Facebook Page
Twitter
RSS
+ Reply to Thread
Results 1 to 3 of 3
  1. MHB Craftsman

    Status
    Offline
    Join Date
    Feb 2012
    Posts
    149
    Thanks
    43 times
    Thanked
    12 times
    #1
    Let $f(x)=-x$ for $-l\le x\le l$ and $f(l)=l.$

    a) Study the pointwise convergence of the Fourier series for $f.$
    b) Compute the series $\displaystyle\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)}.$
    c) Does the Fourier series of $f$ converge uniformly on $\mathbb R$ ?

    -------------

    First I need to compute the Fourier series, so since $f$ is odd, then the Fourier series is just $\displaystyle\sum_{n=1}^\infty b_n\sin\frac{n\pi x}l$ where $b_n=\dfrac 2l\displaystyle\int_0^{l} f(x)\sin\frac{n\pi x}l\,dx$ so I'm getting $\displaystyle-\frac{x}{{{l}^{2}}}=\frac{1}{\pi }\sum\limits_{n=1}^{\infty }{\frac{{{(-1)}^{n}}}{n}\sin \frac{n\pi x}{l}},$ but now I don't know how to proceed with the pointwise convergence, also, how to do part b)?

    Thanks for the help!

  2. # ADS
    Circuit advertisement
    Join Date
    Always
    Posts
    Many
     

  3. MHB Master
    MHB Site Helper
    MHB Math Helper
    MHB Ambassador
    Sudharaka's Avatar
    Status
    Offline
    Join Date
    Feb 2012
    Location
    Canada
    Posts
    1,620
    Thanks
    7,514 times
    Thanked
    2,543 times
    Thank/Post
    1.570
    Awards
    University POTW Award (Jan-June 2013)
    #2
    Quote Originally Posted by Markov View Post
    Let $f(x)=-x$ for $-l\le x\le l$ and $f(l)=l.$

    a) Study the pointwise convergence of the Fourier series for $f.$
    b) Compute the series $\displaystyle\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)}.$
    c) Does the Fourier series of $f$ converge uniformly on $\mathbb R$ ?

    -------------

    First I need to compute the Fourier series, so since $f$ is odd, then the Fourier series is just $\displaystyle\sum_{n=1}^\infty b_n\sin\frac{n\pi x}l$ where $b_n=\dfrac 2l\displaystyle\int_0^{l} f(x)\sin\frac{n\pi x}l\,dx$ so I'm getting $\displaystyle-\frac{x}{{{l}^{2}}}=\frac{1}{\pi }\sum\limits_{n=1}^{\infty }{\frac{{{(-1)}^{n}}}{n}\sin \frac{n\pi x}{l}},$ but now I don't know how to proceed with the pointwise convergence, also, how to do part b)?

    Thanks for the help!
    Hi Markov,

    Firstly the definition of the function \(f\) seem to be erroneous, both \(f(x)=-x\mbox{ for }-l\leq x\leq l\) and \(f(l)=l\) cannot be true. So I shall neglect the latter part: \(f(l)=l\).

    I think the Fourier series that you have obtained is also incorrect. It should be,

    \[-x=\frac{2l}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}\sin\left(\frac{n\pi x}{l}\right)\]

    Since both \(f(x)=-x\) and \(f'(x)=-1\) are continuous on \([-l,\,l]\) the Fourier series converges point-wise on the interval \((-l,\,l)\) (Refer Theorem 5.5 ).

    Substitute \(x=1\mbox{ and }l=2\) and we obtain,

    \begin{eqnarray}

    -\frac{\pi}{4}&=&\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}\sin\left(\frac{n\pi}{2} \right)\\

    &=&\sum_{n=0}^{\infty}\frac{(-1)^{2n+1}}{2n+1}\sin\left(\frac{(2n+1)\pi}{2} \right)\\

    &=&\sum_{n=0}^{\infty}\frac{(-1)^{2n+1}}{2n+1}(-1)^n\\

    &=&\sum_{n=0}^{\infty}\frac{(-1)^{3n+1}}{2n+1}\\

    \therefore\sum_{n=0}^{\infty}\frac{(-1)^{n}}{2n+1}&=&\frac{\pi}{4}

    \end{eqnarray}

    When \(x=l\) we have,

    \[\left|f(l)-\sum_{n=1}^{N}\frac{(-1)^{n}}{n}\sin\left(\frac{n\pi l}{l}\right)\right|=l\]

    Therefore by the definition of uniform convergence (Refer ) it is clear that the Fourier series of \(f\) is not uniformly convergent on \(\Re\).

    Kind Regards,
    Sudharaka.

  4. MHB Journeyman

    Status
    Offline
    Join Date
    Jan 2012
    Posts
    890
    Thanks
    344 times
    Thanked
    1,039 time
    Thank/Post
    1.167
    #3
    Quote Originally Posted by Sudharaka View Post
    Hi Markov,

    Firstly the definition of the function \(f\) seem to be erroneous, both \(f(x)=-x\mbox{ for }-l\leq x\leq l\) and \(f(l)=l\) cannot be true. So I shall neglect the latter part: \(f(l)=l\).
    This is very likely to be intended to indicate the periodic extension: \(f(x)=f(x+2l)\) (or there is a mistake with the value at either \(+l\) or \(-l\), both end points would not normally be included in the domain for a Fourier Series).

    Without the periodic extension the question of uniform convergence is moot, since the Fourier Series is periodic and so does not converge to the function outside of the interval.

    CB

Similar Threads

  1. Fourier series proof
    By Poirot in forum Differential Equations
    Replies: 2
    Last Post: May 2nd, 2012, 02:38
  2. Replies: 5
    Last Post: May 1st, 2012, 18:34
  3. Replies: 9
    Last Post: March 5th, 2012, 15:11
  4. More Fourier series
    By Markov in forum Calculus
    Replies: 3
    Last Post: February 15th, 2012, 16:13
  5. Finding Fourier Series
    By Markov in forum Calculus
    Replies: 4
    Last Post: February 15th, 2012, 15:16

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •  
Math Help Boards