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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










2












$begingroup$


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...










share|cite|improve this question









$endgroup$







  • 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
















2












$begingroup$


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...










share|cite|improve this question









$endgroup$







  • 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














2












2








2


2



$begingroup$


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...










share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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













  • 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











1 Answer
1






active

oldest

votes


















0












$begingroup$

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






share|cite|improve this answer









$endgroup$













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    1 Answer
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    1 Answer
    1






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    active

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    0












    $begingroup$

    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






    share|cite|improve this answer









    $endgroup$

















      0












      $begingroup$

      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






      share|cite|improve this answer









      $endgroup$















        0












        0








        0





        $begingroup$

        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






        share|cite|improve this answer









        $endgroup$



        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







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 18 at 21:02









        mu7zmu7z

        305




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