# Summer courses almost over, "children's" books in the works.

This summer is one I will remember. I taught both differential equations and calculus II for the first time. Seeing these subjects from the instructors side has really opened my eyes to all kinds of details I never noticed before. One of the most striking new insights is how much these two courses have in common. They both rely deeply on sequences and series. Sequences are like a hallway in mathematics, one that connects many many many rooms.

I am working on two math book projects. The first is a Japanese-styled art book on the topic of sine and cosine. It's inspired by many of the lessons I taught this summer.

Japanese-style book about sine and cosine

I want to bring all of the different ways that sine and cosine are presented in elementary and undergraduate mathematics in to one (long) pictorial document. I start with the differential equation, $y''+y=0$ then solved it (using the series method from differential equations) producing $\sin x = \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} x^{2n+1}$ and $\cos x = \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n)!} x^{2n}$

Next I wanted a pictorial way to relate these power series to the unit circle. I have found it in this spiral (the first image shows how it is constructed as an involution):

Constructing The Involute Pinwheel

a sequence of involutes: the vertical and horizontal components will form the power series for sine and cosine respectively.

The vertical and horizontal components will form the power series for sine and cosine respectively. Take the series of vertical line segments: $\sin x = A_1A_2 - A_3A_4 + A_5A_6 - \cdots$ and so on, the segments repeatedly over and under-shoot the accutal value of sine. The full paper by Leo S. Gurin, "A problem", can be found here.

I'm going to incorporate Gurin's spiral in to my book. I want to show the power series literally flying out of it, like they have come to life. I wonder if I can make it like the famous drawing of the sine curve projecting out of the unit circle?

Naturally, I already have planned to put that diagram in my booklet.

Work in progress

The Japanese-style book is perfect for series and periodic functions It's one long continuous piece of paper:

Yet very compact:

I'm also working on a very silly book about hypercycloids (that's the "math" name for the shapes drawn by spirographs, did I mention I collect spirographs?):

I'm trying to make it like a children's book, fun, light, a little silly:

I can't wait to share the final product.

# Almost Convergence.

The title of my last post "Almost Cauchy Sequence" made me wonder if "Almost Cauchy" was already defined in some formal sense. I could not find much on "Almost Cauchy" ... but I did find out about "Almost Convergence." (I think I have ran in to this before... but it is very neat... observe:)

A sequence $S = S_n$ is almost convergent to L if for any $\epsilon > 0$ we can find an integer n such that the average of n or more consecutive terms in the sequence is within $\epsilon$ of L. Formally, $\forall \epsilon > 0 \quad \exists N \ni \quad \displaystyle \left| \frac{1}{n} \sum_{i=0}^{n-1} S_{k-i} -L \right| < \epsilon \quad \forall n > N, \; k \in \mathbb{N}.$ (I took this definition from here, and I wonder why they didn't mention that we need $k \geq n$ or else we are looking at negative terms in the sequence... But, maybe there is an interpretation of something like $S_{-3}$? Or maybe the author thinks that would be obvious? I'm going to ask at the mathematics stack exchange!)

# Several series for powers of pi.

The Weierstrass factorization theorem gives us this identity:

$\displaystyle \prod_{n=1}^{\infty} \left( 1 -\frac{q^2}{n^2} \right) = \frac{\sin(\pi q)}{\pi q}$

Take the left side:

$\left( 1 - \frac{1}{1}q^2 \right) \left( 1 - \frac{1}{4}q^2 \right) \left( 1 - \frac{1}{9}q^2 \right) \cdots$

Thinking of this in a combinatorial way we can expand it. Each even power of q has a sum of combinations of the inverse squares as its coefficient. That is:

$1 + \displaystyle \left( q^2 \sum_{j_1=1}^{\infty} -j_1^{-2} \right) + \left( q^4 \sum_{j_1,j_2=1 \atop j_1 \neq j_2}^{\infty} (j_1j_2)^{-2} \right) + \left( q^6 \sum_{j_1,j_2,j_3=1 \atop j_i \neq j_k} (j_1j_2j_3)^{-2} \right) + \cdots$ $+ \displaystyle q^{2n} \sum_{j_1,...j_n=1 \atop j_i \neq j_k}^{\infty}( -1)^n (j_1j_2...j_n)^{-2} + \cdots$

Now on the right side of
$\displaystyle \prod_{n=1}^{\infty} \left( 1 -\frac{q^2}{n^2} \right) = \frac{\sin(\pi q)}{\pi q}$

Expand using the Taylor series for the sine.

$\frac{\sin(\pi q)}{\pi q} = 1 - \frac{(\pi q)^2}{3!} + \frac{(\pi q)^4}{5!} - \frac{(\pi q)^6}{7!} + \cdots$

Equate the coefficients of the terms with the same exponent.:

• $1 = 1$
• $- \frac{\pi ^2}{3!} = \displaystyle \sum_{j_1=1}^{\infty} -j_1^{-2}$
• $\frac{\pi ^4}{5!} = \displaystyle \sum_{j_1,j_2=1 \atop j_1 \neq j_2}^{\infty} (j_1j_2)^{-2}$
• $- \frac{\pi ^6}{7!} = \displaystyle \sum_{j_1,j_2,j_3=1 \atop j_i \neq j_k} - (j_1j_2j_3)^{-2}$
• $\vdots$
• $\frac{\pi ^{2n}}{(2n+1)!} = \displaystyle \sum_{j_1,...j_n=1 \atop j_i \neq j_k}^{\infty} (j_1j_2...j_n)^{-2}$
• $\vdots$

The second one is famous, more often written:

$\frac{\pi ^2}{6} = 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{25} \cdots$.
And the rest of these series I do not know so well. Do you know them?

# Next step with the interesting difference from last night...

From last night's post:

$\displaystyle \prod_{j=1}^{\infty} (1 +x^j) - \displaystyle \prod_{j=1}^{n} (1 +x^j)$

This difference is the generating function for partitions into distinct parts from the set $\{ n+1, n+2, ... \}$.

I think my next task is to graph the upper bound I found for these coefficients, then see how it differs from the actual number of partitions for this set. The difference should grow pretty quickly. I wonder how mathematica deals with generating functions. I know some of the popular functs are pre-loaded... But, can I give it an expression and have it list exponents for me?

So much to learn!

# Quick thought. (Very rough.)

$\displaystyle \prod_{j=1}^{\infty} (1 +x^j)$

The coefficient of the $x^n$ term will be the number of partitions of n into distinct parts. (It is also the number of partitions in to odd parts as one can prove these are the same using a neat argument with Ferris graphs. )

$\displaystyle \prod_{j=1}^{n} (1 +x^j)$

Now this is the generating function for partitions into distinct parts from the set $\{ 1, 2, 3, ... n \}$. How can I describe it in terms of odd parts?

$\displaystyle \prod_{j=1}^{\infty} (1 +x^j) - \displaystyle \prod_{j=1}^{n} (1 +x^j)$

This will change the coefficients of the first $\frac{n(n+1)}{2}$ terms in $\prod_{j=1}^{\infty} (1 +x^j)$ since $x^{\frac{n(n+1)}{2}}$ is the highest power in $\prod_{j=1}^{n} (1 +x^j)$ Then this difference is the generating function for partitions into distinct parts from the set $\{ n+1, n+2, ... \}$. The first n terms will have a coefficient 0 because the numbers are the smallest in the set of allowable values. The first number with a partition using two terms is $(n+1) + (n+2)$ so $x^{n+1} .. x^{2n+2}$ all have coefficient 1. In the same way $x^{2n+3} .. x^{3n+5}$ have coefficient at most 2. And so $x^{mn+\frac{m(m+1)}{2}} .. x^{(m+1)n+\frac{(m+1)(m+2)}{2}-1}$ have coefficient at most m.

I need to walk through this a few more time to see if what I'm saying really makes sense. And find a better way to write those awful exponents!

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