Fourier Transform of $cos x$ on finite interval












3














Let $f(x)=cos x$ for $-frac{pi}{2}<x<frac{pi}{2}$, and $0$ everywhere else.



Then $tilde f(omega)$ becomes
$$int_{-frac{pi}{2}}^{frac{pi}{2}} cos (x) e^{-i omega x} dx=frac{2cos(frac{pi}{2}omega)}{1-omega^2}$$




There are just some confusion I got for this kind of functions



If we consider the Fourier transform of $f''(x)=- cos x$ for $-frac{pi}{2}<x<frac{pi}{2}$, and $0$ everywhere else. Assuming we can even take the derivatives of such function (defined only on an open interval) to begin with, the Fourier transform of $f''$ becomes $$frac{-omega^22cos(frac{pi}{2}omega)}{1-omega^2}$$
So the Fourier transform of $cos x$ differs by a factor of $omega^2$, but we have defined our Fourier transform to be unique? Are those 2 functions the same after inverse Fourier transform?






fourier-transform







share|cite|improve this question



























    3














    Let $f(x)=cos x$ for $-frac{pi}{2}<x<frac{pi}{2}$, and $0$ everywhere else.



    Then $tilde f(omega)$ becomes
    $$int_{-frac{pi}{2}}^{frac{pi}{2}} cos (x) e^{-i omega x} dx=frac{2cos(frac{pi}{2}omega)}{1-omega^2}$$




    There are just some confusion I got for this kind of functions



    If we consider the Fourier transform of $f''(x)=- cos x$ for $-frac{pi}{2}<x<frac{pi}{2}$, and $0$ everywhere else. Assuming we can even take the derivatives of such function (defined only on an open interval) to begin with, the Fourier transform of $f''$ becomes $$frac{-omega^22cos(frac{pi}{2}omega)}{1-omega^2}$$
    So the Fourier transform of $cos x$ differs by a factor of $omega^2$, but we have defined our Fourier transform to be unique? Are those 2 functions the same after inverse Fourier transform?






    fourier-transform







    share|cite|improve this question

























      3












      3








      3







      Let $f(x)=cos x$ for $-frac{pi}{2}<x<frac{pi}{2}$, and $0$ everywhere else.



      Then $tilde f(omega)$ becomes
      $$int_{-frac{pi}{2}}^{frac{pi}{2}} cos (x) e^{-i omega x} dx=frac{2cos(frac{pi}{2}omega)}{1-omega^2}$$




      There are just some confusion I got for this kind of functions



      If we consider the Fourier transform of $f''(x)=- cos x$ for $-frac{pi}{2}<x<frac{pi}{2}$, and $0$ everywhere else. Assuming we can even take the derivatives of such function (defined only on an open interval) to begin with, the Fourier transform of $f''$ becomes $$frac{-omega^22cos(frac{pi}{2}omega)}{1-omega^2}$$
      So the Fourier transform of $cos x$ differs by a factor of $omega^2$, but we have defined our Fourier transform to be unique? Are those 2 functions the same after inverse Fourier transform?






      fourier-transform







      share|cite|improve this question













      Let $f(x)=cos x$ for $-frac{pi}{2}<x<frac{pi}{2}$, and $0$ everywhere else.



      Then $tilde f(omega)$ becomes
      $$int_{-frac{pi}{2}}^{frac{pi}{2}} cos (x) e^{-i omega x} dx=frac{2cos(frac{pi}{2}omega)}{1-omega^2}$$




      There are just some confusion I got for this kind of functions



      If we consider the Fourier transform of $f''(x)=- cos x$ for $-frac{pi}{2}<x<frac{pi}{2}$, and $0$ everywhere else. Assuming we can even take the derivatives of such function (defined only on an open interval) to begin with, the Fourier transform of $f''$ becomes $$frac{-omega^22cos(frac{pi}{2}omega)}{1-omega^2}$$
      So the Fourier transform of $cos x$ differs by a factor of $omega^2$, but we have defined our Fourier transform to be unique? Are those 2 functions the same after inverse Fourier transform?






      fourier-transform




      fourier-transform






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 29 at 0:12









      Steve

      6941421




      6941421






















          1 Answer
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          2














          Let $f(x)$ be defined as



          $$f(x)=begin{cases}cos(x)&|x|le pi/2\\0&,|x|gepi/2end{cases}$$



          We can write $f(x)=cos(x)text{rect}(x/pi)$ in terms of the rectangular function



          $$text{rect}(x)=begin{cases}
          1&, |x|<1/2\\0&,|x|>1/2
          end{cases}$$



          Then, we have in distribution



          $$f'(x)=-sin(x)text{rect}(x/pi) $$



          and



          $$f''(x)=-cos(x)text{rect}(x/pi)+delta(x-pi/2)+delta(x+pi/2)$$



          So, the Fourier Transform of $f''(x)$ is



          $$begin{align}
          mathscr{F}{f''}(omega)&=mathscr{F}{-f}+e^{-iomega pi/2}+e^{iomega pi/2}\\
          &=-frac{2cos(omega pi/2)}{1-omega^2}+2cos(omega pi/2)\\
          &=-frac{2omega^2 cos(omega pi/2)}{1-omega^2}\\
          &=-omega^2 mathscr{F}{f}(omega)
          end{align}$$



          as expected!






          share|cite|improve this answer























          • Hi, could you please explain how we get the derivatives (I assume by product rule) for $f’$ and $f’’$?
            – Steve
            Nov 29 at 1:49










          • Yes, use the product rule.
            – Mark Viola
            Nov 29 at 2:25










          • What is the derivative of $rect(x/pi)$?
            – Steve
            Nov 29 at 2:27






          • 1




            $$text{rect}'(x/pi)=delta(x+pi/2)-delta(x-pi/2)$$
            – Mark Viola
            Nov 29 at 3:03












          • Thanks! That clears everything up
            – Steve
            Nov 29 at 3:14











          Your Answer





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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          2














          Let $f(x)$ be defined as



          $$f(x)=begin{cases}cos(x)&|x|le pi/2\\0&,|x|gepi/2end{cases}$$



          We can write $f(x)=cos(x)text{rect}(x/pi)$ in terms of the rectangular function



          $$text{rect}(x)=begin{cases}
          1&, |x|<1/2\\0&,|x|>1/2
          end{cases}$$



          Then, we have in distribution



          $$f'(x)=-sin(x)text{rect}(x/pi) $$



          and



          $$f''(x)=-cos(x)text{rect}(x/pi)+delta(x-pi/2)+delta(x+pi/2)$$



          So, the Fourier Transform of $f''(x)$ is



          $$begin{align}
          mathscr{F}{f''}(omega)&=mathscr{F}{-f}+e^{-iomega pi/2}+e^{iomega pi/2}\\
          &=-frac{2cos(omega pi/2)}{1-omega^2}+2cos(omega pi/2)\\
          &=-frac{2omega^2 cos(omega pi/2)}{1-omega^2}\\
          &=-omega^2 mathscr{F}{f}(omega)
          end{align}$$



          as expected!






          share|cite|improve this answer























          • Hi, could you please explain how we get the derivatives (I assume by product rule) for $f’$ and $f’’$?
            – Steve
            Nov 29 at 1:49










          • Yes, use the product rule.
            – Mark Viola
            Nov 29 at 2:25










          • What is the derivative of $rect(x/pi)$?
            – Steve
            Nov 29 at 2:27






          • 1




            $$text{rect}'(x/pi)=delta(x+pi/2)-delta(x-pi/2)$$
            – Mark Viola
            Nov 29 at 3:03












          • Thanks! That clears everything up
            – Steve
            Nov 29 at 3:14
















          2














          Let $f(x)$ be defined as



          $$f(x)=begin{cases}cos(x)&|x|le pi/2\\0&,|x|gepi/2end{cases}$$



          We can write $f(x)=cos(x)text{rect}(x/pi)$ in terms of the rectangular function



          $$text{rect}(x)=begin{cases}
          1&, |x|<1/2\\0&,|x|>1/2
          end{cases}$$



          Then, we have in distribution



          $$f'(x)=-sin(x)text{rect}(x/pi) $$



          and



          $$f''(x)=-cos(x)text{rect}(x/pi)+delta(x-pi/2)+delta(x+pi/2)$$



          So, the Fourier Transform of $f''(x)$ is



          $$begin{align}
          mathscr{F}{f''}(omega)&=mathscr{F}{-f}+e^{-iomega pi/2}+e^{iomega pi/2}\\
          &=-frac{2cos(omega pi/2)}{1-omega^2}+2cos(omega pi/2)\\
          &=-frac{2omega^2 cos(omega pi/2)}{1-omega^2}\\
          &=-omega^2 mathscr{F}{f}(omega)
          end{align}$$



          as expected!






          share|cite|improve this answer























          • Hi, could you please explain how we get the derivatives (I assume by product rule) for $f’$ and $f’’$?
            – Steve
            Nov 29 at 1:49










          • Yes, use the product rule.
            – Mark Viola
            Nov 29 at 2:25










          • What is the derivative of $rect(x/pi)$?
            – Steve
            Nov 29 at 2:27






          • 1




            $$text{rect}'(x/pi)=delta(x+pi/2)-delta(x-pi/2)$$
            – Mark Viola
            Nov 29 at 3:03












          • Thanks! That clears everything up
            – Steve
            Nov 29 at 3:14














          2












          2








          2






          Let $f(x)$ be defined as



          $$f(x)=begin{cases}cos(x)&|x|le pi/2\\0&,|x|gepi/2end{cases}$$



          We can write $f(x)=cos(x)text{rect}(x/pi)$ in terms of the rectangular function



          $$text{rect}(x)=begin{cases}
          1&, |x|<1/2\\0&,|x|>1/2
          end{cases}$$



          Then, we have in distribution



          $$f'(x)=-sin(x)text{rect}(x/pi) $$



          and



          $$f''(x)=-cos(x)text{rect}(x/pi)+delta(x-pi/2)+delta(x+pi/2)$$



          So, the Fourier Transform of $f''(x)$ is



          $$begin{align}
          mathscr{F}{f''}(omega)&=mathscr{F}{-f}+e^{-iomega pi/2}+e^{iomega pi/2}\\
          &=-frac{2cos(omega pi/2)}{1-omega^2}+2cos(omega pi/2)\\
          &=-frac{2omega^2 cos(omega pi/2)}{1-omega^2}\\
          &=-omega^2 mathscr{F}{f}(omega)
          end{align}$$



          as expected!






          share|cite|improve this answer














          Let $f(x)$ be defined as



          $$f(x)=begin{cases}cos(x)&|x|le pi/2\\0&,|x|gepi/2end{cases}$$



          We can write $f(x)=cos(x)text{rect}(x/pi)$ in terms of the rectangular function



          $$text{rect}(x)=begin{cases}
          1&, |x|<1/2\\0&,|x|>1/2
          end{cases}$$



          Then, we have in distribution



          $$f'(x)=-sin(x)text{rect}(x/pi) $$



          and



          $$f''(x)=-cos(x)text{rect}(x/pi)+delta(x-pi/2)+delta(x+pi/2)$$



          So, the Fourier Transform of $f''(x)$ is



          $$begin{align}
          mathscr{F}{f''}(omega)&=mathscr{F}{-f}+e^{-iomega pi/2}+e^{iomega pi/2}\\
          &=-frac{2cos(omega pi/2)}{1-omega^2}+2cos(omega pi/2)\\
          &=-frac{2omega^2 cos(omega pi/2)}{1-omega^2}\\
          &=-omega^2 mathscr{F}{f}(omega)
          end{align}$$



          as expected!







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 29 at 3:19

























          answered Nov 29 at 1:25









          Mark Viola

          130k1273170




          130k1273170












          • Hi, could you please explain how we get the derivatives (I assume by product rule) for $f’$ and $f’’$?
            – Steve
            Nov 29 at 1:49










          • Yes, use the product rule.
            – Mark Viola
            Nov 29 at 2:25










          • What is the derivative of $rect(x/pi)$?
            – Steve
            Nov 29 at 2:27






          • 1




            $$text{rect}'(x/pi)=delta(x+pi/2)-delta(x-pi/2)$$
            – Mark Viola
            Nov 29 at 3:03












          • Thanks! That clears everything up
            – Steve
            Nov 29 at 3:14


















          • Hi, could you please explain how we get the derivatives (I assume by product rule) for $f’$ and $f’’$?
            – Steve
            Nov 29 at 1:49










          • Yes, use the product rule.
            – Mark Viola
            Nov 29 at 2:25










          • What is the derivative of $rect(x/pi)$?
            – Steve
            Nov 29 at 2:27






          • 1




            $$text{rect}'(x/pi)=delta(x+pi/2)-delta(x-pi/2)$$
            – Mark Viola
            Nov 29 at 3:03












          • Thanks! That clears everything up
            – Steve
            Nov 29 at 3:14
















          Hi, could you please explain how we get the derivatives (I assume by product rule) for $f’$ and $f’’$?
          – Steve
          Nov 29 at 1:49




          Hi, could you please explain how we get the derivatives (I assume by product rule) for $f’$ and $f’’$?
          – Steve
          Nov 29 at 1:49












          Yes, use the product rule.
          – Mark Viola
          Nov 29 at 2:25




          Yes, use the product rule.
          – Mark Viola
          Nov 29 at 2:25












          What is the derivative of $rect(x/pi)$?
          – Steve
          Nov 29 at 2:27




          What is the derivative of $rect(x/pi)$?
          – Steve
          Nov 29 at 2:27




          1




          1




          $$text{rect}'(x/pi)=delta(x+pi/2)-delta(x-pi/2)$$
          – Mark Viola
          Nov 29 at 3:03






          $$text{rect}'(x/pi)=delta(x+pi/2)-delta(x-pi/2)$$
          – Mark Viola
          Nov 29 at 3:03














          Thanks! That clears everything up
          – Steve
          Nov 29 at 3:14




          Thanks! That clears everything up
          – Steve
          Nov 29 at 3:14


















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