Difference (or lack thereof) of the minus sign in complex exponentials in Fourier series

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It seems as though the minus sign is left out every so often when writing down Fourier series:



$$f(x) = sum_-infty^infty c_k mathrm e^-ikx$$



versus



$$f(x) = sum_-infty^infty c_k mathrm e^ikx$$



In the first instance,



$$mathrm e^ikx=cos kx +isin kx$$



while



$$mathrm e^-ikx=cos kx -isin kx$$



It makes intuitive sense that the minus sign in the last equation can easily be absorbed into the coefficients.



But what is the reason to use the minus sign? And why are they equivalent?




complex-numbers fourier-analysis




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    up vote
    1
    down vote

    favorite
    1












    It seems as though the minus sign is left out every so often when writing down Fourier series:



    $$f(x) = sum_-infty^infty c_k mathrm e^-ikx$$



    versus



    $$f(x) = sum_-infty^infty c_k mathrm e^ikx$$



    In the first instance,



    $$mathrm e^ikx=cos kx +isin kx$$



    while



    $$mathrm e^-ikx=cos kx -isin kx$$



    It makes intuitive sense that the minus sign in the last equation can easily be absorbed into the coefficients.



    But what is the reason to use the minus sign? And why are they equivalent?




    complex-numbers fourier-analysis




    share|cite|improve this question






















      up vote
      1
      down vote

      favorite
      1









      up vote
      1
      down vote

      favorite
      1






      1





      It seems as though the minus sign is left out every so often when writing down Fourier series:



      $$f(x) = sum_-infty^infty c_k mathrm e^-ikx$$



      versus



      $$f(x) = sum_-infty^infty c_k mathrm e^ikx$$



      In the first instance,



      $$mathrm e^ikx=cos kx +isin kx$$



      while



      $$mathrm e^-ikx=cos kx -isin kx$$



      It makes intuitive sense that the minus sign in the last equation can easily be absorbed into the coefficients.



      But what is the reason to use the minus sign? And why are they equivalent?




      complex-numbers fourier-analysis




      share|cite|improve this question












      It seems as though the minus sign is left out every so often when writing down Fourier series:



      $$f(x) = sum_-infty^infty c_k mathrm e^-ikx$$



      versus



      $$f(x) = sum_-infty^infty c_k mathrm e^ikx$$



      In the first instance,



      $$mathrm e^ikx=cos kx +isin kx$$



      while



      $$mathrm e^-ikx=cos kx -isin kx$$



      It makes intuitive sense that the minus sign in the last equation can easily be absorbed into the coefficients.



      But what is the reason to use the minus sign? And why are they equivalent?





      complex-numbers fourier-analysis





      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Aug 26 at 23:13









      MathAsFun

      2,75821234




      2,75821234




















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          The Fourier transform and the inverse Fourier transform always come in pairs. So you transform from $x$ to $k$ and back. Which one has the negative sign is just a convention.






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            The Fourier transform and the inverse Fourier transform always come in pairs. So you transform from $x$ to $k$ and back. Which one has the negative sign is just a convention.






            share|cite|improve this answer
























              up vote
              0
              down vote













              The Fourier transform and the inverse Fourier transform always come in pairs. So you transform from $x$ to $k$ and back. Which one has the negative sign is just a convention.






              share|cite|improve this answer






















                up vote
                0
                down vote










                up vote
                0
                down vote









                The Fourier transform and the inverse Fourier transform always come in pairs. So you transform from $x$ to $k$ and back. Which one has the negative sign is just a convention.






                share|cite|improve this answer












                The Fourier transform and the inverse Fourier transform always come in pairs. So you transform from $x$ to $k$ and back. Which one has the negative sign is just a convention.







                share|cite|improve this answer












                share|cite|improve this answer



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                answered Aug 27 at 0:22









                Andrei

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