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

Won the lottery - how do I keep the money?

Why do we use the plural of movies in this phrase "We went to the movies last night."?

Preparing Indesign booklet with .psd graphics for print

Contours of a clandestine nature

Why am I allowed to create multiple unique pointers from a single object?

Calculus II Question

Are there any limitations on attacking while grappling?

Would a completely good Muggle be able to use a wand?

Why does the UK parliament need a vote on the political declaration?

What is "(CFMCC)" on an ILS approach chart?

Are there any unintended negative consequences to allowing PCs to gain multiple levels at once in a short milestone-XP game?

Why didn't Khan get resurrected in the Genesis Explosion?

Does it take more energy to get to Venus or to Mars?

Is there a way to save my career from absolute disaster?

Rotate a column

Would this house-rule that treats advantage as a +1 to the roll instead (and disadvantage as -1) and allows them to stack be balanced?

If/When UK leaves the EU, can a future goverment conduct a referendum to join the EU?

Why do professional authors make "consistency" mistakes? And how to avoid them?

How to avoid supervisors with prejudiced views?

Is "for causing autism in X" grammatical?

I believe this to be a fraud - hired, then asked to cash check and send cash as Bitcoin

Why does standard notation not preserve intervals (visually)

What was the first Unix version to run on a microcomputer?

What flight has the highest ratio of time difference to flight time?



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









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function ()
    return StackExchange.using("mathjaxEditing", function ()
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    );
    );
    , "mathjax-editing");

    StackExchange.ready(function()
    var channelOptions =
    tags: "".split(" "),
    id: "69"
    ;
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function()
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled)
    StackExchange.using("snippets", function()
    createEditor();
    );

    else
    createEditor();

    );

    function createEditor()
    StackExchange.prepareEditor(
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader:
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    ,
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    );



    );













    draft saved

    draft discarded


















    StackExchange.ready(
    function ()
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1020231%2ffourier-series-of-real-valued-functions%23new-answer', 'question_page');

    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    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









    $endgroup$

















      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




        305



























            draft saved

            draft discarded
















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid


            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.

            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1020231%2ffourier-series-of-real-valued-functions%23new-answer', 'question_page');

            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            Lowndes Grove History Architecture References Navigation menu32°48′6″N 79°57′58″W / 32.80167°N 79.96611°W / 32.80167; -79.9661132°48′6″N 79°57′58″W / 32.80167°N 79.96611°W / 32.80167; -79.9661178002500"National Register Information System"Historic houses of South Carolina"Lowndes Grove""+32° 48' 6.00", −79° 57' 58.00""Lowndes Grove, Charleston County (260 St. Margaret St., Charleston)""Lowndes Grove"The Charleston ExpositionIt Happened in South Carolina"Lowndes Grove (House), Saint Margaret Street & Sixth Avenue, Charleston, Charleston County, SC(Photographs)"Plantations of the Carolina Low Countrye

            random experiment with two different functions on unit interval Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Random variable and probability space notionsRandom Walk with EdgesFinding functions where the increase over a random interval is Poisson distributedNumber of days until dayCan an observed event in fact be of zero probability?Unit random processmodels of coins and uniform distributionHow to get the number of successes given $n$ trials , probability $P$ and a random variable $X$Absorbing Markov chain in a computer. Is “almost every” turned into always convergence in computer executions?Stopped random walk is not uniformly integrable

            How should I support this large drywall patch? Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?How do I cover large gaps in drywall?How do I keep drywall around a patch from crumbling?Can I glue a second layer of drywall?How to patch long strip on drywall?Large drywall patch: how to avoid bulging seams?Drywall Mesh Patch vs. Bulge? To remove or not to remove?How to fix this drywall job?Prep drywall before backsplashWhat's the best way to fix this horrible drywall patch job?Drywall patching using 3M Patch Plus Primer