Statement of Parseval's theorem for Fourier Transform
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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 Jan 3 at 19:26
Mike LMike L
1
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add a comment |
$begingroup$
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 Jan 3 at 19:26
Mike LMike L
1
$endgroup$
$begingroup$
Sure. Just re-order the sum by replacing $n$ with $-n$.
$endgroup$
– DisintegratingByParts
Jan 3 at 19:33
add a comment |
$begingroup$
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 Jan 3 at 19:26
Mike LMike L
1
$endgroup$
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 Jan 3 at 19:26
Mike LMike L
1
asked Jan 3 at 19:26
Mike LMike L
1
asked Jan 3 at 19:26
Mike LMike L
1
asked Jan 3 at 19:26
Mike LMike L
1
asked Jan 3 at 19:26
Mike LMike L
1
1
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Sure. Just re-order the sum by replacing $n$ with $-n$.
$endgroup$
– DisintegratingByParts
Jan 3 at 19:33
add a comment |
$begingroup$
Sure. Just re-order the sum by replacing $n$ with $-n$.
$endgroup$
– DisintegratingByParts
Jan 3 at 19:33
$begingroup$
Sure. Just re-order the sum by replacing $n$ with $-n$.
$endgroup$
– DisintegratingByParts
Jan 3 at 19:33
$begingroup$
Sure. Just re-order the sum by replacing $n$ with $-n$.
$endgroup$
– DisintegratingByParts
Jan 3 at 19:33
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 Jan 4 at 16:58
Davide GiraudoDavide Giraudo
128k17156268
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1 Answer
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1 Answer
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$begingroup$
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 Jan 4 at 16:58
Davide GiraudoDavide Giraudo
128k17156268
$endgroup$
add a comment |
$begingroup$
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 Jan 4 at 16:58
Davide GiraudoDavide Giraudo
128k17156268
$endgroup$
add a comment |
$begingroup$
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 Jan 4 at 16:58
Davide GiraudoDavide Giraudo
128k17156268
$endgroup$
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 Jan 4 at 16:58
Davide GiraudoDavide Giraudo
128k17156268
answered Jan 4 at 16:58
Davide GiraudoDavide Giraudo
128k17156268
answered Jan 4 at 16:58
Davide GiraudoDavide Giraudo
128k17156268
answered Jan 4 at 16:58
Davide GiraudoDavide Giraudo
128k17156268
128k17156268
add a comment |
add a comment |
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Sure. Just re-order the sum by replacing $n$ with $-n$.
$endgroup$
– DisintegratingByParts
Jan 3 at 19:33