Interpolating an infinite sequence ?

[tommy1729]

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]

oh one more thing.

this is clearly related to continued fractions.

https://en.wikipedia.org/wiki/Euler%27s_...on_formula

regards

tommy1729

[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]

i used to make the " infinite degree newton polynomial " for primes and prime twins.

But without any useful results.

[JmsNxn]

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]

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]

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]

\[
f(z) = A(z)\sin(2 \pi z) \sum_{n=0}^\infty \frac{a_n}{2\pi A(n) (z-n)}\\
\]

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]

\[
f(z) = A(z)\sin(2 \pi z) \sum_{n=0}^\infty \frac{a_n}{2\pi A(n) (z-n)}\\
\]

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]

\[
f(z) = A(z)\sin(2 \pi z) \sum_{n=0}^\infty \frac{a_n}{2\pi A(n) (z-n)}\\
\]

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.