Mean Green Math

Source: https://www.facebook.com/MathematicalMemesLogarithmicallyScaled/photos/a.1605246506167805.1073741827.1605219649503824/2080975208594930/?type=3&theater

by John Quintanilla on April 27, 2018 • Permalink

Posted in Calculus, Humor, Tales from the Classroom

Tagged series

Posted by John Quintanilla on April 27, 2018

https://meangreenmath.com/2018/04/27/my-favorite-one-liners-part-111/

Borwein integrals

When teaching proofs, I always stress to my students that it’s not enough to do a few examples and then extrapolate, because it’s possible that the pattern might break down with a sufficiently large example. Here’s an example of this theme that I recently learned:

No automatic alt text available.

For further reading:

by John Quintanilla on February 5, 2018 • Permalink

Posted in Calculus, Discrete mathematics

Tagged integral, proof, sine

Posted by John Quintanilla on February 5, 2018

https://meangreenmath.com/2018/02/05/borwein-integrals/

Euler’s Equation

This was hands-down my favorite variant of the “distracted boyfriend” meme that went around the internet last year.

Image may contain: 3 people, people smiling, people standing

2 Comments

by John Quintanilla on January 29, 2018 • Permalink

Posted in Calculus, Humor

Tagged complex numbers, Euler's formula

Posted by John Quintanilla on January 29, 2018

https://meangreenmath.com/2018/01/29/eulers-equation/

My Favorite One-Liners: Part 104

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 use today’s quip when discussing the Taylor series expansions for sine and/or cosine:

$\sin x = x - \displaystyle \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} \dots$

$\cos x = 1 - \displaystyle \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} \dots$

To try to convince students that these intimidating formulas are indeed correct, I’ll ask them to pull out their calculators and compute the first three terms of the above expansion for $x=0.2$, and then compute $\sin 0.2$ . The results:

This generates a pretty predictable reaction, “Whoa; it actually works!” Of course, this shouldn’t be a surprise; calculators actually use the Taylor series expansion (and a few trig identity tricks) when calculating sines and cosines. So, I’ll tell my class,

It’s not like your calculator draws a right triangle, takes out a ruler to measure the lengths of the opposite side and the hypotenuse, and divides to find the sine of an angle.

by John Quintanilla on May 15, 2017 • Permalink

Tagged cosine, sine, Taylor series

Posted by John Quintanilla on May 15, 2017

https://meangreenmath.com/2017/05/15/my-favorite-one-liners-part-104/

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.

by John Quintanilla on May 14, 2017 • Permalink

Tagged derivative

Posted by John Quintanilla on May 14, 2017

https://meangreenmath.com/2017/05/14/my-favorite-one-liners-part-103/

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.

by John Quintanilla on May 13, 2017 • Permalink

Tagged derivative

Posted by John Quintanilla on May 13, 2017

https://meangreenmath.com/2017/05/13/my-favorite-one-liners-part-102/

My Favorite One-Liners: Part 101

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 a choice has to be made between two different techniques of approximately equal difficulty. For example:

Calculate $\displaystyle \iint_R e^{-x-2y}$ , where $R$ is the region $\{(x,y): 0 \le x \le y < \infty \}$

There are two reasonable options for calculating this double integral.

Option #1: Integrate with respect to $x$ first:

$\int_0^\infty \int_0^y e^{-x-2y} dx dy$

Option #2: Integrate with respect to $y$ first:

$\int_0^\infty \int_x^\infty e^{-x-2y} dy dx$

Both techniques require about the same amount of effort before getting the final answer. So which technique should we choose? Well, as the instructor, I realize that it really doesn’t matter, so I’ll throw it open for a student vote by asking my class:

Anyone ever read the Choose Your Own Adventure books when you were kids?

After the class decides which technique to use, then we’ll set off on the adventure of computing the double integral.

This quip also works well when finding the volume of a solid of revolution. We teach our students two different techniques for finding such volumes: disks/washers and cylindrical shells. If it’s a toss-up as to which technique is best, I’ll let the class vote as to which technique to use before computing the volume.

by John Quintanilla on May 12, 2017 • Permalink

Tagged integral, proof, solid of revolution, volume

Posted by John Quintanilla on May 12, 2017

https://meangreenmath.com/2017/05/12/my-favorite-one-liners-part-101/

My Favorite One-Liners: Part 99

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.

Today’s quip is a light-hearted one-liner that I’ll use to lighten the mood when in the middle of a complex calculation, like the following limit problem from calculus:

Let $f(x) = 11-4x$ . Find $\delta$ so that $|f(x) - 3| < \epsilon$ whenever $|x-2| < \delta$.

The solution of this problem requires isolating $x$ in the above inequality:

$|(11-4x) - 3| < \epsilon$

$|8-4x| < \epsilon$

$-\epsilon < 8 - 4x < \epsilon$

$-8-\epsilon < -4x < -8 + \epsilon$

At this point, the next step is dividing by $-4$ . So, I’ll ask my class,

When we divide by $-4$ , what happens to the crocodiles?

This usually gets the desired laugh out of the middle-school rule about how the insatiable “crocodiles” of an inequality always point to the larger quantity, leading to the next step:

$2 + \displaystyle \frac{\epsilon}{4} > x > 2 - \displaystyle \frac{\epsilon}{4}$ ,

so that

$\delta = \min \left( \left[ 2 + \displaystyle \frac{\epsilon}{4} \right] - 2, 2 - \left[2 - \displaystyle \frac{\epsilon}{4} \right] \right) = \displaystyle \frac{\epsilon}{4}$ .

Formally completing the proof requires starting with $|x-2| < \displaystyle \frac{\epsilon}{4}$ and ending with $|f(x) - 3| < \epsilon$ .

by John Quintanilla on May 10, 2017 • Permalink

Posted in Calculus, Tales from the Classroom, Theorems

Tagged limit, proof

Posted by John Quintanilla on May 10, 2017

https://meangreenmath.com/2017/05/10/my-favorite-one-liners-part-99/

My Favorite One-Liners: Part 91

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.

Everyone once in a while, a student might make a careless mistake — or just choose an incorrect course of action — that changes what was supposed to be a simple problem into an incredibly difficult problem. For example, here’s a problem that might arise in Calculus I:

Find $f'(x)$ if $f(x) = \displaystyle \int_0^x (1+t^2)^{10} \, dt$

The easy way to do this problem, requiring about 15 seconds to complete, is to use the Fundamental Theorem of Calculus. The hard way is by multiplying out $(1+t^2)^{10}$ — preferably using Pascal’s triangle — taking the integral term-by-term, and then taking the derivative of the result. Naturally, a student who doesn’t see the easy way of doing the problem might get incredibly frustrated by the laborious calculations.

So here’s the advice that I give my students to trying to discourage them from following such rabbit trails:

If you find yourself stuck on what seems to be an incredibly difficult problem, you should ask yourself, “Just how evil do I think my professor is?”

by John Quintanilla on May 2, 2017 • Permalink

Posted in Calculus, Humor, Tales from the Classroom

Tagged binomial series, fundamental theorem of calculus, Pascal's triangle

Posted by John Quintanilla on May 2, 2017

https://meangreenmath.com/2017/05/02/my-favorite-one-liners-part-91/

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.

If $f(x) = c$ , a constant, then $\displaystyle \frac{d}{dx} (c) = 0$ .
If $f(x)$ and $g(x)$ are both differentiable, then $(f+g)'(x) = f'(x) + g'(x)$ .
If $f(x)$ is differentiable and $c$ is a constant, then $(cf)'(x) = c f'(x)$ .
If $f(x) = x^n$ , where $n$ is a nonnegative integer, then $f'(x) = n x^{n-1}$ .
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.