Some Trig

A large portion of the “standard” precalculus curriculum consists of a rather tedious recitation of trigonometric identities. I am largely of the opinion that there are only a couple that one really needs to know—for example, the Pythagorean identity (\(\sin(\theta)^2 + \cos(\theta)^2 = 1\) for all \(\theta\)), one of the angle addition formulæ (e.g. \(\cos(\theta+\varphi) = \cos(\theta)\cos(\varphi) – \sin(\theta)\sin(\varphi)\) for all \(\theta,\varphi\)), and the law of cosines (essentially the Pythagorean theorem with a correction term) immediately come to mind as important results. In the last quarter, while teaching precalculus class, I was introduced to a couple of proofs that I had not seen before. I don’t get the impression that either proof is that unusual or original, but I do think that they are both rather slick. I’m putting them up here for future reference.

Angle Addition Formulæ

Theorem: Given any two angles \(\alpha, \beta\), we have the identities \[\cos(\alpha + \beta) = \cos(\alpha)\cos(\beta) – \sin(\alpha)\sin(\beta) \quad\text{and}\quad \sin(\alpha + \beta) = \cos(\alpha)\sin(\beta) + \sin(\alpha)\cos(\beta).\]

Proof: It is possible to obtain angle addition formulæ for both sines and cosines with one picture. The argument is as follows: first, we need to angles to add together, say \(\alpha\) and \(\beta\). We are going to draw them so that \(\alpha\), \(\beta\), and \(\alpha+\beta\) all live in quadrant I, though the argument can be generalized to other quadrants without too much additional work (exercise: do this).

The sum of two angles on the unit circle.

Note that the point \(B\) is the point on the unit circle where the ray emanating from the origin with angle \(\alpha+\beta\) intersects, hence \[B = (\cos(\alpha+\beta),\sin(\alpha+\beta));\] we’ll come back to this later. Next, we can construct a line through \(B\) that is perpendicular to the ray emanating from the origin with angle \(\alpha\)—call the intersection \(C\).

The segment BC is constructed.

Similarly, construct a segment perpendicular to the \(x\)-axis through \(C\), and a segment perpendicular to the \(y\)-axis through \(B\)—these are the points \(D\) and \(E\), respectively.

Similarly, construct EB and CD.

We finish the figure by extending the segments \(EB\) and \(DC\) until they intersect at a point \(F\). Note that we now have a rectangle \(ODFE\).

Construct F and clean up the mess.

The line \(OB\) is transverse to the parallel lines \(EF\) and \(OD\) (since these are opposite sides of a rectangle), thus \(m\angle EBO = m\angle COD = \alpha+\beta\). Moreover, since \[m\angle FCB + m\angle BCO + m\angle OCD = \pi,\] it follows that \(m\angle FCB = \alpha\). These are labeled in the next figure.

Labeling some angles.

Recall that \(B = (\cos(\alpha+\beta),\sin(\alpha+\beta))\). This gives us the lengths of the segments \(OE\) and \(EB\).

Determining the lengths of two sides.

Via some right triangle trig identities on \(\triangle OCB\) (in particular, the identities given by the mnemonic “SOH CAH TOA”), we have that \(OC = \cos(\beta)\) and \(BC = \sin(\beta)\).

Finding some more sides.

Using the same trig identities, this time applied to \(\triangle ODC\), we have that \(OD = \cos(\alpha)\sin(\beta)\), and \(CD = \sin(\alpha)\cos(\beta)\).

And more sides.

Finally, repeating the same argument on \(\triangle BFC\), we have \(BF = \sin(\alpha)\sin(\beta)\) and \(CF = \cos(\alpha)\sin(\beta)\).

Finish the figure.

But this figure essentially completes the proof. Since opposite sides of a rectangle have the same length, the vertical sides give us \[\sin(\alpha+\beta) = \cos(\alpha)\sin(\beta) + \sin(\alpha)\cos(\beta),\] and the horizontal sides render \[\cos(\alpha+\beta) + \sin(\alpha)\sin(\beta) = \cos(\alpha)\cos(\beta).\] Note that, after moving the \(\sin(\alpha)\sin(\beta)\) term of the second identity to the right-hand side, these are the proposed angle addition formulæ.

Law of Cosines

Theorem: Suppose that an arbitrary triangle has sides of length \(a\), \(b\), and \(c\), with the angle \(\theta\) opposite the side \(c\). Then \[c^2 = a^2 + b^2 – 2ab\cos(\theta).\] Note that if \(\theta\) is a right angle, then \(c\) is the hypotenuse of a right triangle and \(\cos(\theta) = 0\), giving the Pythagorean Theorem. Hence we may regard the Law of Cosines as a generalization of the Pythagorean Theorem with some kind of correction term.

Proof: We may assume that the triangle is oriented such that \(\theta\) is at the origin and one of the two “legs” is along the \(x\)-axis in the Cartesian plane. Labeling the vertices such that vertex \(A\) is opposite the side with length \(a\), \(B\) is opposite \(b\), and \(C\) is opposite \(c\), we obtain the figure shown below.

Finish the figure.

Since the point \(A\) is along the \(x\)-axis, the \(y\)-coordinate of \(A\) is zero, and since the distance from \(A\) to the origin (\(C\)) is the length of the corresponding leg of the triangle, we have that the \(x\)-coordinate of \(A\) is \(b\). That is,\[A = (b,0).\] To determine the coordinates of \(B\), note that the ray from \(C\) through \(B\) must intersect the unit circle at some point \(B’\), as shown below.

Finish the figure.

From the definitions of sine and cosine on the unit circle, we know that the coordinates of \(B’\) are given by \(B’ = (\cos(\theta),\sin(\theta))\). Then, as \(\triangle ABC\) is similar to \(\triangle AB’C\), it follows that \[B = (a\cos(\theta),a\sin(\theta)).\]

On the one hand, we know that \(d(A,B) = c\) (that is, the distance from \(A\) to \(B\) is \(c\) units). On the other hand, the distance formula tells us that \[d(A,B) = \sqrt{(a\cos(\theta)-b)^2 + (a\sin(\theta))^2} = \sqrt{a^2(\cos(\theta)^2 + \sin(\theta)^2) – 2ab\cos(\theta) + b^2}.\] Applying the Pythagorean Identity \(\cos(\theta)^2 + \sin(\theta)^2 = 1\) and squaring, we obtain \[c^2 = d(A,B)^2 = a^2 – 2ab\cos(\theta) + b^2,\] which is the desired result.

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My wife brought home a loaf bread this weekend. The nutrition label reads as follows:

I’m not entirely sure how one loaf of bread can consist of only 4.5 servings when a single serving is one ninth of a loaf. On the other hand, fractions are hard, I guess. :\

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Seminar Talk in FRG

I gave a talk last week before the Fractal Research Group at UCR. The goal was to introduce my colleagues to the Assouad dimension, and to share some of the more interesting and/or surprising results. My notes (typos and all) are available for those that are interested. During my talk, I managed to get through the sections on homogeneity and the Assouad dimension, including most of the examples involving orthogonal sequences. I did not have time to discuss the Laakso graph. As further reading, I suggest James Robinson’s Dimensions, Embeddings, and Attractors—my notes are shamelessly plagarized from chapter 9 of that volume.

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The Sphere is Weakly Dense in the Ball

Another qual question, this one dealing with Hilbert spaces and the weak topology:

Exercise: Let \(\mathcal{H}\) be an infinite dimensional Hilbert space. Show that the unit sphere \(S := \{x\in\mathcal{H} : \|x\| = 1\}\) is weakly dense in the unit ball \(B := \{x\in\mathcal{H} : \|x\| \le 1\}\).

I find this to be a really surprising and counter-intuitive result. What this exercise asks us to prove is that every point inside of the unit ball in an infinite dimensional Hilbert space is, in some sense, arbitrarily close to the boundary of that ball. This is really striking to me—how can the center of a ball be really close to the boundary of that ball? Because this result is so unexpected, I think it is worth understanding. Indeed, the arguments presented below have helped me to build some better intuition about both infinite dimensional Hilbert spaces and the weak topology.

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Nowhere Differentiable Functions

In an undergraduate analysis class, one of the first results that is generally proved after the definition of differentiability is given is the fact that differentiable functions are continuous. We can justifiably ask if the converse holds—are there examples of functions that are continuous but not differentiable? If such examples exist, how “bad” can they be?

Many examples of continuous functions that are not differentiable spring to mind immediately: the absolute value function is not differentiable at zero; a sawtooth wave is not differentiable anywhere that it changes direction; the Cantor function[1] is an example of a continous function that is not differentiable on an uncountable set, though it does remain differentiable “almost everywhere.”

The goal is to show that there exist functions that are continuous, but that are nowhere differentiable. In fact, what we actually show is that the collection of such functions is, in some sense, quite large and that the set of functions that are differentiable—even if only at a single point—is quite small. First, we need a definition and an important result.

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I’ll (hopefully) talk more about this later.
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An Elementary Duality Result

The following problem is a fairly straightforward exercise that seems to come up fairly often on qualifying exams at UCR, and I think that I finally have a nicely argued proof. I am going to put it up here mostly for my own edification. Please note that the following is intended largely for an audience of advanced undergraduate or graduate students. I will be making no effort to give all of the required background, and assume some knowledge of the theory of Lebesgue integration and \(L^p\)-spaces.

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Grad school is keeping me busy. In the meantime, hummingbirds from our balcony feeder:

A hummingbird at our feeder.

A hummingbird at our feeder.

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Fractal Music(?)

In the past, I have treated fractals visually. For instance, in my treatment of the Mandelbrot set, I chose points in the complex plane to represent pixels in an image, applied the iterative function \(z_{n+1} = z_{n}^2 + z_0\) to each of those points until the orbit escaped (or failed to escape after a fixed number of iterations), then colored the pixel based on the number of iterations applied before the orbit escaped. Another visual approach was to compute a large number of orbits and treat their traces like photons hitting a light sensor, thus creating nebulabrot images. I have recently started playing around with an entirely different idea: instead of using the orbits to compute values for pixels in an image, the orbits can be used to generate sound.

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It’s a ξ, See?

While looking for a dead-tree set of Russell and Whitehead’s Principia Mathematica, I came across a scanned version in the University of Michigan’s Historical Math Collection. Not quite what I wanted, but I was highly amused by the annotations on one of the blank pages near the front:

It seems that some poor student in the past was having trouble with his Greek, particularly with the lowercase letter ξ (xi). Since that little guy is the bane of my existence, I couldn’t help but chuckle at the struggles of a another.

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2014 Fields Medal

The International Congress of Mathematicians (ICM) announced the 2014 Fields Medal awardees last night. The Fields Medal is one of (if not the) most prestigious honors in the mathematical community—it is often portrayed as the “Nobel Prize of mathematics.” Of particular note this year is that a woman—Maryam Mirzakhani—has been honored for the first time in the nearly 80 year history of the award. Quanta Magazine has published a nice biographical sketch of Dr. Mirzakhani (as well as the other winners).

Mathematics is a field that has been historically dominated by men. This has been a self fulfilling prophecy—our society often tells young women that math isn’t for women (see, for instance, the Barbie doll who states “Math class is tough!” or the innumerable stories about women being advised to “consider ‘easier’ programs”) leading to women pursuing other courses of study, which in turn leads to an underrepresentation of women in mathematics, thereby reinforcing the perception that women can’t do math. Each of these factors needs to be addressed and I hope that Dr. Mirzakhani’s being honored in this manner will help with the misconception that women cannot do well in mathematics.

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