tommy's "linear" summability method

[Caleb]

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]

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]

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]

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

What 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]

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

What 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]

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.