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Fourier series, pointwise convergence, series computation

Markov

Member
Feb 1, 2012
149
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!
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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 here).

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 this) it is clear that the Fourier series of \(f\) is not uniformly convergent on \(\Re\).

Kind Regards,
Sudharaka.
 

CaptainBlack

Well-known member
Jan 26, 2012
890
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