Finding the Fourier series of a piecewise function












0












$begingroup$


I'm s little confused about Fourier series of functions that are piecewise. Here’s an example of such a function:
$$f(x) = begin{cases}
x & -fracpi2 < x < fracpi2 \[5pt]
pi - x & fracpi2 < x < frac{3pi}2
end{cases}$$
Please can you explain this example to me?






fourier-analysis fourier-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    There is a standard formula to compute coefficients $a_0, a_n, b_n$. Your question is how these coefficients are derived or how they get computed?
    $endgroup$
    – thanasissdr
    Sep 28 '14 at 18:25












  • $begingroup$
    yes,in other word in don't know the interval of integral for obtaining coefficients.
    $endgroup$
    – Panda
    Sep 28 '14 at 19:43






  • 2




    $begingroup$
    Remember that you're not computing coefficients for two different functions - you're computing the coefficients of one function, except you will have two integrals when computing the Fourier coefficients due to the function being piecewise across the period.
    $endgroup$
    – Eweler
    Sep 28 '14 at 20:59
















0












$begingroup$


I'm s little confused about Fourier series of functions that are piecewise. Here’s an example of such a function:
$$f(x) = begin{cases}
x & -fracpi2 < x < fracpi2 \[5pt]
pi - x & fracpi2 < x < frac{3pi}2
end{cases}$$
Please can you explain this example to me?






fourier-analysis fourier-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    There is a standard formula to compute coefficients $a_0, a_n, b_n$. Your question is how these coefficients are derived or how they get computed?
    $endgroup$
    – thanasissdr
    Sep 28 '14 at 18:25












  • $begingroup$
    yes,in other word in don't know the interval of integral for obtaining coefficients.
    $endgroup$
    – Panda
    Sep 28 '14 at 19:43






  • 2




    $begingroup$
    Remember that you're not computing coefficients for two different functions - you're computing the coefficients of one function, except you will have two integrals when computing the Fourier coefficients due to the function being piecewise across the period.
    $endgroup$
    – Eweler
    Sep 28 '14 at 20:59














0












0








0


2



$begingroup$


I'm s little confused about Fourier series of functions that are piecewise. Here’s an example of such a function:
$$f(x) = begin{cases}
x & -fracpi2 < x < fracpi2 \[5pt]
pi - x & fracpi2 < x < frac{3pi}2
end{cases}$$
Please can you explain this example to me?






fourier-analysis fourier-series







share|cite|improve this question











$endgroup$




I'm s little confused about Fourier series of functions that are piecewise. Here’s an example of such a function:
$$f(x) = begin{cases}
x & -fracpi2 < x < fracpi2 \[5pt]
pi - x & fracpi2 < x < frac{3pi}2
end{cases}$$
Please can you explain this example to me?






fourier-analysis fourier-series




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 Sep 28 '14 at 21:06









alexwlchan

1,9351118




1,9351118










asked Sep 28 '14 at 17:50









PandaPanda

4141618




4141618












  • $begingroup$
    There is a standard formula to compute coefficients $a_0, a_n, b_n$. Your question is how these coefficients are derived or how they get computed?
    $endgroup$
    – thanasissdr
    Sep 28 '14 at 18:25












  • $begingroup$
    yes,in other word in don't know the interval of integral for obtaining coefficients.
    $endgroup$
    – Panda
    Sep 28 '14 at 19:43






  • 2




    $begingroup$
    Remember that you're not computing coefficients for two different functions - you're computing the coefficients of one function, except you will have two integrals when computing the Fourier coefficients due to the function being piecewise across the period.
    $endgroup$
    – Eweler
    Sep 28 '14 at 20:59


















  • $begingroup$
    There is a standard formula to compute coefficients $a_0, a_n, b_n$. Your question is how these coefficients are derived or how they get computed?
    $endgroup$
    – thanasissdr
    Sep 28 '14 at 18:25












  • $begingroup$
    yes,in other word in don't know the interval of integral for obtaining coefficients.
    $endgroup$
    – Panda
    Sep 28 '14 at 19:43






  • 2




    $begingroup$
    Remember that you're not computing coefficients for two different functions - you're computing the coefficients of one function, except you will have two integrals when computing the Fourier coefficients due to the function being piecewise across the period.
    $endgroup$
    – Eweler
    Sep 28 '14 at 20:59
















$begingroup$
There is a standard formula to compute coefficients $a_0, a_n, b_n$. Your question is how these coefficients are derived or how they get computed?
$endgroup$
– thanasissdr
Sep 28 '14 at 18:25






$begingroup$
There is a standard formula to compute coefficients $a_0, a_n, b_n$. Your question is how these coefficients are derived or how they get computed?
$endgroup$
– thanasissdr
Sep 28 '14 at 18:25














$begingroup$
yes,in other word in don't know the interval of integral for obtaining coefficients.
$endgroup$
– Panda
Sep 28 '14 at 19:43




$begingroup$
yes,in other word in don't know the interval of integral for obtaining coefficients.
$endgroup$
– Panda
Sep 28 '14 at 19:43




2




2




$begingroup$
Remember that you're not computing coefficients for two different functions - you're computing the coefficients of one function, except you will have two integrals when computing the Fourier coefficients due to the function being piecewise across the period.
$endgroup$
– Eweler
Sep 28 '14 at 20:59




$begingroup$
Remember that you're not computing coefficients for two different functions - you're computing the coefficients of one function, except you will have two integrals when computing the Fourier coefficients due to the function being piecewise across the period.
$endgroup$
– Eweler
Sep 28 '14 at 20:59










1 Answer
1






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oldest

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3












$begingroup$

Your function is defined on the interval $left(-dfrac pi 2, dfrac pi 2 right)cup left( dfrac pi 2 , dfrac {3pi} 2right) $. That means the length of the interval is $boxed{L= 2pi}$.



Now, how to compute the coefficients:



$a_0=displaystyle dfrac 1 L cdotint_{-frac pi 2}^ frac {3pi} 2 f(x) , dx=dfrac 1 Lcdot bigg(int_{-frac pi 2}^ frac {pi} 2 x , dx +int_{frac pi 2}^ frac {3pi} {2} (pi- x) , dxbigg)$



$a_n=displaystyle dfrac 2 L cdotint_{-frac pi 2}^ frac {3pi} 2 f(x)cdot cosleft(dfrac {2npi x} {L}right) , dx$



$b_n=displaystyle dfrac 2 L cdotint_{-frac pi 2}^ frac {3pi} 2 f(x)cdot sinleft(dfrac {2npi x} {L}right) , dx$



The Fourier series of $f$ is:



$$ displaystyle a_0+sum_{n=1}^infty Big[a_ncdot cosleft(dfrac {2npi x} {L}right)+b_n cdot sinleft(dfrac {2npi x} {L}right)Big]$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    but we know for obtaining coefficients we have to integrate function from [-T/2,T/2] and intervals are Symmetric but you didn't write that.I have been confused now.
    $endgroup$
    – Panda
    Sep 29 '14 at 14:31










  • $begingroup$
    I don't think this is necessary to be always true. In general, $a_n=displaystyleint_{x_0}^{x_0+L} f(x)cdot cosbigg(dfrac { 2 n pi x } {L}bigg), dx$ , where $L$ stands for the period of the periodic function.
    $endgroup$
    – thanasissdr
    Sep 29 '14 at 15:21














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1 Answer
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1 Answer
1






active

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active

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3












$begingroup$

Your function is defined on the interval $left(-dfrac pi 2, dfrac pi 2 right)cup left( dfrac pi 2 , dfrac {3pi} 2right) $. That means the length of the interval is $boxed{L= 2pi}$.



Now, how to compute the coefficients:



$a_0=displaystyle dfrac 1 L cdotint_{-frac pi 2}^ frac {3pi} 2 f(x) , dx=dfrac 1 Lcdot bigg(int_{-frac pi 2}^ frac {pi} 2 x , dx +int_{frac pi 2}^ frac {3pi} {2} (pi- x) , dxbigg)$



$a_n=displaystyle dfrac 2 L cdotint_{-frac pi 2}^ frac {3pi} 2 f(x)cdot cosleft(dfrac {2npi x} {L}right) , dx$



$b_n=displaystyle dfrac 2 L cdotint_{-frac pi 2}^ frac {3pi} 2 f(x)cdot sinleft(dfrac {2npi x} {L}right) , dx$



The Fourier series of $f$ is:



$$ displaystyle a_0+sum_{n=1}^infty Big[a_ncdot cosleft(dfrac {2npi x} {L}right)+b_n cdot sinleft(dfrac {2npi x} {L}right)Big]$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    but we know for obtaining coefficients we have to integrate function from [-T/2,T/2] and intervals are Symmetric but you didn't write that.I have been confused now.
    $endgroup$
    – Panda
    Sep 29 '14 at 14:31










  • $begingroup$
    I don't think this is necessary to be always true. In general, $a_n=displaystyleint_{x_0}^{x_0+L} f(x)cdot cosbigg(dfrac { 2 n pi x } {L}bigg), dx$ , where $L$ stands for the period of the periodic function.
    $endgroup$
    – thanasissdr
    Sep 29 '14 at 15:21


















3












$begingroup$

Your function is defined on the interval $left(-dfrac pi 2, dfrac pi 2 right)cup left( dfrac pi 2 , dfrac {3pi} 2right) $. That means the length of the interval is $boxed{L= 2pi}$.



Now, how to compute the coefficients:



$a_0=displaystyle dfrac 1 L cdotint_{-frac pi 2}^ frac {3pi} 2 f(x) , dx=dfrac 1 Lcdot bigg(int_{-frac pi 2}^ frac {pi} 2 x , dx +int_{frac pi 2}^ frac {3pi} {2} (pi- x) , dxbigg)$



$a_n=displaystyle dfrac 2 L cdotint_{-frac pi 2}^ frac {3pi} 2 f(x)cdot cosleft(dfrac {2npi x} {L}right) , dx$



$b_n=displaystyle dfrac 2 L cdotint_{-frac pi 2}^ frac {3pi} 2 f(x)cdot sinleft(dfrac {2npi x} {L}right) , dx$



The Fourier series of $f$ is:



$$ displaystyle a_0+sum_{n=1}^infty Big[a_ncdot cosleft(dfrac {2npi x} {L}right)+b_n cdot sinleft(dfrac {2npi x} {L}right)Big]$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    but we know for obtaining coefficients we have to integrate function from [-T/2,T/2] and intervals are Symmetric but you didn't write that.I have been confused now.
    $endgroup$
    – Panda
    Sep 29 '14 at 14:31










  • $begingroup$
    I don't think this is necessary to be always true. In general, $a_n=displaystyleint_{x_0}^{x_0+L} f(x)cdot cosbigg(dfrac { 2 n pi x } {L}bigg), dx$ , where $L$ stands for the period of the periodic function.
    $endgroup$
    – thanasissdr
    Sep 29 '14 at 15:21
















3












3








3





$begingroup$

Your function is defined on the interval $left(-dfrac pi 2, dfrac pi 2 right)cup left( dfrac pi 2 , dfrac {3pi} 2right) $. That means the length of the interval is $boxed{L= 2pi}$.



Now, how to compute the coefficients:



$a_0=displaystyle dfrac 1 L cdotint_{-frac pi 2}^ frac {3pi} 2 f(x) , dx=dfrac 1 Lcdot bigg(int_{-frac pi 2}^ frac {pi} 2 x , dx +int_{frac pi 2}^ frac {3pi} {2} (pi- x) , dxbigg)$



$a_n=displaystyle dfrac 2 L cdotint_{-frac pi 2}^ frac {3pi} 2 f(x)cdot cosleft(dfrac {2npi x} {L}right) , dx$



$b_n=displaystyle dfrac 2 L cdotint_{-frac pi 2}^ frac {3pi} 2 f(x)cdot sinleft(dfrac {2npi x} {L}right) , dx$



The Fourier series of $f$ is:



$$ displaystyle a_0+sum_{n=1}^infty Big[a_ncdot cosleft(dfrac {2npi x} {L}right)+b_n cdot sinleft(dfrac {2npi x} {L}right)Big]$$






share|cite|improve this answer











$endgroup$



Your function is defined on the interval $left(-dfrac pi 2, dfrac pi 2 right)cup left( dfrac pi 2 , dfrac {3pi} 2right) $. That means the length of the interval is $boxed{L= 2pi}$.



Now, how to compute the coefficients:



$a_0=displaystyle dfrac 1 L cdotint_{-frac pi 2}^ frac {3pi} 2 f(x) , dx=dfrac 1 Lcdot bigg(int_{-frac pi 2}^ frac {pi} 2 x , dx +int_{frac pi 2}^ frac {3pi} {2} (pi- x) , dxbigg)$



$a_n=displaystyle dfrac 2 L cdotint_{-frac pi 2}^ frac {3pi} 2 f(x)cdot cosleft(dfrac {2npi x} {L}right) , dx$



$b_n=displaystyle dfrac 2 L cdotint_{-frac pi 2}^ frac {3pi} 2 f(x)cdot sinleft(dfrac {2npi x} {L}right) , dx$



The Fourier series of $f$ is:



$$ displaystyle a_0+sum_{n=1}^infty Big[a_ncdot cosleft(dfrac {2npi x} {L}right)+b_n cdot sinleft(dfrac {2npi x} {L}right)Big]$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Apr 17 '15 at 21:53

























answered Sep 28 '14 at 20:51









thanasissdrthanasissdr

5,57111325




5,57111325












  • $begingroup$
    but we know for obtaining coefficients we have to integrate function from [-T/2,T/2] and intervals are Symmetric but you didn't write that.I have been confused now.
    $endgroup$
    – Panda
    Sep 29 '14 at 14:31










  • $begingroup$
    I don't think this is necessary to be always true. In general, $a_n=displaystyleint_{x_0}^{x_0+L} f(x)cdot cosbigg(dfrac { 2 n pi x } {L}bigg), dx$ , where $L$ stands for the period of the periodic function.
    $endgroup$
    – thanasissdr
    Sep 29 '14 at 15:21




















  • $begingroup$
    but we know for obtaining coefficients we have to integrate function from [-T/2,T/2] and intervals are Symmetric but you didn't write that.I have been confused now.
    $endgroup$
    – Panda
    Sep 29 '14 at 14:31










  • $begingroup$
    I don't think this is necessary to be always true. In general, $a_n=displaystyleint_{x_0}^{x_0+L} f(x)cdot cosbigg(dfrac { 2 n pi x } {L}bigg), dx$ , where $L$ stands for the period of the periodic function.
    $endgroup$
    – thanasissdr
    Sep 29 '14 at 15:21


















$begingroup$
but we know for obtaining coefficients we have to integrate function from [-T/2,T/2] and intervals are Symmetric but you didn't write that.I have been confused now.
$endgroup$
– Panda
Sep 29 '14 at 14:31




$begingroup$
but we know for obtaining coefficients we have to integrate function from [-T/2,T/2] and intervals are Symmetric but you didn't write that.I have been confused now.
$endgroup$
– Panda
Sep 29 '14 at 14:31












$begingroup$
I don't think this is necessary to be always true. In general, $a_n=displaystyleint_{x_0}^{x_0+L} f(x)cdot cosbigg(dfrac { 2 n pi x } {L}bigg), dx$ , where $L$ stands for the period of the periodic function.
$endgroup$
– thanasissdr
Sep 29 '14 at 15:21






$begingroup$
I don't think this is necessary to be always true. In general, $a_n=displaystyleint_{x_0}^{x_0+L} f(x)cdot cosbigg(dfrac { 2 n pi x } {L}bigg), dx$ , where $L$ stands for the period of the periodic function.
$endgroup$
– thanasissdr
Sep 29 '14 at 15:21




















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