2009 February 21 « Foxmaths! 2.0

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,