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 Another way to continuum sum! JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 08/29/2013, 04:05 PM Well if one way to continuum sum wasn't enough, I found another one! Let's start: $\mathcal{L} f(x) = \frac{1}{\Gamma(x)}\int_0^\infty e^{-t} f(t) t^{x-1}dt$ $\mathcal{L}^{-1} f(x) = \frac{e^x}{2 \pi i} \int_{c - i \infty}^{c+ i \infty} \Gamma(t) f(t) x^{-t} dt$ If $\frac{\Delta f(x)}{\Delta x} = f(x) - f(x-1)$ then: $\frac{\Delta}{\Delta x} \mathcal{L} f(x) = \mathcal{L} \frac{d f(x)}{dx}$ The method for continuum sum follows from this! $\sum_a^b f(y) \, \Delta y = (\mathcal{L} \int \mathcal{L}^{-1} f(x))\,|_{x=a}^{x=b}$ tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 08/29/2013, 10:04 PM I like your enthousiasm but I hoped you would be a bit more skeptical after you found your " latest paradoxes ". Not that math is easy but did you even test with say x^7 and 2^x ? regards tommy1729 JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 08/29/2013, 11:50 PM (This post was last modified: 08/29/2013, 11:52 PM by JmsNxn.) It works perfectly for polynomials, which follows from Mellin transforms. I'm thinking that it should work for functions that don't grow too fast (polynomially, less than exponential) as the imaginary part goes to \pm infinity. If that's satisfied then I'm pretty sure for a large class of functions the result should hold. It starts with pretty much the same mellin inverse integral, so I'm thinking it works on the same class of functions as the first continuum sum method since the mellin inversion transformation guarantees that the rest will converge. On exponentials it should work as well, I haven't tested, but I'm confident it does. Functions that blow up at \pm imaginary infinity like $e^{-x^2}$ don't work. I'll have to think about that one, maybe solving for $e^{x^2}$ and then making a change of argument in the continuum sum so you count across imaginary values (that should work!) I'm really confident about this way and I think it may be more effective. My next mission is to define contour summation. tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 06/05/2014, 10:57 PM Do you still believe in this ? regards tommy1729 MphLee Fellow Posts: 95 Threads: 7 Joined: May 2013 06/06/2014, 11:21 AM (This post was last modified: 06/06/2014, 11:40 AM by MphLee.) Just for curiosity, but in other words the operator $\mathcal {L}$ conjugates the differential operator and the difference operator? That is the same as $\Delta \circ \mathcal {L}=\mathcal {L}$$\circ {D}$ So $\mathcal {L}$ is like a the solution of an "abel functional equation for operators"... Maybe if you can find an operator $\mathcal {V}$ such that $\Sigma \circ \mathcal {V}=\mathcal {V}$$\circ {D}$ where $\Sigma f(x):=f(S(f^{-1}(x)))$ (aka an operator that conujugates the differential operator and the subfunction operator) you could conjugate the fractional differentiation of a function by $\mathcal {V}$ and obtain a fractional iteration of the subfunction operator: $\Sigma^{\circ\sigma} \circ \mathcal {V}=\mathcal {V}$$\circ {D^{\circ\sigma}}$ $\Sigma^{\circ\sigma}=\mathcal {V}$$\circ {D^{\circ\sigma}} \circ \mathcal {V^{-1}}$ Do you think it is possible to find shuch $\mathcal {V}$? MathStackExchange account:MphLee JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 06/06/2014, 04:00 PM (This post was last modified: 06/06/2014, 04:01 PM by JmsNxn.) (06/06/2014, 11:21 AM)MphLee Wrote: Just for curiosity, but in other words the operator $\mathcal {L}$ conjugates the differential operator and the difference operator? That is the same as $\Delta \circ \mathcal {L}=\mathcal {L}$$\circ {D}$ So $\mathcal {L}$ is like a the solution of an "abel functional equation for operators"... Maybe if you can find an operator $\mathcal {V}$ such that $\Sigma \circ \mathcal {V}=\mathcal {V}$$\circ {D}$ where $\Sigma f(x):=f(S(f^{-1}(x)))$ (aka an operator that conujugates the differential operator and the subfunction operator) you could conjugate the fractional differentiation of a function by $\mathcal {V}$ and obtain a fractional iteration of the subfunction operator: $\Sigma^{\circ\sigma} \circ \mathcal {V}=\mathcal {V}$$\circ {D^{\circ\sigma}}$ $\Sigma^{\circ\sigma}=\mathcal {V}$$\circ {D^{\circ\sigma}} \circ \mathcal {V^{-1}}$ Do you think it is possible to find shuch $\mathcal {V}$? Yep, but it's not linear and it requires a lot more sophistication to it than in the form I have. And before you get ahead of yourself, it looks as though it WILL NOT work on hyper operators. Unless some very magical theorems fall from the air. It's hard enough getting it to work for tetration let alone all functions. MphLee Fellow Posts: 95 Threads: 7 Joined: May 2013 06/06/2014, 05:09 PM (06/06/2014, 04:00 PM)JmsNxn Wrote: it looks as though it WILL NOT work on hyper operators. Can you summarize the hints that makes you think this? I'm only courious because I don't know nothing about differentiation (and fractional differentiation) so I wanted to know what is the landscape that you can see with your bigger knowledge of the subject. MathStackExchange account:MphLee « Next Oldest | Next Newest »

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