Prove the equality of Parseval
$begingroup$
Let $f$ a continuous function in $[-L,L]$ with $f'$ piecewise continuous. Suppose $f(-L)=f(L)$, then the coefficients of fourier of $f$ in $[-L,L]$ satisfy:
$$2a_0+sum_{n=1}^infty (a_n^2+b_n^2)=frac{1}{L}int_{-L}^{L}f^2(x)dx.$$
My attempt
As $f$ is integrable in $[-L,L]$ the coefficients of fourier serie is
$$a_0=frac{1}{2L}intlimits_{-L}^{L}f(x)dx$$
$$a_n=frac{1}{L}intlimits_{-L}^{L}f(x)cos(frac{npi x}{L})dx$$
$$b_n=frac{1}{L}intlimits_{-L}^{L}f(x)sin(frac{npi x}{L})dx.$$
Let the N-esima partial sum of fourier serie of $f$
$$S_N(x)=a_0+sumlimits_{n=1}^{N}[a_ncos(frac{npi x}{L})+b_nsin(frac{npi x}{L})].$$
I know by Bessel inequality
$$2a_0+sum_{n=1}^infty (a_n^2+b_n^2)leqfrac{1}{L}int_{-L}^{L}f^2(x)dx.$$
Here I'm stuck. Can someone help me?
calculus fourier-analysis harmonic-analysis
edited Dec 10 '18 at 16:59
rafa11111
1,1932417
asked Dec 10 '18 at 16:14
Bvss12Bvss12
1,821619
$endgroup$
add a comment |
$begingroup$
Let $f$ a continuous function in $[-L,L]$ with $f'$ piecewise continuous. Suppose $f(-L)=f(L)$, then the coefficients of fourier of $f$ in $[-L,L]$ satisfy:
$$2a_0+sum_{n=1}^infty (a_n^2+b_n^2)=frac{1}{L}int_{-L}^{L}f^2(x)dx.$$
My attempt
As $f$ is integrable in $[-L,L]$ the coefficients of fourier serie is
$$a_0=frac{1}{2L}intlimits_{-L}^{L}f(x)dx$$
$$a_n=frac{1}{L}intlimits_{-L}^{L}f(x)cos(frac{npi x}{L})dx$$
$$b_n=frac{1}{L}intlimits_{-L}^{L}f(x)sin(frac{npi x}{L})dx.$$
Let the N-esima partial sum of fourier serie of $f$
$$S_N(x)=a_0+sumlimits_{n=1}^{N}[a_ncos(frac{npi x}{L})+b_nsin(frac{npi x}{L})].$$
I know by Bessel inequality
$$2a_0+sum_{n=1}^infty (a_n^2+b_n^2)leqfrac{1}{L}int_{-L}^{L}f^2(x)dx.$$
Here I'm stuck. Can someone help me?
calculus fourier-analysis harmonic-analysis
edited Dec 10 '18 at 16:59
rafa11111
1,1932417
asked Dec 10 '18 at 16:14
Bvss12Bvss12
1,821619
$endgroup$
1
$begingroup$
I think the first term is wrong. Just plug in $f(x)=C$.
$endgroup$
– Andrei
Dec 10 '18 at 16:34
$begingroup$
Yes sorry @Andrei is $2L$
$endgroup$
– Bvss12
Dec 10 '18 at 16:35
$begingroup$
There should be a square too.
$endgroup$
– Andrei
Dec 10 '18 at 16:42
add a comment |
$begingroup$
Let $f$ a continuous function in $[-L,L]$ with $f'$ piecewise continuous. Suppose $f(-L)=f(L)$, then the coefficients of fourier of $f$ in $[-L,L]$ satisfy:
$$2a_0+sum_{n=1}^infty (a_n^2+b_n^2)=frac{1}{L}int_{-L}^{L}f^2(x)dx.$$
My attempt
As $f$ is integrable in $[-L,L]$ the coefficients of fourier serie is
$$a_0=frac{1}{2L}intlimits_{-L}^{L}f(x)dx$$
$$a_n=frac{1}{L}intlimits_{-L}^{L}f(x)cos(frac{npi x}{L})dx$$
$$b_n=frac{1}{L}intlimits_{-L}^{L}f(x)sin(frac{npi x}{L})dx.$$
Let the N-esima partial sum of fourier serie of $f$
$$S_N(x)=a_0+sumlimits_{n=1}^{N}[a_ncos(frac{npi x}{L})+b_nsin(frac{npi x}{L})].$$
I know by Bessel inequality
$$2a_0+sum_{n=1}^infty (a_n^2+b_n^2)leqfrac{1}{L}int_{-L}^{L}f^2(x)dx.$$
Here I'm stuck. Can someone help me?
calculus fourier-analysis harmonic-analysis
edited Dec 10 '18 at 16:59
rafa11111
1,1932417
asked Dec 10 '18 at 16:14
Bvss12Bvss12
1,821619
$endgroup$
Let $f$ a continuous function in $[-L,L]$ with $f'$ piecewise continuous. Suppose $f(-L)=f(L)$, then the coefficients of fourier of $f$ in $[-L,L]$ satisfy:
$$2a_0+sum_{n=1}^infty (a_n^2+b_n^2)=frac{1}{L}int_{-L}^{L}f^2(x)dx.$$
My attempt
As $f$ is integrable in $[-L,L]$ the coefficients of fourier serie is
$$a_0=frac{1}{2L}intlimits_{-L}^{L}f(x)dx$$
$$a_n=frac{1}{L}intlimits_{-L}^{L}f(x)cos(frac{npi x}{L})dx$$
$$b_n=frac{1}{L}intlimits_{-L}^{L}f(x)sin(frac{npi x}{L})dx.$$
Let the N-esima partial sum of fourier serie of $f$
$$S_N(x)=a_0+sumlimits_{n=1}^{N}[a_ncos(frac{npi x}{L})+b_nsin(frac{npi x}{L})].$$
I know by Bessel inequality
$$2a_0+sum_{n=1}^infty (a_n^2+b_n^2)leqfrac{1}{L}int_{-L}^{L}f^2(x)dx.$$
Here I'm stuck. Can someone help me?
calculus fourier-analysis harmonic-analysis
calculus fourier-analysis harmonic-analysis
edited Dec 10 '18 at 16:59
rafa11111
1,1932417
asked Dec 10 '18 at 16:14
Bvss12Bvss12
1,821619
edited Dec 10 '18 at 16:59
rafa11111
1,1932417
asked Dec 10 '18 at 16:14
Bvss12Bvss12
1,821619
edited Dec 10 '18 at 16:59
rafa11111
1,1932417
edited Dec 10 '18 at 16:59
rafa11111
1,1932417
edited Dec 10 '18 at 16:59
rafa11111
1,1932417
1,1932417
asked Dec 10 '18 at 16:14
Bvss12Bvss12
1,821619
asked Dec 10 '18 at 16:14
Bvss12Bvss12
1,821619
asked Dec 10 '18 at 16:14
Bvss12Bvss12
1,821619
1,821619
1
$begingroup$
I think the first term is wrong. Just plug in $f(x)=C$.
$endgroup$
– Andrei
Dec 10 '18 at 16:34
$begingroup$
Yes sorry @Andrei is $2L$
$endgroup$
– Bvss12
Dec 10 '18 at 16:35
$begingroup$
There should be a square too.
$endgroup$
– Andrei
Dec 10 '18 at 16:42
add a comment |
1
$begingroup$
I think the first term is wrong. Just plug in $f(x)=C$.
$endgroup$
– Andrei
Dec 10 '18 at 16:34
$begingroup$
Yes sorry @Andrei is $2L$
$endgroup$
– Bvss12
Dec 10 '18 at 16:35
$begingroup$
There should be a square too.
$endgroup$
– Andrei
Dec 10 '18 at 16:42
1
1
$begingroup$
I think the first term is wrong. Just plug in $f(x)=C$.
$endgroup$
– Andrei
Dec 10 '18 at 16:34
$begingroup$
I think the first term is wrong. Just plug in $f(x)=C$.
$endgroup$
– Andrei
Dec 10 '18 at 16:34
$begingroup$
Yes sorry @Andrei is $2L$
$endgroup$
– Bvss12
Dec 10 '18 at 16:35
$begingroup$
Yes sorry @Andrei is $2L$
$endgroup$
– Bvss12
Dec 10 '18 at 16:35
$begingroup$
There should be a square too.
$endgroup$
– Andrei
Dec 10 '18 at 16:42
$begingroup$
There should be a square too.
$endgroup$
– Andrei
Dec 10 '18 at 16:42
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint: Just consider it intuitively. Suppose that
$$f=sum_{n=1}^{N}b_nsinleft(frac{pi n x}{L}right)$$
then
$$f^2=sum_{n,k=1}^{N}b_{n}b_{k}sinleft(frac{pi n x}{L}right)sinleft(frac{pi k x}{L}right).$$
Since
$$int_{-L}^L sinleft(frac{pi n x}{L}right)sinleft(frac{pi k x}{L}right)dx=Ldelta_{n,k}$$
we have that in this case
$$int_{-L}^L f^2 dx=Lsum_{n,k=1}^{N}b_{n}b_{k}delta_{n,k}=Lsum_{n=1}^N b_n^2.$$
The square of the function will sum over all possible pairs, and pairs that aren't the same will integrate to zero. How can you extend this intuition for finite sums to infinite sums? The argument is also the same when you toss in cosines and a constant term.
answered Dec 10 '18 at 16:42
Will FisherWill Fisher
4,05311132
$endgroup$
$begingroup$
In this case i need $f=a_0+sum_{n=1}^{N}[a_ncos (frac{npi x}{L}+b_nsin(frac{pi n x}{L})]$ and the square of this. no?
$endgroup$
– Bvss12
Dec 10 '18 at 16:59
$begingroup$
Yes, but it is the same argument when you also add those terms in because $$int_{-L}^Lsinleft(frac{pi n x}{L}right)cosleft(frac{pi k x}{L}right)dx = 0.$$
$endgroup$
– Will Fisher
Dec 10 '18 at 17:02
$begingroup$
By the way, Parsevel's theorem gives $$frac{1}{L}int_{-L}^L f^2 dx=frac{1}{2}a_0^2+sum_{nge 1}(a_n^2+b_n^2).$$ Your $a_0$ term is off.
$endgroup$
– Will Fisher
Dec 10 '18 at 17:03
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Hint: Just consider it intuitively. Suppose that
$$f=sum_{n=1}^{N}b_nsinleft(frac{pi n x}{L}right)$$
then
$$f^2=sum_{n,k=1}^{N}b_{n}b_{k}sinleft(frac{pi n x}{L}right)sinleft(frac{pi k x}{L}right).$$
Since
$$int_{-L}^L sinleft(frac{pi n x}{L}right)sinleft(frac{pi k x}{L}right)dx=Ldelta_{n,k}$$
we have that in this case
$$int_{-L}^L f^2 dx=Lsum_{n,k=1}^{N}b_{n}b_{k}delta_{n,k}=Lsum_{n=1}^N b_n^2.$$
The square of the function will sum over all possible pairs, and pairs that aren't the same will integrate to zero. How can you extend this intuition for finite sums to infinite sums? The argument is also the same when you toss in cosines and a constant term.
answered Dec 10 '18 at 16:42
Will FisherWill Fisher
4,05311132
$endgroup$
$begingroup$
In this case i need $f=a_0+sum_{n=1}^{N}[a_ncos (frac{npi x}{L}+b_nsin(frac{pi n x}{L})]$ and the square of this. no?
$endgroup$
– Bvss12
Dec 10 '18 at 16:59
$begingroup$
Yes, but it is the same argument when you also add those terms in because $$int_{-L}^Lsinleft(frac{pi n x}{L}right)cosleft(frac{pi k x}{L}right)dx = 0.$$
$endgroup$
– Will Fisher
Dec 10 '18 at 17:02
$begingroup$
By the way, Parsevel's theorem gives $$frac{1}{L}int_{-L}^L f^2 dx=frac{1}{2}a_0^2+sum_{nge 1}(a_n^2+b_n^2).$$ Your $a_0$ term is off.
$endgroup$
– Will Fisher
Dec 10 '18 at 17:03
add a comment |
$begingroup$
Hint: Just consider it intuitively. Suppose that
$$f=sum_{n=1}^{N}b_nsinleft(frac{pi n x}{L}right)$$
then
$$f^2=sum_{n,k=1}^{N}b_{n}b_{k}sinleft(frac{pi n x}{L}right)sinleft(frac{pi k x}{L}right).$$
Since
$$int_{-L}^L sinleft(frac{pi n x}{L}right)sinleft(frac{pi k x}{L}right)dx=Ldelta_{n,k}$$
we have that in this case
$$int_{-L}^L f^2 dx=Lsum_{n,k=1}^{N}b_{n}b_{k}delta_{n,k}=Lsum_{n=1}^N b_n^2.$$
The square of the function will sum over all possible pairs, and pairs that aren't the same will integrate to zero. How can you extend this intuition for finite sums to infinite sums? The argument is also the same when you toss in cosines and a constant term.
answered Dec 10 '18 at 16:42
Will FisherWill Fisher
4,05311132
$endgroup$
$begingroup$
In this case i need $f=a_0+sum_{n=1}^{N}[a_ncos (frac{npi x}{L}+b_nsin(frac{pi n x}{L})]$ and the square of this. no?
$endgroup$
– Bvss12
Dec 10 '18 at 16:59
$begingroup$
Yes, but it is the same argument when you also add those terms in because $$int_{-L}^Lsinleft(frac{pi n x}{L}right)cosleft(frac{pi k x}{L}right)dx = 0.$$
$endgroup$
– Will Fisher
Dec 10 '18 at 17:02
$begingroup$
By the way, Parsevel's theorem gives $$frac{1}{L}int_{-L}^L f^2 dx=frac{1}{2}a_0^2+sum_{nge 1}(a_n^2+b_n^2).$$ Your $a_0$ term is off.
$endgroup$
– Will Fisher
Dec 10 '18 at 17:03
add a comment |
$begingroup$
Hint: Just consider it intuitively. Suppose that
$$f=sum_{n=1}^{N}b_nsinleft(frac{pi n x}{L}right)$$
then
$$f^2=sum_{n,k=1}^{N}b_{n}b_{k}sinleft(frac{pi n x}{L}right)sinleft(frac{pi k x}{L}right).$$
Since
$$int_{-L}^L sinleft(frac{pi n x}{L}right)sinleft(frac{pi k x}{L}right)dx=Ldelta_{n,k}$$
we have that in this case
$$int_{-L}^L f^2 dx=Lsum_{n,k=1}^{N}b_{n}b_{k}delta_{n,k}=Lsum_{n=1}^N b_n^2.$$
The square of the function will sum over all possible pairs, and pairs that aren't the same will integrate to zero. How can you extend this intuition for finite sums to infinite sums? The argument is also the same when you toss in cosines and a constant term.
answered Dec 10 '18 at 16:42
Will FisherWill Fisher
4,05311132
$endgroup$
Hint: Just consider it intuitively. Suppose that
$$f=sum_{n=1}^{N}b_nsinleft(frac{pi n x}{L}right)$$
then
$$f^2=sum_{n,k=1}^{N}b_{n}b_{k}sinleft(frac{pi n x}{L}right)sinleft(frac{pi k x}{L}right).$$
Since
$$int_{-L}^L sinleft(frac{pi n x}{L}right)sinleft(frac{pi k x}{L}right)dx=Ldelta_{n,k}$$
we have that in this case
$$int_{-L}^L f^2 dx=Lsum_{n,k=1}^{N}b_{n}b_{k}delta_{n,k}=Lsum_{n=1}^N b_n^2.$$
The square of the function will sum over all possible pairs, and pairs that aren't the same will integrate to zero. How can you extend this intuition for finite sums to infinite sums? The argument is also the same when you toss in cosines and a constant term.
answered Dec 10 '18 at 16:42
Will FisherWill Fisher
4,05311132
edited Dec 10 '18 at 16:51
answered Dec 10 '18 at 16:42
Will FisherWill Fisher
4,05311132
answered Dec 10 '18 at 16:42
Will FisherWill Fisher
4,05311132
answered Dec 10 '18 at 16:42
Will FisherWill Fisher
4,05311132
4,05311132
$begingroup$
In this case i need $f=a_0+sum_{n=1}^{N}[a_ncos (frac{npi x}{L}+b_nsin(frac{pi n x}{L})]$ and the square of this. no?
$endgroup$
– Bvss12
Dec 10 '18 at 16:59
$begingroup$
Yes, but it is the same argument when you also add those terms in because $$int_{-L}^Lsinleft(frac{pi n x}{L}right)cosleft(frac{pi k x}{L}right)dx = 0.$$
$endgroup$
– Will Fisher
Dec 10 '18 at 17:02
$begingroup$
By the way, Parsevel's theorem gives $$frac{1}{L}int_{-L}^L f^2 dx=frac{1}{2}a_0^2+sum_{nge 1}(a_n^2+b_n^2).$$ Your $a_0$ term is off.
$endgroup$
– Will Fisher
Dec 10 '18 at 17:03
add a comment |
$begingroup$
In this case i need $f=a_0+sum_{n=1}^{N}[a_ncos (frac{npi x}{L}+b_nsin(frac{pi n x}{L})]$ and the square of this. no?
$endgroup$
– Bvss12
Dec 10 '18 at 16:59
$begingroup$
Yes, but it is the same argument when you also add those terms in because $$int_{-L}^Lsinleft(frac{pi n x}{L}right)cosleft(frac{pi k x}{L}right)dx = 0.$$
$endgroup$
– Will Fisher
Dec 10 '18 at 17:02
$begingroup$
By the way, Parsevel's theorem gives $$frac{1}{L}int_{-L}^L f^2 dx=frac{1}{2}a_0^2+sum_{nge 1}(a_n^2+b_n^2).$$ Your $a_0$ term is off.
$endgroup$
– Will Fisher
Dec 10 '18 at 17:03
$begingroup$
In this case i need $f=a_0+sum_{n=1}^{N}[a_ncos (frac{npi x}{L}+b_nsin(frac{pi n x}{L})]$ and the square of this. no?
$endgroup$
– Bvss12
Dec 10 '18 at 16:59
$begingroup$
In this case i need $f=a_0+sum_{n=1}^{N}[a_ncos (frac{npi x}{L}+b_nsin(frac{pi n x}{L})]$ and the square of this. no?
$endgroup$
– Bvss12
Dec 10 '18 at 16:59
$begingroup$
Yes, but it is the same argument when you also add those terms in because $$int_{-L}^Lsinleft(frac{pi n x}{L}right)cosleft(frac{pi k x}{L}right)dx = 0.$$
$endgroup$
– Will Fisher
Dec 10 '18 at 17:02
$begingroup$
Yes, but it is the same argument when you also add those terms in because $$int_{-L}^Lsinleft(frac{pi n x}{L}right)cosleft(frac{pi k x}{L}right)dx = 0.$$
$endgroup$
– Will Fisher
Dec 10 '18 at 17:02
$begingroup$
By the way, Parsevel's theorem gives $$frac{1}{L}int_{-L}^L f^2 dx=frac{1}{2}a_0^2+sum_{nge 1}(a_n^2+b_n^2).$$ Your $a_0$ term is off.
$endgroup$
– Will Fisher
Dec 10 '18 at 17:03
$begingroup$
By the way, Parsevel's theorem gives $$frac{1}{L}int_{-L}^L f^2 dx=frac{1}{2}a_0^2+sum_{nge 1}(a_n^2+b_n^2).$$ Your $a_0$ term is off.
$endgroup$
– Will Fisher
Dec 10 '18 at 17:03
add a comment |
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1
$begingroup$
I think the first term is wrong. Just plug in $f(x)=C$.
$endgroup$
– Andrei
Dec 10 '18 at 16:34
$begingroup$
Yes sorry @Andrei is $2L$
$endgroup$
– Bvss12
Dec 10 '18 at 16:35
$begingroup$
There should be a square too.
$endgroup$
– Andrei
Dec 10 '18 at 16:42