Statement of Parseval's theorem for Fourier Transform
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
asked 2 days ago
Mike L
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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
asked 2 days ago
Mike L
1
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
add a comment |
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
asked 2 days ago
Mike L
1
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
asked 2 days ago
Mike L
1
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.
asked 2 days ago
Mike L
1
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.
asked 2 days ago
Mike L
1
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.
asked 2 days ago
Mike L
1
asked 2 days ago
Mike L
1
1
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.
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.
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
add a comment |
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
add a comment |
1 Answer
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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.
answered yesterday
Davide Giraudo
125k16150260
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1 Answer
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oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
answered yesterday
Davide Giraudo
125k16150260
add a comment |
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.
answered yesterday
Davide Giraudo
125k16150260
add a comment |
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.
answered yesterday
Davide Giraudo
125k16150260
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.
answered yesterday
Davide Giraudo
125k16150260
answered yesterday
Davide Giraudo
125k16150260
answered yesterday
Davide Giraudo
125k16150260
answered yesterday
Davide Giraudo
125k16150260
125k16150260
add a comment |
add a comment |
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Sure. Just re-order the sum by replacing $n$ with $-n$.
– DisintegratingByParts
2 days ago