Statement of Parseval's theorem for Fourier Transform












0














the following is the statement of Parseval's theorem from Wikipedia,



Suppose that $A(x)$ and $B(x)$ are two square integrable (with respect to the Lebesgue measure), complex-valued functions on $mathbb{R}$ of period $2pi$ with Fourier series
$$A(x) = sum_{n=-infty}^{infty} a_n e^{inx} $$
and
$$B(x) = sum_{n=-infty}^{infty} b_n e^{inx} $$
respectively. Then
$$sum_{n=-infty}^{infty}a_n overline{b_n} = frac{1}{2pi} int_{-pi}^{pi}A(x) overline{B(x)} dx$$
where $i$ is the imaginary unit and horizontal bars indicate complex conjugation.



I would like to know if the above statement still hold when $A(x) = sum_{n=-infty}^{infty} a_n e^{-inx} $ and $B(x) = sum_{n=-infty}^{infty} b_n e^{-inx} $. The only changes is the negative in exponential.






fourier-analysis parsevals-identity







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  • Sure. Just re-order the sum by replacing $n$ with $-n$.
    – DisintegratingByParts
    2 days ago
















0














the following is the statement of Parseval's theorem from Wikipedia,



Suppose that $A(x)$ and $B(x)$ are two square integrable (with respect to the Lebesgue measure), complex-valued functions on $mathbb{R}$ of period $2pi$ with Fourier series
$$A(x) = sum_{n=-infty}^{infty} a_n e^{inx} $$
and
$$B(x) = sum_{n=-infty}^{infty} b_n e^{inx} $$
respectively. Then
$$sum_{n=-infty}^{infty}a_n overline{b_n} = frac{1}{2pi} int_{-pi}^{pi}A(x) overline{B(x)} dx$$
where $i$ is the imaginary unit and horizontal bars indicate complex conjugation.



I would like to know if the above statement still hold when $A(x) = sum_{n=-infty}^{infty} a_n e^{-inx} $ and $B(x) = sum_{n=-infty}^{infty} b_n e^{-inx} $. The only changes is the negative in exponential.






fourier-analysis parsevals-identity







share|cite|improve this question







New contributor




Mike L is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • Sure. Just re-order the sum by replacing $n$ with $-n$.
    – DisintegratingByParts
    2 days ago














0












0








0







the following is the statement of Parseval's theorem from Wikipedia,



Suppose that $A(x)$ and $B(x)$ are two square integrable (with respect to the Lebesgue measure), complex-valued functions on $mathbb{R}$ of period $2pi$ with Fourier series
$$A(x) = sum_{n=-infty}^{infty} a_n e^{inx} $$
and
$$B(x) = sum_{n=-infty}^{infty} b_n e^{inx} $$
respectively. Then
$$sum_{n=-infty}^{infty}a_n overline{b_n} = frac{1}{2pi} int_{-pi}^{pi}A(x) overline{B(x)} dx$$
where $i$ is the imaginary unit and horizontal bars indicate complex conjugation.



I would like to know if the above statement still hold when $A(x) = sum_{n=-infty}^{infty} a_n e^{-inx} $ and $B(x) = sum_{n=-infty}^{infty} b_n e^{-inx} $. The only changes is the negative in exponential.






fourier-analysis parsevals-identity







share|cite|improve this question







New contributor




Mike L is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











the following is the statement of Parseval's theorem from Wikipedia,



Suppose that $A(x)$ and $B(x)$ are two square integrable (with respect to the Lebesgue measure), complex-valued functions on $mathbb{R}$ of period $2pi$ with Fourier series
$$A(x) = sum_{n=-infty}^{infty} a_n e^{inx} $$
and
$$B(x) = sum_{n=-infty}^{infty} b_n e^{inx} $$
respectively. Then
$$sum_{n=-infty}^{infty}a_n overline{b_n} = frac{1}{2pi} int_{-pi}^{pi}A(x) overline{B(x)} dx$$
where $i$ is the imaginary unit and horizontal bars indicate complex conjugation.



I would like to know if the above statement still hold when $A(x) = sum_{n=-infty}^{infty} a_n e^{-inx} $ and $B(x) = sum_{n=-infty}^{infty} b_n e^{-inx} $. The only changes is the negative in exponential.






fourier-analysis parsevals-identity




fourier-analysis parsevals-identity






share|cite|improve this question







New contributor




Mike L is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




Mike L is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









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share|cite|improve this question






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asked 2 days ago









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Mike L is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Mike L is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • Sure. Just re-order the sum by replacing $n$ with $-n$.
    – DisintegratingByParts
    2 days ago


















  • Sure. Just re-order the sum by replacing $n$ with $-n$.
    – DisintegratingByParts
    2 days ago
















Sure. Just re-order the sum by replacing $n$ with $-n$.
– DisintegratingByParts
2 days ago




Sure. Just re-order the sum by replacing $n$ with $-n$.
– DisintegratingByParts
2 days ago










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Yes: a possibility is to work with the functions $widetilde{A}colon xmapsto Aleft(-xright)$ and $widetilde{B}colon xmapsto Bleft(-xright)$ then do the substitution $y=-x$ in the integral.






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    Yes: a possibility is to work with the functions $widetilde{A}colon xmapsto Aleft(-xright)$ and $widetilde{B}colon xmapsto Bleft(-xright)$ then do the substitution $y=-x$ in the integral.






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      Yes: a possibility is to work with the functions $widetilde{A}colon xmapsto Aleft(-xright)$ and $widetilde{B}colon xmapsto Bleft(-xright)$ then do the substitution $y=-x$ in the integral.






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        Yes: a possibility is to work with the functions $widetilde{A}colon xmapsto Aleft(-xright)$ and $widetilde{B}colon xmapsto Bleft(-xright)$ then do the substitution $y=-x$ in the integral.






        share|cite|improve this answer












        Yes: a possibility is to work with the functions $widetilde{A}colon xmapsto Aleft(-xright)$ and $widetilde{B}colon xmapsto Bleft(-xright)$ then do the substitution $y=-x$ in the integral.







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        Davide Giraudo

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