The antiderivative of 1/(x^4+1): Part 2

This antiderivative has arguable the highest ratio of “really hard to compute” to “really easy to write”:

\displaystyle \int \frac{1}{x^4 + 1} dx

To compute this integral, I will use the technique of partial fractions. This requires factoring the denominator over the real numbers, which can be accomplished by finding the roots of the denominator. In other words, I need to solve

x^4 + 1 = 0,

or

z^4 = -1.

I switched to the letter z since the roots will be complex. The four roots of this quartic equation can be found with De Moivre’s Theorem by writing

z = r (\cos \theta + i \sin \theta),

where r is a real number, and

-1 + 0i = 1(\cos \pi + \i \sin \pi)

By De Moivre’s Theorem, I obtain

r^4 (\cos 4\theta + i \sin 4 \theta) = 1 (\cos \pi + i \sin \pi).

Matching terms, I obtain the two equations

r^4 = 1 and 4\theta = \pi + 2\pi n

or

r = 1 and \theta = \displaystyle \frac{\pi}{4} + \displaystyle \frac{\pi n}{2}

or

r = 1 and \theta = \displaystyle \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}.

This yields the four solutions

z = 1 \left[ \cos \displaystyle \frac{\pi}{4} + i \sin \frac{\pi}{4} \right] = \displaystyle \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}

z = 1 \left[ \cos \displaystyle \frac{3\pi}{4} + i \sin \frac{3\pi}{4} \right] = -\displaystyle \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}

z = 1 \left[ \cos \displaystyle \frac{5\pi}{4} + i \sin \frac{5\pi}{4} \right] = -\displaystyle \frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2}

z = 1 \left[ \cos \displaystyle \frac{7\pi}{4} + i \sin \frac{7\pi}{4} \right] = \displaystyle \frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2}

Therefore, the denominator x^4 + 1 can be written as the following product of linear factors over the complex plane:

\displaystyle \left(x - \left[ \displaystyle \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \right] \right)\left(x - \left[ \displaystyle \frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2} \right] \right) \left(x - \left[ -\displaystyle \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \right] \right) \left(x - \left[ - \displaystyle \frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2} \right] \right)

or

\displaystyle \left(\left[x - \displaystyle \frac{\sqrt{2}}{2} \right] - i \frac{\sqrt{2}}{2} \right)\left( \left[ x - \displaystyle \frac{\sqrt{2}}{2} \right] + i \frac{\sqrt{2}}{2} \right) \left( \left[ x + \displaystyle \frac{\sqrt{2}}{2} \right] - i \frac{\sqrt{2}}{2} \right) \left( \left[ x + \displaystyle \frac{\sqrt{2}}{2} \right] + i \frac{\sqrt{2}}{2} \right)

or

\displaystyle \left(\left[x - \displaystyle \frac{\sqrt{2}}{2} \right]^2 - \left[ i \frac{\sqrt{2}}{2} \right]^2 \right) \left( \left[ x + \displaystyle \frac{\sqrt{2}}{2} \right]^2 - \left[i \frac{\sqrt{2}}{2} \right]^2 \right)

or

\displaystyle \left(x^2 - x \sqrt{2} + \displaystyle \frac{1}{2} + \displaystyle \frac{1}{2}\right) \left(x^2 + x \sqrt{2} + \displaystyle \frac{1}{2} + \displaystyle \frac{1}{2}\right)

or

\displaystyle \left(x^2 - x \sqrt{2} + 1 \right) \left(x^2 + x \sqrt{2} + 1\right).

We have thus factored the denominator over the real numbers:

\displaystyle \int \frac{dx}{x^4 + 1} = \displaystyle \int \frac{dx}{\left(x^2 - x \sqrt{2} + 1 \right) \left(x^2 + x \sqrt{2} + 1\right)} ,

and the technique of partial fractions can be applied.

I’ll continue the calculation of this integral with tomorrow’s post.

The antiderivative of 1/(x^4+1): Part 1

Here’s an innocuous looking integral:

\displaystyle \int \frac{1}{x^4 + 1} dx

This integral arguably has the highest ratio of “really hard to compute” to “really easy to write” of any indefinite integral, since it is merely a rational function without any powers with non-integer exponents, trigonometric functions, exponential functions, or logarithms. Furthermore, the numerator is a constant while the denominator has only two terms. It doesn’t look that hard.

But this integral is really hard to compute. Indeed, in my experience, this integral is often held as the gold standard for Calculus II (or AP Calculus) students who are learning the various techniques of integration. In this series, I will discuss the various methods that have to be employed to find this antiderivative.

I’ll begin this tomorrow. In the meantime, I’ll leave a thought bubble if you’d like to think about how to compute this integral.

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How to Avoid Thinking in Math Class: Index

I’m doing something that I should have done a long time ago: collecting a series of posts into one single post. Recently, Math With Bad Drawings had a terrific series on how students try to avoid thinking in math class.

Part 1: Introduction: “In teaching math, I’ve come across a whole taxonomy of insidious strategies for avoiding thinking. Albeit for understandable reasons, kids employ an arsenal of time-tested ways to short-circuit the learning process, to jump to right answers and good test scores without putting in the cognitive heavy lifting. I hope to classify and illustrate these academic maladies: their symptoms, their root causes, and (with any luck) their cures.”

Part 2: Students’ natural desire to mindlessly plug numbers into a formula without conceptual understanding.

Part 3: The importance of both computational proficiency and conceptual understanding.

Part 4: Fears of word problems.

Part 5: What happens when students get stuck getting started on a problem.

Part 6: Is only getting the right answer important?

 

Engaging students: Deriving the Pythagorean theorem

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Belle Duran. Her topic, from Geometry: deriving the Pythagorean theorem.

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How can technology be used to effectively engage students with this topic?

Using the video in which the scarecrow from The Wizard of Oz “explains” the Pythagorean theorem, I can get the students to review what the definition of it is. Since the scarecrow’s definition was wrong, I can ask the students what was wrong with his phrasing (he said isosceles, when the Pythagorean theorem pertains to right triangles). Thus, I can ask why it only relates to right triangles, starting the proof for the Pythagorean theorem.

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What interesting things can you say about the people who contributed to the discovery and/or the development of this topic?

While Pythagoras is an important figure in the development of mathematics, little is truly known about him since he was the leader of a half religious, half scientific cult-like society who followed a code of secrecy and often presented Pythagoras as a god-like figure. These Pythagoreans believed that “number rules the universe” and thus gave numerical values to many objects and ideas; these numerical values were endowed with mystical and spiritual qualities. Numbers were an obsession for these people, so much so that they put to death a member of the cult who founded the idea of irrational numbers through finding that if we take the legs of measure 1 of an isosceles right triangle, then the hypotenuse would be equal to sqrt(2). The most interesting of all, is the manner in which Pythagoras died. It all roots back to Pythagoras’ vegetarian diet. He had a strong belief in the transmigration of souls after death, so he obliged to become a vegetarian to avoid the chance of eating a relative or a friend. However, not only did he abstain from eating meat, but also beans since he believed that humans and beans were spawned from the same source, hence the human fetal shape of the bean. In a nutshell, he refused access to the Pythagorean Brotherhood to a wealthy man who grew vengeful and thus, unleashed a mob to go after the Brotherhood. Most of the members were killed, save for a few including Pythagoras (his followers created a human bridge to help him out of a burning building). He was meters ahead from the mob, and was about to run into safety when he froze, for before him stretched a vast bean field. Refusing to trample over a single bean, his pursuers caught up and immediately ended his life.

 

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How has this topic appeared in the news?

Dallas Cowboys coach, Jason Garrett recently made it mandatory for his players to know the Pythagorean theorem. He wants his players to understand that “’if you’re running straight from the line of scrimmage, six yards deep…it takes you a certain amount of time…If you’re doing it from ten yards inside and running to that same six yards, that’s the hypotenuse of the right triangle’” (NBC Sports). Also, recently the Museum of Mathematics (MoMath) and about 500 participants recently proved that New York’s iconic Flatiron building is indeed a right triangle. They measured the sides of the building by first handing out glow sticks for the participants to hold from end to end, then by counting while handing out the glow sticks, MoMath was able to estimate the length of the building in terms of glow sticks.

The lengths came out to be 75^2 + 180^2 = 38,025. After showing their Pythagorean relationship, MoMath projected geometric proofs on the side of the Flatiron building.

 

References

http://www.youtube.com/watch?v=DUCZXn9RZ9s

http://www.youtube.com/watch?v=X1E7I7_r3Cw

http://www.geom.uiuc.edu/~demo5337/Group3/hist.html

http://profootballtalk.nbcsports.com/2013/07/24/jason-garrett-wants-the-cowboys-to-know-the-pythagorean-theorem/

http://www.businessinsider.com/500-math-enthusiasts-prove-the-flatiron-building-is-a-right-triangle-2013-12

Different ways of computing a limit (Part 5)

One of my colleagues placed the following problem on an exam for his Calculus II course…

\displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2+1}}{x}

and was impressed by the variety of correct responses that he received. I thought it would be fun to discuss some of the different ways that this limit can be computed.

Method #5. Another geometric approach. The numbers x and \sqrt{x^2+1} can be viewed as two sides of a right triangle with legs 1 and x and hypotenuse \sqrt{x^2+1}. Therefore, the length of the hypotenuse must be larger than the length of one leg but less than the sum of the lengths of the two legs. In other words,

x < \sqrt{x^2+1} < x+1,

or

1 < \displaystyle \frac{\sqrt{x^2+1}}{x} < \displaystyle 1+\frac{1}{x}.

 Clearly \displaystyle \lim_{x \to \infty} 1 = 1 and \displaystyle \lim_{x \to \infty} \left( 1 + \frac{1}{x} \right) = 1. Therefore, by the Sandwich Theorem, we can conclude that \displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2+1}}{x} = 1.

Different ways of computing a limit (Part 4)

One of my colleagues placed the following problem on an exam for his Calculus II course…

\displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2+1}}{x}

and was impressed by the variety of correct responses that he received. I thought it would be fun to discuss some of the different ways that this limit can be computed.

Method #4. The geometric approach. The numbers x and \sqrt{x^2+1} can be viewed as two sides of a right triangle with legs 1 and x and hypotenuse \sqrt{x^2+1}. So as x gets larger and larger, the longer leg x will get closer and closer in length to the length of the hypotenuse. Therefore, the ratio of the length of the hypotenuse to the length of the longer leg must be 1.

 

Different ways of computing a limit (Part 3)

One of my colleagues placed the following problem on an exam for his Calculus II course…

\displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2+1}}{x}

and was impressed by the variety of correct responses that he received. I thought it would be fun to discuss some of the different ways that this limit can be computed.

Method #3. A trigonometric identity. When we see \sqrt{x^2+1} inside of an integral, one kneejerk reaction is to try the trigonometric substitution x = \tan \theta. So let’s use this here. Also, since x \to \infty, we can change the limit to be \theta \to \pi/2:

\displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2+1}}{x} = \displaystyle \lim_{\theta \to \pi/2} \frac{\sqrt{\tan^2 \theta+1}}{\tan \theta}

= \displaystyle \lim_{\theta \to \pi/2} \frac{\sqrt{\sec^2 \theta}}{\tan \theta}

= \displaystyle \lim_{\theta \to \pi/2} \frac{ \sec \theta}{\tan \theta}

= \displaystyle \lim_{\theta \to \pi/2} \frac{ ~~\displaystyle \frac{1}{\cos \theta} ~~}{ ~~ \displaystyle \frac{\sin \theta}{\cos \theta} ~~ }

= \displaystyle \lim_{\theta \to \pi/2} \frac{ 1}{\sin \theta}

= 1.

Different ways of computing a limit (Part 2)

One of my colleagues placed the following problem on an exam for his Calculus II course…

\displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2+1}}{x}

and was impressed by the variety of correct responses that he received. I thought it would be fun to discuss some of the different ways that this limit can be computed.

Method #2. Using L’Hopital’s Rule. The limit has the indeterminant form \infty/\infty, and so I can differentiate the top and the bottom with respect to x:

\displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2+1}}{x} = \displaystyle \lim_{x \to \infty} \frac{ \displaystyle \frac{d}{dx} \left( \sqrt{x^2+1} \right) }{\displaystyle \frac{d}{dx} \left( x \right)}

= \displaystyle \lim_{x \to \infty} \frac{ \displaystyle \frac{1}{2} \left( x^2+1 \right)^{-1/2} \cdot 2x }{1}

= \displaystyle \lim_{x \to \infty} \frac{x}{\sqrt{x^2+1}}.

Oops… it looks like I just got the reciprocal of the original limit! Indeed, if I use L’Hopital’s Rule again, I’ll just return back to the original limit.

So that doesn’t look very helpful… except it is. If I define the value of this limit to be equal to L, then I’ve just shown that L = 1/L (assuming that the limit exists in the first place, of course). That means that L = 1 or L = -1. Well, clearly the limit of this nonnegative function can’t be negative, and so we conclude that the limit is equal to 1.