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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add a comment |
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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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What's the domain of the function?
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– Botond
Dec 3 '18 at 17:13
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Why don't you compute the coefficients by evaluating the integrals?
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– copper.hat
Dec 3 '18 at 17:21
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where are you getting stuck?
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– Doug M
Dec 3 '18 at 17:43
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Since $n$ is integer, $e^{i npi} = (e^{ipi})^n = (-1)^n$. Was that your problem?
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– Paul Sinclair
Dec 3 '18 at 22:55
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@Botond $mathbb{R}$.
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– Felix Marin
Dec 15 '18 at 3:11
add a comment |
$begingroup$
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
$endgroup$
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
edited Dec 3 '18 at 17:49
Rabih Assaf
asked Dec 3 '18 at 17:05
Rabih AssafRabih Assaf
474
474
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What's the domain of the function?
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– 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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
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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.
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
edited Dec 14 '18 at 23:35
answered Dec 14 '18 at 23:22
PoujhPoujh
590516
590516
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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