsibility for my children’s learning than I ever have before.  It’s been a lot of work, but it’s also been fun.  In fact, I’ve been surprised at how much I’ve enjoyed it.

One of these enjoyable aspects has been introducing my children to some mathematical concepts that are more advanced than they would normally get in third or sixth grade.  My sixth grader in particular is ready for some basic number theory, such as the representation of numbers in bases other than 10.

Here’s a problem I posed for him a few weeks ago, after making sure he understood the conversion concept.

Take the number 42178 and convert it to base 2.

Dutifully, he began converting 42178 to base 10.  It took him a minute or two, but he got the correct answer of 219110.  Then he started working on the conversion from base 10 to base 2.  I told him to tell me when he finished the calculation but not tell me what the answer is.  After another couple of minutes, he did so.  I then quickly wrote down the answer of 1000100011112 off the top of my head.  His eyes bugged and his jaw dropped – a response that is always gratifying to see from a middle-schooler. 🙂

I didn’t keep him in suspense long, though.  Since 8 is a power of 2, there’s a fast way to convert between those two bases.  In particular, 23 = 8, so you can convert the digits in the base-8 representation of a number in groups of three.  For the example of 42178, we have 48 = 1002, 28 = 0102, 18 = 0012, and 78 = 1112.  (All of these base-2 representations I had in my head.)  String those four together to get

42178 = 1000100011112.

This process goes in the other direction, too.  And let’s convert from binary to base 16, just to work with a different number than 8.  Thus, for example,

1101010001112 = D4716,

as 01112 = 716, 01002 = 416, and 11012 = D16.  (Note that we have to do the conversion starting with the least significant digit; i.e., from right to left.)

This process works when converting between any two bases where one base is a positive integer power of the other.

A Coin-Flipping Problem

One problem that I’ve assigned when discussing Markov chains is to calculate the expected number of flips required for a particular pattern to appear.  (Here I mean a pattern such as heads, heads, heads, or HHH.)  In this post I’m going to discuss another approach – one that doesn’t use Markov chains – to solve this problem.

Suppose we want to find the expected number of flips required for the pattern HHH to appear.  Call this X.  We can calculate X by conditioning on the various patterns we might achieve that do or don’t give us HHH.  For example, for our first few flips we could observe T, HT, HHT, or HHH.  These cover the entire outcome space and have probability 1/2, 1/4, 1/8, and 1/8, respectively.

• If we observe T, then we effectively have to start the entire process over, and we’ve used one flip to get to that point.  So, if we observe T, then the average number of flips required is $1+X$.
• If we observe HT, then we also have to start the entire process over, and we’ve used two flips to get there.  For this outcome, the average number of flips required is $2+X$.
• If we observe HHT, then we have to start the entire process over, and we’ve used three flips.  The average number of flips required for this scenario is $3+X$.
• Finally, if we observe HHH, then we have achieved our goal.  The number of flips required is 3.

All together, then, the average number of flips required satisfies the equation

$\displaystyle X = 1/2(1+X) + 1/4(2+X) + 1/8(3+X) + 1/8(3).$

Solving for X, we obtain $X = 14$.  So it takes 14 flips, on average, to obtain three consecutive heads.

What if we have a more complicated pattern, though?  Let’s look at HHT as an example.  Let X be the expected number of flips required for HHT to appear for the first time.

Once again, we can condition on the various patterns we might achieve that do or don’t give us HHT.  These are T, HT, HHH, and HHT.  As in the previous example, these cover the entire outcome space and have probability 1/2, 1/4, 1/8, and 1/8, respectively.

• If we observe T, then we have to start over.  The average number of flips required is $1+X$.
• If we observe HT, then we have to start over.  The average number of flips required is $2+X$.
• If we observe HHH, then we don’t have to start over.  It’s entirely possible that the HH at the end of HHH could be followed by a T, and then we would have achieved our pattern!  Mathematically, this means that things get a bit more complicated.
• Let $E_{HH}$ denote the expected number of flips required for HHT to appear given that we currently have HH.
• Thus if we start with HHH, then the average number of flips required to obtain HHT is $3 + E_{HH}$.
• Now we need to determine $E_{HH}$.
• If we currently have HH, then the next flip is either a T, which completes our pattern, or an H, we means we must start over in our quest to complete HHT given that we currently have HH.
• Thus $E_{HH} = 1/2(1) + 1/2(1 + E_{HH})$.
• Solving this yields $E_{HH} = 2$.
• Thus if we observe HHH, then the average number of flips required to obtain HHT is 3 + 2 = 5.
• Finally, if we observe HHT, then we have achieved our goal in 3 flips.

Therefore, $X = 1/2(1 + X) + 1/4(2 + X) + 1/8(5) + 1/8(3)$.  Solving this equation for X gives us $X = 8$ flips.

Notice that it takes fewer flips, on average, to achieve HHT than it does HHH.  This is because we don’t always have to start over every time the sequence fails to match our goal sequence.

The interested reader is invited to find the expected number of flips for the other sequences.

A Request for a Proof of a Binomial Identity

A few weeks ago I received an email from Professor Steve Drekic at the University of Waterloo. He asked if I knew of a way to prove the following binomial identity:

$\displaystyle \sum_{k=1}^n \frac{1}{2k-1} \binom{2k}{k} \binom{2n-2k}{n-k} = \binom{2n}{n}.$

(He told me later that he wanted it to help prove that the random walk on the integers with transition probability $p = 1/2$ is null recurrent.)

It’s an interesting binomial identity, in that it has a nontrivial but not overly complicated sum on the left side and a simple expression on the right. Yet I had not seen it before. I was able to find a proof that uses generating functions, and I thought I would reproduce it here.

Lemma 1.

$\displaystyle \binom{1/2}{k} = \frac{-1}{2k-1} \binom{-1/2}{k}$.

Proof.
By Identity 17 in my book The Art of Proving Binomial Identities [1],

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

where in the second step we use properties of the falling factorial $x^{\underline{k}}$.

Next, we need the following generating function.

Lemma 2.

Proof.
$\displaystyle - \sqrt{1-4x} = \sum_{k=0}^{\infty} \frac{1}{2k-1} \binom{2k}{k} x^k.$

By Newton’s binomial series (Identity 18 in my book [1]),
$\displaystyle -\sqrt{1-4x} = -(1-4x)^{1/2} = -\sum_{k=0}^{\infty} \binom{1/2}{k} (-4x)^k \\ = -\sum_{k=0}^{\infty} \frac{-1}{2k-1} \binom{-1/2}{k} (-4x)^k, \text{ by Lemma 1} \\ = \sum_{k=0}^{\infty} \frac{1}{2k-1} \left(\frac{-1}{4}\right)^k \binom{2k}{k} (-4x)^k, \text{ by Identity 30 in [1]} \\ = \sum_{k=0}^{\infty} \frac{1}{2k-1} \binom{2k}{k} x^k.$

Finally, we’re ready to prove the identity.

Identity 1.

$\displaystyle \sum_{k=1}^n \frac{1}{2k-1} \binom{2k}{k} \binom{2n-2k}{n-k} = \binom{2n}{n}.$

Proof.
Let $a_k = \binom{2k}{k}/(2k-1)$, and let $b_k = \binom{2k}{k}$. Since the generating function for $(a_k)$ is $-\sqrt{1-4x}$ (Lemma 2), and the generating function for $(b_k)$ is $1/\sqrt{1-4x}$ (Identity 150 in [1]), the generating function for their convolution,

$\displaystyle c_n = \sum_{k=0}^n \frac{1}{2k-1} \binom{2k}{k} \binom{2n-2k}{n-k},$

is, by the convolution property for generating functions (see Theorem 13 in [1]),

$\displaystyle - \frac{\sqrt{1-4x}}{\sqrt{1-4x}} = -1.$

Since $-1$ just generates the sequence $-1, 0, 0, 0, \ldots$, this means that

$\displaystyle \sum_{k=0}^n \frac{1}{2k-1} \binom{2k}{k} \binom{2n-2k}{n-k} = \begin{cases} -1, & \: n = 0; \\ 0, & \: n \geq 1.\end{cases}$

Therefore, when $n \geq 1$, we have
$\displaystyle \sum_{k=0}^n \frac{1}{2k-1} \binom{2k}{k} \binom{2n-2k}{n-k} = 0 \\ \implies \sum_{k=1}^n \frac{1}{2k-1} \binom{2k}{k} \binom{2n-2k}{n-k} - \binom{2n}{n} = 0 \\ \implies \sum_{k=1}^n \frac{1}{2k-1} \binom{2k}{k} \binom{2n-2k}{n-k} = \binom{2n}{n}.$

References

1. Michael Z. Spivey, The Art of Proving Binomial Identities, CRC Press, 2019.

Strong Induction Wasn’t Needed After All

Lately when I’ve taught the second principle of mathematical induction – also called “strong induction” – I’ve used the following example to illustrate why we need it.

Prove that you can make any amount of postage of 12 cents or greater using only 4-cent and 5-cent stamps.

At this point in a class we would have just done several examples using regular induction, and so we would be naturally inclined to try to prove this postage stamp problem using that technique.  The base case is easy: Just use three 4-cent stamps to make 12 cents of postage.  The induction hypothesis is as usual, too: Suppose, for some $k \geq 12$, that we can make k cents’ worth of postage using only 4-cent and 5-cent stamps.  But regular induction fails at the induction step: To prove that you can make $k+1$ cents using only 4-cent and 5-cent stamps, knowing that you can make k cents isn’t helpful, since you can’t add either a 4-cent or a 5-cent stamp to k cents’ worth of postage to generate $k+1$ cents’ worth of postage.  Instead, you need to know that you can make, say, $k-3$ cents’ worth of postage.  Then you can add a 4-cent stamp to that amount to produce your $k+1$ cents.  In other words, you need to assume in the induction hypothesis not that you can make k cents but that you can make $12, 13, \ldots, k$ cents.  And that’s the essence of strong induction.

However, before we get to this realization my students will often suggest that we can produce $k+1$ cents from k cents by simply removing one of the 4-cent stamps used to produce k cents and replacing it with a 5-cent stamp.  This is a good idea.  However, it assumes that we actually used at least one 4-cent stamp to produce k cents, and that’s a faulty assumption.  Sometimes we don’t need a 4-cent stamp, such as if we use three 5-cent stamps to produce 15 cents.  (In fact, for 15 cents we must use only 5-cent stamps.)  If we don’t use any 4-cent stamps then we can’t generate $k+1$ cents from k cents by replacing a 4-cent stamp with a 5-cent one.

Normally the students are a little chagrined that this idea fails.  And that happened again this term.  After discussing the idea and why it doesn’t quite work I was about ready to move on and introduce strong induction when one of my students interjected, “I’ve got it!  I’ve got it!”  He pointed out that if we’re going to use only 5-cent stamps to generate k cents, when $k \geq 12$, then we’ll need at least three of them.  That’s true, I said.  Then, he added, if we have at least three 5-cent stamps to make k cents, we can replace them with four 4-cent stamps to yield $k+1$ cents’ worth of postage.  That covers the remaining case.  I was impressed enough that I applauded the class for their collective work in solving the problem.

To recap: You don’t actually need strong induction to solve the stamp problem described above.  Use regular induction, and break the induction step into two cases: (1) If you use at least one 4-cent stamp to make k cents, replace it with a 5-cent stamp to make $k+1$ cents.  (2) If you only use 5-cent stamps to make k cents, you’ll need at least three of them.  Replace those three with four 4-cent stamps to generate $k+1$ cents’ worth of postage.

Well done, Spring 2020 Math 210 students!

Arguments for 0.9999… Being Equal to 1

Recently I tried to explain to my 11-year-old son why 0.9999… equals 1.  The standard arguments for $0.9999... = 1$ (at least the ones I’ve seen) assume more math background than he has.  So I tried another couple of arguments, and they seemed to convince him.

The first argument.  The usual claim you get from people who aren’t yet convinced that $0.9999... = 1$ is that 0.9999… is the real number just before 1 on the number line.  Let’s suppose this is true.  What is the average of 0.9999… and 1?  Stop and think about that for a bit.

If the average exists, it must be larger than 0.9999…, and it must be smaller than 1.  But if 0.9999… is just before 1 on the number line, there can’t be such a number.  So either (1) the averaging operation doesn’t apply to 0.9999… and 1, (2) there are numbers between 0.9999… and 1 on the number line, or (3) 0.9999… equals 1.  But (1) immediately leads to the question “Why not?”, which has no obvious answer, and (2) leads to the question of what their decimal representations would be, which also has no obvious answer.  Explanation (3), that 0.9999… equals 1, starts to look more plausible.

The second argument.  Again, let’s assume that 0.9999… is the real number just before 1 on the number line.  If this is true, then what is the difference of 1 and 0.9999…?  Again, stop and think about that for a bit.

If it’s not zero, then you could just add half of that difference to 0.9999… to get a new number between 0.9999… and 1, which not only contradicts our assumption but also forces us to come up with the decimal representation of such a number.  If it is 0, then you have that $0.9999... = 1$.  And if you try to argue that you can’t subtract 0.9999… from 1, then you need to explain why that operation is not allowed for those two real numbers.  (This second argument is a lot like the first one, really.)  The most reasonable of the three options is that the difference is 0, which means that 0.9999… is actually equal to 1.

Final comments and the standard algebra argument.  Both arguments are reductio ad absurdum arguments; that is, they assume that 0.9999… is not equal to 1 and then reason to a contradiction.  The other arguments that I’ve seen are all direct arguments; i.e., they reason from basic mathematical principles to the conclusion that 0.9999… equals 1.

For example, here’s the standard argument via algebra.  We know that 0.9999… must be equal to some number, so let’s call that number x.  Multiplying by 10 yields $10x = 9.9999...$.  Subtracting the first equation from the second leaves us $9x = 9.0000...$, which implies that $x = 1$.  Thus $0.9999... = 1$.

The algebraic argument is a great one, provided you know algebra.  But for my preteen with a pre-algebra background, these two reductio ad absurdum arguments seemed to be enough to convince him.