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 applying continuum sum to interpolate any sequence. JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 08/17/2013, 11:00 PM (This post was last modified: 08/17/2013, 11:02 PM by JmsNxn.) Well I've always liked the idea of having a sequence $a_n$ and finding an analytic continuation $a(s)$ such that it returns $a_n$ at integer values. Recently I have found a method of continuum summing http://math.eretrandre.org/tetrationforu...hp?tid=801 This is one of the exploits we have. We will be using two theorems, my continuum sum theorem and a result thanks to Grunwald and Letnikov. $\frac{d^{-s}f(t)}{dt^{-s}} = \lim_{n\to \infty} n^{s}\sum_{k=0}^{\infty} \frac{s \cdot \cdot \cdot (s + k -1)}{k!} f(t - \frac{k}{n})$ Which reduces to the natural derivative and integral for integer values of s We take our sequence $a_n$ and create an auxillary function $\vartheta(s) = \sum_{n=0}^{\infty} a_n \frac{s^n}{n!}$ We are clear that $\frac{d^n}{dt^n}\vartheta(t)|_{t=0} = a_n$ And therefore if we create the following function: $\phi(s) = \int_0^\infty e^{-t} \frac{d^{-s}\vartheta(t)}{dt^{-s}}dt$ We have, thanks to my continuum sum theorem: $\phi(-n) - \phi(-n-1) = a_n$ Therefore the function: $\phi(-s) - \phi(-s-1) = a(s)$ and interpolates the sequence $a_n$ And even more beautifully: $\phi(-n) - \phi(-n-k) = \sum_{j=n}^{n+k-1} a_j$ and so we have also a continuum sum from phi. This is a nifty little way of creating interpolation functions. Thought it was neat. Allows us to interpolate something like the prime number sequence and get an entire function that returns prime numbers at integer values. this is because the primenumber sequence doesn't grow too fast and so our auxillary function is entire. tommy1729 Ultimate Fellow Posts: 1,372 Threads: 336 Joined: Feb 2009 08/18/2013, 08:55 PM (08/17/2013, 11:00 PM)JmsNxn Wrote: Well I've always liked the idea of having a sequence $a_n$ and finding an analytic continuation $a(s)$ such that it returns $a_n$ at integer values. Recently I have found a method of continuum summing http://math.eretrandre.org/tetrationforu...hp?tid=801 There are many interpolations methods. But for tetration we are intrested in continuum sums yes. Quote:We will be using two theorems, my continuum sum theorem and a result thanks to Grunwald and Letnikov. $\frac{d^{-s}f(t)}{dt^{-s}} = \lim_{n\to \infty} n^{s}\sum_{k=0}^{\infty} \frac{s \cdot \cdot \cdot (s + k -1)}{k!} f(t - \frac{k}{n})$ I think that should be : $\frac{d^{-s}f(t)}{dt^{-s}} = \lim_{n\to \infty} n^{s}\sum_{k=0}^{n} \frac{s \cdot \cdot \cdot (s + k -1)}{k!} f(t - \frac{k}{n})$ Not ? Grunwald and Letnikov hmm , probably my ignorance but if you told me Newton I might have believed that too. Should I know these 2 gentlemen ? Quote:Thought it was neat. Allows us to interpolate something like the prime number sequence and get an entire function that returns prime numbers at integer values. this is because the primenumber sequence doesn't grow too fast and so our auxillary function is entire. Why do you want to interpolate the prime numbers ? Its not like you will get a nice elementary function or a zeta function. You might as well take F(x) = the product over primes of (1-x/p) ( and use weierstrass factorization theorem to show its entire ) or such and then count the number of zero's with a contour integral , why interpolate ? We already have explicit forms for the prime counting function ( multiple actually alhtough some are tricky to prove or read about ) in terms of the Riemann zeta function nontrivial zero's. Dont get me wrong , I like Number Theory , the Continuum sum and your post. But at this point I only see " aeshtetic " value in this interpolation of the prime numbers unless you have a nice form ( big reductions/simplifications in the integrals ). regards tommy1729 « Next Oldest | Next Newest »

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