$$$$

Project

The project in this class is an opportunity to both dive deeper into a topic and to be creative with it - presenting your work in a means that is fun for you (not just a writeup of disconnected topics on a homework assignment). The goal of this is to be an enjoyable way to do some independent research - so if you can think of something that would make this even more fun for you (that isn’t listed below) just ask!

Below are two possible projects, both of which ask you to take what you have learned in our analysis course and apply them to the study of special functions: logarithms on the one hand and trigonometric on the other. Both require you to prove some things, but the focus of the first is more on historical context important to analysis, the second on a careful, rigorous development of classical material.

Unlike a homework assignment, the result of a project will be a research paper, neatly typed including introductory material, commentary and connecting paragraphs in addition to whatever examples or proofs you complete. There is no specific page length or rules on how you must write: you are all in your later years of university and these are the sort of choices that its important to learn to make yourself. However, to help you out, in each of the project proposals below, I walk through an outline of the minimal topics you need to cover. But do not treat this like a checklist of things to knock out one after the other and call it good enough. Do you think potential readers would benefit from seeing another example? Add one! Would it be helpful to explain why a certain theorem needs oen of the conditions in its assumptions? Explain it! Did you learn a cool anecdote along the way? Share it!

Remember this project is optional: its certainly a challenge, and is here for those of you who are looking for some serious mathematical work. I am more than happy to help you out along the way, if you have questions, want me to look at a proof in office hours, or want to discuss your writing. And for anyone considering graduate school, here’s some motivation to give it a try: some graduate schools in math (and surely other fields) ask for a writing sample with your application; I’ve had several past students tell me they submitted a project from one my classes like this one, and were glad they took the time to produce something like this before application season!

Finally, some administrative details: this project is due at the end of Finals Week, Friday May 16. To submit, just email it to me then (or before!).

I: The History of Logarithms

Logarithms came about for very practical purposes centuries ago, and have a fascinating conceptual and calculational history. If you choose this project, your goal is to learn this history, and write a thoroughly researched paper introducing the topic to other mathematics students. This is a big story, so you’ll probably learn more in the research phase than will end up in a final paper, so part of the problem is to sort through what you learned and figure out how to tell a coherent story. You should think about where you would like to include examples, which historical figures / details / dates are most relevant to the story you want to tell. To provide some guidance:

• Your intended readership is other advanced mathematics undergraduates. You can expect your reader to have a solid grasp of an introductory to proofs class, but you should review / recall / introduce any definitions and theorems from real analysis that you need.
• You should include a discussion of why logarithms helped people speed up calculation by hand.
• You should include an example of how to use a log table, and compare the number of steps in calculating the product of two large numbers to the number of steps needed to get the same answer, if you use a log table to convert multiplication to addition.
• The biggest component: you should learn how people computed log tables, and explain it to your reader. Once you understand them, you should do an example and approximate the value of some logarithms.
• Using your knowledge of real analysis then try to phrase this logarithm computing technique in rigorous language (using sequences and limits?) You don’t have to prove the technique works or converges, but instead just make a precise and modern-understandable rigorous statement of the procedure. (Though of course I encourage you to try and prove it converges!!)

In terms of choices of historical approach, there are several techniques people used to approximate logarithms. The first was due to Napier and is kind of difficult / confusing (as first ideas often are, in the history of math and science) as he in fact computed the functin ${10}^7\ln (x/{10}^7)$. A better technique was invented by Henry Briggs who calculated log base $10$. I would recommend looking into that technique, and figuring out how to use it!

Resources

Here are some resources that might be useful in researching the history of logarithms, and answering the questions above.

• Here is a magazine article on the history of logarithms by the Mathematical Association of America. Each section of the article is on a different webpage, accessed by the various hyperlinks on the bottom of this introduction.

Here are some additional papers that can help you get started: you’ll certainly need to do your own independent literature search to find out all that you would like to write about, but these may help you get going. Paper I Paper II Paper II Paper IV

II: Deriving Trigonometry

The goal of this project is to prove the existence of the trigonometric functions $\sin$ and $\cos$ and derive their familiar algebraic properties (trigonometric identities) as well as calculus properties.

Trigonometric Functions

We define the trigonometric functions like we did the exponential, by giving an equation they ought to satisfy: here we pick the angle difference identities

Definition 1 (Angle Identities) A pair of two functions $(c,s)$ are trigonometric if they are a continuous nonconstant solution to the angle identities $$s(x-y)=s(x)c(y)-c(x)s(y)$$ $$c(x-y)=c(x)c(y)+s(x)s(y)$$

It may seem strange at first: is this really enough to fully nail down trigonometry? It turns out it is: if $s,c$ satisfy these identities then they actually satisfy all the usual trigonometric identities! Its good practice working with functional equations to confirm some of this: I’ll start

Lemma 1 (Values at Zero) If $s,c$ are trigonometric, then we can calculate their values at $0$: $$s(0)=0c(0)=1$$

Proof. Setting $x=y$ in the first immediately gives the first claim $$s(0)=s(x-x)=s(x)c(x)-c(x)s(x)=0$$

Evaluating the second functional equation also at $x=y$ $$c(0)=c(x-x)=c(x)c(x)+s(x)s(x)=c(x{)}^2+s(x{)}^2$$

From this we can see that $c(0)\ne 0$, as if it were, we would have $c(x{)}^2+s(x{)}^2=0$: since both $c(x{)}^2$ and $s(x{)}^2$ are nonnegative this implies each are zero, and so we would have $c(x)=s(x)=0$ are constant, contradicting the definition. Now, plug in $0$ to what we’ve derived, and use that we know $s(0)=0$

$$c(0)=c(0{)}^2+s(0{)}^2=c(0{)}^2$$

Finally, since $c(0)$ is nonzero we may divide by it, which gives $c(0)=1$ as claimed.

An important corollary showed up during the proof here, when we observed that $c(0)=c(x{)}^2+s(x{)}^2$: now that we know $c(0)=1$, we see that $(c,s)$ satisfy the Pythagorean identity!

Corollary 1 (Pythagorean Identity) If $s,c$ are trigonometric, then for every $x\in \mathbb{R}$ $$s(x{)}^2+c(x{)}^2=1$$

Continuing this way, we can prove many other trigonometric identities: for instance, the double angle identity (which will be useful to us later)

Lemma 2 (Evenness and Oddness) If $s,$ are trigonometric, then $s$ is odd and $c$ is even: $$s(-x)=-s(x)c(-x)=c(x)$$

Lemma 3 (Angle Sums) If $s,c$ are trigonometric, then for every $x\in \mathbb{R}$ $$s(x+y)=c(x)s(y)+s(x)c(y)$$ $$c(x+y)=c(x)c(y)-s(x)s(y)$$

Corollary 2 (Double Angles) If $s,c$ satisfy the angle sum identities, then for any $x\in \mathbb{R}$, $$s(2x)=2s(x)c(x)$$

Another useful identity we’ll need is the ‘Half Angle Identites’:

Lemma 4 If $s,c$ are trigonometric functions, then $$c(x{)}^2=\frac{1+c(2x)}{2}$$

Proof. Using the angle sum identity we see $$c(2x)=c(x)c(x)-s(x)s(x)=c(x{)}^2-s(x{)}^2$$ Then applying the pythagorean identity $$\begin{array}{rl}c(2x)& =c(x{)}^2-s(x{)}^2\\ & =c(x{)}^2-(1-c(x{)}^2)\\ & =2c(x{)}^2-1\end{array}$$

Re-arranging yields the claimed identity.

Lemma 5 If $s,c$ are trigonometric functions then $$s(x{)}^2=\frac{1-c(2x)}{2}$$

Existence of Trigonometry

We’ve proven several things of the form if$ $s,c$ are trigonometric, then…. So we know a lot about what trig functions are like if they exist, but we so far don’t know the angle sum identities actually have any solutions!

Thus the next goal of this project is to prove the existence of trigonometric functions, so that we can use them in our further arguments. There are many routes to doing so, but the path we follow here meanders through some particularly beautiful mathematics in its own right. We will – as a corollary of proving that trigonometry exists – discover the famous identity of Euler $e^{i\pi }=-1$.

Definition 2 (The Series $\cos (x)$ and $\sin (x)$) Define the following two series $$\cos (x)=\sum _{n\ge 0}\frac{(-1{)}^n}{(2n)!}{x}^{2n}\sin (x)=\sum _{n\ge 0}\frac{(-1{)}^n}{(2n+1)!}{x}^{2n+1}$$

Proposition 1 The functions $\cos (x)$ and $\sin (x)$ are absolutely convergent on the entire real line. And thus, they are continuous on the entire real line.

At the moment we do not know these functions are actually trigonometric (other than we’ve used suggestive and familiar names for them). They are just some infinite series! We need to show they satisfy the angle difference identities. There are many possible arguments here. The one we’ll take uses some complex numbers, but does not require any complex analysis. The only fact needed is that $i$ is a number where $i^2=-1$, which allows complex multiplication to be defined by the distributive property:

$$(a+bi)(c+di)=ac+adi+bci+bd{i}^2=(ac-bd)+(ad+bc)i$$

We can use the fact that a complex number $a+bi$ has a real component ($a$) and an imaginary component $b$ to define convergence: a complex series converges if both its real and imaginary parts converge.

Definition 3 (The Function $\mathrm{CIS}(x)$) Define the function $\mathrm{CIS}(x)$ as $$\mathrm{CIS}(x)=\cos (x)+i\sin (x)$$

Using the series for $\cos (x)$ and $\sin (x)$ we can produce a series for $\mathrm{CIS}(x)$

Proposition 2 $$\mathrm{CIS}(x)=\sum _{n\ge 0}\frac{1}{n!}(ix{)}^n$$

But this looks awfully familiar: we are acquainted with the power series $E(x)=\sum _{n\ge 0}{x}^n/n!$: this defines the exponential function!

Corollary 3 $$CIS(x)=e^{ix}$$

From here, it is easy work to show that the the functions $C$ and $S$ satisfy the angle identities: this is just the law of exponents, which was the defining property of any exponential $E(x+y)=E(x)E(y)$

Proposition 3 The functions $\cos$ and $\sin$ satisfy the angle difference identities.

Thus, they are trigonometric, and trigonometric functions exist!

Periodicity and $\pi$

Because $\sin$ and $\cos$ have been defined with power series that converge everywhere, its very easy to do calculus with them.
We’ve proven (or will prove, depending on when you are reading this) that within a power series’ radius of convergence it is differentiable, and you can differentiate it term - by - term.

Lemma 6 Show that for all $x\in \mathbb{R}$ $$\sin (x{)}^{\prime }=\cos (x)$$ $$\cos (x{)}^{\prime }=-\sin (x)$$

This is of course useful for many things - but for our first application we will use it to prove a rather mundane sounding fact: that the cosine function has a zero! This relies on a corollary that comes from combining the above with the pythagorean identity:

Corollary 4 The $n^{th}$ derivative of cosine is bounded to lie between $-1$ and $1$ for all $n$.

Theorem 1 The cosine function has a zero, somewhere between $x=0$ and $x=2$.

Proving this uses some great material from our course:

• Use the fact that power series are continuous where they converge.
• Use the Taylor Error Bounds to prove that $\cos (1)$ is positive but $\cos (2)$ is negative:
  • Compute approximate values of each using partial sums, showing the approximate value is pos/neg respectively.
  • Show the error bounds for your chosen approximations are small enough that the interval around each is fully positive, and fully negative respectively.
• Use the intermediate value theorem to conclude the existence of a zero in between.

The existence of this zero is a big deal, when combined with the angle sum identity we proved previously:

Proposition 4 Let $z$ be the first positive zero of the cosine function. Then both sine and cosine are periodic, with period $4z$.

The fact that these functions are periodic is a big deal, and it would be sensible to give a name to their period. However, for historical reasons it turns out we have instead decided to reserve the special name for their half period: this is our rigorous definition of $\pi$!

Definition 4 $\pi$ is the half period of the trigonometric functions $\sin (x),\cos (x)$.

Proposition 5 Equivalently, $\pi$ is the first positive zero of the sine function.

Now with a definition of $\pi$ in hand, we can really pull together all of the things we’ve done and prove Euler’s identity, often called the most beautiful equation in mathematics

Theorem 2 $$e^{i\pi }+1=0$$

Pi and the Circle

Euler’s identity is indeed beautiful, but much of the beauty comes from the fact that it somehow relates the numbers $1,0$ from arithmetic to the number $\pi$ of geometry (from the unit circle) and the number $e$ from calculus. But our story thus far is missing a bit of that beauty: we’ve merely shown that $0,1,$ and $e$ are related to the first positive zero of the power series $\sin (x)=\sum (-1{)}^n{x}^{2n+1}/(2n+1)!$. As a final step, we make this connection and show that the first zero of $\sin (x)$ is actually the area of the unit circle. There are no exercises here for you to do - I’ve worked it out for you, using material we will cover at the end of the semester (but you know from your calculus course: $u$-substitution!)

Since we have defined area rigorously with integration, and the unit circle is given by $x^2+y^2=1$, we can write its area as

$$\mathrm{Area}=2{\int }_{[-1,1]}\sqrt{1-x^2}$$

Now we compute this integral with our newfound integration techniques (substitution), and show it equals the half-period of our trigonometric functions in natural units.

Theorem 3 $$2{\int }_{[-1,1]}\sqrt{1-x^2}=\pi$$

Proof. By subsitution, we see that the following two integrals are equal $${\int }_{[0,1]}\sqrt{1-x^2}={\int }_I\sqrt{1-(\sin (t){)}^2}(\sin (t){)}^{\prime }$$ Where $I=[a,b]$ is the interval such that $[\sin (a),\sin (b)]=[0,1]$. Since $\sin (0)=0$ and $\sin (\pi /2)=1$ we see $I=[0,\pi /2]$. Now we focus on simplifying the integrand:

By the Pythagorean identity, $1-{\sin }^2(t)={\cos }^2(t)$, thus by ?@exm-sqrt-x-squared, $$\sqrt{1-{\sin }^2(t)}=\sqrt{{\cos }^2(t)}=|\cos (t)|$$ and by definition we recall $(\sin t{)}^{\prime }=\cos t$. Thus $$\begin{array}{rl}{\int }_{[0,\pi /2]}& ={\int }_{[0,\pi /2]}|\cos (t)|\cos (t)\\ & ={\int }_{[0,\pi /2]}{\cos }^2(t)\end{array}$$ Where we can drop the absolute value as $\cos$ is nonnegative on $[0,\pi /2]$ (its first zero is at half the period, so $\pi$). We can simplify this using the “half angle formula” ${\cos }^2(x)=(1+\cos (2x))/2$ $${\int }_{[0,\pi /2]}{\cos }^2(t)={\int }_{[0,\pi /2]}\frac{1+\cos (2t)}{2}$$ Using the linearity of the integral, this reduces to

$$\begin{array}{rl}{\int }_{[0,\pi /2]}{\cos }^2(t)& =\frac{1}{2}{\int }_{[0,\pi /2]}1+\frac{1}{2}{\int }_{[0,\pi /2]}\cos (2t)\\ & =\frac{\pi }{4}+\frac{1}{2}{\int }_{[0,\pi /2]}\cos (2t)\end{array}$$

The first of these integrals could be immediately evaluated as the integral of a constant, but the second requires us to do another substitution. If $u=2t$ then $${\int }_{[0,\pi /2]}\cos (2t)=\frac{1}{2}{\int }_{[0,\pi ]}\cos u$$

We recall again that by definition $\cos u=(\sin u{)}^{\prime }$, so by the first fundamental theorem

$${\int }_{[0,\pi ]}\cos u={\int }_{[0,\pi ]}{(\sin u)}^{\prime }=\sin u{|}_{[0,\pi /]}$$

But, $\sin$ is equal to $0$ both at $0$ and $\pi$! So after all this work, this integral evaluates to zero. Thus

$$\begin{array}{rl}{\int }_{[0,1]}\sqrt{1-x^2}& ={\int }_{[0,\pi /2]}{\cos }^{2}t\\ & =\frac{\pi }{4}+\frac{1}{2}{\int }_{[0,\pi ]}\cos (2t)\\ & =\frac{\pi }{4}+0\end{array}$$

Now, we are ready to assemble the pieces. Because $x^2$ is an even function so is $\sqrt{1-x^2}$, and so its integral over $[-1,1]$ is twice its integral over $[0,1]$. Thus

$$\mathrm{Area}=2{\int }_{[-1,1]}\sqrt{1-x^2}=4{\int }_{[0,1]}\sqrt{1-x^2}=4\frac{\pi }{4}=\pi$$