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?

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1 Comment »

  1. Just found your blog. Kind of interesting. Constructive criticism: learn some lessons from Creative Writing. (And more aesthetic LaTeX, if that’s what you’re using.)

    Anyway, I scribbled a few things down and found a generalization.

    Let the sequence be defined by:

    a_{n+1} = q a_n + r

    Then the function you define satisfies the following differential equation:

    f^\prime (x) = x^{a_1 - 1} + x^{r - 1} ( f(x^q) - 1 )

    Comment by Brad — May 17, 2009 @ 1:01 am

    Reply

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