tommy's "linear" summability method
#13
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
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Messages In This Thread
RE: tommy's "linear" summability method - by tommy1729 - 02/08/2023, 01:25 PM

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