general sums

[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.