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The Riemann Sum of Sin(x)


Date: 02/07/2001 at 16:05:26
From: kaldoo
Subject: Riemann sum

Hello all. The problem is to evaluate 

      pi
     INT [(sin x)dx]
      0

by Riemann sum. I got approximately 2.04 using the midpoint formula. 
But by integrating you get the answer 2, so how do you prove that you 
can also use Riemann sums to get 2?


Date: 02/07/2001 at 16:57:14
From: Doctor Anthony
Subject: Re: Riemann sum

The Riemann sum is   

      n
     SUM[(pi/n)sin(r*(pi/n))] =
     r=1

     (pi/n)[sin(pi/n) + sin(2pi/n) + sin(3pi/n) + ... + sin((n)pi/n)]

This is equal to the imaginary part of the following by Euler's 
formula, e^(ix) = cos x + i*sin x.

     (pi/n)[e^(i*pi/n) + e^(i*2pi/n) + ... + e^(i*(n)pi/n)]

     = (pi/n)*e^(i*pi/n)[1+e^(i*pi/n)+e^(i*2pi/n)+...+e^(i*(n-1)pi/n)]

The expression in brackets can be replaced, because 
1+x+x^2+...+x^(n-1) = (1-x^n)/(1-x).

     = (pi/n)*e^(i*pi/n)[1 - e^(i*n*pi/n)]/[1-e^(i*pi/n)]

                            e^(i*pi/2)[e^(-i*pi/2) - e^(i*pi/2)]
     = (pi/n)*e^(i*pi/n)---------------------------------------------
                        e^(i*pi/(2n))[e^(-i*pi/(2n)) - e^(i*pi/(2n))]

     = (pi/n)*e^(i*pi(n+1)/(2n))*[sin(pi/2)/sin(pi/(2n))]

Expand e^(i*pi(n+1)/(2n)) with Euler's formula:

     = (pi/n)[cos(pi(n+1)/(2n)) + i*sin(pi(n+1)/(2n))]/sin(pi/(2n))

The Riemann sum is equal to the imaginary part of the above:

     (pi/n)sin(pi(n+1)/(2n))/sin(pi/(2n))

Take the limit as n goes to infinity. Since x = lim[sin x] as x->0, 
the limit of the above expression is

     (pi/n)sin(pi/2)
     ---------------   =  2
         pi/(2n)

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/   


Date: 02/14/2001 at 13:44:22
From: Khaldoon Barakat
Subject: Re: Riemann sum

Could you explain with a little more detail? For example, where did 
the imaginary part come from, and how did you get rid of the sine in 
the denominator at the end of the proof?

Thanks.


Date: 02/14/2001 at 18:46:30
From: Doctor Anthony
Subject: Re: Riemann sum

We have e^(ix) = cos(x) + i.sin(x), and so if you have a series of 
sine terms you can replace it with the imaginary part of terms in 
e^(ix)

At the last stage of the calculation the denominator was  

     sin(pi/(2n))

Now if n -> infinity, then pi/(2n) -> 0, and we use the fact that as 
x -> 0, sin(x) -> x. So I replaced sin(pi/(2n)) with pi/(2n).

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Calculus

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