Posted by: Dave Richeson | August 27, 2014

Tangent lines to the sine function with rational slope

Today I was wondering the following thing (I won’t bore you with how I ended up with this question):

Are there any rational values of {m} for which the line {y=mx} is tangent to the graph of {y=\sin x?}

Clearly the answer is yes: {m=1.} But my gut feeling was that this was the only such {m.} After some head scratching, I obtained the following proof.

Suppose they are tangent at {x=x_{0}}. Then the point of tangency is {(x_{0},mx_{0})=(x_{0},\sin(x_{0})).} Moreover, the slope of the tangent line is {\cos(x_{0})}. Thus, {x_{0}} must satisfy the two equations:

{\sin(x_{0})=mx_{0},} and

{\cos(x_{0})=m}.

Using a trig identity,

{1=\sin^{2}(x_{0})+\cos^{2}(x_{0})=m^{2}x_{0}^{2}+m^{2}}.

So,

{x_{0}=\sqrt{\frac{1-m^{2}}{m^{2}}}.}

Note that {x_{0}} is an algebraic number—it is the root of a polynomial with integer coefficients.

In 1882 Ferdinand von Lindemann famously proved that {e^{a}} is transcendental (that is, non-algebraic) for every non-zero algebraic number {a,} and a similar proof holds for the sine and cosine functions.

Thus, because {\cos(x_{0})=m,} and {m} is algebraic, it must be the case that {x_{0}=0}. In particular, {m=1.}

This concludes the proof.

 

Posted by: Dave Richeson | March 25, 2014

Gabriel’s paper horn

I just returned from the eleventh Gathering for Gardner. One of the many special things about this unusual conference is that the attendees are strongly encouraged to participate in the “gift exchange.” We were each asked to bring a physical exchange item (one for each of the 350 conference-goers) or to submit a written contribution.

The bag of gifts that I received at the end of the conference was amazing. It contained exotic dice, clever puzzles, unique playing cards, mathematical earrings, 3d printed objects, cool pencils, puzzle books, artwork, etc. Nearly all of the gifts were handmade. We haven’t received the written gifts yet. What a fantastic idea!

gabrielshorn

My gift was a set of instructions and templates for making Gabriel’s horn out of paper cones. Gabriel’s horn is the surface obtained by revolving the curve y=1/x (x\ge 1/2) about the x-axis. Mathematics professors wow introductory calculus students by sharing its paradoxical properties: it has finite volume, but infinite surface area. As they say, “you can fill it with paint, but you can’t paint it.”

Photos of the paper horn (in white and in rainbow) and a photo of the cones are shown below. I’m linking to a pdf copy of the article here in case you want to make one.

hornphoto2hornphoto3cones

Note: my inspiration was this blog post by Dan Walsh in which he makes a pseudosphere out of cones.

 

Posted by: Dave Richeson | February 18, 2014

Undergraduate Math Bloggers

I was interested seeing how undergraduate math students used blogs (and related platforms, like Tumblr). So I posted a call on Google+ and Twitter:

I received quite a few links. I’m looking forward to reading them. It seemed like there was a lot of interest, so I’m posting the unedited list here.

If you know of any others, post them in the comments.

Posted by: Dave Richeson | September 9, 2013

2013: the year of pi

A couple days ago I saw this tweet.

Pretty cool! Let’s see why

\arctan(2)+\arctan(0)+\arctan(1)+\arctan(3)=\pi.

Two terms are easy to deal with:

\arctan(0)=0 and

\arctan(1)=\pi/4.

But why is \arctan(2)+\arctan(3)=3\pi/4?

One way to prove this is via the inverse trigonometric identity

\displaystyle \arctan(a)+\arctan(b)=\arctan\left(\frac{a+b}{1-ab}\right)\, (\text{mod }\pi).

In this case

\arctan(2)+\arctan(3)=\arctan(-1)\, (\text{mod }\pi).

Because \arctan(2) and \arctan(3) are between 0 and \pi/2,  \arctan(a)+\arctan(b)=3\pi/4.

I didn’t use this approach on my first attempt to solve the problem.  (To be honest, I didn’t know this identity existed before finding it online.) I used geometry and trigonometry. We are interested in the angle \theta in the diagram below.

triangle

The law of cosines tells us that

(2+3)^2=(\sqrt{5})^2+(\sqrt{10})^2-2\sqrt{5}\sqrt{10}\cos\theta.

This implies that \cos\theta=-\frac{1}{\sqrt{2}}, and hence \theta=3\pi/4.

(I wonder if there is a year of tau coming up sooner than the year 112233.)

Update: I just saw that Cut-the-Knot has a page devoted to this topic too.

Posted by: Dave Richeson | July 4, 2013

Using a kayak to measure the perimeter of a lake

I’m on vacation this week on a lake in northern Michigan (hold up your right hand, palm toward you, point at the first knuckle of your middle finger—that’s where I am). Yesterday I paddled around the perimeter of the lake in a kayak. On a whim I brought my GPS-enabled phone. My route is shown below. Note that the cartographer didn’t do a perfect job—despite what you see, I can assure you that I paddled only on water.

When I got back, the GPS said that I had paddled for 2.74 miles (14470 feet). Let’s say that I stayed exactly 50 feet off the shore for the entire trip. What is the perimeter of the lake?

We should probably make some assumptions about the shoreline and about my path. For simplicity, let’s say that my path is differentiable, that it bounds a region X, and that the lake is the set {N_{50}(X)}, where N_{r}(X)
is the radius {r} neighborhood of {X}. Intuitively, the boundary of N_r(X) is always r units off the port side of the boat. (See Ravi Vakil’s article “The Mathematics of Doodling” for more about this set.)

You may not think you have enough information to solve the problem, but you do! The clue is found in a well-known puzzler. Suppose we have an electrical wire running round the equator of the earth. We want to get it up off the ground and put it at the top of 50-foot electrical poles. How much longer must the wire be? [If you haven't seen this puzzler before, stop now and think about it.] If the radius of the Earth is {R} feet, then the original wire must be {2\pi R} feet long. The new wire is a circle with a radius 50 feet longer than before, so its length must be {2\pi (R+50)=2\pi R+100\pi} feet. That is, it must be {100\pi\approx 314} feet longer.

It turns out that this argument can be generalized in various ways (again, see Vakil’s article for more information). First of all, if {X} is any convex set with a polygonal boundary, then, just like in the puzzle, the perimeter of {N_{r}(X)} is {2\pi r} units longer than the perimeter of {X}. Vakil illustrates this with the picture below.

It turns out that the exact same result holds if {X} is nonconvex—such as the region enclosed by my kayak trip. As the kayak goes in coves and around peninsulas, the “extra” regions of convexity and nonconvexity cancel each other out. In the end it is the same as going around a circle one time. Thus the perimeter of the lake is {2\pi\cdot 50\approx 314} feet longer than my kayak path: {14470+314=14784} feet.

Posted by: Dave Richeson | April 16, 2013

Countability of the rationals drawn using TikZ

I’m continuing my exploration of TikZ (here is my first post about TikZ). I will be showing my Discrete Math class how to “count” the positive rational numbers. (See this old blog post for more information about countable sets.) I used TikZ to create the picture below.

Screen Shot 2013-04-16 at 9.23.14 PM

Here is the source code for this figure. If you click on the link you can get an editable copy of the document in WriteLatex

Posted by: Dave Richeson | April 12, 2013

Greatest living mathematician and expositor?

On Twitter I posed the following question:

I got a great repsonse. Here is the complete—unedited—list of names (in alphabetical order).

If you have other suggestions, add them in the comments.

Posted by: Dave Richeson | April 9, 2013

Bubble diagrams for functions in LaTeX using TikZ

I am teaching Discrete Math this semester (our intro-to-proof course). One of the topics is functions. Not surprisingly my students and I have to draw “bubble diagrams” for functions between finite sets—and we have to include them in LaTeX documents. Rather than simply sketching them in Adobe Illustrator and importing them as graphics, I decided to try creating them in TikZ. After a lot of tinkering I came up with something pretty nice (see below). Moreover, I set it up so that all the complicated TikZ and LaTeX commands are in the header of the document. So my students and I can easily generate new bubble diagrams.

Screen shot 2013-04-09 at 3.53.26 PM

I’ve put a sample document on WriteLatex. If you click on this link you can get an editable copy of the document. You can edit it and it won’t change my copy—so go wild with it! (This cool feature of WriteLatex is described on their WriteLatex for Education page.)

Posted by: Dave Richeson | March 5, 2013

Circular reasoning: who first proved that C/d is a constant?

I just uploaded an article “Circular reasoning: who first proved that C/d is a constant?” to the arXiv. The abstract is below. It is on a topic that I’ve been thinking about and reading about off-and-on for the last year and a half. I’d be happy to hear people’s thoughts, reactions, and impressions.

Abstract. We answer the question: who first proved that C/d is a constant? We argue that Archimedes proved that the ratio of the circumference of a circle to its diameter is a constant independent of the circle and that the circumference constant equals the area constant (C/d=A/r^{2}). He stated neither result explicitly, but both are implied by his work. His proof required the addition of two axioms beyond those in Euclid’s Elements; this was the first step toward a rigorous theory of arc length. We also discuss how Archimedes’s work coexisted with the 2000-year belief—championed by scholars from Aristotle to Descartes—that it is impossible to find the ratio of a curved line to a straight line.

Posted by: Dave Richeson | February 26, 2013

Interview on Wild about Math!

I had a very nice conversation with Sol Lederman (of the Wild about Math! blog) on his “Inspired by Math” podcast series. Check it out and be sure to check out his other podcast episodes. 

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