Notebook Collection

There is nothing quite as exciting as a fresh blank notebook full of possibilities, well except, perhaps a full worn notebook filled with new mathematics! I can admit that I'm addicted to notebooks. Whenever I see one, with a new kind of color or material I buy it. As such, a shelf in our house is totally devoted to blank notebooks.

The Notebook Shelf

The Notebook Shelf

It was only a few months ago when I put them all in one place that it hit me how many I'd accumulated. These are all blank. I go through about one or two notebooks per week (of the Moleskine Cahier size) So, I have almost a whole bookcase full of full notebooks as well. I use Rhodia, Moleskine, Whitelines, Writer's block and other big brands quite often, In this post I'll focus on the more rare and unique stuff, the notebooks that make my hand tremble with excitement as I get ready to make that first mark on the unblemished page! The notebooks that I cherish, they are almost too good to write in, yet, I await the day that I mark them with such pent-up desire!

Okay, maybe I'm getting a little too excited here, on with the notebooks!

I like large notebooks best, they are best suited for lesson plans and proofs.

Although I like large notebooks best I do have a few small ones that I could not resist. Here we have: Quattero, a generic red notebook, an antique 1950s notebook, Writer's block "Dots" mini notebook and "Tidbit" Free Cut memo.

Quattero notebooks come in all sizes, they are a little pricy, but the paper is so smooth and thick it's worth it, the only thing I don't like about them is that the backs of the papers are blank, you only get the grid on one side, considering that the paper is bleed-proof this seems like a missed opportunity.

Here's what they look like inside: Writerblock "dots", Quattro, and the little red one.

Writersblok is one of the most underrated notebooks around. They are a less expensive version of the "Moleskine Cahier" but, even if they cost more than Moleskine I'd still take them over the Moleskine any day! Why? DOTS. Dots are the ideal method of ruling the blank page, grid paper is too distracting, lined paper looks awful if you stop to make drawings and graphs as often as I do. So dotted paper is one of the main things I look for in a notebook. The dots in the Writersblok notebooks are the best, they are whisper light, so that you'll almost forget they are there. Yet when you look at your pages you'll see that the lines are straighter, graphs are more accurate and the text is evenly spaced. I've been seeing more and more kinds of dot paper, I really hope it catches on!

Antique notebooks fascinate me. Especially those that have survived totally blank and untouched like this one for decades. I love that this little memo book has an even smaller notebook (not an address book!) tucked inside. If I ever think of the perfect use for it I will use it, but for now it remains untouched, it is one of my most cherished notebooks. (*sniff*)

Notebook within a notebook...

Notebook within a notebook...

Ever since I started loving blank books and shopping for them I have starting HATING address books, planners and photo albums. Why? Often I'll see a notebook on the shelf, the perfect size! The perfect color! I reach for it and.... it's a PHOTO ALBUM! ugh. So annoying. I end up wondering who are all of the people who like photo albums and address books so much, have they not heard of Flikr and gmail? I suppose they must be wondering what's with all of the blank books... who needs paper? I have an iPad.

(I will continue sharing my collection in the next post.)

Of yarn and math.

A collection of hand-spun yarn and my spindle. I did not spin all of this yarn, some is from etsy.com

Yarn has been on my mind. It's on my mind because I love to knit, but also because of this post I read some weeks ago on the mathematics stack exchange. The author presents a paper about an "optimal yarn ball" that looks like this:

via. http://arxiv.org/abs/1005.4609

Very cool, but it would be impossible to do that with real yarn! (Though, I doubt that was their goal here.) When wrapping yarn around a ball you cannot make sharp turns or the yarn will fall off. If we think of the yarn as a curve on the surface of the sphere, we would say it must have curvature less than some small constant m. (meaning the best fit tangental circle must have radius greater than \frac 1m)

The yarn should be wrapped evenly: it should not overlap itself too often. (This would lead to the yarn ball taking on a non-spherical shape over time. No good!) The distribution of the self-intersections should be of nearly consistent density over the surface of the sphere. The yarn will divide the surface of the sphere in to regions. If A_i, A_j are the areas of any two of these regions then for some small constant F, |A_i- A_j| < F, \forall i, j.

As the curve is extended further and further (wrapped around more times) the density of intersections should increase and A_i \rightarrow 0 for all of the regions.

I know one answer is a randomly generated curve that deviates from a straight path. Overtime it produces a perfect yarn ball. That's how I make my yarn balls, randomness produces even inward tension which forms a sphere.

But, I want to know if there is a non-random answer to this question.

First Thoughts:
In trying to solving this myself I thought it might be a good idea to project the sphere on to the plane, as a Riemann surface. But, since so many of the requirements focus on uniformity, this makes the problem a little strage. Observe: The area of the regions will need to *increase* as we move away from the unit circle toward infinity. If the curve ever passes through the north pole, then the plane version would shoot off to infinity. What I found even more weird is what would happen if the curve simply passed very near, but not through, the north pole. Then, on the plane, the curve would go very far from the origin and then loop back. By making the curve near enough to the north pole the loops can be as large as we please.

Since wrapping increases the density uniformly, if we observe the wrapping pattern projected on to the plane it would be like loopy knot that, over time, keeps casting bigger and bigger outlier loops, it would always grow and be un-bounded. (even without any intersection with the north pole!)
The requirement about curvature, would mean that the knot could be more curvy inside of the unit circle, and then grow less curvy as we move away from the origin.

Next to find such a curve... Maybe if I projected the following polar plot back on to the Riemann sphere?

The graph of the form r= \theta^2 \sin (k \theta) produced good-looking results for 0 < k <1... but all of the graphs are far too dense near the origin. Also, wrapping would start in a lop-sided manner, filling in the bottom of the sphere first then moving up... and never reaching the north pole! But, maybe there is some way to tweak this?

(This graph shows the kind of non-random pattern I have in mind for a solution. In a practical sense, a solution to this problem could be used to wind spherical balls of ropes, yarns, or cords in a factory setting. Now, I will think about how I would instruct a robot to wind a yarn ball... if the robot is unable to do things randomly. What would you tell the robot to do?)

This post grew out of a question I asked at the Stack Exchange.

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!)

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!

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