Results 1 to 5 of 5

Thread: integral inner product

  1. #1
    Member
    Joined
    Dec 2009
    Posts
    225

    integral inner product



    Here's the solution provided:


    I don't understand how they got that answer! This is my own attempt:

    Since $\displaystyle f$ is a piece-wise function I divided the interval to two and calculated the inner product like this:

    $\displaystyle \left\langle f, sin(x) \right\rangle = \int^{0}_{-\pi} -sin(x)+\int^{\pi}_{0} sin(x)=$$\displaystyle cos(0)-cos(\pi)-(-cos(\pi)-cos(0))=-2$

    So, why is my method and my answer wrong? And could anyone please explain the method used in the model answers?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    10
    $\displaystyle -\int_{-\pi}^{0}sin(x)dx+\int_{0}^{\pi}sin(x)dx=cos(x)]_{-\pi}^{0}-cos(x)]_{0}^{\pi}$
    $\displaystyle =cos(0)-cos(-\pi)-cos(\pi)+cos(0)=1-(-1)-(-1)+1=1+1+1+1=4$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2010
    Posts
    36
    The method used in the mode proof is this: If $\displaystyle a \geq 0$ and $\displaystyle g : [-a,a] \to \mathbb{R}$ is integrable then:
    $\displaystyle \int_{-a}^a g(x) \mathrm{d} \! x = \begin{cases} 2 \int_0^a g(x) \mathrm{d} \! x & \text{if } g \text{ is even.} \\ 0 & \text{if } g \text{ is odd.} \end{cases}$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Dec 2009
    Posts
    225
    Quote Originally Posted by Giraffro View Post
    The method used in the mode proof is this: If $\displaystyle a \geq 0$ and $\displaystyle g : [-a,a] \to \mathbb{R}$ is integrable then:
    $\displaystyle \int_{-a}^a g(x) \mathrm{d} \! x = \begin{cases} 2 \int_0^a g(x) \mathrm{d} \! x & \text{if } g \text{ is even.} \\ 0 & \text{if } g \text{ is odd.} \end{cases}$
    Thanks. But how did they figure out that the given function $\displaystyle f(x)$ is odd? Could you please explain that?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Mar 2010
    Posts
    36
    Quote Originally Posted by demode View Post
    Thanks. But how did they figure out that the given function $\displaystyle f(x)$ is odd? Could you please explain that?
    Ah, I seem to have glanced over something. Well $\displaystyle \forall x \in (0,\pi], f(x) = -f(-x)$ and integrating over a point doesn't contribute to any integrals, so you could swap $\displaystyle f$ for:
    $\displaystyle f(x) := \begin{cases} -1 & \text{if } -\pi \leq x < 0 \\ 0 & \text{if } x = 0 \\ 1 & \text{if } 0 < x \leq \pi \end{cases}$

    Then $\displaystyle f$ would be odd and so $\displaystyle f(x) \sin(x)$ is odd, since $\displaystyle \sin(x)$ is even and the product of 2 odd functions is even. But explicit calculation is just as fine.
    Last edited by Giraffro; Apr 17th 2010 at 12:22 AM. Reason: Mistake
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Integral of product legendre functions
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: Mar 17th 2010, 02:29 AM
  2. Wierd product contour integral
    Posted in the Calculus Forum
    Replies: 0
    Last Post: Apr 5th 2009, 10:06 PM
  3. Replies: 13
    Last Post: Jan 5th 2009, 12:15 PM
  4. Multiple integral of a product
    Posted in the Calculus Forum
    Replies: 0
    Last Post: Feb 1st 2007, 07:42 AM
  5. Integral of a product
    Posted in the Calculus Forum
    Replies: 5
    Last Post: Jan 23rd 2007, 02:56 AM

Search Tags


/mathhelpforum @mathhelpforum