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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/