Page 1 of 2 12 LastLast
Results 1 to 15 of 22

Math Help - Exponential Fourier Series - Trouble solving

  1. #1
    Junior Member
    Joined
    Mar 2008
    From
    Sweden
    Posts
    61

    Exponential Fourier Series - Trouble solving

    Hello,

    I have troubling simplifying the following problem. It is the second integral solving for c(k) of f involving [i] that is confusing.

    I also was wondering if there is simpler way to solve for the exponetial fourier series for the given problem, than the way I done it. I would greatly appriciate guidance.

    Thank you
    Attached Thumbnails Attached Thumbnails Exponential Fourier Series - Trouble solving-untitled.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Part of your problem is that for the c_{k} integral, you can't use half the interval and multiply by two like you could for the c_{0} case. You can only do that, in general, when your complete integrand is even. But the presence of that particular exponential function in the integrand precludes that. So you have to keep the full [-\pi,\pi] interval for the limits in the c_{k} integration, I'm afraid.

    I don't know of any way to integrate x^{2}e^{-ikx} other than by parts twice. I'm not sure I would recommend breaking the integrals up into the trig functions, because then you're going to have to do integration by parts twice on two integrals. There's no need, when working with the exponential Fourier series, to look at sin and cos individually. It's an easier integration with just the exponential in there.

    So I would carry these changes through, and then we'll see what happens at the end. Sound good?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2008
    From
    Sweden
    Posts
    61
    Quote Originally Posted by Ackbeet View Post
    Part of your problem is that for the c_{k} integral, you can't use half the interval and multiply by two like you could for the c_{0} case. You can only do that, in general, when your complete integrand is even. But the presence of that particular exponential function in the integrand precludes that. So you have to keep the full [-\pi,\pi] interval for the limits in the c_{k} integration, I'm afraid.

    I don't know of any way to integrate x^{2}e^{-ikx} other than by parts twice. I'm not sure I would recommend breaking the integrals up into the trig functions, because then you're going to have to do integration by parts twice on two integrals. There's no need, when working with the exponential Fourier series, to look at sin and cos individually. It's an easier integration with just the exponential in there.

    So I would carry these changes through, and then we'll see what happens at the end. Sound good?
    I got the following expression after integrating the integral [-\pi,\pi]:

    c_{k} = \frac{1}{2\pi} ({-\frac{\pi^{2}}{ik}}e^{-ik\pi}}+{\frac{2\pi}{k^{2}}e^{-ik\pi}+{\frac{2}{ik^{3}}e^{-ik\pi}+\frac{\pi^{2}}{ik}}e^{ik\pi}+{\frac{2\pi}{k  ^{2}}e^{ik\pi}-{\frac{2}{ik^{3}}e^{ik\pi})

    and I know that

    e^{ik\pi}=(-1)^{k}

    \frac{1}{2\pi}(\frac{\pi^{2}}{ik}}e^{ik\pi}{-\frac{\pi^{2}}{ik}}e^{-ik\pi}})=0

    \frac{1}{2\pi}({\frac{2}{ik^{3}}e^{-ik\pi}-{\frac{2}{ik^{3}}e^{ik\pi})=0

    \frac{1}{2\pi}({\frac{2\pi}{k^{2}}e^{ik\pi}+{\frac  {2\pi}{k^{2}}e^{-ik\pi})={\frac{1}{k^{2}}e^{ik\pi}+{\frac{1}{k^{2}}  e^{-ik\pi}

    c_{k} suppose to be {\frac{{2}(-1^{k})}{k^{2}}

    How do I get:

    ({\frac{1}{k^{2}}e^{ik\pi}+{\frac{1}{k^{2}}e^{-ik\pi})

    to be equal {\frac{{2}(-1^{k})}{k^{2}}

    I would appriciate a response. Thank you
    Follow Math Help Forum on Facebook and Google+

  4. #4
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Your k exponent should be outside the parentheses (otherwise it only applies to the 1). Otherwise, it looks fine.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Mar 2008
    From
    Sweden
    Posts
    61
    Quote Originally Posted by Ackbeet View Post
    Your k exponent should be outside the parentheses (otherwise it only applies to the 1). Otherwise, it looks fine.
    Where does the 2 in {\frac{{2}(-1)^{k}}{k^{2}} comes from. How do I get that. I just got it from the answer in the book but don't know where it comes from.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    ({\frac{1}{k^{2}}e^{ik\pi}+{\frac{1}{k^{2}}e^{-ik\pi})=\frac{1}{k^{2}}(e^{ikx}+e^{-ikx})=\frac{1}{k^{2}}(\cos(k\pi)+i\sin(k\pi)+\cos(  k\pi)-i\sin(k\pi))
    =\frac{2}{k^{2}}\,\cos(k\pi).

    But now \cos(k\pi)=(-1)^{k} for all k\in\mathbb{Z}.

    The answer follows.

    Does that help? It's basically all from the Euler formula, and the even-ness of cosine and the odd-ness of sine:

    e^{i\theta}=\cos(\theta)+i\sin(\theta).
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Mar 2008
    From
    Sweden
    Posts
    61
    I have a question regarding solving for the exponetial Fourier series for a given function. The general formula is the following (the way I understood):

    \displaystyle f(x)=\displaystyle \sum_{k=-\infty}^{\infty}c_{k}e^{ik\Omega x}

    there \Omega T={2\pi}

    Now, to solve for the exponetial Fourier series for the given problem above :

     \displaystyle f(x) = {x^{2}} ,\,\,-\pi < x \leq \pi

    Why do we solve for c_{0} ? Its not in the general fomula.
    --------------------------------------------------------------------------
    Where as I have a different problem where I have to solve for the exponetial Fourier series for a {2\pi} - periodic function for the following given function:

    \displaystyle f(x)=e^{-x}},\,\,0 < x < {2\pi}

    In this case we just solve for c_{k} only and not for c_{0}. Why is that?

    I do appriciate for a explanation. Thank you
    Follow Math Help Forum on Facebook and Google+

  8. #8
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Why do we solve for c_{0}? It's not in the general formula.
    I don't understand. If you have the sum

    \displaystyle{f(x)=\sum_{k=-\infty}^{\infty}c_{k}e^{ik\Omega x}},

    then the k=0 term has the expression c_{0} in it. Does that qualify as "being in the general formula"?

    In the problem you have there, I would think you would solve for c_{0}. Both functions you're dealing with have a nonzero average value. As I recall, the expression for the c_{0} term looks a lot like the average value of the function over its period. *looking it up* In fact, it is the average value of the function over a period. Now if you have a function like f(x)=x over the interval (-\pi,\pi), and you can tell at a glance that its average value is zero, well then, you have no need to explicitly compute it. But such examples should not be considered to be the usual.

    Make sense?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Mar 2008
    From
    Sweden
    Posts
    61
    Alright, it makes sense. Thanks.

    How about for this function:

    \displaystyle f(x)=e^{-x}},\,\,0 < x < {2\pi}

    Do I need to compute c_{0} ? If not, why?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Well, do you think the average value of the function over that interval is nonzero?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Junior Member
    Joined
    Mar 2008
    From
    Sweden
    Posts
    61
    Quote Originally Posted by Ackbeet View Post
    Well, do you think the average value of the function over that interval is nonzero?
    I guess I am confused after all and don't understand.

    Do you mean that:

    \displaystyle{f(x)=\sum_{k=-\infty}^{\infty}c_{k}e^{ik\Omega x}}\iff {f(x)=c_{0} + \sum_ {k\neq 0}c_{k}e^{ik\Omega x}}
    Follow Math Help Forum on Facebook and Google+

  12. #12
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Your double implication is correct, if the understood range of k values is the integers. All you're doing there is splitting off one term from the sum.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Junior Member
    Joined
    Mar 2008
    From
    Sweden
    Posts
    61
    Quote Originally Posted by Ackbeet View Post
    Now if you have a function like f(x)=x over the interval (-\pi,\pi), and you can tell at a glance that its average value is zero, well then, you have no need to explicitly compute it. But such examples should not be considered to be the usual.
    If I compute c_{0} for this function:

    f(x)=x over the interval (-\pi,\pi)

    I would get the following:

    \displaystyle c_{0} =\frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) dx =\frac{2}{2\pi}\int_{0}^{\pi}x dx =\frac{\pi}{2} the value is {\neq 0}
    Follow Math Help Forum on Facebook and Google+

  14. #14
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    No, the formula you used is valid for an even integrand. You have an odd integrand.
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Junior Member
    Joined
    Mar 2008
    From
    Sweden
    Posts
    61
    Quote Originally Posted by Ackbeet View Post
    No, the formula you used is valid for an even integrand. You have an odd integrand.
    Which formula do I use for an odd integrand? By the way thanks for your response.
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: August 13th 2011, 02:29 AM
  2. Fourier series - complex exponential form
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: September 10th 2010, 07:58 AM
  3. Complex Fourier Series & Full Fourier Series
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 9th 2009, 05:39 AM
  4. Exponential Fourier-series expansion
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 23rd 2009, 06:48 PM
  5. Replies: 0
    Last Post: October 12th 2008, 10:09 PM

/mathhelpforum @mathhelpforum