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 Very curious question JmsNxn Long Time Fellow Posts: 565 Threads: 94 Joined: Dec 2010 08/18/2013, 11:14 PM (This post was last modified: 08/19/2013, 05:26 PM by JmsNxn.) 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? JmsNxn Long Time Fellow Posts: 565 Threads: 94 Joined: Dec 2010 08/19/2013, 05:39 PM (This post was last modified: 08/19/2013, 05:40 PM by JmsNxn.) 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? mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 08/19/2013, 09:41 PM (This post was last modified: 08/19/2013, 11:51 PM by mike3.) (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. JmsNxn Long Time Fellow Posts: 565 Threads: 94 Joined: Dec 2010 08/20/2013, 08:56 PM 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 « Next Oldest | Next Newest »

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