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Full Version: Very curious question
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What If I told you I can find infinite functions that equal their own derivative?

Take some fractional differentiation method $\frac{d^t}{ds^t}f(s)$ which differentiates f across s, t times. Now assume that:

$\frac{d^t}{ds^t} f(s) < e^{-t^2}$ for some s in some set $D \subset \mathbb{C}$, which can be easily constructed using some theorems I have.

Then:

$\phi(s) = \int_{-\infty}^{\infty} \cos(2 \pi t) \frac{d^{t}}{ds^{t}} f(s) dt$

If you differentiate $\phi$ by the continuity of this improper integral $\frac{d}{ds} \phi(s) = \phi(s)$

What does this mean? How did I get this? Where is the mistake?
Let's make another function that equals its own derivative. I'm very curious as to why this is happening!

$g(s) = \sum_{n=-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(y+n)^2} \frac{s^y}{\Gamma(y+1)}dy$

Differentiate and watch for your self!

Does this mean the function cannot converge? I know the integral converges, not sure about the summation though.

Using the other method I can easily create a function that converges for some domain... What's going on?
(08/19/2013, 05:39 PM)JmsNxn Wrote: [ -> ]Let's make another function that equals its own derivative. I'm very curious as to why this is happening!

$g(s) = \sum_{n=-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(y+n)^2} \frac{s^y}{\Gamma(y+1)}dy$

Differentiate and watch for your self!

Does this mean the function cannot converge? I know the integral converges, not sure about the summation though.

Using the other method I can easily create a function that converges for some domain... What's going on?

The summation does not look like it converges. Try graphing the integrand for s = 1 and look what happens as n increases.

Also, using a numerical integration from $-8-n$ to $8-n$ (roughly centers around the "peak", at least for relatively small n), one can approximate the integral and see the divergence:

n = 1, s = 1: 0.38446
n = 2, s = 1: 0.042752
n = 3, s = 1: -0.082158
n = 4, s = 1: 0.26084
n = 5, s = 1: -0.83652
n = 6, s = 1: 2.2210
n = 7, s = 1: 2.4999
n = 8, s = 1: -149.51

So the sum of these values approaches no limit. While the values do shrink for negative $n$, the sum also includes the problematic positive values.

Note that this numerical test is not a proof of divergence, but it strongly indicates that is what is happening.
Aww thank you mike. I've been coming across a lot of these functions and I've yet to see one that converges so I think I'm not doing anything too wrong.

Btw, you should look at my continuum sum thread, I know you were looking into the method earlier, I found a way using fractional calculus, but I'm a little mirky on some of the formal fine tunings, help would be greatly appreciated