Use orthogonality to proof Parseval's identity for the general Fourier series written as the power spectrum
$begingroup$
I need to show that $$int_{-pi}^{pi}left|frac{a_0}{2}+sum_{n=1}^{infty}alpha_ncos(nx-theta_n)right|^2dx=2pileft(frac{a_0^2}{4}+frac12sum_{n=1}^{infty}alpha_n^2right)tag{1}$$
Just for reference the trigonometric Fourier series is
$$f(x)= frac{a_0}{2} + sum_{n=1}^{infty}left(a_ncos(nx)+b_nsin(nx)right)$$
and the connection between the trigonometric Fourier series and the power spectrum is given by
$$a_n=alpha_ncostheta_n$$
$$b_n=alpha_nsintheta_n$$
$$alpha_n^2=a_n^2+b_n^2$$
$$tantheta_n=frac{b_n}{a_n}$$
So I start by expanding the LHS of $(mathrm{1})$
$$int_{-pi}^{pi}left{frac{a_0^2}{4}+a_0sum_{n=1}^{infty}alpha_ncos(nx-theta_n)+left[sum_{n=1}^{infty}alpha_ncos(nx-theta_n)right]^2right}dx$$
$$=
int_{-pi}^{pi}frac{a_0^2}{4}dx+a_0int_{-pi}^{pi}sum_{n=1}^{infty}alpha_ncos(nx-theta_n)dx+int_{-pi}^{pi}left[sum_{n=1}^{infty}alpha_ncos(nx-theta_n)right]^2dx$$
$$= 2pifrac{a_0^2}{4}+a_0int_{-pi}^{pi}sum_{n=1}^{infty}alpha_ncos(nx-theta_n)dx$$
$$+int_{-pi}^{pi}sum_{n=1}^{infty}alpha_ncos(nx-theta_n)sum_{m=1}^{infty}alpha_mcos(mx-theta_m)dxtag{2}$$
I don't know how to proceed any further with this but do I know that for integer $nne m$
$$langlecos(nx)|cos(mx)rangle=0$$
but I am struggling to apply the same logic to $(mathrm{2})$ as the cosines have different phase offsets, I am also confused about how to deal with the 2 sums in the second integral.
The answer just states that:
Does anyone have any advice on how I can complete this proof?
summation fourier-analysis fourier-series
asked Nov 26 '18 at 17:22
BLAZEBLAZE
6,076112755
$endgroup$
add a comment |
$begingroup$
I need to show that $$int_{-pi}^{pi}left|frac{a_0}{2}+sum_{n=1}^{infty}alpha_ncos(nx-theta_n)right|^2dx=2pileft(frac{a_0^2}{4}+frac12sum_{n=1}^{infty}alpha_n^2right)tag{1}$$
Just for reference the trigonometric Fourier series is
$$f(x)= frac{a_0}{2} + sum_{n=1}^{infty}left(a_ncos(nx)+b_nsin(nx)right)$$
and the connection between the trigonometric Fourier series and the power spectrum is given by
$$a_n=alpha_ncostheta_n$$
$$b_n=alpha_nsintheta_n$$
$$alpha_n^2=a_n^2+b_n^2$$
$$tantheta_n=frac{b_n}{a_n}$$
So I start by expanding the LHS of $(mathrm{1})$
$$int_{-pi}^{pi}left{frac{a_0^2}{4}+a_0sum_{n=1}^{infty}alpha_ncos(nx-theta_n)+left[sum_{n=1}^{infty}alpha_ncos(nx-theta_n)right]^2right}dx$$
$$=
int_{-pi}^{pi}frac{a_0^2}{4}dx+a_0int_{-pi}^{pi}sum_{n=1}^{infty}alpha_ncos(nx-theta_n)dx+int_{-pi}^{pi}left[sum_{n=1}^{infty}alpha_ncos(nx-theta_n)right]^2dx$$
$$= 2pifrac{a_0^2}{4}+a_0int_{-pi}^{pi}sum_{n=1}^{infty}alpha_ncos(nx-theta_n)dx$$
$$+int_{-pi}^{pi}sum_{n=1}^{infty}alpha_ncos(nx-theta_n)sum_{m=1}^{infty}alpha_mcos(mx-theta_m)dxtag{2}$$
I don't know how to proceed any further with this but do I know that for integer $nne m$
$$langlecos(nx)|cos(mx)rangle=0$$
but I am struggling to apply the same logic to $(mathrm{2})$ as the cosines have different phase offsets, I am also confused about how to deal with the 2 sums in the second integral.
The answer just states that:
Does anyone have any advice on how I can complete this proof?
summation fourier-analysis fourier-series
asked Nov 26 '18 at 17:22
BLAZEBLAZE
6,076112755
$endgroup$
1
$begingroup$
You have to argue why the infinite sum can be exchanged with the integral, and then, as $cos$ is periodic, we get the same integrals for each $n$ if we drop $theta_n$..
$endgroup$
– Berci
Nov 26 '18 at 17:40
$begingroup$
@Berci Thanks for your reply, I'm not sure why the infinite sum can be exchanged with the integral. In fact, I don't even understand what you mean by 'exchange'. Could you please elaborate on this in an answer?
$endgroup$
– BLAZE
Nov 26 '18 at 19:37
add a comment |
$begingroup$
I need to show that $$int_{-pi}^{pi}left|frac{a_0}{2}+sum_{n=1}^{infty}alpha_ncos(nx-theta_n)right|^2dx=2pileft(frac{a_0^2}{4}+frac12sum_{n=1}^{infty}alpha_n^2right)tag{1}$$
Just for reference the trigonometric Fourier series is
$$f(x)= frac{a_0}{2} + sum_{n=1}^{infty}left(a_ncos(nx)+b_nsin(nx)right)$$
and the connection between the trigonometric Fourier series and the power spectrum is given by
$$a_n=alpha_ncostheta_n$$
$$b_n=alpha_nsintheta_n$$
$$alpha_n^2=a_n^2+b_n^2$$
$$tantheta_n=frac{b_n}{a_n}$$
So I start by expanding the LHS of $(mathrm{1})$
$$int_{-pi}^{pi}left{frac{a_0^2}{4}+a_0sum_{n=1}^{infty}alpha_ncos(nx-theta_n)+left[sum_{n=1}^{infty}alpha_ncos(nx-theta_n)right]^2right}dx$$
$$=
int_{-pi}^{pi}frac{a_0^2}{4}dx+a_0int_{-pi}^{pi}sum_{n=1}^{infty}alpha_ncos(nx-theta_n)dx+int_{-pi}^{pi}left[sum_{n=1}^{infty}alpha_ncos(nx-theta_n)right]^2dx$$
$$= 2pifrac{a_0^2}{4}+a_0int_{-pi}^{pi}sum_{n=1}^{infty}alpha_ncos(nx-theta_n)dx$$
$$+int_{-pi}^{pi}sum_{n=1}^{infty}alpha_ncos(nx-theta_n)sum_{m=1}^{infty}alpha_mcos(mx-theta_m)dxtag{2}$$
I don't know how to proceed any further with this but do I know that for integer $nne m$
$$langlecos(nx)|cos(mx)rangle=0$$
but I am struggling to apply the same logic to $(mathrm{2})$ as the cosines have different phase offsets, I am also confused about how to deal with the 2 sums in the second integral.
The answer just states that:
Does anyone have any advice on how I can complete this proof?
summation fourier-analysis fourier-series
asked Nov 26 '18 at 17:22
BLAZEBLAZE
6,076112755
$endgroup$
I need to show that $$int_{-pi}^{pi}left|frac{a_0}{2}+sum_{n=1}^{infty}alpha_ncos(nx-theta_n)right|^2dx=2pileft(frac{a_0^2}{4}+frac12sum_{n=1}^{infty}alpha_n^2right)tag{1}$$
Just for reference the trigonometric Fourier series is
$$f(x)= frac{a_0}{2} + sum_{n=1}^{infty}left(a_ncos(nx)+b_nsin(nx)right)$$
and the connection between the trigonometric Fourier series and the power spectrum is given by
$$a_n=alpha_ncostheta_n$$
$$b_n=alpha_nsintheta_n$$
$$alpha_n^2=a_n^2+b_n^2$$
$$tantheta_n=frac{b_n}{a_n}$$
So I start by expanding the LHS of $(mathrm{1})$
$$int_{-pi}^{pi}left{frac{a_0^2}{4}+a_0sum_{n=1}^{infty}alpha_ncos(nx-theta_n)+left[sum_{n=1}^{infty}alpha_ncos(nx-theta_n)right]^2right}dx$$
$$=
int_{-pi}^{pi}frac{a_0^2}{4}dx+a_0int_{-pi}^{pi}sum_{n=1}^{infty}alpha_ncos(nx-theta_n)dx+int_{-pi}^{pi}left[sum_{n=1}^{infty}alpha_ncos(nx-theta_n)right]^2dx$$
$$= 2pifrac{a_0^2}{4}+a_0int_{-pi}^{pi}sum_{n=1}^{infty}alpha_ncos(nx-theta_n)dx$$
$$+int_{-pi}^{pi}sum_{n=1}^{infty}alpha_ncos(nx-theta_n)sum_{m=1}^{infty}alpha_mcos(mx-theta_m)dxtag{2}$$
I don't know how to proceed any further with this but do I know that for integer $nne m$
$$langlecos(nx)|cos(mx)rangle=0$$
but I am struggling to apply the same logic to $(mathrm{2})$ as the cosines have different phase offsets, I am also confused about how to deal with the 2 sums in the second integral.
The answer just states that:
Does anyone have any advice on how I can complete this proof?
summation fourier-analysis fourier-series
summation fourier-analysis fourier-series
asked Nov 26 '18 at 17:22
BLAZEBLAZE
6,076112755
asked Nov 26 '18 at 17:22
BLAZEBLAZE
6,076112755
edited Nov 26 '18 at 19:51
BLAZE
asked Nov 26 '18 at 17:22
BLAZEBLAZE
6,076112755
asked Nov 26 '18 at 17:22
BLAZEBLAZE
6,076112755
asked Nov 26 '18 at 17:22
BLAZEBLAZE
6,076112755
6,076112755
1
$begingroup$
You have to argue why the infinite sum can be exchanged with the integral, and then, as $cos$ is periodic, we get the same integrals for each $n$ if we drop $theta_n$..
$endgroup$
– Berci
Nov 26 '18 at 17:40
$begingroup$
@Berci Thanks for your reply, I'm not sure why the infinite sum can be exchanged with the integral. In fact, I don't even understand what you mean by 'exchange'. Could you please elaborate on this in an answer?
$endgroup$
– BLAZE
Nov 26 '18 at 19:37
add a comment |
1
$begingroup$
You have to argue why the infinite sum can be exchanged with the integral, and then, as $cos$ is periodic, we get the same integrals for each $n$ if we drop $theta_n$..
$endgroup$
– Berci
Nov 26 '18 at 17:40
$begingroup$
@Berci Thanks for your reply, I'm not sure why the infinite sum can be exchanged with the integral. In fact, I don't even understand what you mean by 'exchange'. Could you please elaborate on this in an answer?
$endgroup$
– BLAZE
Nov 26 '18 at 19:37
1
1
$begingroup$
You have to argue why the infinite sum can be exchanged with the integral, and then, as $cos$ is periodic, we get the same integrals for each $n$ if we drop $theta_n$..
$endgroup$
– Berci
Nov 26 '18 at 17:40
$begingroup$
You have to argue why the infinite sum can be exchanged with the integral, and then, as $cos$ is periodic, we get the same integrals for each $n$ if we drop $theta_n$..
$endgroup$
– Berci
Nov 26 '18 at 17:40
$begingroup$
@Berci Thanks for your reply, I'm not sure why the infinite sum can be exchanged with the integral. In fact, I don't even understand what you mean by 'exchange'. Could you please elaborate on this in an answer?
$endgroup$
– BLAZE
Nov 26 '18 at 19:37
$begingroup$
@Berci Thanks for your reply, I'm not sure why the infinite sum can be exchanged with the integral. In fact, I don't even understand what you mean by 'exchange'. Could you please elaborate on this in an answer?
$endgroup$
– BLAZE
Nov 26 '18 at 19:37
add a comment |
1 Answer
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$begingroup$
$$cos(nx-theta_n)=cos(nx)cos(theta_n)+sin(nx)sin(theta_n)$$
Therefore
$$int_{-pi}^pi cos(nx-theta_n)cos(mx-theta_m)dx={cos(theta_n)cos(theta_m)int_{-pi}^pi left(cos(nx)cos(mx)right)dx quadtext{etc.}}$$
For $nne m$, $$int_{-pi}^pi cos(nx)cos(mx)dx=0$$ and in general $$int_{-pi}^pi cos(nx)sin(mx)dx=0$$ Meanwhile $$int_{-pi}^pi cos^2(nx)dx=pi$$ This will allow you to complete $(2)$.
edited Nov 26 '18 at 19:46
BLAZE
6,076112755
answered Nov 26 '18 at 17:49
herb steinbergherb steinberg
2,5582310
$endgroup$
$begingroup$
I don't know who edited my answer. Moreover I don't see the point of it.
$endgroup$
– herb steinberg
Nov 26 '18 at 20:48
$begingroup$
It was me that edited the format of your answer. Do you not agree that it is easier to read now? If not you can revert back to the original.
$endgroup$
– BLAZE
Nov 27 '18 at 15:01
$begingroup$
It is just more spread out. I have found from past experience that reverting to the original is too troublesome, so leave it.
$endgroup$
– herb steinberg
Nov 27 '18 at 16:54
add a comment |
Your Answer
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$begingroup$
$$cos(nx-theta_n)=cos(nx)cos(theta_n)+sin(nx)sin(theta_n)$$
Therefore
$$int_{-pi}^pi cos(nx-theta_n)cos(mx-theta_m)dx={cos(theta_n)cos(theta_m)int_{-pi}^pi left(cos(nx)cos(mx)right)dx quadtext{etc.}}$$
For $nne m$, $$int_{-pi}^pi cos(nx)cos(mx)dx=0$$ and in general $$int_{-pi}^pi cos(nx)sin(mx)dx=0$$ Meanwhile $$int_{-pi}^pi cos^2(nx)dx=pi$$ This will allow you to complete $(2)$.
edited Nov 26 '18 at 19:46
BLAZE
6,076112755
answered Nov 26 '18 at 17:49
herb steinbergherb steinberg
2,5582310
$endgroup$
$begingroup$
I don't know who edited my answer. Moreover I don't see the point of it.
$endgroup$
– herb steinberg
Nov 26 '18 at 20:48
$begingroup$
It was me that edited the format of your answer. Do you not agree that it is easier to read now? If not you can revert back to the original.
$endgroup$
– BLAZE
Nov 27 '18 at 15:01
$begingroup$
It is just more spread out. I have found from past experience that reverting to the original is too troublesome, so leave it.
$endgroup$
– herb steinberg
Nov 27 '18 at 16:54
add a comment |
$begingroup$
$$cos(nx-theta_n)=cos(nx)cos(theta_n)+sin(nx)sin(theta_n)$$
Therefore
$$int_{-pi}^pi cos(nx-theta_n)cos(mx-theta_m)dx={cos(theta_n)cos(theta_m)int_{-pi}^pi left(cos(nx)cos(mx)right)dx quadtext{etc.}}$$
For $nne m$, $$int_{-pi}^pi cos(nx)cos(mx)dx=0$$ and in general $$int_{-pi}^pi cos(nx)sin(mx)dx=0$$ Meanwhile $$int_{-pi}^pi cos^2(nx)dx=pi$$ This will allow you to complete $(2)$.
edited Nov 26 '18 at 19:46
BLAZE
6,076112755
answered Nov 26 '18 at 17:49
herb steinbergherb steinberg
2,5582310
$endgroup$
$begingroup$
I don't know who edited my answer. Moreover I don't see the point of it.
$endgroup$
– herb steinberg
Nov 26 '18 at 20:48
$begingroup$
It was me that edited the format of your answer. Do you not agree that it is easier to read now? If not you can revert back to the original.
$endgroup$
– BLAZE
Nov 27 '18 at 15:01
$begingroup$
It is just more spread out. I have found from past experience that reverting to the original is too troublesome, so leave it.
$endgroup$
– herb steinberg
Nov 27 '18 at 16:54
add a comment |
$begingroup$
$$cos(nx-theta_n)=cos(nx)cos(theta_n)+sin(nx)sin(theta_n)$$
Therefore
$$int_{-pi}^pi cos(nx-theta_n)cos(mx-theta_m)dx={cos(theta_n)cos(theta_m)int_{-pi}^pi left(cos(nx)cos(mx)right)dx quadtext{etc.}}$$
For $nne m$, $$int_{-pi}^pi cos(nx)cos(mx)dx=0$$ and in general $$int_{-pi}^pi cos(nx)sin(mx)dx=0$$ Meanwhile $$int_{-pi}^pi cos^2(nx)dx=pi$$ This will allow you to complete $(2)$.
edited Nov 26 '18 at 19:46
BLAZE
6,076112755
answered Nov 26 '18 at 17:49
herb steinbergherb steinberg
2,5582310
$endgroup$
$$cos(nx-theta_n)=cos(nx)cos(theta_n)+sin(nx)sin(theta_n)$$
Therefore
$$int_{-pi}^pi cos(nx-theta_n)cos(mx-theta_m)dx={cos(theta_n)cos(theta_m)int_{-pi}^pi left(cos(nx)cos(mx)right)dx quadtext{etc.}}$$
For $nne m$, $$int_{-pi}^pi cos(nx)cos(mx)dx=0$$ and in general $$int_{-pi}^pi cos(nx)sin(mx)dx=0$$ Meanwhile $$int_{-pi}^pi cos^2(nx)dx=pi$$ This will allow you to complete $(2)$.
edited Nov 26 '18 at 19:46
BLAZE
6,076112755
answered Nov 26 '18 at 17:49
herb steinbergherb steinberg
2,5582310
edited Nov 26 '18 at 19:46
BLAZE
6,076112755
edited Nov 26 '18 at 19:46
BLAZE
6,076112755
edited Nov 26 '18 at 19:46
BLAZE
6,076112755
6,076112755
answered Nov 26 '18 at 17:49
herb steinbergherb steinberg
2,5582310
answered Nov 26 '18 at 17:49
herb steinbergherb steinberg
2,5582310
answered Nov 26 '18 at 17:49
herb steinbergherb steinberg
2,5582310
2,5582310
$begingroup$
I don't know who edited my answer. Moreover I don't see the point of it.
$endgroup$
– herb steinberg
Nov 26 '18 at 20:48
$begingroup$
It was me that edited the format of your answer. Do you not agree that it is easier to read now? If not you can revert back to the original.
$endgroup$
– BLAZE
Nov 27 '18 at 15:01
$begingroup$
It is just more spread out. I have found from past experience that reverting to the original is too troublesome, so leave it.
$endgroup$
– herb steinberg
Nov 27 '18 at 16:54
add a comment |
$begingroup$
I don't know who edited my answer. Moreover I don't see the point of it.
$endgroup$
– herb steinberg
Nov 26 '18 at 20:48
$begingroup$
It was me that edited the format of your answer. Do you not agree that it is easier to read now? If not you can revert back to the original.
$endgroup$
– BLAZE
Nov 27 '18 at 15:01
$begingroup$
It is just more spread out. I have found from past experience that reverting to the original is too troublesome, so leave it.
$endgroup$
– herb steinberg
Nov 27 '18 at 16:54
$begingroup$
I don't know who edited my answer. Moreover I don't see the point of it.
$endgroup$
– herb steinberg
Nov 26 '18 at 20:48
$begingroup$
I don't know who edited my answer. Moreover I don't see the point of it.
$endgroup$
– herb steinberg
Nov 26 '18 at 20:48
$begingroup$
It was me that edited the format of your answer. Do you not agree that it is easier to read now? If not you can revert back to the original.
$endgroup$
– BLAZE
Nov 27 '18 at 15:01
$begingroup$
It was me that edited the format of your answer. Do you not agree that it is easier to read now? If not you can revert back to the original.
$endgroup$
– BLAZE
Nov 27 '18 at 15:01
$begingroup$
It is just more spread out. I have found from past experience that reverting to the original is too troublesome, so leave it.
$endgroup$
– herb steinberg
Nov 27 '18 at 16:54
$begingroup$
It is just more spread out. I have found from past experience that reverting to the original is too troublesome, so leave it.
$endgroup$
– herb steinberg
Nov 27 '18 at 16:54
add a comment |
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You have to argue why the infinite sum can be exchanged with the integral, and then, as $cos$ is periodic, we get the same integrals for each $n$ if we drop $theta_n$..
$endgroup$
– Berci
Nov 26 '18 at 17:40
$begingroup$
@Berci Thanks for your reply, I'm not sure why the infinite sum can be exchanged with the integral. In fact, I don't even understand what you mean by 'exchange'. Could you please elaborate on this in an answer?
$endgroup$
– BLAZE
Nov 26 '18 at 19:37