# Convexity and Orthogonality at Saddle Points

Today, the Texas Section of the Mathematical Association of America is holding its annual conference. Like many other professional conferences these days, this conference will be held virtually, and so my contribution to the conference is saved on YouTube and is available to the public.

Here’s the abstract of my talk: “At a saddle point (like the middle of a Pringles potato chip), the directions of maximum upward concavity and maximum downward concavity are perpendicular. The usual proof requires a fair amount of linear algebra: eigenvectors of different eigenvalues of a real symmetric matrix, like the Hessian, must be orthogonal. For this reason, the orthogonality of these two directions is not often stated in calculus textbooks, let alone proven, when the Second Partial Derivative Test for identifying local extrema and saddle points is discussed. In this talk, we present an elementary proof of the orthogonality of these two directions that requires only ideas from Calculus III and trigonometry. Not surprisingly, this proof can be connected to the usual proof from linear algebra.”

If you have 12 minutes to spare, here’s the talk.

# Adding by a Form of 0: Index

I’m doing something that I should have done a long time ago: collecting a series of posts into one single post. The following links comprised my series on adding by a form of 0 (analogous to multiplying by a form of 1).

Part 1: Introduction.

Part 2: The Product and Quotient Rules from calculus.

Part 3: A formal mathematical proof from discrete mathematics regarding equality of sets.

Part 4: Further thoughts on adding by a form of 0 in the above proof.

# A Clean Calculus Joke

The change of position over time is velocity.

The change of velocity over time is acceleration.

The change of acceleration over time is jerk.

And the change of jerk over time is an election.d

# Adding by a Form of 0 (Part 2)

Often intuitive appeals for the proof of the Product Rule rely on pictures like the following:

The above picture comes from https://mrchasemath.com/2017/04/02/the-product-rule/, which notes the intuitive appeal of the argument but also its lack of rigor.

My preferred technique is to use the above rectangle picture but make it more rigorous. Assuming that the functions $f$ and $g$ are increasing, the difference $f(x+h) g(x+h) - f(x) g(x)$ is exactly equal to the sum of the green and blue areas in the figure below.

In other words,

$f(x+h) g(x+h) - f(x) g(x) = f(x+h) [g(x+h) - g(x)] + [f(x+h) - f(x)] g(x)$,

or

$f(x+h) g(x+h) - f(x+h) g(x) + f(x+h) g(x) - f(x) g(x)$.

This gives a geometrical way of explaining this otherwise counterintuitive step for students not used to adding by a form of 0. I make a point of noting that we took one term, $f(x+h)$, from the first product $f(x+h) g(x+h)$, while the second term, $g(x)$, came from the second product $f(x) g(x)$. From this, the usual proof of the Product Rule follows:

$[(fg)(x)]' = \displaystyle \lim_{h \to 0} \frac{f(x+h) g(x+h) - f(x) g(x)}{h}$

$\displaystyle = \lim_{h \to 0} \frac{f(x+h) [g(x+h) - g(x)]}{h} + \lim_{h\ to 0} \frac{[f(x+h) - f(x)] g(x)}{h}$

$\displaystyle = \lim_{h \to 0} \frac{f(x+h) [g(x+h) - g(x)]}{h} + \lim_{h\ to 0} \frac{[f(x+h) - f(x)] g(x)}{h}$

$\displaystyle = \lim_{h \to 0} f(x+h) \frac{g(x+h) - g(x)}{h} + \lim_{h\ to 0} \frac{f(x+h) - f(x) }{h} g(x)$

$= f(x)g'(x) + f'(x) g(x)$

For what it’s worth, a Google Images search for proofs of the Product Rule yielded plenty of pictures like the one at the top of this post but did not yield any pictures remotely similar to the green and blue rectangles above. This suggests to me that the above approach of motivating this critical step of this derivation might not be commonly known.

Once students have been introduced to the idea of adding by a form of 0, my experience is that the proof of the Quotient Rule is much more palatable. I’m unaware of a geometric proof that I would be willing to try with students (a description of the best attempt I’ve seen can be found here), and so adding by a form of 0 becomes unavoidable. The proof begins

$\left[\left( \displaystyle \frac{f}{g} \right)(x) \right]' = \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{f(x+h)}{ g(x+h)} - \frac{f(x)}{ g(x)}}{h}$

$= \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{f(x+h) g(x) - f(x) g(x+h)}{ g(x) g(x+h)}}{h}$

$= \displaystyle \lim_{h \to 0} \frac{f(x+h) g(x) - f(x) g(x+h)}{ h g(x) g(x+h)}$.

At this point, I ask my students what we should add and subtract this time to complete the derivation. Given the previous experience with the Product Rule, students are usually quick to chose one factor from the first term and another factor from the second term, usually picking $f(x) g(x)$. In fact, they usually find this step easier than the analogous step in the Product Rule because this expression is more palatable than the slightly more complicated $f(x+h) g(x)$. From here, the rest of the proof follows:

$[(fg)(x)]' = \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{f(x+h) g(x) - f(x) g(x+h)}{h }}{g(x) g(x+h)}$

$= \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{f(x+h) g(x) - f(x)g(x) + f(x)g(x) - f(x) g(x+h)}{h }}{g(x) g(x+h)}$

$= \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{f(x+h) g(x) - f(x)g(x)}{h} + \frac{f(x)g(x) - f(x) g(x+h)}{h }}{g(x) g(x+h)}$

$= \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{f(x+h) g(x) - f(x)g(x)}{h} - \frac{f(x) g(x+h) - f(x)g(x)}{h }}{g(x) g(x+h)}$

$= \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{[f(x+h) - f(x)] g(x)}{h} - \frac{f(x) [g(x+h) - g(x)]}{h }}{g(x) g(x+h)}$

$= \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{f(x+h) - f(x) }{h} g(x) - f(x) \frac{ g(x+h) - g(x)}{h }}{g(x) g(x+h)}$

$= \displaystyle \frac{ f'(x) g(x) - f(x) g'(x)}{g(x)^2}$

P.S.

• The website https://mrchasemath.com/2017/04/02/the-product-rule/ also suggests an interesting pedagogical idea: before giving the formal proof of the Product Rule, use a particular function and the limit definition of a derivative so that students can intuitively guess the form of the rule. For example, if $g(x) = x^2$:

# Adding by a Form of 0 (Part 1)

Adding by a form of 0, or adding and subtracting the same quantity, is a common technique in mathematical proofs. For example, this technique is used in the second step of the standard proof of the Product Rule in calculus:

$[(fg)(x)]' = \displaystyle \lim_{h \to 0} \frac{f(x+h) g(x+h) - f(x) g(x)}{h}$

$\displaystyle = \lim_{h \to 0} \frac{f(x+h) g(x+h) - f(x+h) g(x) + f(x+h) g(x) - f(x) g(x)}{h}$

$\displaystyle = \lim_{h \to 0} \left[ \frac{f(x+h) g(x+h) - f(x+h) g(x)}{h} + \frac{f(x+h) g(x) - f(x) g(x)}{h} \right]$

$\displaystyle = \lim_{h \to 0} \frac{f(x+h) g(x+h) - f(x+h) g(x)}{h} + \lim_{h\ to 0} \frac{f(x+h) g(x) - f(x) g(x)}{h}$

$\displaystyle = \lim_{h \to 0} \frac{f(x+h) [g(x+h) - g(x)]}{h} + \lim_{h\ to 0} \frac{[f(x+h) - f(x)] g(x)}{h}$

$\displaystyle = \lim_{h \to 0} f(x+h) \frac{g(x+h) - g(x)}{h} + \lim_{h\ to 0} \frac{f(x+h) - f(x) }{h} g(x)$

$= f(x)g'(x) + f'(x) g(x)$

Or the proof of the Quotient Rule:

$\left[\left( \displaystyle \frac{f}{g} \right)(x) \right]' = \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{f(x+h)}{ g(x+h)} - \frac{f(x)}{ g(x)}}{h}$

$= \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{f(x+h) g(x) - f(x) g(x+h)}{ g(x) g(x+h)}}{h}$

$= \displaystyle \lim_{h \to 0} \frac{f(x+h) g(x) - f(x) g(x+h)}{ h g(x) g(x+h)}$

$= \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{f(x+h) g(x) - f(x) g(x+h)}{h }}{g(x) g(x+h)}$

$= \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{f(x+h) g(x) - f(x)g(x) + f(x)g(x) - f(x) g(x+h)}{h }}{g(x) g(x+h)}$

$= \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{f(x+h) g(x) - f(x)g(x)}{h} + \frac{f(x)g(x) - f(x) g(x+h)}{h }}{g(x) g(x+h)}$

$= \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{f(x+h) g(x) - f(x)g(x)}{h} - \frac{f(x) g(x+h) - f(x)g(x)}{h }}{g(x) g(x+h)}$

$= \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{[f(x+h) - f(x)] g(x)}{h} - \frac{f(x) [g(x+h) - g(x)]}{h }}{g(x) g(x+h)}$

$= \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{f(x+h) - f(x) }{h} g(x) - f(x) \frac{ g(x+h) - g(x)}{h }}{g(x) g(x+h)}$

$= \displaystyle \frac{ f'(x) g(x) - f(x) g'(x)}{g(x)^2}$

This is a technique that we expect math majors to add to their repertoire of techniques as they progress through the curriculum. I forget the exact proof, but I remember that, when I was a student in honors calculus, we had some theorem that required an argument of the form

$|x - y| = |x - A + A - B + B - C + C - D + D - E + E - F + F - y|$

$\le |x - A| + |A - B| + |B - C| + |C - D| + |D - E| + |E - F| + |F - y|$

$\le \displaystyle \frac{\epsilon}{7} + \frac{\epsilon}{7} +\frac{\epsilon}{7} +\frac{\epsilon}{7} +\frac{\epsilon}{7} +\frac{\epsilon}{7} +\frac{\epsilon}{7}$

$= \epsilon$

But while this is a technique that expect students to master, there’s no doubt that this looks utterly foreign to a student first encountering this technique. After all, in high school algebra, students would simplify something like $x - A + A - B + B - C + C - D + D - E + E - F + F - y$ into $x-y$. If they were to convert $x-y$ into something more complicated like $x - A + A - B + B - C + C - D + D - E + E - F + F - y$, they would most definitely get points taken off.

In this brief series, I’d like to give some thoughts on getting students comfortable with this technique.

# My Favorite One-Liners: Part 103

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.

I’ll use today’s one-liner to give students my expectations about simplifying incredibly complicated answers. For example,

Find $f'(x)$ if $f(x) = \displaystyle \frac{\sqrt{x} \csc^5 (\sqrt{x} )}{x^2+1}$.

Using the rules for differentiation,

$f(x) = \displaystyle \frac{[\sqrt{x} \csc^5 (\sqrt{x} )]'(x^2+1) -[\sqrt{x} \csc^5 (\sqrt{x} )](x^2+1)' }{(x^2+1)^2}$

$= \displaystyle \frac{[(\sqrt{x})' \csc^5 (\sqrt{x} ) + \sqrt{x} (\csc^5(\sqrt{x}))'](x^2+1) - \sqrt{x} \csc^5 (\sqrt{x} )](2x) }{(x^2+1)^2}$

$= \displaystyle \frac{[\frac{1}{2\sqrt{x}} \csc^5 (\sqrt{x} ) + 5 \sqrt{x} \csc^4(\sqrt{x}) [-\csc(\sqrt{x})\cot(\sqrt{x})]\frac{1}{2\sqrt{x}}(x^2+1) - \sqrt{x} \csc^5 (\sqrt{x} )](2x) }{(x^2+1)^2}$

With some effort, this simplifies somewhat:

$f'(x) = -\displaystyle \frac{\left(5 x^{5/2} \cot \left(\sqrt{x}\right)+3 x^2+5 \sqrt{x} \cot \left(\sqrt{x}\right)-1\right) \csc ^5\left(\sqrt{x}\right)}{2 \sqrt{x} \left(x^2+1\right)^2}$

Still, the answer is undeniably ugly, and students have been well-trained by their previous mathematical education to think the final answers are never that messy. So, if they want to try to simplify it further, I’ll give them this piece of wisdom:

You can lipstick on a pig, but it remains a pig.

# My Favorite One-Liners: Part 102

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.

I’ll use today’s one-liner when the final answer is a hideous mess. For example,

Find $f'(x)$ if $f(x) = \displaystyle \frac{\sqrt{x} \csc^5 (\sqrt{x} )}{x^2+1}$.

The answer isn’t pretty:

$f'(x) = -\displaystyle \frac{\left(5 x^{5/2} \cot \left(\sqrt{x}\right)+3 x^2+5 \sqrt{x} \cot \left(\sqrt{x}\right)-1\right) \csc ^5\left(\sqrt{x}\right)}{2 \sqrt{x} \left(x^2+1\right)^2}$

This leads to the only possible response:

As all the King’s horses and all the King’s men said when discovering Humpty Dumpty… yuck.

# My Favorite One-Liners: Part 88

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.

In the first few weeks of my calculus class, after introducing the definition of a derivative,

$\displaystyle \frac{dy}{dx} = y' = f'(x) = \lim_{h \to 0} \displaystyle \frac{f(x+h) - f(x)}{h}$,

I’ll use the following steps to guide my students to find the derivatives of polynomials.

1. If $f(x) = c$, a constant, then $\displaystyle \frac{d}{dx} (c) = 0$.
2. If $f(x)$ and $g(x)$ are both differentiable, then $(f+g)'(x) = f'(x) + g'(x)$.
3.  If $f(x)$ is differentiable and $c$ is a constant, then $(cf)'(x) = c f'(x)$.
4. If $f(x) = x^n$, where $n$ is a nonnegative integer, then $f'(x) = n x^{n-1}$.
5. If $f(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$ is a polynomial, then $f'(x) = n a_n x^{n-1} + (n-1) a_{n-1} x^{n-2} + a_1$.

After doing a few examples to help these concepts sink in, I’ll show the following two examples with about 3-4 minutes left in class.

Example 1. Let $A(r) = \pi r^2$. Notice I’ve changed the variable from $x$ to $r$, but that’s OK. Does this remind you of anything? (Students answer: the area of a circle.)

What’s the derivative? Remember, $\pi$ is just a constant. So $A'(r) = \pi \cdot 2r = 2\pi r$.

Does this remind you of anything? (Students answer: Whoa… the circumference of a circle.)

Generally, students start waking up even though it’s near the end of class. I continue:

Example 2. Now let’s try $V(r) = \displaystyle \frac{4}{3} \pi r^3$. Does this remind you of anything? (Students answer: the volume of a sphere.)

What’s the derivative? Again, $\displaystyle \frac{4}{3} \pi$ is just a constant. So $V'(r) = \displaystyle \frac{4}{3} \pi \cdot 3r^2 = 4\pi r^2$.

Does this remind you of anything? (Students answer: Whoa… the surface area of a sphere.)

By now, I’ve really got my students’ attention with this unexpected connection between these formulas from high school geometry. If I’ve timed things right, I’ll say the following with about 30-60 seconds left in class:

Hmmm. That’s interesting. The derivative of the area of a circle is the circumference of the circle, and the derivative of the area of a sphere is the surface area of the sphere. I wonder why this works. Any ideas? (Students: stunned silence.)

This is what’s known as a cliff-hanger, and I’ll give you the answer at the start of class tomorrow. (Students groan, as they really want to know the answer immediately.) Class is dismissed.

If you’d like to see the answer, see my previous post on this topic.

# My Favorite One-Liners: Part 87

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.

When teaching Calculus I, I use the following mantra throughout the semester. I heard this from my calculus instructor back in 1984, and I repeat it for my own students:

There are two themes of calculus: approximating curved things by straight things, and passing to limits.

For example, to find a derivative, we approximate a curved function by a straight tangent line and then pass to a limit. Later in the semester, to find a definite integral, we approximate the area under a curve by the sum of a bunch of straight rectangles and then pass to a limit.

For further reading, I’ll refer to this series of posts on what I typically do on the first day of my calculus class.