Foxmaths! 2.0

February 22, 2009

Clearing My Head

Filed under: Maths — Fox @ 5:16 am

Sometimes, when I get caught up in maths, I just need to write something down so I can stop thinking about it, and get some real work done.

So consider the differential equation,

f'(x) = f(x)

With the initial condition, f(0) = 1. Relatively straight forward, the solution is f(x) = e^x . But, I would like to express that in a slightly different form, the usual Taylor Series for e^x ,

f(x) = \sum_{n = 0}^\infty \frac{x^n}{n!}

But, we can write this in a slightly more evocative form.

f(x) = 1 + \sum_{n = 1}^\infty \frac{x^n}{\prod_{k=1}^n k}

Why is this more interesting?

Consider the differential equation,

f'(x) = f(x^2)

with the initial condition, f(0) = 1. Now, with the careful application of power series (a trick I learned from you, Doc!), you can show that the solution function can be expressed as a summation in the following way.

f(x) = 1 + \sum_{n=1}^\infty \frac{x^{2^n-1}}{ \prod_{k=1}^n 2^k-1 }

Which bears more than a passing resemblance to the previous formula.

So, for various sequences a_n , what in general can be said about summations of the form

f(x) = 1 + \sum_{n = 1}^\infty \frac{x^{a_n}}{ \prod_{k=1}^n a_k }

Thoughts?

February 21, 2009

A Pair of Problems

Filed under: Maths — Fox @ 5:46 am

I work too hard, I really do. That’s my main excuse for the lack of maths lately. I’ve been working far too hard!

For instance, consider the following integral.

\int_{0}^\infty (\frac{a^2 e^{a x}}{(e^{a x}-1)^2} - \frac{1}{x^2}) dx = -\frac{a}{2}

Things like that fascinate me. Either part of the sum, integrated over that interval, diverges, but the difference of the two does not. It balances out just so. Anti-differentiating it really isn’t too hard, and then evaluating at the limits of integration took several applications of l’Hoptial. Time consuming. And, as an added point of interest, Mathematica fails to perform the integral in any reasonable amount of time.

But the point is, for the assignment this occurred in, the appropriate application of approximations and relevant limits, that integral, and the rest of the page and a half of calculations I wrote out, were completely unnecessary to achieve the same result. Working hard, not smart. Story of my life.

But, I do like that integral.

The second thing to consider is the following differential equation. Differential equations have a lot of potential for fun and excitement. My PDE teacher actually stopped his lecture the other day, and asked us what we hoped to get out of that class. His point was that, while we are learning (in theory) how to solve PDE’s, all the really interesting differential equations cannot be solved at all. In general, it really doesn’t take much to render a differential equation almost completely unmanageable.

So, consider a function f(x) such that

f'(x) = f(f(x))

Now, I like to view such a thing as simply a statement, meant to inspire questions. Of course, the kind of question that sort of thing generally inspires is, can we solve for f(x). And I’m fairly certain the answer is no, beyond the trivial solution f(x) = 0. But there are many different ways one might go about solving such a thing. One way is to solve for a specific function f(x). Another way would be to use the differential equation to solve for the values of the n-th derivatives of f, and use those to construct the Taylor Series of the solution function. That is possibly doable, though nothing I’ve seen suggests anything easy.

Another interesting question to ask would be, can f(x) be re-expressed in another way? Possibly one that doesn’t have the nested function call?

This late at night, I don’t really have too much more to offer except for this interesting calculation.

Let f_n(x) be f, applied to itself n times. So, f_3(x) = f(f(f(x))) . Notice then, we can write f_{n+1}(x) = f( f_n(x)) . We can then do the following,

f_{n+1}(x) = f(f_n(x))

f_{n+1}'(x) = f_n'(x)*f'(f_n(x))

But then, applying the differential equation, f’(x) = f(f(x)), that becomes

f_{n+1}'(x) = f_n'(x)*f(f(f_n(x)))

f_{n+1}'(x) = f_n'(x)*f_{n+2}(x)

But, applying the above relation, we also have

f_{n}'(x) = f_{n-1}'(x)*f_{n+1}(x)

Which can then be substituted back to yield

f_{n+1}'(x) = f_{n-1}'(x)*f_{n+1}(x)*f_{n+2}(x)

Continuing that substitution, and noting that f_1'(x) = f'(x) = f(f(x)) = f_2(x) , yields the following formula.

f_n'(x) = \prod_{i = 2}^{n+1} f_i(x)

I have no idea what that formula means, but it seems relatively slick.

What are your thoughts?

February 10, 2009

Math: A Different Perspective

Filed under: Maths — Fox @ 5:32 am

First, I would like to draw your attention to this article, about a processor that uses probabilistic calculations, and as such, runs 7 times as fast and at 1/30th the power of conventional hardware. The basic idea is, what information do you -really- need?

Which brings me to the point at hand. I am a math person, first and foremost. But for the last three years or so, I’ve also been pursuing physics a great deal. For the most part, if you know math, then physics is relatively easy. But, physics also provides a different perspective on math that I think is very important. In math, there is a tendency to get lost in symbolic manipulations, algorithms, problems of a specific type, and forget what the math is ‘actually’ saying. You can become entrenched. Blinded, almost. In physics, math is often treated as almost an obstacle to be dealt with. Something to be worked around to get at the ‘real’ answer. At least, the relevant answer. It’s fascinating to me. The solution is often to not do the problem at all. Doing by not doing. Math without math. And yet, at the same time, it gets straight to the matter of what the math is trying to express, and in the processes, helps you develop an intuition for it.

And if none of that made sense, here is an example. Consider the expression, for a positive value of x,

\sum_{j = 0}^\infty (2j + 1)\ e^{-j(j+1)x}

In general, we want to understand the behavior of that function with respect to x. Now, as a mathematician, I jump in and think, hmmm. What does that summation converge to? Can we express it in a nicer formula? Things start to jump out, generating functions, possible Taylor summations, etc, any one of which could be used to turn that summation into something recognizable. Usable.

But then, staring at the summation a good long while … nothing forms. The usual power series formulas are no help, due to the quadratic nature of the exponent. And the factor of (2j+1) in front of the exponent certainly complicates things. The form of the summation fits to nothing I recall seeing, ever. Asking Mathematica, it can’t find any convenient expression for the sum either. General maths suggest no reasonable way of turning this summation into something tractable. And, if we wanted to do more interesting things to that summation, such as take the log, and then various derivatives, by keeping the entire summation, formulas would rapidly turn quite nasty.

So, what is there to be done? How best to do that summation? How can we really appreciate what this function is? The answer is, to not do the summation.
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