general sums

[tommy1729]

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.

[mike3]

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 \( e^{ux} \), the basis of Fourier series, we have

\( \Delta e^{ux} = \Sigma^{-1} e^{ux} = (e^u - 1) e^{ux} \).
\( \Delta^2 e^{ux} = \Delta \Delta e^{ux} = \Delta (e^u - 1) e^{ux} = (e^u - 1) \Delta e^{ux} = (e^u - 1)(e^u - 1) e^{ux} = (e^u - 1)^2 e^{ux} \).

Induction shows that

\( \Delta^n e^{ux} = (e^u - 1)^n e^{ux} \).

Thus we have the (indefinite!) continuum sum \( \frac{e^{ux}}{e^u - 1} \) by setting \( n = -1 \), and we can "formally" continuously iterate the summation and difference operator by setting fractional, real, and complex values for \( n \). 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,

\( f(x) = \sum_{n=-\infty}^{\infty} a_n e^{nux} \),

the fractional forward difference is

\( \Delta^t f(x) = \sum_{n=-\infty}^{\infty} a_n (e^{nu} - 1)^t e^{nux} \)

which only gives iterations of the formal continuum sum if \( a_0 = 0 \). If \( a_0 \ne 0 \), then we get \( 0^t \) 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.

[tommy1729]

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

[kobi_78]

There exists a formula for iterated sums.

If

\( Sum_n({ a_n }) = S_n \)
\( S_0 = 0 \)
\( S_n = S_{n - 1} + a_n \)

Then

\( Sum_n^k({ a_n }) = \sum_{j=1}^{n} { {k + (n - j) - 1}\choose{k - 1} } \cdot a_j \)

[tommy1729]

There exists a formula for iterated sums.

If

\( Sum_n({ a_n }) = S_n \)
\( S_0 = 0 \)
\( S_n = S_{n - 1} + a_n \)

Then

\( Sum_n^k({ a_n }) = \sum_{j=1}^{n} { {k + (n - j) - 1}\choose{k - 1} } \cdot a_j \)

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

[kobi_78]

Hi, I discovered it myself

To derive it, use the rule that the multiplication of two power series is the discrete convolution of their coefficients and that \( (1 - x)^{-1} = \sum_{n=0}^{\infty}{x^n} \).

Now try to multiply a formal power series with \( (1 - x)^{-1} = \sum_{n=0}^{\infty}{x^n} \).

Repeat it k times and use the formula of \( (1 - x)^{-k} \) to derive the formula.

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