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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 which differentiates f across s, t times. Now assume that:

for some s in some set , which can be easily constructed using some theorems I have.

Then:



If you differentiate by the continuity of this improper integral

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!




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!




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 to (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 , 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 Smile