Reimann sums, okay. How about a Reimann product ?

1. Nov 6, 2003

pellman

Reimann sums, okay. How about a "Reimann product"?

An integral is a sort of "continuous sum". Very roughly, the sum &Sigma;k f(xk) &Delta;x goes over to the integral &int;f(x)dx when the number of terms becomes infinite while &Delta;x goes to zero.

What about a similar "continuous product"? If we have the product

&Pi;k [f(xk)] Pk

and let the exponents Pk go to zero while the number of products goes to infinite (and the range of xk is fixed as in an integral) what sort of an animal do we get?

I realize that we can turn this product into a sum by taking the anti-logarithm. Then the continuous limit of this product is e raised to an integral with ln(f) in the integrand. However, I'm curious if the properties of these sorts of entities are known without resorting to integrals. And is there a notation for them?

2. Nov 6, 2003

Ambitwistor

Re: Reimann sums, okay. How about a "Reimann product"?

I don't think you want to let the exponents go to zero; then each factor in the product will approach 1.

I don't know if it's really what you're looking for, but something a little analogous to a "continuous product" is the exponentiation of an element of a Lie algebra to get an element of a Lie group.

As a simple, scalar example, consider the n-fold product of (1+x/n), which is &Pi;n (1+x/n)n. The limit as n goes to infinity is just the function exp(x).

3. Nov 7, 2003

pellman

Thanks, Ambitwistor.

Having thought it over, I see now that these products have probably not been studied in themselves since they can always be reduced to an integral. Like so, ..

&Pi;k [f(xk)] Pk
= exp[ ln ( &Pi;k [f(xk)] Pk ) ]
= exp[ &Sigma;k Pkf(xk) ]
= exp[ &Sigma;k &alpha;(xk)f(xk) &Delta;x ] , Pk = &alpha;(xk)&Delta;x
--> exp[ &int; &alpha;(x) f(x) dx ]

which shows why I said the exponents go to zero.