From: Lance <a@b.c>
Subject: Integral as the limit of a sum
Date: Thu, 29 Jun 2000 12:58:57 +0200
Newsgroups: sci.math.symbolic
Summary: [missing]

Hi

I know that if I have a summation of the form :
/sum_{i=1}^{n}[ f(x_{i})dx_{i}] then in the limit as n-->infinty and
dx-->0, this becomes the definite integral of f(x)

However, I now have an expression of the form :
/sum_{i=1}^{n} [f(x_{i})exp(adx_{i})] and wish to write this as an
integral in the limit.  Is it possible to do this ? How would I go about
doing this ?

Thanks
Lance
==============================================================================

From: israel@math.ubc.ca (Robert Israel)
Subject: Re: Integral as the limit of a sum
Date: 29 Jun 2000 17:43:49 GMT
Newsgroups: sci.math.symbolic

In article <395B2BF1.B2499650@b.c>, Lance  <a@b.c> wrote:

>However, I now have an expression of the form :
>/sum_{i=1}^{n} [f(x_{i})exp(adx_{i})] and wish to write this as an
>integral in the limit.  Is it possible to do this ? How would I go about
>doing this ?


Use the "tangent-line approximation" to the exponential:

exp(a dx_i) = 1 + a dx_i + O(dx_i^2)

so the sum becomes 
sum_{i=1}^n f(x_i) + a sum_{i=1}^n f(x_i) dx_i + ...

The first sum will only have a finite limit under 
rather special circumstances, the second (under 
rather more general conditions) becomes an 
integral, and the "..." (under these same 
conditions) should go to 0. 

Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            
Vancouver, BC, Canada V6T 1Z2
============================================================================== 

From: martin.kahlert@keksy.mchp.siemens.de (Martin Kahlert)
Subject: Re: Integral as the limit of a sum
Date: 29 Jun 2000 12:21:11 GMT
Newsgroups: sci.math.num-analysis

In article <395B2BED.1960DA3E@b.c>,
	Lance <a@b.c> writes:
> Hi
> 
> I know that if I have a summation of the form :
> /sum_{i=1}^{n}[ f(x_{i})dx_{i}] then in the limit as n-->infinty and
> dx-->0, this becomes the definite integral of f(x)
> 
> However, I now have an expression of the form :
> /sum_{i=1}^{n} [f(x_{i})exp(adx_{i})] and wish to write this as an
> integral in the limit.  Is it possible to do this ? How would I go about
> doing this ?

I am not sure, that i understand, what you want:
If you really mean this
/sum_{i=1}^{n} [f(x_{i}) exp(a*dx_{i})],
you will get a limit of infinity (dx_i --> 0 ==> exp(a*dx_i) --> 1)
and you will get more and more nearly identical terms f(x_{i})
(assuming a continous f).

If you mean /sum_{i=1}^{n} [f(x_{i}) exp(a*d*x_{i})], it will be 
basically the same.

What you *can* do is to transform something like /sum_{i=1}^n f(i)
into an integral and a rest sum, which could be quite fast convergent.
(see the Euler Mc. Laurin summation formula)

Perhaps it would help, if you told us more about f and the exact
formula you meant.

Have fun,
Martin.

-- 
The early bird gets the worm. If you want something else for       
breakfast, get up later.
==============================================================================

From: spellucci@mathematik.tu-darmstadt.de (Peter Spellucci)
Subject: Re: Integral as the limit of a sum
Date: 29 Jun 2000 18:28:27 GMT
Newsgroups: sci.math.num-analysis

In article <395B2BED.1960DA3E@b.c>,
 Lance <a@b.c> writes:
|> Hi
|> 
|> I know that if I have a summation of the form :
|> /sum_{i=1}^{n}[ f(x_{i})dx_{i}] then in the limit as n-->infinty and
|> dx-->0, this becomes the definite integral of f(x)
|> 
|> However, I now have an expression of the form :
|> /sum_{i=1}^{n} [f(x_{i})exp(adx_{i})] and wish to write this as an
                           ^^^^^^^^^^^^^^^^^^^^^^^  this cannot be the case 
                           if dx{i} -> 0 this goes to one hence you measure
                           arbirarily small intervals by one

|> integral in the limit.  Is it possible to do this ? How would I go about
|> doing this ?

I think you intended what is known as the Riemann-Stieltjes-Integral:
    \sum_{i=1}^n [f(x_i)(g(x_{i+1})-g(x_i))  
with g(x)=exp(ax) in your case. If g is continuously differentiable you get 
the limit
   \int_a^b f(x)g'(x)dx
but the limit exists for every g of bounded variation. 
here a=x_1, b=x_{n+1} for all n.

hope that helps
peter