Tag Archives: mean value theorem

“On Functions Whose Mean Value Abscissas are Midpoints, with Connections to Harmonic Functions” (with Paul Carter)

This is joint work with Paul Carter. Humorously, we completed this while on a cross-country drive as we moved the newly minted Dr. Carter from Brown to Arizona.

I’ve had a longtime fascination with the standard mean value theorem of calculus.

Mean Value Theorem
Suppose $f$ is a differentiable function. Then there is some $c \in (a,b)$ such that
\frac{f(b) – f(a)}{b-a} = f'(c).

The idea for this project started with a simple question: what happens when we interpret the mean value theorem as a differential equation and try to solve it? As stated, this is too broad. To narrow it down, we might specify some restriction on the $c$, which we refer to as the mean value abscissa, guaranteed by the Mean Value Theorem.

So I thought to try to find functions satisfying
\frac{f(b) – f(a)}{b-a} = f’ \left( \frac{a + b}{2} \right)
for all $a$ and $b$ as a differential equation. In other words, let’s try to find all functions whose mean value abscissas are midpoints.

This looks like a differential equation, which I only know some things about. But my friend and colleague Paul Carter knows a lot about them, so I thought it would be fun to ask him about it.

He very quickly told me that it’s essentially impossible to solve this from the perspective of differential equations. But like a proper mathematician with applied math leanings, he thought we should explore some potential solutions in terms of their Taylor expansions. Proceeding naively in this way very quickly leads to the answer that those (assumed smooth) solutions are precisely quadratic polynomials.

It turns out that was too simple. It was later pointed out to us that verifying that quadratic polynomials satisfy the midpoint mean value property is a common exercise in calculus textbooks, including the one we use to teach from at Brown. Digging around a bit reveals that this was even known (in geometric terms) to Archimedes.

So I thought we might try to go one step higher, and see what’s up with
\frac{f(b) – f(a)}{b-a} = f’ (\lambda a + (1-\lambda) b), \tag{1}
where $\lambda \in (0,1)$ is a weight. So let’s find all functions whose mean value abscissas are weighted averages. A quick analysis with Taylor expansions show that (assumed smooth) solutions are precisely linear polynomials, except when $\lambda = \frac{1}{2}$ (in which case we’re looking back at the original question).

That’s a bit odd. It turns out that the midpoint itself is distinguished in this way. Why might that be the case?

It is beneficial to look at the mean value property as an integral property instead of a differential property,
\frac{1}{b-a} \int_a^b f'(t) dt = f’\big(c(a,b)\big).
We are currently examining cases when $c = c_\lambda(a,b) = \lambda a + (1-\lambda b)$. We can see the right-hand side is differentiable by differentiating the left-hand side directly. Since any point can be a weighted midpoint, one sees that $f$ is at least twice-differentiable. One can actually iterate this argument to show that any $f$ satisfying one of the weighted mean value properties is actually smooth, justifying the Taylor expansion analysis indicated above.

An attentive eye might notice that the midpoint mean value theorem, written as the integral property
\frac{1}{b-a} \int_a^b f'(t) dt = f’ \left( \frac{a + b}{2} \right)
is exactly the one-dimensional case of the harmonic mean value property, usually written
\frac{1}{\lvert B_h \rvert} = \int_{B_h(x)} g(t) dV = g(x).
Here, $B_h(x)$ is the ball of radius $h$ and center $x$. Any harmonic function satisfies this mean value property, and any function satisfying this mean value property is harmonic.

From this viewpoint, functions satisfying our original midpoint mean value property~\eqref{eq:original_midpoint} have harmonic derivatives. But the only one-dimensional harmonic functions are affine functions $g(x) = cx + d$. This gives immediately that the set of solutions to~\eqref{eq:original_midpoint} are quadratic polynomials.

The weighted mean value property can also be written as an integral property. Trying to connect it similarly to harmonic functions led us to consider functions satisfying
\frac{1}{\lvert B_h \rvert} = \int_{B_h(x)} g(t) dV = g(c_\lambda(x,h)),
where $c_\lambda(x,h)$ should be thought of as some distinguished point in the ball $B_h(x)$ with a weight parameter $\lambda$. More specifically,

Are there weighted harmonic functions corresponding to a weighted harmonic mean value property?
In one dimension, the answer is no, as seen above. But there are many more multivariable harmonic functions [in fact, I’ve never thought of harmonic functions on $\mathbb{R}^1$ until this project, as they’re too trivial]. So maybe there are weighted harmonic functions in higher dimensions?

This ends up being the focus of the latter half of our paper. Unexpectedly (to us), an analogous methodology to our approach in the one-dimensional case works, with only a few differences.

It turns out that no, there are no weighted harmonic functions on $\mathbb{R}^n$ other than trivial extensions of harmonic functions from $\mathbb{R}^{n-1}$.

Harmonic functions are very special, and even more special than we had thought. The paper is a fun read, and can be found on the arxiv now. It has been accepted and will appear in American Mathematical Monthly.

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Notes from a talk on the Mean Value Theorem

1. Introduction

When I first learned the Mean Value Theorem and the Intermediate Value Theorem, I thought they were both intuitively obvious and utterly useless. In one of my courses in analysis, I was struck when, after proving the Mean Value Theorem, my instructor said that all of calculus was downhill from there. But it was a case of not being able to see the forest for the trees, and I missed the big picture.

I have since come to realize that almost every major (and often, minor) result of calculus is a direct and immediate consequence of the Mean Value Theorem and the Intermediate Value Theorem. In this talk, we will focus on the forest, the big picture, and see the Mean Value Theorem for what it really is: the true Fundamental Theorem of Calculus.


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Continuity of the Mean Value

1. Introduction

When I first learned the mean value theorem as a high schooler, I was thoroughly unimpressed. Part of this was because it’s just like Rolle’s Theorem, which feels obvious. But I think the greater part is because I thought it was useless. And I continued to think it was useless until I began my first proof-oriented treatment of calculus as a second year at Georgia Tech. Somehow, in the interceding years, I learned to value intuition and simple statements.

I have since completely changed my view on the mean value theorem. I now consider essentially all of one variable calculus to be the Mean Value Theorem, perhaps in various forms or disguises. In my earlier note An Intuitive Introduction to Calculus, we state and prove the Mean Value Theorem, and then show that we can prove the Fundamental Theorem of Calculus with the Mean Value Theorem and the Intermediate Value Theorem (which also felt silly to me as a high schooler, but which is not silly).

In this brief note, I want to consider one small aspect of the Mean Value Theorem: can the “mean value” be chosen continuously as a function of the endpoints? To state this more clearly, first recall the theorem:

Suppose $latex {f}$ is a differentiable real-valued function on an interval $latex {[a,b]}$. Then there exists a point $latex {c}$ between $latex {a}$ and $latex {b}$ such that $$ \frac{f(b) – f(a)}{b – a} = f'(c), \tag{1}$$
which is to say that there is a point where the slope of $latex {f}$ is the same as the average slope from $latex {a}$ to $latex {b}$.

What if we allow the interval to vary? Suppose we are interested in a differentiable function $latex {f}$ on intervals of the form $latex {[0,b]}$, and we let $latex {b}$ vary. Then for each choice of $latex {b}$, the mean value theorem tells us that there exists $latex {c_b}$ such that $$ \frac{f(b) – f(0)}{b} = f'(c_b). $$
Then the question we consider today is, as a function of $latex {b}$, can $latex {c_b}$ be chosen continuously? We will see that we cannot, and we’ll see explicit counterexamples. This, after the fold.


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An Intuitive Overview of Taylor Series

This is a note written for my fall 2013 Math 100 class, but it was not written “for the exam,” nor does anything on here subtly hint at anything on any exam. But I hope that this will be helpful for anyone who wants to get a basic understanding of Taylor series. What I want to do is try to get some sort of intuitive grasp on Taylor series as approximations of functions. By intuitive, I mean intuitive to those with a good grasp of functions, the basics of a first semester of calculus (derivatives, integrals, the mean value theorem, and the fundamental theorem of calculus) – so it’s a mathematical intuition. In this way, this post is a sort of follow-up of my earlier note, An Intuitive Introduction to Calculus.

PLEASE NOTE that my math compiler and my markdown compiler sometimes compete, and sometimes repeated derivatives are too high or too low by one pixel.

We care about Taylor series because they allow us to approximate other functions in predictable ways. Sometimes, these approximations can be made to be very, very, very accurate without requiring too much computing power. You might have heard that computers/calculators routinely use Taylor series to calculate things like $latex {e^x}$ (which is more or less often true). But up to this point in most students’ mathematical development, most mathematics has been clean and perfect; everything has been exact algorithms yielding exact answers for years and years. This is simply not the way of the world.

Here’s a fundamental fact to both mathematics and life: almost anything worth doing is probably pretty hard and pretty messy.

For a very recognizable example, let’s think about finding zeroes of polynomials. Finding roots of linear polynomials is very easy. If we see $latex {5 + x = 0}$, we see that $latex {-5}$ is the zero. Similarly, finding roots of quadratic polynomials is very easy, and many of us have memorized the quadratic formula to this end. Thus $latex {ax^2 + bx + c = 0}$ has solutions $latex {x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}}$. These are both nice, algorithmic, and exact. But I will guess that the vast majority of those who read this have never seen a “cubic polynomial formula” for finding roots of cubic polynomials (although it does exist, it is horrendously messy – look up Cardano’s formula). There is even an algorithmic way of finding the roots of quartic polynomials. But here’s something amazing: there is no general method for finding the exact roots of 5th degree polynomials (or higher degree).

I don’t mean We haven’t found it yet, but there may be one, or even You’ll have to use one of these myriad ways – I mean it has been shown that there is no general method of finding exact roots of degree 5 or higher polynomials. But we certainly can approximate them arbitrarily well. So even something as simple as finding roots of polynomials, which we’ve been doing since we were in middle school, gets incredibly and unbelievably complicated.

So before we hop into Taylor series directly, I want to get into the mindset of approximating functions with other functions.

1. Approximating functions with other functions

We like working with polynomials because they’re so easy to calculate and manipulate. So sometimes we try to approximate complicated functions with polynomials, a problem sometimes called
polynomial interpolation”.

Suppose we wanted to approximate $latex {\sin(x)}$. The most naive approximation that we might do is see that $latex {\sin(0) = 0}$, so we might approximate $latex {\sin(x)}$ by $latex {p_0(x) = 0}$. We know that it’s right at least once, and since $latex {\sin(x)}$ is periodic, it’s going to be right many times. I write $latex {p_0}$ to indicate that this is a degree $latex {0}$ polynomial, that is, a constant polynomial. Clearly though, this is a terrible approximation, and we can do better.


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Math 90: Week 8

Today, we had a set of problems as usual, and a quiz! (And I didn’t tell you about the quiz, even though others did, so I’m going to pretend that it was a pop quiz)!. Below, you’ll find the three problems, their solutions, and a worked-out quiz.


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