# Almost Cauchy Sequence, unfinished business.

I just remembered a question from beginning analysis that my professor mentioned in class:

"Find a sequence $A$ that is not Cauchy, yet any consecutive terms $A_n, A_{n+1}$ will become arbitrarily close."

It was one of those questions that he mentioned in class for us to do on our own, but I never quite got to it. I was thinking about convergence and complete metric spaces today (and about unfinished business, as I've made a lot of my own since I started teaching.) And the question came back to me.

We seek a sequence, $A$, where $\forall \; \epsilon > 0 \quad \exists N \in \mathbb{N} \; \ni \; \forall n > N \quad |A_n - A_{n+1}| < \epsilon$ and $\exists \epsilon > 0 \; \ni \quad \forall N \in \mathbb{N} \quad \exists m,n > N \quad |A_m - A_n| > \epsilon$. (Writing the negation of what it means for a sequence to be Cauchy was a little more tricky than I thought...)

The natural log function is the most obvious answer. For large vales of n $\ln (n) - \ln (n+1)$ is small. That is, $\displaystyle \lim_{n \rightarrow \infty} \ln \frac{n}{n+1} = 0.$ This is true becuase after taking EXP of both sides we have: $\displaystyle \lim_{n \rightarrow \infty} \frac{n}{n+1} = 1$

Next, it seems obvious that the natural log can't be Cauchy, but just to be certian if we let $\epsilon = \frac{1}{2}$ then, $\forall N \in \mathbb{N}$ if we choose some $n > N$ and let $m= ne$ (so, $n,m > N$) then

$| \ln m - \ln n | = |\ln \frac{m}{n}| = |\ln \frac{ne}{n}| = 1 > \frac{1}{2} = \epsilon$.

NOT Cauchy at all!

Ok, I imagine my professor was thinking of the natural log... but, what other functions also have this property? $\sqrt[n]{x}$? So, I guess this is really asking when $\displaystyle \lim_{x \rightarrow \infty} f(x) - f(x+1) = 0$ while $\displaystyle \lim_{x \rightarrow \infty} f(x) = \pm \infty$. It is necesairy but not sufficent for $f'(x) > 0 \; \forall x$ and $f''(x) < 0 \; \forall x$ , both will have the x-axis as a horizontal asymptote.

Nice to have finally gotten around to doing this problem! I think it will be good practice to find more unfinished business in my notes.

# Introducing the multi-zeta function!

Contemplating:

$\frac{\pi ^{2n}}{(2n+1)!} = \displaystyle \sum_{j_1,...j_n=1 \atop j_i \neq j_k}^{\infty} (j_1j_2...j_n)^{-2}$

My friend, and fellow math blogger, Owen, tried to tell me I was dealing with the multi-zeta function the other day, but the very general definition on WolframMathworld left me feeling a little mystified. I could see how the identities I was playing with fit in to it, but what was all of that other stuff (like $\sigma_1,...,\sigma_k$? I guess in my equation they are all 1? Why isn't it dealing with powers of the active variable in the product in the denominator? Why the extra level of subscripts?) So, I asked about it on stackexchange to gain more insight, and, thanks to Marni, I was promptly pointed to this lovely paper:

Now, I can see easily that the function is:

$A(i_1, i_2, ... , i_k) = \sum_{n_1 > n_2 > \cdots > n_k > 1} \frac{1}{n_1^{i_1}n_2^{i_2} \cdots n_k^{i_k}}$

With $i_1, i_2, ... , i_k = 2$.

More on this later.

Meaning one may repeat values,

# 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!

s

# Using LaTeX on the iPad: A review of the best and worst apps for writing mathematics.

LaTeX  is a powerful markup language that allows you to generate beautiful equations like this:

$\left\{\begin{matrix} n \\ k \end{matrix}\right\} =\sum_{j=1}^k (-1)^{k-j} \frac{j^{n-1}}{(j-1)!(k-j)!} =\frac{1}{k!}\sum_{j=0}^{k}(-1)^{k-j}{k \choose j} j^n$

Without cumbersome "equation editors." You simply type the TeX code for the equation and it appears. (And it looks as good as what you'd find in any research paper since LaTeX is what most publishers use for their typesetting.)

Naturally, if you do math and if you have an iPad you want to find ways to use LaTeX on your iPad. As it turns out there are a few great options, though each leaves a little more to be desired. At this time there is no full LaTeX engine that runs on the iPad that will let you use every typesetting and graphics command that LaTeX offers. But there are a few work-arounds that give you access to that full engine. And there are a myriad of small apps that let you typeset equations. (I often forget that there is far more to LaTeX than just writing equations!) So, here is the round up:

## MathBot

Overall: A bare-bones equation editor that shows your LaTeX equations in real-time as you edit. Most of the important keys for writing mathematics are handy, but, annoyingly, the numbers (0,1, 2, 3,...) are on a a separate screen so typing equations can be quite tedious.  Once you have your equation ready you can export it as an image to email, to the photo album or to other apps.  It also has an "export as TeX" option that lets you move the code in to another editing environment as text. (Though copy/past would work just as well for this, so I don't really see the point.) If you are new to LaTeX you'll find the short-cut keys for most of the basic math symbols  helpful, and the way the keyboard shifts from one view to another is smooth and delightful, fitting well within the iPad user interface.  There are no folders and it is slightly cumbersome to rename files. This app is really for people who are just looking to write one or two quick equations to paste in to an email or document. Naturally, it has no full LaTeX engine and can't make pdfs or postscript files.

Ease of Use: Short-cut buttons are great for beginners. Real-time editing gives instant feedback. However, switching screens to type numbers is annoying.
Customization: Only one font size. Few options for advanced users.
Aesthetics and Flow: Simple clean design, intuitive sliding shortcuts on the keyboard. Classy, no distractions. Designed for the iPad. Uses the full screen.
Price: Free!

## Tex Touch

Overall: Tex Touch is the only LaTeX app (available right now) that will allow you to type documents of any size in LaTeX and generate pdfs as if you were using your desktop... sounds amazing right? Well, the catch is that you ARE using your desktop! The app uses Dropbox (an excellent, free, "cloud" style remote hard-drive) to send .tex files to your desktop computer where a second application installed (and always running) will pick up the file and compile your pdf. Hence, you can create and edit exams, research papers and even books from your iPad. But, as the review implies, the instillation and set-up process, along with the complexity of the program may be too much for some. Once running, it works like a charm and I use it often to revise my exams, (though typing an entire exam in LaTeX on the iPad is perhaps too much even for an enthusiast like myself.) I give it high marks for being the only "real" LaTeX application. But, I and many others, look forward to the day when we can compile our pdfs directly on the iPad without an internet connection, and without leaving a helper application running on our home computer.

Ease of Use: Hard to set up. Not for novices. Requires multiple applications on multiple computers and an account with Dropbox. However, if you were able to install TeX on your iMac you should have no trouble getting this set up. Once running it works like a charm. (Though you must remember to turn the helper application on or you're out of luck.)
Customization: You have access to everything LaTeX can do including things like the picture environment, that lets you use LaTeX to draw graphics on the fly. If it's on your home computer you'll have access to it with TeX Touch.
Aesthetics and Flow: Well designed for editing .tex files, though it would be vastly improved if the code were color coded.

## Formula

Overall: I want to like this app, but there are so many little infuriating things about it that I find it unusable. First the interface design: Much of your screen is taken up with cheesy greek columns, it feels like a bad CD rom from the 90s. But, once you get past that, there are a few nice touches: equations appear in real time, copious short-cuts will make this app appealing to the LaTeX beginner.  However, whoever designed it did not test it for writing actual equations. Most  equations contain both numbers and letters, but on the main keyboard you are given access to only a numeric key pad. Try writing something simple like:  $x= \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ and you will end up switching keyboards 7 times! It's very tedious. The app is packed with useless flashy features like voice recognition. In theory you say the name of the symbol and the application will provide the correct short-cut keys... To say this feature "dosen't work" is an understatement. It doesn't work and it is an incredible waste of time!  I have decided that this app is not for people who do math, but rather for people who would like to look like they are doing math.  It's very serious and impressive to look at, and nothing says 'mathematician' like loudly shouting Greek letters into your iPad's built-in microphone. You will be pleased with this app until you try to use it to get any kind of work done. Stay away.

Ease of Use: Looks fancy but poorly thought out.
Customization: You have no control over font size, there are no maros.
Aesthetics and Flow: 1990s CD ROM look. Feels junky.
Price: \$3.99

# Counting using products.

Today I proved the binomial theorem for my class. I did an informal proof showing them how, if we consider the set containing the alphabet: $S = \{ a,b ,c, d, ..., z \}$ and the following product.

$\displaystyle \prod_{s \in S} (1+s_i) = (1+a)(1+b)(1+c)\cdots (1+z)$

expanding this produces a sum with every combination of the letter a-z:

$\displaystyle \prod_{s \in S} (1+s_i) = 1 + a + b + \cdots + z + ab + ac + ad + \cdots + az + bd + be + \cdots + bz + \cdots + abc + abd + \cdots + abcd + \cdots + abcdefghijklmnopqrstuvwxyz$

if we count all of the terms in this sum with, say three elements, we have the number combinations of 26 items taken 3 at a time or $26 \choose 3$. Next, we make every letter x:

$(1+x)^{26} = 1 + x + x + \cdots + x + xx + xx + xx + \cdots + xx + xx + xx + \cdots + xx + \cdots + xxx + xxx + \cdots + xxxx + \cdots + xxxxxxxxxxxxxxxxxxxxxxxxx$

(Pretty silly looking I know!) But, now we can see that the coefficients of the terms in the expansion are given by the same formula used to count combinations in probability.

That was today's lesson. What I've noticed is that this manner of counting is the same basic idea as what I've been looking at with integer partitions. (There we use a different product and coefficents to count.)

I wonder if this type of process has a name, or if it is further formalized in any contexts?

# Looks can be deceiving!

Look at the graph in the previous post. I was right! They don't intersect at a common point, rather the limit of the consecutive intersections is $\left(1, \frac{1}{6} \right)$.

Close up of the intersections of y^n + x^n -6xy = 0 for the first ten integers.

This is just numeric, from Mathematica ... I'll try to prove it later...

By the way an answer to the question of fining a curve orthogonal to the follium of Descartes is $x^3 +y^3 +6xy=$ the reflection in the line $y=-x$. Do you know of any other trivial answers?

Goodnight! Tomorrow: Back to the generating functions, Owen has been bugging me!

Orthogonal Curves.