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i was thinking about " general sums ".

that means continuum iterations of continuum sums.

i believe q-analogues and fourrier series play an important role in this.

its still vague , but i wanted to throw it on the table.
(06/22/2010, 10:29 PM)tommy1729 Wrote: [ -> ]i was thinking about " general sums ".

that means continuum iterations of continuum sums.

i believe q-analogues and fourrier series play an important role in this.

its still vague , but i wanted to throw it on the table.

What do you mean? You mean continuous iteration of the sum operator?

Hmmmmm... Well, for , the basis of Fourier series, we have

.
.

Induction shows that

.

Thus we have the (indefinite!) continuum sum by setting , and we can "formally" continuously iterate the summation and difference operator by setting fractional, real, and complex values for . Iteration of the difference operator seems to have been studied before -- look up "fractional finite differences". The generalization above may remind one of generalizing the derivative to non-integer order.

For a Fourier/exp-series,

,

the fractional forward difference is



which only gives iterations of the formal continuum sum if . If , then we get which is undefined for negative t and even t = 0, meaning we can't even apply the operator 0 times. I'm not sure how to extend it in those cases.
an intresting paper related to continuum sum ( but not continuum iterations of it ) is this :

http://www.math.tu-berlin.de/~mueller/HowToAdd.pdf

especially " 3. Basic Algebraic Identities " where the geometric part is what mike3 uses (together with fourrier expansion) to get his continuum sum.

the idea of ' removing the period ' is also known and the origin of this 'geometric part equation' is as old as " q-math " ( q-series and q-analogues and fourrier series )

i knew id seen it before ... in fact i used it myself even way before that paper was written , although probably similar papers have been written much earlier.

not to mention eulers example given in the paper.

intresting is the continuum product

product x ; sin(x) + 5/4.

or equivalent the continuum sum

sum x ; ln(sin(x) + 5/4).

and the question if these sums resp products are periodic themselves.

and the question if these sums resp products are divergent ( lim x -> oo does not equal +/-oo or 0)

( it is known that integral 0,2pi log(sin(x) + 5/4) = 0 )

regards

tommy1729
There exists a formula for iterated sums.

If





Then

(06/24/2010, 06:53 PM)kobi_78 Wrote: [ -> ]There exists a formula for iterated sums.

If





Then


thats great !

now all we need to do is take the continuum sum of that.

where did you get that formula btw ? newton ? gauss ?

thanks

regards

tommy1729
Hi, I discovered it myself Smile

To derive it, use the rule that the multiplication of two power series is the discrete convolution of their coefficients and that .

Now try to multiply a formal power series with .

Repeat it k times and use the formula of to derive the formula.

It however reminds Cauchy's formula for repeated integration (with all umbral calculus variations).