C is for constant of integration, and that’s good enough for me

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Derivative of 1/x



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Richard Feynman’s Integral Trick

“I had learned to do integrals by various methods shown in a book that my high school physics teacher Mr. Bader had given me. [It] showed how to differentiate parameters under the integral sign — it’s a certain operation. It turns out that’s not taught very much in the universities; they don’t emphasize it. But I caught on how to use that method, and I used that one damn tool again and again. [If] guys at MIT or Princeton had trouble doing a certain integral, [then] I come along and try differentiating under the integral sign, and often it worked. So I got a great reputation for doing integrals, only because my box of tools was different from everybody else’s, and they had tried all their tools on it before giving the problem to me.” (Surely you’re Joking, Mr. Feynman!)

I read Surely You’re Joking, Mr. Feynman! dozens of times when I was a teenager, and I was always curious about exactly what this integration technique actually was. So I enjoyed reading this article about the Leibniz Integration Rule: https://medium.com/dialogue-and-discourse/richard-feynmans-integral-trick-e7afae85e25c

Powers Great and Small

I enjoyed this reflective piece from Math with Bad Drawings about determining whether a^b or b^a is larger. The final answer, involving the number e, was a complete surprise to me.

Short story: e is the unique number so that e^x > x^e for all positive x.

Powers Great and Small

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.


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.