tommy's "linear" summability method Caleb Junior Fellow Posts: 42 Threads: 5 Joined: Feb 2023 02/07/2023, 03:38 AM Also, I'm interested in the continuum summation object you've mentioned. About a year and a half I thought up something similar, but since then I haven't really thought about it in details, so I'm interested to see what perspectives other's have on the idea JmsNxn Ultimate Fellow Posts: 1,176 Threads: 123 Joined: Dec 2010 02/08/2023, 03:43 AM (This post was last modified: 02/08/2023, 03:44 AM by JmsNxn.) Hey, Caleb, since you've talked about Fractional calculus, and indefinite sums, I reference this paper by me from a long time ago! https://arxiv.org/abs/1503.06211 This is about fractional calculus approaches to indefinite sums! I think you will find it very similar to what you are talking about here . Please don't mind the language, and the rough nature. This was the first paper I ever wrote; part of an undergrad thesis on Fractional Calculus. tommy1729 Ultimate Fellow Posts: 1,859 Threads: 402 Joined: Feb 2009 02/08/2023, 01:25 PM Well I guess I will explain my " interpretation " of continuum sum. Notice the continuumsum is actually unique ( up to a constant ) as the solution to F(x+1) - F(X) = f(x) or so.   So all methods agree somewhat when they converge. Wiki already has integral representations due to Riemann etc. All major posters here ( till 2022 ) have devoted time on continuum sums. I believe mike3 came up first with fourier series by letting the period go to oo. and then summing like I do here :  https://math.stackexchange.com/questions...f-m-frac54 which works because we get exp sums. Notice how this looks similar to how I do my summability method , but I said that before I guess. I think getting the period to oo is wasting time and we can do it directly by my interpretation. Although that fact is nice to understand four analysis and related integrals ! Let c be a real constant. Let CS stand for continuum sum. Then  CS f(x) = CS g( exp(x) - c ) This implies g(x) = f( ln(x + c) ) Now expand g(x) as a taylor :  g(x) = g0 + g1 x + g2 x^2 + ... Then we get CS f(x) = CS g( exp(x) - c ) = CS ( g0 + g1 (exp(x) - c) + g2 (exp(x) - c)^2 + ... ) By using newtons binomium : CS ( h0 + h1 exp(x) + h2 exp(x)^2 + ... ) and linearity = Constant + CS (h0) + h1 CS (exp(x)) + h2  CS( exp(2x) ) + ... then using the q-identity 1 + q + q^2 + q^3 + ... + q^n = (q^(n+1) - 1)/(q-1) We finally get an expression for the continuum sum. As long as everything converges the choice of c is not important and the method works. I hope that is clear. regards tommy1729 Caleb Junior Fellow Posts: 42 Threads: 5 Joined: Feb 2023 02/09/2023, 04:27 AM (02/08/2023, 01:25 PM)tommy1729 Wrote: Well I guess I will explain my " interpretation " of continuum sum. Notice the continuumsum is actually unique ( up to a constant ) as the solution to F(x+1) - F(X) = f(x) or so.   So all methods agree somewhat when they converge. Wiki already has integral representations due to Riemann etc. All major posters here ( till 2022 ) have devoted time on continuum sums. I believe mike3 came up first with fourier series by letting the period go to oo. and then summing like I do here :  https://math.stackexchange.com/questions...f-m-frac54 which works because we get exp sums. Notice how this looks similar to how I do my summability method , but I said that before I guess. I think getting the period to oo is wasting time and we can do it directly by my interpretation. Although that fact is nice to understand four analysis and related integrals ! Let c be a real constant. Let CS stand for continuum sum. Then  CS f(x) = CS g( exp(x) - c ) This implies g(x) = f( ln(x + c) ) Now expand g(x) as a taylor :  g(x) = g0 + g1 x + g2 x^2 + ... Then we get CS f(x) = CS g( exp(x) - c ) = CS ( g0 + g1 (exp(x) - c) + g2 (exp(x) - c)^2 + ... ) By using newtons binomium : CS ( h0 + h1 exp(x) + h2 exp(x)^2 + ... ) and linearity = Constant + CS (h0) + h1 CS (exp(x)) + h2  CS( exp(2x) ) + ... then using the q-identity 1 + q + q^2 + q^3 + ... + q^n = (q^(n+1) - 1)/(q-1) We finally get an expression for the continuum sum. As long as everything converges the choice of c is not important and the method works. I hope that is clear. regards tommy1729What is this uniqueness you are talking about?  $$F(x+1)-F(x) = f(x)$$ is definitely not unique, just add in any 1-periodic function so that $$G(x+1)-G(x) = 0$$ And then $$(F+G)(x+1)-(F+G)(x) = f(x)$$ is a new solution.  If I remember correct, Ramanujan even had issues with uniqueness for his summation method, which was also based around solutions to $F(x+1)-F(x) = f(x)$. The usual way I think of trying to achieve uniqueness now-a-days is to think about the operator  $$(e^D-1) F(x) = f(x) \implies F(x) = \frac{1}{e^D-1} f(x)$$ which I think leads naturally into the E-M formula and is similar to Ramanujan's approach. Do you have something like this in mind when you are talking about uniqueness? tommy1729 Ultimate Fellow Posts: 1,859 Threads: 402 Joined: Feb 2009 02/09/2023, 12:54 PM (02/09/2023, 04:27 AM)Caleb Wrote: (02/08/2023, 01:25 PM)tommy1729 Wrote: Well I guess I will explain my " interpretation " of continuum sum. Notice the continuumsum is actually unique ( up to a constant ) as the solution to F(x+1) - F(X) = f(x) or so.   So all methods agree somewhat when they converge. Wiki already has integral representations due to Riemann etc. All major posters here ( till 2022 ) have devoted time on continuum sums. I believe mike3 came up first with fourier series by letting the period go to oo. and then summing like I do here :  https://math.stackexchange.com/questions...f-m-frac54 which works because we get exp sums. Notice how this looks similar to how I do my summability method , but I said that before I guess. I think getting the period to oo is wasting time and we can do it directly by my interpretation. Although that fact is nice to understand four analysis and related integrals ! Let c be a real constant. Let CS stand for continuum sum. Then  CS f(x) = CS g( exp(x) - c ) This implies g(x) = f( ln(x + c) ) Now expand g(x) as a taylor :  g(x) = g0 + g1 x + g2 x^2 + ... Then we get CS f(x) = CS g( exp(x) - c ) = CS ( g0 + g1 (exp(x) - c) + g2 (exp(x) - c)^2 + ... ) By using newtons binomium : CS ( h0 + h1 exp(x) + h2 exp(x)^2 + ... ) and linearity = Constant + CS (h0) + h1 CS (exp(x)) + h2  CS( exp(2x) ) + ... then using the q-identity 1 + q + q^2 + q^3 + ... + q^n = (q^(n+1) - 1)/(q-1) We finally get an expression for the continuum sum. As long as everything converges the choice of c is not important and the method works. I hope that is clear. regards tommy1729What is this uniqueness you are talking about?  $$F(x+1)-F(x) = f(x)$$ is definitely not unique, just add in any 1-periodic function so that $$G(x+1)-G(x) = 0$$ And then $$(F+G)(x+1)-(F+G)(x) = f(x)$$ is a new solution.  If I remember correct, Ramanujan even had issues with uniqueness for his summation method, which was also based around solutions to $F(x+1)-F(x) = f(x)$. The usual way I think of trying to achieve uniqueness now-a-days is to think about the operator  $$(e^D-1) F(x) = f(x) \implies F(x) = \frac{1}{e^D-1} f(x)$$ which I think leads naturally into the E-M formula and is similar to Ramanujan's approach. Do you have something like this in mind when you are talking about uniqueness? Oh sorry I did not mention it , By uniqueness I also mean that the continuum sum of a polynomial is a polynomial. so the CS x = x(x+1)/2 But you have a point such a 1 periodic function might exist. But not when  the method gives  CS x = x(x+1)/2 Or so I believe ... worth consideration regards tommy1729 JmsNxn Ultimate Fellow Posts: 1,176 Threads: 123 Joined: Dec 2010 02/10/2023, 03:55 AM (This post was last modified: 02/10/2023, 04:47 AM by JmsNxn.) The uniqueness is Exponential space. If: $$f(s) = O(e^{\rho |\Re(s)| +\tau|\Im(s)|})$$ for $$0 < \tau < \pi/2$$ then there is a unique continuum sum $$F(s)$$ that also belongs to this space. That's the central thesis of my paper. No 1-periodic function is in this space. No function which satisfies $$g(s+1) - g(s) = 0$$ in this space other than a constant... And yes, this is Tommy's continuum sum, sending polynomials to polynomials. « Next Oldest | Next Newest »

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