Interpolating an infinite sequence ? tommy1729 Ultimate Fellow     Posts: 1,906 Threads: 409 Joined: Feb 2009 06/12/2022, 03:39 PM Consider an infinite sequence of positive reals :  f(n) := a_1 , a_2 , ... Now we want to interpolate to define f(x) for all real x >= 1. Many things are written about interpolation , extrapolation , curve fitting etc. But they usually deal with a finite sequence or finite interval. And adding data changes the entire interpolation function. But I want a stable interpolation of an infinite sequence. ** So when i get 1 , 4 , 9 , 16 , 25 then the interpolation is trivial. But when I am givin complicated sequences and not functions ( like n^3 or taylors : 2 + 3 n + 0.5 n^4 + ... " for integer imput " ) then this way does not work. Im also not looking for best fit , but an actual match. ** I have issues with fractional derivatives and find contour integrals too hard for this. So I came  up with this : a_i = sum  b_n * (i)_n where (i)_n is a kind of falling factorial. In other words : a_1 = b_1 * 1 = b_1 a_2 = b_1 * 2 + b_2 * 2 * 1 = 2 b_1 + 2 b_2 = 2 a_1 + 2 b_2. a_3 = b_1 * 3  + b_2 * 3 * 2 + b_3 * 3 * 2 * 1 = 3 b_1 + 6 b_2 + 6 b_3. etc a_i = b_1 i + b_2 i (i-1) + b_3 i (i-1)(i-2) + b_4 i (i-1)(i-2)(i-3) + ... Notice how the b_n are solvable when the a_i are given. This reduces to linear algebra. This resembles ideas from newton and lagrange. I want to better understand this ( and use it for tetration ). notice that  1 i + 2 i (i-1) + 4 i (i-1) (i-2) + 8 i(i-1)(i-2)(i-3)  + ... does not converge for non-integer i !! So that is problematic as a solution for interpolation. So this creates questions and problems. should be invert the sequence ( replace a_i by 1/a_i ) in case of divergeance and then after interpolation invert again ? Another question is summability methods and ramanujan master theorem. How do they relate ? And ofcourse this falling factorial interpolation is a taylor series in disguise.  So that requires research too.  In fact where does this converge ? It is clearly not within a radius. And how does this relate to other interpolation methods ?? does n^3 interpolate as x^3 ? does f(n) interpolate to f(x) as a continuum sum ; f(x) = sum_0^x  f(x) - f(x-1) or something like that ? And if not , how do they relate ?? We do have the additive property. Vandermonde matrices are related. This all looks very familiar. I even wonder ; how many interpolation methods are there ? How many are interesting ? And how do they relate to dynamical systems ? Finally i want to write : f(x) = v_1 x/2! + v_2 x(x-1)/4! + v_3 x(x-1)(x-2)/6! + v_4 x(x-1)(x-2)(x-3)/8! + ... which converges for bounded v_n. And thus f(x) is an entire function and a consistant interpolation of "something". As mentioned above , we probably wont be able to interpolate 2^^n directly with such ideas but we could perhaps interpolate 1 / 2^^n with this and then take the multiplicative inverse. But we know our method is linear but not how it related to things like multiplicative inverse , summability methods , continuum sum etc. Maybe this is just my lack of a deep understanding of interpolation or linear algebra. Or my memory is getting old. But right now Im puzzled. One more thing  suppose a_i converges to a constant. Can we then use this interpolation as a fixpoint method for dynamical systems ?? regards tommy1729 tommy1729 Ultimate Fellow     Posts: 1,906 Threads: 409 Joined: Feb 2009 06/12/2022, 03:44 PM (This post was last modified: 06/12/2022, 03:45 PM by tommy1729.) oh one more thing. this is clearly related to continued fractions. https://en.wikipedia.org/wiki/Euler%27s_...on_formula regards tommy1729 tommy1729 Ultimate Fellow     Posts: 1,906 Threads: 409 Joined: Feb 2009 06/12/2022, 03:57 PM (This post was last modified: 06/12/2022, 04:13 PM by tommy1729.) ofcourse im aware of newton's divided differences and the newton polynomial what is basicly the same idea. And perhaps this is useful : https://math.stackexchange.com/questions...a-sequence But that does not answer all my questions. regards tommy1729 tommy1729 Ultimate Fellow     Posts: 1,906 Threads: 409 Joined: Feb 2009 06/12/2022, 04:01 PM i used to make the " infinite degree newton polynomial " for primes and prime twins. But without any useful results. JmsNxn Ultimate Fellow     Posts: 1,214 Threads: 126 Joined: Dec 2010 06/12/2022, 09:56 PM Interpolating is actually pretty easy. It's when you ask for a functional equation that it's difficult. Assume $a_n \to \infty$ and $b_n\to\infty$ and we want to find $f(b_n) = a_n$. Define a Weierstrass function $W(z)$, such that $W(b_n) = 0$. Then define: $f(z) = W(z) \sum_{n=0}^\infty \frac{a_n}{W'(b_n)(z-b_n)}\\$ You can choose $W$ such that the series converges, and that's pretty much it. This is an exercise in John B Conway's complex analysis, if you're looking for a source. tommy1729 Ultimate Fellow     Posts: 1,906 Threads: 409 Joined: Feb 2009 06/12/2022, 11:31 PM (06/12/2022, 09:56 PM)JmsNxn Wrote: Interpolating is actually pretty easy. It's when you ask for a functional equation that it's difficult. Assume $a_n \to \infty$ and $b_n\to\infty$ and we want to find $f(b_n) = a_n$. Define a Weierstrass function $W(z)$, such that $W(b_n) = 0$. Then define: $f(z) = W(z) \sum_{n=0}^\infty \frac{a_n}{W'(b_n)(z-b_n)}\\$ You can choose $W$ such that the series converges, and that's pretty much it. This is an exercise in John B Conway's complex analysis, if you're looking for a source. hmm Is sin(x) = W(x) valid ? then we get  $f(z) = sin(z) \sum_{n=0}^\infty \frac{a_n}{cos(2 \pi n)(z-2 \pi n)}\\$ I guess I made a mistake there ... JmsNxn Ultimate Fellow     Posts: 1,214 Threads: 126 Joined: Dec 2010 06/13/2022, 08:29 PM (06/12/2022, 11:31 PM)tommy1729 Wrote: (06/12/2022, 09:56 PM)JmsNxn Wrote: Interpolating is actually pretty easy. It's when you ask for a functional equation that it's difficult. Assume $a_n \to \infty$ and $b_n\to\infty$ and we want to find $f(b_n) = a_n$. Define a Weierstrass function $W(z)$, such that $W(b_n) = 0$. Then define: $f(z) = W(z) \sum_{n=0}^\infty \frac{a_n}{W'(b_n)(z-b_n)}\\$ You can choose $W$ such that the series converges, and that's pretty much it. This is an exercise in John B Conway's complex analysis, if you're looking for a source. hmm Is sin(x) = W(x) valid ? then we get  $f(z) = sin(z) \sum_{n=0}^\infty \frac{a_n}{cos(2 \pi n)(z-2 \pi n)}\\$ I guess I made a mistake there ... You'd need to choose a specific zero function depending on $a_n$. Such that we have $W'(b_n)$ is large enough to force the series to converge. In your cause, you would need to find a function with zeroes at $n$ but has a derivative at $n$ which causes the series to converge. For example, use: $f(z) = A(z)\sin(2 \pi z) \sum_{n=0}^\infty \frac{a_n}{2\pi A(n) (z-n)}\\$ This satisfies: $f(n) = a_n\\$ And you can force convergence of the series by letting  $A(n)$ be as large as possible. So for example $A(z) = e^z$ works. tommy1729 Ultimate Fellow     Posts: 1,906 Threads: 409 Joined: Feb 2009 06/13/2022, 10:14 PM (This post was last modified: 06/13/2022, 10:16 PM by tommy1729.) (06/13/2022, 08:29 PM)JmsNxn Wrote: $f(z) = A(z)\sin(2 \pi z) \sum_{n=0}^\infty \frac{a_n}{2\pi A(n) (z-n)}\\$ This satisfies: $f(n) = a_n\\$ And you can force convergence of the series by letting  $A(n)$ be as large as possible. So for example $A(z) = e^z$ works. How does f(5) = a_5 follow from  $f(z) = \Exp(z)\sin(2 \pi z) \sum_{n=0}^\infty \frac{a_n}{2\pi \Exp(n) (z-n)}\\$ or  $f(5) = \Exp(5)\sin(2 \pi 5) \sum_{n=0}^\infty \frac{a_n}{2\pi \Exp(n) (z-n)}\\$ ?? regards tommy1729 tommy1729 Ultimate Fellow     Posts: 1,906 Threads: 409 Joined: Feb 2009 06/13/2022, 10:18 PM (06/13/2022, 10:14 PM)tommy1729 Wrote: (06/13/2022, 08:29 PM)JmsNxn Wrote: $f(z) = A(z)\sin(2 \pi z) \sum_{n=0}^\infty \frac{a_n}{2\pi A(n) (z-n)}\\$ This satisfies: $f(n) = a_n\\$ And you can force convergence of the series by letting  $A(n)$ be as large as possible. So for example $A(z) = e^z$ works. How does f(5) = a_5 follow from  $f(z) = \Exp(z)\sin(2 \pi z) \sum_{n=0}^\infty \frac{a_n}{2\pi \Exp(n) (z-n)}\\$ or  $f(5) = \Exp(5)\sin(2 \pi 5) \sum_{n=0}^\infty \frac{a_n}{2\pi \Exp(n) (z-n)}\\$ ?? regards tommy1729 not sure why tex fails  slightly better How does f(5) = a_5 follow from  $f(z) = Exp(z)\sin(2 \pi z) \sum_{n=0}^\infty \frac{a_n}{2\pi Exp(n) (z-n)}\\$ or  $f(5) = Exp(5)\sin(2 \pi 5) \sum_{n=0}^\infty \frac{a_n}{2\pi Exp(n) (z-n)}\\$ ?? regards tommy1729 tommy1729 Ultimate Fellow     Posts: 1,906 Threads: 409 Joined: Feb 2009 06/13/2022, 10:19 PM (06/13/2022, 10:18 PM)tommy1729 Wrote: (06/13/2022, 10:14 PM)tommy1729 Wrote: (06/13/2022, 08:29 PM)JmsNxn Wrote: $f(z) = A(z)\sin(2 \pi z) \sum_{n=0}^\infty \frac{a_n}{2\pi A(n) (z-n)}\\$ This satisfies: $f(n) = a_n\\$ And you can force convergence of the series by letting  $A(n)$ be as large as possible. So for example $A(z) = e^z$ works. How does f(5) = a_5 follow from  $f(z) = \Exp(z)\sin(2 \pi z) \sum_{n=0}^\infty \frac{a_n}{2\pi \Exp(n) (z-n)}\\$ or  $f(5) = \Exp(5)\sin(2 \pi 5) \sum_{n=0}^\infty \frac{a_n}{2\pi \Exp(n) (z-n)}\\$ ?? regards tommy1729 not sure why tex fails  slightly better How does f(5) = a_5 follow from  $f(z) = Exp(z)\sin(2 \pi z) \sum_{n=0}^\infty \frac{a_n}{2\pi Exp(n) (z-n)}\\$ or  $f(5) = Exp(5)\sin(2 \pi 5) \sum_{n=0}^\infty \frac{a_n}{2\pi Exp(n) (z-n)}\\$ ?? regards tommy1729 site crashed in my browser ... maybe that relates. tex looked different without changing it. « Next Oldest | Next Newest »

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