Fourier series with complex coefficients












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Hello i need to find the complex coffecients of the fouriers series of the $2pi$ perriodic function:
$$f(x)= frac{pi-x}{2}, forall xin]-pi;pi]$$
I stock in the computation of complex coefficients $$c_n=frac{1}{2pi}int_{-pi}^{pi}{f(x)e^{-inx}}dx$$. Precisely in what doing with $e^{inpi}$ in the result.






real-analysis analysis fourier-analysis fourier-series







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  • $begingroup$
    What's the domain of the function?
    $endgroup$
    – Botond
    Dec 3 '18 at 17:13










  • $begingroup$
    Why don't you compute the coefficients by evaluating the integrals?
    $endgroup$
    – copper.hat
    Dec 3 '18 at 17:21










  • $begingroup$
    where are you getting stuck?
    $endgroup$
    – Doug M
    Dec 3 '18 at 17:43










  • $begingroup$
    Since $n$ is integer, $e^{i npi} = (e^{ipi})^n = (-1)^n$. Was that your problem?
    $endgroup$
    – Paul Sinclair
    Dec 3 '18 at 22:55










  • $begingroup$
    @Botond $mathbb{R}$.
    $endgroup$
    – Felix Marin
    Dec 15 '18 at 3:11
















0












$begingroup$


Hello i need to find the complex coffecients of the fouriers series of the $2pi$ perriodic function:
$$f(x)= frac{pi-x}{2}, forall xin]-pi;pi]$$
I stock in the computation of complex coefficients $$c_n=frac{1}{2pi}int_{-pi}^{pi}{f(x)e^{-inx}}dx$$. Precisely in what doing with $e^{inpi}$ in the result.






real-analysis analysis fourier-analysis fourier-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    What's the domain of the function?
    $endgroup$
    – Botond
    Dec 3 '18 at 17:13










  • $begingroup$
    Why don't you compute the coefficients by evaluating the integrals?
    $endgroup$
    – copper.hat
    Dec 3 '18 at 17:21










  • $begingroup$
    where are you getting stuck?
    $endgroup$
    – Doug M
    Dec 3 '18 at 17:43










  • $begingroup$
    Since $n$ is integer, $e^{i npi} = (e^{ipi})^n = (-1)^n$. Was that your problem?
    $endgroup$
    – Paul Sinclair
    Dec 3 '18 at 22:55










  • $begingroup$
    @Botond $mathbb{R}$.
    $endgroup$
    – Felix Marin
    Dec 15 '18 at 3:11














0












0








0


1



$begingroup$


Hello i need to find the complex coffecients of the fouriers series of the $2pi$ perriodic function:
$$f(x)= frac{pi-x}{2}, forall xin]-pi;pi]$$
I stock in the computation of complex coefficients $$c_n=frac{1}{2pi}int_{-pi}^{pi}{f(x)e^{-inx}}dx$$. Precisely in what doing with $e^{inpi}$ in the result.






real-analysis analysis fourier-analysis fourier-series







share|cite|improve this question











$endgroup$




Hello i need to find the complex coffecients of the fouriers series of the $2pi$ perriodic function:
$$f(x)= frac{pi-x}{2}, forall xin]-pi;pi]$$
I stock in the computation of complex coefficients $$c_n=frac{1}{2pi}int_{-pi}^{pi}{f(x)e^{-inx}}dx$$. Precisely in what doing with $e^{inpi}$ in the result.






real-analysis analysis fourier-analysis fourier-series




real-analysis analysis fourier-analysis fourier-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 3 '18 at 17:49







Rabih Assaf

















asked Dec 3 '18 at 17:05









Rabih AssafRabih Assaf

474




474












  • $begingroup$
    What's the domain of the function?
    $endgroup$
    – Botond
    Dec 3 '18 at 17:13










  • $begingroup$
    Why don't you compute the coefficients by evaluating the integrals?
    $endgroup$
    – copper.hat
    Dec 3 '18 at 17:21










  • $begingroup$
    where are you getting stuck?
    $endgroup$
    – Doug M
    Dec 3 '18 at 17:43










  • $begingroup$
    Since $n$ is integer, $e^{i npi} = (e^{ipi})^n = (-1)^n$. Was that your problem?
    $endgroup$
    – Paul Sinclair
    Dec 3 '18 at 22:55










  • $begingroup$
    @Botond $mathbb{R}$.
    $endgroup$
    – Felix Marin
    Dec 15 '18 at 3:11


















  • $begingroup$
    What's the domain of the function?
    $endgroup$
    – Botond
    Dec 3 '18 at 17:13










  • $begingroup$
    Why don't you compute the coefficients by evaluating the integrals?
    $endgroup$
    – copper.hat
    Dec 3 '18 at 17:21










  • $begingroup$
    where are you getting stuck?
    $endgroup$
    – Doug M
    Dec 3 '18 at 17:43










  • $begingroup$
    Since $n$ is integer, $e^{i npi} = (e^{ipi})^n = (-1)^n$. Was that your problem?
    $endgroup$
    – Paul Sinclair
    Dec 3 '18 at 22:55










  • $begingroup$
    @Botond $mathbb{R}$.
    $endgroup$
    – Felix Marin
    Dec 15 '18 at 3:11
















$begingroup$
What's the domain of the function?
$endgroup$
– Botond
Dec 3 '18 at 17:13




$begingroup$
What's the domain of the function?
$endgroup$
– Botond
Dec 3 '18 at 17:13












$begingroup$
Why don't you compute the coefficients by evaluating the integrals?
$endgroup$
– copper.hat
Dec 3 '18 at 17:21




$begingroup$
Why don't you compute the coefficients by evaluating the integrals?
$endgroup$
– copper.hat
Dec 3 '18 at 17:21












$begingroup$
where are you getting stuck?
$endgroup$
– Doug M
Dec 3 '18 at 17:43




$begingroup$
where are you getting stuck?
$endgroup$
– Doug M
Dec 3 '18 at 17:43












$begingroup$
Since $n$ is integer, $e^{i npi} = (e^{ipi})^n = (-1)^n$. Was that your problem?
$endgroup$
– Paul Sinclair
Dec 3 '18 at 22:55




$begingroup$
Since $n$ is integer, $e^{i npi} = (e^{ipi})^n = (-1)^n$. Was that your problem?
$endgroup$
– Paul Sinclair
Dec 3 '18 at 22:55












$begingroup$
@Botond $mathbb{R}$.
$endgroup$
– Felix Marin
Dec 15 '18 at 3:11




$begingroup$
@Botond $mathbb{R}$.
$endgroup$
– Felix Marin
Dec 15 '18 at 3:11










1 Answer
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The integral with $frac{pi}{2}$ vanishes as you get $frac{sin(n pi)}{pi n}$ with $n$ being an integer as result. For $frac{-1}{2 pi}int_{-pi}^{pi}xe^{-inx}dx$, I get $-frac{i(-1)^n}{n}$ as result using integration by parts. For $c_0$, I get $frac{pi}{2}$ as result. For $c_0$, it's the $-x$ which doesn't contribute to the result as you get $-pi - (-pi)=0$ for the variable $x$. So $c_0$ is $frac{pi}{2}$ and $c_n$ is $frac{-i(-1)^n}{n}$ if I made no mistake. Maybe you were troubled to get $0$ for the $frac{pi}{2}$ of the $c_n$ ?



Hope this helps. If I made any mistake, please say it in the comments as soon as possible.






share|cite|improve this answer











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    $begingroup$

    The integral with $frac{pi}{2}$ vanishes as you get $frac{sin(n pi)}{pi n}$ with $n$ being an integer as result. For $frac{-1}{2 pi}int_{-pi}^{pi}xe^{-inx}dx$, I get $-frac{i(-1)^n}{n}$ as result using integration by parts. For $c_0$, I get $frac{pi}{2}$ as result. For $c_0$, it's the $-x$ which doesn't contribute to the result as you get $-pi - (-pi)=0$ for the variable $x$. So $c_0$ is $frac{pi}{2}$ and $c_n$ is $frac{-i(-1)^n}{n}$ if I made no mistake. Maybe you were troubled to get $0$ for the $frac{pi}{2}$ of the $c_n$ ?



    Hope this helps. If I made any mistake, please say it in the comments as soon as possible.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      The integral with $frac{pi}{2}$ vanishes as you get $frac{sin(n pi)}{pi n}$ with $n$ being an integer as result. For $frac{-1}{2 pi}int_{-pi}^{pi}xe^{-inx}dx$, I get $-frac{i(-1)^n}{n}$ as result using integration by parts. For $c_0$, I get $frac{pi}{2}$ as result. For $c_0$, it's the $-x$ which doesn't contribute to the result as you get $-pi - (-pi)=0$ for the variable $x$. So $c_0$ is $frac{pi}{2}$ and $c_n$ is $frac{-i(-1)^n}{n}$ if I made no mistake. Maybe you were troubled to get $0$ for the $frac{pi}{2}$ of the $c_n$ ?



      Hope this helps. If I made any mistake, please say it in the comments as soon as possible.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        The integral with $frac{pi}{2}$ vanishes as you get $frac{sin(n pi)}{pi n}$ with $n$ being an integer as result. For $frac{-1}{2 pi}int_{-pi}^{pi}xe^{-inx}dx$, I get $-frac{i(-1)^n}{n}$ as result using integration by parts. For $c_0$, I get $frac{pi}{2}$ as result. For $c_0$, it's the $-x$ which doesn't contribute to the result as you get $-pi - (-pi)=0$ for the variable $x$. So $c_0$ is $frac{pi}{2}$ and $c_n$ is $frac{-i(-1)^n}{n}$ if I made no mistake. Maybe you were troubled to get $0$ for the $frac{pi}{2}$ of the $c_n$ ?



        Hope this helps. If I made any mistake, please say it in the comments as soon as possible.






        share|cite|improve this answer











        $endgroup$



        The integral with $frac{pi}{2}$ vanishes as you get $frac{sin(n pi)}{pi n}$ with $n$ being an integer as result. For $frac{-1}{2 pi}int_{-pi}^{pi}xe^{-inx}dx$, I get $-frac{i(-1)^n}{n}$ as result using integration by parts. For $c_0$, I get $frac{pi}{2}$ as result. For $c_0$, it's the $-x$ which doesn't contribute to the result as you get $-pi - (-pi)=0$ for the variable $x$. So $c_0$ is $frac{pi}{2}$ and $c_n$ is $frac{-i(-1)^n}{n}$ if I made no mistake. Maybe you were troubled to get $0$ for the $frac{pi}{2}$ of the $c_n$ ?



        Hope this helps. If I made any mistake, please say it in the comments as soon as possible.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 14 '18 at 23:35

























        answered Dec 14 '18 at 23:22









        PoujhPoujh

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        590516






























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