limit to infinity of Left Riemann Sum

Hi there,

Lets say I have created a function Sn=LeftRiemannSum(f,left,right,N) ,that computes the left riemann sum over the interval left to right with N subdomains.i.e.:

Sn=sum(f(xi)*h) for all subdomains i=0 to N-1. f is my function and xi=left+i*h , so the input arguments left=x0 and right=xN.

Let f be my anonymous function (ex f=@(x)(x.*log(1+x)) .

I also estimated the Sn for varying N, from N=10 to 100000.

Now, I simply want to compute the value of the series Sn when N -> infinity. Inside the function I have a for loop [ for i=0:(N-1)] so I will have endless loop ..

Can I pass the function somehow to the 'limit' command? Any clues?

Thanx!

PS: The main part of the code of my function LeftRiemann Sum is the following:

if true
  for i=0:(N-1)
    x=x0+i.*h;
    y=f(x);
    A=y.*h;
    S=S+A
  end

   Sn=S

end

Answer 1

[Zoltán Csáti]

If I am not mistaken, you try to determine the limit of the left Riemann-sum for the value of the definite integral. Of course you can't take infinite members. If you want to use this approach, I recommend you to use a large number for N and also estimate the right Riemann-sum. If the two sums are approximately equal, then there is hope that this is the approximate value of the integral.

A comment: it can easily be vectorized:

N = 1000;
a = 1;
b = 2;
h = (b-a)/N;
x = x0+(0:N)*h;
fx = f(x);
sum(fx*h);

Answer 2

[Torsten]

Use

S_inf = integral(f,left,right)

to get the limit.

Best wishes

Torsten.

[Zoltán Csáti]

Yes, it is straightforward. But I think it is homework problem for better understanding of limit and MATLAB.

[Stelina]

Yes, I should have clarified. I already integrated my function analytically with

syms x;f=x.*log(1+x);
F=int(f)  %integrate
a=0;b=1;
area=int(f,a,b)

, so I know the exact result.

Though, I still don't quite get whats the point of this homework exercise. For example, even if I use a relatively not that large value for N (I tried N=10000000) to substitute "inf", I get the exact same number, meaning my numerical approximation is perfect.

Besides, I guess MATLAB uses sth like that to calculate int "analytically" right?

Lol, I should ask MATLAB.

[Torsten]

If you use "int", MATLAB tries to analytically determine a function F such that F'=f.

In your case, F(x)=0.25*(2*(x^2-1)*log(1+x)-(x-2)*x).

Thus in the limit you get F(1)-F(0)=0.25.

Best wishes

Torsten.