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+--- Thread: Very curious question (/showthread.php?tid=812)
Very curious question - JmsNxn - 08/18/2013
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?
RE: Very curious question - JmsNxn - 08/19/2013
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?
RE: Very curious question - mike3 - 08/19/2013
(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.
RE: Very curious question - JmsNxn - 08/20/2013
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
