Results 1 to 5 of 5

Math Help - change of variables

  1. #1
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3

    change of variables

    The problem is to show that  \int^{x+ct+L}_{x-ct-L} g(s) \ ds = - \int^{x+ct}_{x-ct} g(s) \ ds

    where g is an odd 2L periodic function and g(L-x)=g(x)


    I don't understand the following change of variables:  \int^{x+ct+L}_{x-ct-L} g(s) \ ds = \int^{x+ct}_{x-ct} g(s+L) \ ds

    But if if the above is true, then it's easy to see that \int^{x+ct}_{x-ct} g(s+L) \ ds = -\int^{x+ct}_{x-ct} g(-s-L) \ ds = - \int^{x+ct}_{x-ct} g(-s+L) \ ds  = -\int^{x+ct}_{x-ct} g(s) \ ds
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by Random Variable View Post
    The problem is to show that  \int^{x+ct+L}_{x-ct-L} g(s) \ ds = - \int^{x+ct}_{x-ct} g(s) \ ds

    where g is an odd 2L periodic function and g(L-x)=g(x)


    I don't understand the following change of variables:  \int^{x+ct+L}_{x-ct-L} g(s) \ ds = \int^{x+ct}_{x-ct} g(s+L) \ ds

    But if if the above is true, then it's easy to see that \int^{x+ct}_{x-ct} g(s+L) \ ds = -\int^{x+ct}_{x-ct} g(-s-L) \ ds = - \int^{x+ct}_{x-ct} g(-s+L) \ ds  = -\int^{x+ct}_{x-ct} g(s) \ ds

    Let u=s-L \implies du=ds using this we get

     <br />
\int^{x+ct+L}_{x-ct-L} g(s) \ ds =\int g(u+L)<br />

    To figure out the new limits of integration plug them into the above eqation for s. So we get

    u=\underbrace{(x+ct+L)}_{s}-L=x+ct

    Now just do the same thing for the bottom limit.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
    But then the bottom limit would be  x-ct-2L .

    What am I not understanding?


    Wait. If g is 2L periodic, then then an antiderivative of g is also 2L periodic.
    Last edited by Random Variable; February 20th 2010 at 07:58 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by Random Variable View Post
    But then the bottom limit would be  x-ct-2L .

    What am I not understanding?
    The function is 2L periodic so you can can add any multiple of 2L as many times as you want

    It is exactly like adding 2\pi to this integral

    \int_{0}^{2\pi}sin(x)dx=\int_{2\pi}^{4\pi}\sin(x)d  x

    I hope this helps
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
    Wow. That was a brain fart of epic proportions.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Change of variables
    Posted in the Pre-Calculus Forum
    Replies: 0
    Last Post: January 27th 2011, 02:10 AM
  2. Help with Change of Variables
    Posted in the Calculus Forum
    Replies: 5
    Last Post: January 19th 2011, 06:09 PM
  3. Change of Variables
    Posted in the Calculus Forum
    Replies: 0
    Last Post: October 20th 2010, 04:13 AM
  4. Change of variables
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 14th 2010, 05:18 AM
  5. PDE; change of variables
    Posted in the Calculus Forum
    Replies: 4
    Last Post: October 5th 2008, 06:27 AM

Search Tags


/mathhelpforum @mathhelpforum