Fourier Series of Real-valued Functions The Next CEO of Stack OverflowReal-valued 2D Fourier series?Is my Fourier Series computation done correctly?Understanding the indices in a Fourier seriesReal-valued Fourier series representationShow that Fourier series arising in solution of differential eqn. converges uniformlyA Fourier cosine series of type $f(x)=fraca_02+sum_k=1^infty a_k cos(2^k x)$.On the equivalence of representations of Fourier seriesShow that the coefficients of the Fourier Series of $f in C^1$ go to zerovanishing fourier seriesIntegrating a Fourier Series to Derive Another Fourier Series
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Fourier Series of Real-valued Functions
The Next CEO of Stack OverflowReal-valued 2D Fourier series?Is my Fourier Series computation done correctly?Understanding the indices in a Fourier seriesReal-valued Fourier series representationShow that Fourier series arising in solution of differential eqn. converges uniformlyA Fourier cosine series of type $f(x)=fraca_02+sum_k=1^infty a_k cos(2^k x)$.On the equivalence of representations of Fourier seriesShow that the coefficients of the Fourier Series of $f in C^1$ go to zerovanishing fourier seriesIntegrating a Fourier Series to Derive Another Fourier Series
$begingroup$
Context: For a $2pi$-periodic bounded function $f:mathbbRtomathbbC$, we define the complex Fourier coefficients of $f$ by
$$
hatf_k:=frac12piint_0^2pif(x)e^-ikx,dx.
$$
We call Fourier series of $f$ the formal series
$$
sum_k=-infty^inftyhatf_ke^ikx.tag1
$$
Now, it is easily shown that
$$
overlinehatf_k=hatoverlinef_-k.
$$
Hence if $f$ is real-valued then
$$
overlinehatf_k=hatf_-k.tag2
$$
Reading some notes, it is said that if $f$ is a real-valued function, then we can write
$$
hatf_k=frac12piint_0^2pif(x)cos(kx),dx-ifrac12piint_0^2pif(x)sin(kx),dx.tag3
$$
Putting
$$
a_k:=frac1piint_0^2pif(x)cos(kx),dx,\
b_k:=frac1piint_0^2pif(x)sin(kx),dx,
$$
where $a_k,b_k$ are called the real Fourier coefficients of $f$, it is said that in view of $(2)$, we have the relations
beginalign
hatf_0&=frac12a_0,\
hatf_k&=frac12(a_k-ib_k),\
a_k&=hatf_k+hatf_-k,tag4\
b_k&=i(hatf_k-hatf_-k),\
a_-k&=a_k,\
b_-k&=-b_k.
endalign
Finally it is said that if, once again, $f$ is real-valued, then we can write $(1)$ as the trigonometric series
$$
frac12a_0+sum_k=1^inftya_kcos(kx)+b_ksin(kx).tag5
$$
Question: Is it really essential that $f$ be real-valued for $(3)$, $(4)$ and $(5)$ to hold? It seems to me that everything holds even if $f$ is complex-valued...
fourier-analysis fourier-series
$endgroup$
add a comment |
$begingroup$
Context: For a $2pi$-periodic bounded function $f:mathbbRtomathbbC$, we define the complex Fourier coefficients of $f$ by
$$
hatf_k:=frac12piint_0^2pif(x)e^-ikx,dx.
$$
We call Fourier series of $f$ the formal series
$$
sum_k=-infty^inftyhatf_ke^ikx.tag1
$$
Now, it is easily shown that
$$
overlinehatf_k=hatoverlinef_-k.
$$
Hence if $f$ is real-valued then
$$
overlinehatf_k=hatf_-k.tag2
$$
Reading some notes, it is said that if $f$ is a real-valued function, then we can write
$$
hatf_k=frac12piint_0^2pif(x)cos(kx),dx-ifrac12piint_0^2pif(x)sin(kx),dx.tag3
$$
Putting
$$
a_k:=frac1piint_0^2pif(x)cos(kx),dx,\
b_k:=frac1piint_0^2pif(x)sin(kx),dx,
$$
where $a_k,b_k$ are called the real Fourier coefficients of $f$, it is said that in view of $(2)$, we have the relations
beginalign
hatf_0&=frac12a_0,\
hatf_k&=frac12(a_k-ib_k),\
a_k&=hatf_k+hatf_-k,tag4\
b_k&=i(hatf_k-hatf_-k),\
a_-k&=a_k,\
b_-k&=-b_k.
endalign
Finally it is said that if, once again, $f$ is real-valued, then we can write $(1)$ as the trigonometric series
$$
frac12a_0+sum_k=1^inftya_kcos(kx)+b_ksin(kx).tag5
$$
Question: Is it really essential that $f$ be real-valued for $(3)$, $(4)$ and $(5)$ to hold? It seems to me that everything holds even if $f$ is complex-valued...
fourier-analysis fourier-series
$endgroup$
1
$begingroup$
Regarding $(3)$: No, that's just $e^-ikx = cos(kx) - isin(kx)$ plus linearity of the integral. The others should also be fine but remember that the $a_k, b_k$ will become complex, so that doesn't ease anything.
$endgroup$
– AlexR
Nov 13 '14 at 16:13
add a comment |
$begingroup$
Context: For a $2pi$-periodic bounded function $f:mathbbRtomathbbC$, we define the complex Fourier coefficients of $f$ by
$$
hatf_k:=frac12piint_0^2pif(x)e^-ikx,dx.
$$
We call Fourier series of $f$ the formal series
$$
sum_k=-infty^inftyhatf_ke^ikx.tag1
$$
Now, it is easily shown that
$$
overlinehatf_k=hatoverlinef_-k.
$$
Hence if $f$ is real-valued then
$$
overlinehatf_k=hatf_-k.tag2
$$
Reading some notes, it is said that if $f$ is a real-valued function, then we can write
$$
hatf_k=frac12piint_0^2pif(x)cos(kx),dx-ifrac12piint_0^2pif(x)sin(kx),dx.tag3
$$
Putting
$$
a_k:=frac1piint_0^2pif(x)cos(kx),dx,\
b_k:=frac1piint_0^2pif(x)sin(kx),dx,
$$
where $a_k,b_k$ are called the real Fourier coefficients of $f$, it is said that in view of $(2)$, we have the relations
beginalign
hatf_0&=frac12a_0,\
hatf_k&=frac12(a_k-ib_k),\
a_k&=hatf_k+hatf_-k,tag4\
b_k&=i(hatf_k-hatf_-k),\
a_-k&=a_k,\
b_-k&=-b_k.
endalign
Finally it is said that if, once again, $f$ is real-valued, then we can write $(1)$ as the trigonometric series
$$
frac12a_0+sum_k=1^inftya_kcos(kx)+b_ksin(kx).tag5
$$
Question: Is it really essential that $f$ be real-valued for $(3)$, $(4)$ and $(5)$ to hold? It seems to me that everything holds even if $f$ is complex-valued...
fourier-analysis fourier-series
$endgroup$
Context: For a $2pi$-periodic bounded function $f:mathbbRtomathbbC$, we define the complex Fourier coefficients of $f$ by
$$
hatf_k:=frac12piint_0^2pif(x)e^-ikx,dx.
$$
We call Fourier series of $f$ the formal series
$$
sum_k=-infty^inftyhatf_ke^ikx.tag1
$$
Now, it is easily shown that
$$
overlinehatf_k=hatoverlinef_-k.
$$
Hence if $f$ is real-valued then
$$
overlinehatf_k=hatf_-k.tag2
$$
Reading some notes, it is said that if $f$ is a real-valued function, then we can write
$$
hatf_k=frac12piint_0^2pif(x)cos(kx),dx-ifrac12piint_0^2pif(x)sin(kx),dx.tag3
$$
Putting
$$
a_k:=frac1piint_0^2pif(x)cos(kx),dx,\
b_k:=frac1piint_0^2pif(x)sin(kx),dx,
$$
where $a_k,b_k$ are called the real Fourier coefficients of $f$, it is said that in view of $(2)$, we have the relations
beginalign
hatf_0&=frac12a_0,\
hatf_k&=frac12(a_k-ib_k),\
a_k&=hatf_k+hatf_-k,tag4\
b_k&=i(hatf_k-hatf_-k),\
a_-k&=a_k,\
b_-k&=-b_k.
endalign
Finally it is said that if, once again, $f$ is real-valued, then we can write $(1)$ as the trigonometric series
$$
frac12a_0+sum_k=1^inftya_kcos(kx)+b_ksin(kx).tag5
$$
Question: Is it really essential that $f$ be real-valued for $(3)$, $(4)$ and $(5)$ to hold? It seems to me that everything holds even if $f$ is complex-valued...
fourier-analysis fourier-series
fourier-analysis fourier-series
asked Nov 13 '14 at 16:08
GuestGuest
2,87811530
2,87811530
1
$begingroup$
Regarding $(3)$: No, that's just $e^-ikx = cos(kx) - isin(kx)$ plus linearity of the integral. The others should also be fine but remember that the $a_k, b_k$ will become complex, so that doesn't ease anything.
$endgroup$
– AlexR
Nov 13 '14 at 16:13
add a comment |
1
$begingroup$
Regarding $(3)$: No, that's just $e^-ikx = cos(kx) - isin(kx)$ plus linearity of the integral. The others should also be fine but remember that the $a_k, b_k$ will become complex, so that doesn't ease anything.
$endgroup$
– AlexR
Nov 13 '14 at 16:13
1
1
$begingroup$
Regarding $(3)$: No, that's just $e^-ikx = cos(kx) - isin(kx)$ plus linearity of the integral. The others should also be fine but remember that the $a_k, b_k$ will become complex, so that doesn't ease anything.
$endgroup$
– AlexR
Nov 13 '14 at 16:13
$begingroup$
Regarding $(3)$: No, that's just $e^-ikx = cos(kx) - isin(kx)$ plus linearity of the integral. The others should also be fine but remember that the $a_k, b_k$ will become complex, so that doesn't ease anything.
$endgroup$
– AlexR
Nov 13 '14 at 16:13
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Yes they hold assuming $a_k,b_k$ are complex, but the point is that they are real valued and thus by using (2) we can show that
$$
hat f_k e^ikx + hat f_-k e^-ikx = hat f_k e^ikx + (hat f_k e^ikx)^* = 2 mathrmRe f_k e^ikx = 2 mathrmRe Big frac12 (a_k-ib_k)(cos(kx)+ isin(kx)) Big = a_kcos(kx) + b_k sin(kx)
$$
We can also show that same thing holds if $a_k,b_k$ are complex:
beginalign
hat f_k e^ikx + hat f_-k e^-ikx &= frac12(a_k - ib_k) e^ikx + frac12(a_-k - ib_-k) e^-ikx\&= frac12(a_k - ib_k) e^ikx + frac12(a_k + ib_k) e^-ikx \&= frac12 a_k (e^ikx+e^-ikx) - frac12i b_k( e^ikx - e^-ikx) \&= a_k cos(kx) + b_ksin(kx)
endalign
$endgroup$
add a comment |
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$begingroup$
Yes they hold assuming $a_k,b_k$ are complex, but the point is that they are real valued and thus by using (2) we can show that
$$
hat f_k e^ikx + hat f_-k e^-ikx = hat f_k e^ikx + (hat f_k e^ikx)^* = 2 mathrmRe f_k e^ikx = 2 mathrmRe Big frac12 (a_k-ib_k)(cos(kx)+ isin(kx)) Big = a_kcos(kx) + b_k sin(kx)
$$
We can also show that same thing holds if $a_k,b_k$ are complex:
beginalign
hat f_k e^ikx + hat f_-k e^-ikx &= frac12(a_k - ib_k) e^ikx + frac12(a_-k - ib_-k) e^-ikx\&= frac12(a_k - ib_k) e^ikx + frac12(a_k + ib_k) e^-ikx \&= frac12 a_k (e^ikx+e^-ikx) - frac12i b_k( e^ikx - e^-ikx) \&= a_k cos(kx) + b_ksin(kx)
endalign
$endgroup$
add a comment |
$begingroup$
Yes they hold assuming $a_k,b_k$ are complex, but the point is that they are real valued and thus by using (2) we can show that
$$
hat f_k e^ikx + hat f_-k e^-ikx = hat f_k e^ikx + (hat f_k e^ikx)^* = 2 mathrmRe f_k e^ikx = 2 mathrmRe Big frac12 (a_k-ib_k)(cos(kx)+ isin(kx)) Big = a_kcos(kx) + b_k sin(kx)
$$
We can also show that same thing holds if $a_k,b_k$ are complex:
beginalign
hat f_k e^ikx + hat f_-k e^-ikx &= frac12(a_k - ib_k) e^ikx + frac12(a_-k - ib_-k) e^-ikx\&= frac12(a_k - ib_k) e^ikx + frac12(a_k + ib_k) e^-ikx \&= frac12 a_k (e^ikx+e^-ikx) - frac12i b_k( e^ikx - e^-ikx) \&= a_k cos(kx) + b_ksin(kx)
endalign
$endgroup$
add a comment |
$begingroup$
Yes they hold assuming $a_k,b_k$ are complex, but the point is that they are real valued and thus by using (2) we can show that
$$
hat f_k e^ikx + hat f_-k e^-ikx = hat f_k e^ikx + (hat f_k e^ikx)^* = 2 mathrmRe f_k e^ikx = 2 mathrmRe Big frac12 (a_k-ib_k)(cos(kx)+ isin(kx)) Big = a_kcos(kx) + b_k sin(kx)
$$
We can also show that same thing holds if $a_k,b_k$ are complex:
beginalign
hat f_k e^ikx + hat f_-k e^-ikx &= frac12(a_k - ib_k) e^ikx + frac12(a_-k - ib_-k) e^-ikx\&= frac12(a_k - ib_k) e^ikx + frac12(a_k + ib_k) e^-ikx \&= frac12 a_k (e^ikx+e^-ikx) - frac12i b_k( e^ikx - e^-ikx) \&= a_k cos(kx) + b_ksin(kx)
endalign
$endgroup$
Yes they hold assuming $a_k,b_k$ are complex, but the point is that they are real valued and thus by using (2) we can show that
$$
hat f_k e^ikx + hat f_-k e^-ikx = hat f_k e^ikx + (hat f_k e^ikx)^* = 2 mathrmRe f_k e^ikx = 2 mathrmRe Big frac12 (a_k-ib_k)(cos(kx)+ isin(kx)) Big = a_kcos(kx) + b_k sin(kx)
$$
We can also show that same thing holds if $a_k,b_k$ are complex:
beginalign
hat f_k e^ikx + hat f_-k e^-ikx &= frac12(a_k - ib_k) e^ikx + frac12(a_-k - ib_-k) e^-ikx\&= frac12(a_k - ib_k) e^ikx + frac12(a_k + ib_k) e^-ikx \&= frac12 a_k (e^ikx+e^-ikx) - frac12i b_k( e^ikx - e^-ikx) \&= a_k cos(kx) + b_ksin(kx)
endalign
answered Mar 18 at 21:02
mu7zmu7z
305
305
add a comment |
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$begingroup$
Regarding $(3)$: No, that's just $e^-ikx = cos(kx) - isin(kx)$ plus linearity of the integral. The others should also be fine but remember that the $a_k, b_k$ will become complex, so that doesn't ease anything.
$endgroup$
– AlexR
Nov 13 '14 at 16:13