Proving theorems and special cases (Part 15): The Mean Value Theorem

In a recent class with my future secondary math teachers, we had a fascinating discussion concerning how a teacher should respond to the following question from a student:

Usually, the answer is no. In this series of posts, we’ve seen that a conjecture could be true for the first 40 cases or even the first $10^{316}$ cases yet not always be true. We’ve also explored the computational evidence for various unsolved problems in mathematics, noting that even this very strong computational evidence, by itself, does not provide a proof for all possible cases.

However, there are plenty of examples in mathematics where it is possible to prove a theorem by first proving a special case of the theorem. For the remainder of this series, I’d like to list, in no particular order, some common theorems used in secondary mathematics which are typically proved by first proving a special case.

6. Theorem (Mean Value Theorem). If $f$ is a continuous function on the interval $[a,b]$ which is differentiable on the interior $(a,b)$ , then there is a point $c \in (a,b)$ so that

$f'(c) = \displaystyle \frac{f(b)-f(a)}{b-a}$

In other words, there is a point $c$ in $(a,b)$ so that the slope of the tangent line at $c$ is the same as the slope of the line segment connecting the endpoints.

This is a consequence of the following lemma.

Lemma (Rolle’s Theorem). If $f$ is a continuous function on the interval $[a,b]$ which is differentiable on the interior $(a,b)$ so that $f(a) = 0$ and $f(b) = 0$ , then there is a point $c \in (a,b)$ so that $f'(c) = 0$ .

Notice that Rolle’s Theorem is really a special case of the Mean Value Theorem: if $f(a) = 0$ and $f(b) = 0$ , then the right-hand side of the conclusion of the Mean Value Theorem becomes

$\displaystyle \frac{f(b)-f(a)}{b-a} = \displaystyle \frac{0-0}{b-a} = 0$ ,

thus matching the conclusion of Rolle’s Theorem.

I won’t type out the proofs of Rolle’s Theorem and the Mean Value Theorem here, since Wikipedia has already done that very well. Suffice it to say that Rolle’s Theorem logically comes first, and then the Mean Value Theorem can be proven using Rolle’s Theorem. The main idea is to assume that the function $f$ satisfies the hypotheses of the Mean Value Theorem and then define

$g(x) = f(x) - f(a) - \displaystyle \frac{f(b)-f(a)}{b-a} (x-a)$

It’s straightforward to show that $g$ satisfies the hypotheses of Rolle’s Theorem and conclude that there must be a point so that $g'(c) = 0$ , from which we obtain the conclusion of the Mean Value Theorem.

Proving theorems and special cases (Part 14): The Power Law of differentiation

In a recent class with my future secondary math teachers, we had a fascinating discussion concerning how a teacher should respond to the following question from a student:

5. Theorem. For any rational number $r$ , we have $\displaystyle \frac{d}{dx} x^r = r x^{r-1}$ .

This theorem is typically proven using the Chain Rule (in the guise of implicit differentiation) and the following lemma:

Lemma. For any integer $n$ , we have $\displaystyle \frac{d}{dx} x^n = n x^{n-1}$ .

Clearly, the lemma is a special case of the main theorem. However, the lemma can be proven without using the main theorem:

Proof of Lemma (Case 1). If $n$ is a positive integer, then

$\displaystyle \frac{d}{dx} x^n = \displaystyle \lim_{h \to 0} \frac{(x+h)^n - x^n}{h}$

$= \displaystyle \lim_{h \to 0} \frac{x^n + n x^{n-1} h + \frac{1}{2} n(n-1) x^{n-2} + \dots + h^n - x^n}{h}$

$= \displaystyle \lim_{h \to 0} \left[ n x^{n-1} + \frac{1}{2} n(n-1) x^{n-2} h + \dots + h^{n-1} \right]$

$= n x^{n-1} + 0 + \dots + 0$

$= n x^{n-1}$

Case 1 can also be proven using the Product Rule and mathematical induction.

Proof of Lemma (Case 2). If $n = 0$ , then the theorem is trivially true since $x^0 = 1$ , and the derivative of a constant is zero.

Proof of Lemma (Case 3). If $n$ is a negative integer, then write $n = -m$ , where $m$ is a positive integer. Then, using the Quotient Rule,

$\displaystyle \frac{d}{dx} x^n = \displaystyle \frac{d}{dx} \left( x^{-m} \right)$

$= \displaystyle \frac{d}{dx} \left( \frac{1}{x^m} \right)$

$= \displaystyle \frac{0 \cdot x^m - 1 \cdot m x^{m-1}}{x^{2m}}$

$= -m x^{-m - 1}$

$= n x^{n-1}$

QED

Now that the lemma has been proven, the main theorem can be proven using the lemma.

Proof of Theorem. Suppose that $r = p/q$ , where $p$ and $q$ are integers. Suppose that $y = x^r = x^{p/q}$ . Then:

$y = x^{p/q}$

$y^q = \displaystyle \left[ x^{p/q} \right]^q$

$y^q = x^p$

Let’s now differentiate with respect to $x$ :

$q y^{q-1} \displaystyle \frac{dy}{dx} = p x^{p-1}$

$\displaystyle \frac{dy}{dx} = \displaystyle \frac{p x^{p-1}}{q y^{q-1}}$

$\displaystyle \frac{dy}{dx} = \displaystyle \frac{p}{q} \frac{x^{p-1}}{[x^{p/q}]^{q-1}}$

$\displaystyle \frac{dy}{dx} = \displaystyle \frac{p}{q} \frac{x^{p-1}}{x^{p - p/q}}$

$\displaystyle \frac{dy}{dx} = \displaystyle \frac{p}{q} x^{p-1 - (p-p/q)}$

$\displaystyle \frac{dy}{dx} = \displaystyle \frac{p}{q} x^{p - 1 - p + p/q}$

$\displaystyle \frac{dy}{dx} = \displaystyle \frac{p}{q} x^{p/q - 1}$

$\displaystyle \frac{dy}{dx} = r x^{r-1}$

QED

Proving theorems and special cases (Part 13): Uniqueness of logarithms

In a recent class with my future secondary math teachers, we had a fascinating discussion concerning how a teacher should respond to the following question from a student:

The next theorem is needed in calculus to show that $\ln x = \displaystyle \int_1^x \frac{dt}{t}$ .

4. Theorem. Let $a \in \mathbb{R}^+ \setminus \{1\}$ . Suppose that $f: \mathbb{R}^+ \rightarrow \mathbb{R}$ has the following four properties:

$f(1) = 0$
$f(a) = 1$
$f(xy) = f(x) + f(y)$ for all $x, y \in \mathbb{R}^+$
$f$ is continuous

Then $f(x) = \log_a x$ for all $x \in \mathbb{R}^+$ .

In other blog posts, I went through the full proof of this theorem, which is divided — actually, scaffolded — into cases:

Case 1. $f(x) = \log_a x$ if $x$ is a positive integer.

Case 2. $f(x) = \log_a x$ if $x$ is a positive rational number.

Case 3. $f(x) = \log_a x$ if $x$ is a negative rational number.

Case 4. $f(x) = \log_a x$ if $x$ is a real number.

Clearly, Case 1 is a subset of Case 2, and Case 3 is a subset of Case 4. Once again, a special case of a theorem is used to prove the full theorem.

Proving theorems and special cases (Part 12): The sum and difference formulas for sine

In a recent class with my future secondary math teachers, we had a fascinating discussion concerning how a teacher should respond to the following question from a student:

3. Theorem 1. $\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha + \sin \beta$

Theorem 2. $\sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$

For angles that are not acute, these theorems can be proven using a unit circle and the following four lemmas:

Lemma 1. $\cos(x - y) = \cos x \cos y + \sin x \sin y$

Lemma 2. $\cos(x + y) = \cos x \cos y - \sin x \sin y$

Lemma 3. $\sin(\pi/2 - x) = \cos x$

Lemma 4. $\cos(\pi/2 - x) = \sin x$

Specifically, assuming Lemmas 1-4, then:

$\sin(\alpha + \beta) = \cos(\pi/2 - [\alpha + \beta])$ by Lemma 4

$= \cos([\pi/2 - \alpha] - \beta)$

$= \cos(\pi/2 - \alpha) \cos \beta + \sin(\pi/2 - \alpha) \sin \beta$ by Lemma 1

$= \sin \alpha \cos \beta + \cos \alpha \sin \beta$ by Lemmas 3 and 4.

Also,

$\sin(\alpha - \beta) = \cos(\pi/2 - [\alpha - \beta])$ by Lemma 4

$= \cos([\pi/2 - \alpha] + \beta)$

$= \cos(\pi/2 - \alpha) \cos \beta - \sin(\pi/2 - \alpha) \sin \beta$ by Lemma 2

$= \sin \alpha \cos \beta - \cos \alpha \sin \beta$ by Lemmas 3 and 4.

However, we see that what I’ve called Lemma 3, often called a cofunction identity, can be considered a special case of Theorem 2. However, this is not circular logic since the cofunction identities can be proven without appealing to Theorems 1 and 2.

Proving theorems and special cases (Part 11): The Law of Cosines

In a recent class with my future secondary math teachers, we had a fascinating discussion concerning how a teacher should respond to the following question from a student:

2. Theorem. In $\triangle ABC$ where $a = BC$ , $b = AC$ , and $c = AB$ , we have $c^2 = a^2 + b^2 - 2 a b \cos (m \angle C)$ .

This is typically proven using the Pythagorean theorem:

Lemma. In right triangle $\triangle ABC$ , where $\angle C$ is a right angle, we have $c^2 = a^2 + b^2$ .

Though it usually isn’t thought of this way, the Pythagorean theorem is a special case of the Law of Cosines since $\cos 90^\circ = 0$ .

There are well over 100 different proofs of the Pythagorean theorem that do not presuppose the Law of Cosines. The standard proof of the Law of Cosines then uses the Pythagorean theorem. In other words, a special case of the Law of Cosines is used to prove the Law of Cosines.

Proving theorems and special cases (Part 10): Angles in a convex n-gon

In a recent class with my future secondary math teachers, we had a fascinating discussion concerning how a teacher should respond to the following question from a student:

1. Theorem. The sum of the angles in a convex n-gon is $180(n-2)$ degrees.

This theorem is typically proven after first proving the following lemma:

Lemma. The sum of the angles in a triangle is $180$ degrees.

Clearly the lemma is a special case of the main theorem: for a triangle, $n=3$ and so $180(n-2) = 180 \times 1 = 180$ . The proof of the lemma uses alternate interior angles and the convention that the angle of a straight line is 180 degrees.

Using this, the main theorem follows by using diagonals to divide a convex n-gon into $n-2$ triangles. (For example, drawing a diagonal divides a quadrilateral into two triangles.) The sum of the angles of the n-gon must equal the sum of the angles of the $n-2$ triangles.

So it is possible to prove a theorem by proving a special case of the theorem. Using the sum of the angles of a triangle to prove the formula for the sum of the angles of a convex n-gon is qualitatively different than the previous computational examples seen earlier in this series.

Proving theorems and special cases (Part 9): The Riemann hypothesis

In a recent class with my future secondary math teachers, we had a fascinating discussion concerning how a teacher should respond to the following question from a student:

Usually, the answer is no. In this series of posts, we’ve already seen that a conjecture could be true for the first 40 cases or even the first $10^{316}$ yet ultimately prove false for all cases.

For the next few posts, I thought I’d share a few of the most famous unsolved problems in mathematics… and just how much computational work has been done to check for a counterexample.

4. The Riemann hypothesis (see here, here, and here for more information) is perhaps the outstanding unsolved problem in pure mathematics, and a prize of $1 million has been offered for its proof.

The Riemann zeta function is defined by

$\zeta(s) = \displaystyle \sum_{n=1}^\infty \frac{1}{n^s}$

for complex numbers $s$ with real part greater than 1. For example,

$\zeta(2) = \displaystyle \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \dots = \displaystyle \frac{\pi^2}{6}$

and

$\zeta(4) = \displaystyle \frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \dots = \displaystyle \frac{\pi^4}{90}$

The definition of the Riemann zeta function can be extended to all complex numbers (except a pole at $s = 1$ ) by the integral

$\zeta(s) = \displaystyle \frac{1}{\Gamma(s)} \int_0^\infty \frac{x^{s-1}}{e^s - 1} dx$

and also analytic continuation.

The Riemann hypothesis conjectures to all solutions of the equation $\zeta(s) = 0$ other than negative even integers occur on the line $s = \frac{1}{2} + i t$ . At present, it is known that the first 10 trillion solutions are on this line, so that every solution with $t < 2.4 \times 10^{12}$ is on this line. Of course, that’s not a proof that all solutions are on this line.

A full description of known results concerning the Riemann hypothesis requires much more than a single post. I’ll refer the interested reader to the links above from MathWorld, Wikipedia, and Claymath and the references embedded in those links. An excellent book for the layman concerning the Riemann hypothesis is Prime Obsession: Bernhard Riemann and the Greatest Unsolved Problem in Mathematics, by John Derbyshire.

Proving theorems and special cases (Part 8): The Collatz conjecture

In a recent class with my future secondary math teachers, we had a fascinating discussion concerning how a teacher should respond to the following question from a student:

Usually, the answer is no. In this series of posts, we’ve already seen that a conjecture could be true for the first 40 cases or even the first $10^{316}$ cases yet ultimately prove false for all cases.

For the next few posts, I thought I’d share a few of the most famous unsolved problems in mathematics… and just how much computational work has been done to check for a counterexample.

3. The Collatz conjecture (see here and here for more information) is an easily stated unsolved problem that can be understood by most fourth and fifth graders. Restated from a previous post:

Here’s the statement of the problem.

Start with any positive integer.
If the integer is even, divide it by $2$ . If it’s odd, multiply it by $3$ and then add $1$ .
Repeat until (and if) you reach $1$ .

Here’s the question: Does this sequence eventually reach $1$ no matter the starting value? Or is there a number out there that you could use as a starting value that has a sequence that never reaches $1$ ?

For every integer less than $5 \times 2^{60} = 5,764,607,523,034,234,880$ , this sequence returns to 1. Of course, this is not a proof that the conjecture will hold for every integer.

Proving theorems and special cases (Part 7): The twin prime conjecture

In a recent class with my future secondary math teachers, we had a fascinating discussion concerning how a teacher should respond to the following question from a student:

For the next few posts, I thought I’d share a few of the most famous unsolved problems in mathematics… and just how much computational work has been done to check for a counterexample.

2. The twin prime conjecture (see here and here for more information) asserts that there are infinitely many primes that have a difference of 2. For example:

3 and 5 are twin primes;

5 and 7 are twin primes;

11 and 13 are twin primes;

17 and 19 are twin primes;

29 and 31 are twin primes; etc.

While most mathematicians believe the twin prime conjecture is correct, an explicit proof has not been found. The largest known twin primes are

$3,756,801,695,685 \times 2^{666,669} \pm 1$ ,

numbers which have 200,700 decimal digits. Also, there are 808,675,888,577,436 twin prime pairs less than $10^{18}$ .

Most mathematicians also believe that there are infinitely many cousin primes, which differ by 4:

3 and 7 are cousin primes;

7 and 11 are cousin primes;

13 and 17 are cousin primes;

19 and 23 are cousin primes;

37 and 41 are cousin primes; etc.

Most mathematicians also believe that there are infinitely many sexy primes (no, I did not make that name up), which differ by 6:

5 and 11 are sexy primes;

7 and 13 are sexy primes;

11 and 17 are sexy primes;

13 and 19 are sexy primes;

17 and 23 are sexy primes; etc.

(Parenthetically, a “sexy” prime is probably the most unfortunate name in mathematics ever since Paul Dirac divided a bracket into a “bra” and a “ket,” thereby forever linking women’s underwear to quantum mechanics.)

While it is unknown if there are infinitely many twin primes, it was recently shown — in 2013 — that, for some integer $N$ that is less than 70 million, there are infinitely many pairs of primes that differ by $N$ . In 2014, this upper bound was reduced to 246. Furthermore, if a certain other conjecture is true, the bound has been reduced to 6. In other words, there are infinitely many twin primes or cousin primes or sexy primes… but, at this moment, we don’t know which one (or ones) is infinite.

In November 2014, Dr. Terence Tao of the UCLA Department of Mathematics was interviewed on the Colbert Report to discuss the twin prime conjecture… and he does a good job explaining to Stephen Colbert how we can know one of the three categories is infinite without knowing which category it is.

From the Colbert Report: http://thecolbertreport.cc.com/videos/6wtwlg/terence-tao

Proving theorems and special cases (Part 6): The Goldbach conjecture

In a recent class with my future secondary math teachers, we had a fascinating discussion concerning how a teacher should respond to the following question from a student:

For the next few posts, I thought I’d share a few of the most famous unsolved problems in mathematics… and just how much computational work has been done to check for a counterexample.

1. The Goldbach conjecture (see here and here for more information) claims that every even integer greater than 4 can be written as the sum of two prime numbers. For example,

4 = 2 + 2,

6 = 3 + 3,

8 = 3 + 5,

10 = 3 + 7,

12 = 5 + 7,

14 = 3 + 11, etc.

This has been verified for all even numbers less than $4 \times 10^{18} = 4,000,000,000,000,000,000$ . A proof for all even numbers, however, has not been found yet.

Here are some results related to the Goldbach conjecture that are known:

1. Any integer greater than 4 is the sum of at most six primes.

2. Every sufficiently large even number can be written as the sum or two primes or the sum of a prime and the product of two primes.

3. Every sufficiently large even number can be written as the sum of two primes and at most 8 powers of 2.

4. Every sufficiently large odd number can be written as the sum of three primes.