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Interpolating an infinite sequence ? - tommy1729 - 06/12/2022
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
RE: Interpolating an infinite sequence ? - tommy1729 - 06/12/2022
oh one more thing.
this is clearly related to continued fractions.
https://en.wikipedia.org/wiki/Euler%27s_continued_fraction_formula
regards
tommy1729
RE: Interpolating an infinite sequence ? - tommy1729 - 06/12/2022
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/54559/how-to-interpolate-a-sequence
But that does not answer all my questions.
regards
tommy1729
RE: Interpolating an infinite sequence ? - tommy1729 - 06/12/2022
i used to make the " infinite degree newton polynomial " for primes and prime twins.
But without any useful results.
RE: Interpolating an infinite sequence ? - JmsNxn - 06/12/2022
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.
RE: Interpolating an infinite sequence ? - tommy1729 - 06/12/2022
(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 ...
RE: Interpolating an infinite sequence ? - JmsNxn - 06/13/2022
(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.
RE: Interpolating an infinite sequence ? - tommy1729 - 06/13/2022
(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
RE: Interpolating an infinite sequence ? - tommy1729 - 06/13/2022
(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
RE: Interpolating an infinite sequence ? - tommy1729 - 06/13/2022
(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.