Show ${int_{0}^{1}|e^{-2picdot int} f(t)| dt} = {int_{0}^{1}|f(t)| dt}$











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The expressions below are from page 36 of Brad Osgood's "The Fourier Transform and its Applications". I don't understand how to establish the equality between the second and third expressions below (the text doesn't expand on this).



Some inequalities Fourier coefficient to function
For the inequality between the first and second expressions, I can tell that it is using the idea that the absolute value of an integral is less than or equal to the integral of the absolute value of the same integrand. I also recognize the first expression as the absolute value of the n-th Fourier coefficient. The context of the previous pages was on the Cauchy-Schwartz inequality and the Triangle Inequality; however, it doesn't look like either of these inequalities are being used to establish this equality. Any hints on how to get at this?










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    up vote
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    down vote

    favorite












    The expressions below are from page 36 of Brad Osgood's "The Fourier Transform and its Applications". I don't understand how to establish the equality between the second and third expressions below (the text doesn't expand on this).



    Some inequalities Fourier coefficient to function
    For the inequality between the first and second expressions, I can tell that it is using the idea that the absolute value of an integral is less than or equal to the integral of the absolute value of the same integrand. I also recognize the first expression as the absolute value of the n-th Fourier coefficient. The context of the previous pages was on the Cauchy-Schwartz inequality and the Triangle Inequality; however, it doesn't look like either of these inequalities are being used to establish this equality. Any hints on how to get at this?










    share|cite|improve this question


























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      The expressions below are from page 36 of Brad Osgood's "The Fourier Transform and its Applications". I don't understand how to establish the equality between the second and third expressions below (the text doesn't expand on this).



      Some inequalities Fourier coefficient to function
      For the inequality between the first and second expressions, I can tell that it is using the idea that the absolute value of an integral is less than or equal to the integral of the absolute value of the same integrand. I also recognize the first expression as the absolute value of the n-th Fourier coefficient. The context of the previous pages was on the Cauchy-Schwartz inequality and the Triangle Inequality; however, it doesn't look like either of these inequalities are being used to establish this equality. Any hints on how to get at this?










      share|cite|improve this question















      The expressions below are from page 36 of Brad Osgood's "The Fourier Transform and its Applications". I don't understand how to establish the equality between the second and third expressions below (the text doesn't expand on this).



      Some inequalities Fourier coefficient to function
      For the inequality between the first and second expressions, I can tell that it is using the idea that the absolute value of an integral is less than or equal to the integral of the absolute value of the same integrand. I also recognize the first expression as the absolute value of the n-th Fourier coefficient. The context of the previous pages was on the Cauchy-Schwartz inequality and the Triangle Inequality; however, it doesn't look like either of these inequalities are being used to establish this equality. Any hints on how to get at this?







      fourier-analysis fourier-series






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      edited Nov 26 at 2:37









      Tianlalu

      2,9801936




      2,9801936










      asked Nov 26 at 1:59









      hgil

      226




      226






















          2 Answers
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          accepted










          Note that



          $$|e^{it}|=1 hspace{4mm}forall tin mathbb{R}.$$



          It is because



          $$|e^{it}|=|cos(t)+isin(t)|=sqrt{cos^2(t)+sin^2(t)}=1.$$



          Thus,



          $$|e^{-2pi i n t}f(t)|=|e^{-2pi i n t}||f(t)|=|e^{(-2pi n t)i}||f(t)|=1|f(t)|=|f(t)|.$$






          share|cite|improve this answer





















          • Perfect! Thanks!
            – hgil
            Nov 26 at 2:13


















          up vote
          1
          down vote













          For any two complex numbers $z_1$ and $z_2$ we have $$|z_1z_2|=|z_1||z_2|$$ so here
          $$left|e^{-2pi i n t}f(t)right|=left|e^{-2pi i n t}right|left|f(t)right|=left|f(t)right|$$
          if $ntin mathbb{R}$.






          share|cite|improve this answer





















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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            1
            down vote



            accepted










            Note that



            $$|e^{it}|=1 hspace{4mm}forall tin mathbb{R}.$$



            It is because



            $$|e^{it}|=|cos(t)+isin(t)|=sqrt{cos^2(t)+sin^2(t)}=1.$$



            Thus,



            $$|e^{-2pi i n t}f(t)|=|e^{-2pi i n t}||f(t)|=|e^{(-2pi n t)i}||f(t)|=1|f(t)|=|f(t)|.$$






            share|cite|improve this answer





















            • Perfect! Thanks!
              – hgil
              Nov 26 at 2:13















            up vote
            1
            down vote



            accepted










            Note that



            $$|e^{it}|=1 hspace{4mm}forall tin mathbb{R}.$$



            It is because



            $$|e^{it}|=|cos(t)+isin(t)|=sqrt{cos^2(t)+sin^2(t)}=1.$$



            Thus,



            $$|e^{-2pi i n t}f(t)|=|e^{-2pi i n t}||f(t)|=|e^{(-2pi n t)i}||f(t)|=1|f(t)|=|f(t)|.$$






            share|cite|improve this answer





















            • Perfect! Thanks!
              – hgil
              Nov 26 at 2:13













            up vote
            1
            down vote



            accepted







            up vote
            1
            down vote



            accepted






            Note that



            $$|e^{it}|=1 hspace{4mm}forall tin mathbb{R}.$$



            It is because



            $$|e^{it}|=|cos(t)+isin(t)|=sqrt{cos^2(t)+sin^2(t)}=1.$$



            Thus,



            $$|e^{-2pi i n t}f(t)|=|e^{-2pi i n t}||f(t)|=|e^{(-2pi n t)i}||f(t)|=1|f(t)|=|f(t)|.$$






            share|cite|improve this answer












            Note that



            $$|e^{it}|=1 hspace{4mm}forall tin mathbb{R}.$$



            It is because



            $$|e^{it}|=|cos(t)+isin(t)|=sqrt{cos^2(t)+sin^2(t)}=1.$$



            Thus,



            $$|e^{-2pi i n t}f(t)|=|e^{-2pi i n t}||f(t)|=|e^{(-2pi n t)i}||f(t)|=1|f(t)|=|f(t)|.$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 26 at 2:11









            LeB

            938217




            938217












            • Perfect! Thanks!
              – hgil
              Nov 26 at 2:13


















            • Perfect! Thanks!
              – hgil
              Nov 26 at 2:13
















            Perfect! Thanks!
            – hgil
            Nov 26 at 2:13




            Perfect! Thanks!
            – hgil
            Nov 26 at 2:13










            up vote
            1
            down vote













            For any two complex numbers $z_1$ and $z_2$ we have $$|z_1z_2|=|z_1||z_2|$$ so here
            $$left|e^{-2pi i n t}f(t)right|=left|e^{-2pi i n t}right|left|f(t)right|=left|f(t)right|$$
            if $ntin mathbb{R}$.






            share|cite|improve this answer

























              up vote
              1
              down vote













              For any two complex numbers $z_1$ and $z_2$ we have $$|z_1z_2|=|z_1||z_2|$$ so here
              $$left|e^{-2pi i n t}f(t)right|=left|e^{-2pi i n t}right|left|f(t)right|=left|f(t)right|$$
              if $ntin mathbb{R}$.






              share|cite|improve this answer























                up vote
                1
                down vote










                up vote
                1
                down vote









                For any two complex numbers $z_1$ and $z_2$ we have $$|z_1z_2|=|z_1||z_2|$$ so here
                $$left|e^{-2pi i n t}f(t)right|=left|e^{-2pi i n t}right|left|f(t)right|=left|f(t)right|$$
                if $ntin mathbb{R}$.






                share|cite|improve this answer












                For any two complex numbers $z_1$ and $z_2$ we have $$|z_1z_2|=|z_1||z_2|$$ so here
                $$left|e^{-2pi i n t}f(t)right|=left|e^{-2pi i n t}right|left|f(t)right|=left|f(t)right|$$
                if $ntin mathbb{R}$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 26 at 2:13









                M1183

                943




                943






























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