Increasingly difficult integrals

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My Favorite One-Liners: Part 111

I tried a new wise-crack in class recently, and it was a rousing success. My math majors had trouble recalling basic facts about tests for convergent and divergent series, and so I projected onto the front screen the Official Repository of all Knowledge (www.google.com) and searched for “divergent series” to “help” them recall their prior knowledge.

Worked like a charm.

https://www.google.com/search?q=divergent+series

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.

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For further reading:

Euler’s Equation

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

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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.

 

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 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.

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.

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?”