Finding a series of cosines to represent the absolute value function
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Question: Prove the following identity:
$|x|=frac{pi}{2}+sum_{n=1}^{infty}frac{cos(2n-1)x}{(2n-1)^2}$. Our current topic is Fourier analysis, but I am truly lost here. I tried to compute the Fourier series, but it seemed extremely different than the wanted identity.
fourier-analysis
asked Dec 13 '18 at 17:00
Uri George PeterzilUri George Peterzil
949
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add a comment |
$begingroup$
Question: Prove the following identity:
$|x|=frac{pi}{2}+sum_{n=1}^{infty}frac{cos(2n-1)x}{(2n-1)^2}$. Our current topic is Fourier analysis, but I am truly lost here. I tried to compute the Fourier series, but it seemed extremely different than the wanted identity.
fourier-analysis
asked Dec 13 '18 at 17:00
Uri George PeterzilUri George Peterzil
949
$endgroup$
1
$begingroup$
Just a comment. Note that the right hand side is periodic while the left hand side is not (as it is written). What you are doing is in fact the Fourier series of a sawtooth-like function. So the result depends on which interval you pick. Common choices are $[-pi,pi]$ or $[0,2pi]$. Since the right hand side is an even function, your choice is the former. This may be the source of confusion.
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– lcv
Dec 13 '18 at 18:47
add a comment |
$begingroup$
Question: Prove the following identity:
$|x|=frac{pi}{2}+sum_{n=1}^{infty}frac{cos(2n-1)x}{(2n-1)^2}$. Our current topic is Fourier analysis, but I am truly lost here. I tried to compute the Fourier series, but it seemed extremely different than the wanted identity.
fourier-analysis
asked Dec 13 '18 at 17:00
Uri George PeterzilUri George Peterzil
949
$endgroup$
Question: Prove the following identity:
$|x|=frac{pi}{2}+sum_{n=1}^{infty}frac{cos(2n-1)x}{(2n-1)^2}$. Our current topic is Fourier analysis, but I am truly lost here. I tried to compute the Fourier series, but it seemed extremely different than the wanted identity.
fourier-analysis
fourier-analysis
asked Dec 13 '18 at 17:00
Uri George PeterzilUri George Peterzil
949
asked Dec 13 '18 at 17:00
Uri George PeterzilUri George Peterzil
949
asked Dec 13 '18 at 17:00
Uri George PeterzilUri George Peterzil
949
asked Dec 13 '18 at 17:00
Uri George PeterzilUri George Peterzil
949
asked Dec 13 '18 at 17:00
Uri George PeterzilUri George Peterzil
949
949
1
$begingroup$
Just a comment. Note that the right hand side is periodic while the left hand side is not (as it is written). What you are doing is in fact the Fourier series of a sawtooth-like function. So the result depends on which interval you pick. Common choices are $[-pi,pi]$ or $[0,2pi]$. Since the right hand side is an even function, your choice is the former. This may be the source of confusion.
$endgroup$
– lcv
Dec 13 '18 at 18:47
add a comment |
1
$begingroup$
Just a comment. Note that the right hand side is periodic while the left hand side is not (as it is written). What you are doing is in fact the Fourier series of a sawtooth-like function. So the result depends on which interval you pick. Common choices are $[-pi,pi]$ or $[0,2pi]$. Since the right hand side is an even function, your choice is the former. This may be the source of confusion.
$endgroup$
– lcv
Dec 13 '18 at 18:47
1
1
$begingroup$
Just a comment. Note that the right hand side is periodic while the left hand side is not (as it is written). What you are doing is in fact the Fourier series of a sawtooth-like function. So the result depends on which interval you pick. Common choices are $[-pi,pi]$ or $[0,2pi]$. Since the right hand side is an even function, your choice is the former. This may be the source of confusion.
$endgroup$
– lcv
Dec 13 '18 at 18:47
$begingroup$
Just a comment. Note that the right hand side is periodic while the left hand side is not (as it is written). What you are doing is in fact the Fourier series of a sawtooth-like function. So the result depends on which interval you pick. Common choices are $[-pi,pi]$ or $[0,2pi]$. Since the right hand side is an even function, your choice is the former. This may be the source of confusion.
$endgroup$
– lcv
Dec 13 '18 at 18:47
add a comment |
2 Answers
2
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You can evaluate it using complex fourier coefficients $c_k$. You get: $frac{1}{2pi}$($int_{0}^{-pi}xe^{-ikx}dx$ + $int_{0}^{pi}xe^{-ikx}dx$) which results after integration by parts in $frac{1}{pi k^2}((-1)^k-1)=c_k$ Now, we use $a_k=c_k+c_{-k}$ and $b_k=i(c_k-c_{-k})$ which is $0$.
$a_k$ becomes $frac{-4}{pi k^2}$ where k is odd,i.e. $k=2n-1$ and $a_0$, after evaluation, becomes $pi$
So now, you have : $$f(x)=frac{a_0}{2}+sum_{k=1}^{infty}a_kcos(kx), k=(2n-1)$$ So after plugging in all values, we get $$f(x)=|x|=frac{pi}{2}+frac{-4}{pi} sum_{n=1}^{infty} frac{1}{(2n-1)^2} cos((2n-1)x)$$
Hope this helps !
answered Dec 13 '18 at 19:29
PoujhPoujh
616516
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It should be
$$
|x|=frac{pi}{2}-frac{4}{pi}sum_{n=1}^{infty}frac{cos(2n-1)x}{(2n-1)^2}.
$$
Since $|x|$ is even, there are no sine terms, and the coefficient of $cos(n,x)$ is
$$
frac{1}{pi}int_{-pi}^pi|x|cos(n,x),dx=frac{2}{pi}int_0^pi xcos(n,x),dx,quad nge0.
$$
This integral easily computed using integration by parts.
answered Dec 13 '18 at 18:41
Julián AguirreJulián Aguirre
69.1k24096
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can evaluate it using complex fourier coefficients $c_k$. You get: $frac{1}{2pi}$($int_{0}^{-pi}xe^{-ikx}dx$ + $int_{0}^{pi}xe^{-ikx}dx$) which results after integration by parts in $frac{1}{pi k^2}((-1)^k-1)=c_k$ Now, we use $a_k=c_k+c_{-k}$ and $b_k=i(c_k-c_{-k})$ which is $0$.
$a_k$ becomes $frac{-4}{pi k^2}$ where k is odd,i.e. $k=2n-1$ and $a_0$, after evaluation, becomes $pi$
So now, you have : $$f(x)=frac{a_0}{2}+sum_{k=1}^{infty}a_kcos(kx), k=(2n-1)$$ So after plugging in all values, we get $$f(x)=|x|=frac{pi}{2}+frac{-4}{pi} sum_{n=1}^{infty} frac{1}{(2n-1)^2} cos((2n-1)x)$$
Hope this helps !
answered Dec 13 '18 at 19:29
PoujhPoujh
616516
$endgroup$
add a comment |
$begingroup$
You can evaluate it using complex fourier coefficients $c_k$. You get: $frac{1}{2pi}$($int_{0}^{-pi}xe^{-ikx}dx$ + $int_{0}^{pi}xe^{-ikx}dx$) which results after integration by parts in $frac{1}{pi k^2}((-1)^k-1)=c_k$ Now, we use $a_k=c_k+c_{-k}$ and $b_k=i(c_k-c_{-k})$ which is $0$.
$a_k$ becomes $frac{-4}{pi k^2}$ where k is odd,i.e. $k=2n-1$ and $a_0$, after evaluation, becomes $pi$
So now, you have : $$f(x)=frac{a_0}{2}+sum_{k=1}^{infty}a_kcos(kx), k=(2n-1)$$ So after plugging in all values, we get $$f(x)=|x|=frac{pi}{2}+frac{-4}{pi} sum_{n=1}^{infty} frac{1}{(2n-1)^2} cos((2n-1)x)$$
Hope this helps !
answered Dec 13 '18 at 19:29
PoujhPoujh
616516
$endgroup$
add a comment |
$begingroup$
You can evaluate it using complex fourier coefficients $c_k$. You get: $frac{1}{2pi}$($int_{0}^{-pi}xe^{-ikx}dx$ + $int_{0}^{pi}xe^{-ikx}dx$) which results after integration by parts in $frac{1}{pi k^2}((-1)^k-1)=c_k$ Now, we use $a_k=c_k+c_{-k}$ and $b_k=i(c_k-c_{-k})$ which is $0$.
$a_k$ becomes $frac{-4}{pi k^2}$ where k is odd,i.e. $k=2n-1$ and $a_0$, after evaluation, becomes $pi$
So now, you have : $$f(x)=frac{a_0}{2}+sum_{k=1}^{infty}a_kcos(kx), k=(2n-1)$$ So after plugging in all values, we get $$f(x)=|x|=frac{pi}{2}+frac{-4}{pi} sum_{n=1}^{infty} frac{1}{(2n-1)^2} cos((2n-1)x)$$
Hope this helps !
answered Dec 13 '18 at 19:29
PoujhPoujh
616516
$endgroup$
You can evaluate it using complex fourier coefficients $c_k$. You get: $frac{1}{2pi}$($int_{0}^{-pi}xe^{-ikx}dx$ + $int_{0}^{pi}xe^{-ikx}dx$) which results after integration by parts in $frac{1}{pi k^2}((-1)^k-1)=c_k$ Now, we use $a_k=c_k+c_{-k}$ and $b_k=i(c_k-c_{-k})$ which is $0$.
$a_k$ becomes $frac{-4}{pi k^2}$ where k is odd,i.e. $k=2n-1$ and $a_0$, after evaluation, becomes $pi$
So now, you have : $$f(x)=frac{a_0}{2}+sum_{k=1}^{infty}a_kcos(kx), k=(2n-1)$$ So after plugging in all values, we get $$f(x)=|x|=frac{pi}{2}+frac{-4}{pi} sum_{n=1}^{infty} frac{1}{(2n-1)^2} cos((2n-1)x)$$
Hope this helps !
answered Dec 13 '18 at 19:29
PoujhPoujh
616516
answered Dec 13 '18 at 19:29
PoujhPoujh
616516
answered Dec 13 '18 at 19:29
PoujhPoujh
616516
answered Dec 13 '18 at 19:29
PoujhPoujh
616516
616516
add a comment |
add a comment |
$begingroup$
It should be
$$
|x|=frac{pi}{2}-frac{4}{pi}sum_{n=1}^{infty}frac{cos(2n-1)x}{(2n-1)^2}.
$$
Since $|x|$ is even, there are no sine terms, and the coefficient of $cos(n,x)$ is
$$
frac{1}{pi}int_{-pi}^pi|x|cos(n,x),dx=frac{2}{pi}int_0^pi xcos(n,x),dx,quad nge0.
$$
This integral easily computed using integration by parts.
answered Dec 13 '18 at 18:41
Julián AguirreJulián Aguirre
69.1k24096
$endgroup$
add a comment |
$begingroup$
It should be
$$
|x|=frac{pi}{2}-frac{4}{pi}sum_{n=1}^{infty}frac{cos(2n-1)x}{(2n-1)^2}.
$$
Since $|x|$ is even, there are no sine terms, and the coefficient of $cos(n,x)$ is
$$
frac{1}{pi}int_{-pi}^pi|x|cos(n,x),dx=frac{2}{pi}int_0^pi xcos(n,x),dx,quad nge0.
$$
This integral easily computed using integration by parts.
answered Dec 13 '18 at 18:41
Julián AguirreJulián Aguirre
69.1k24096
$endgroup$
add a comment |
$begingroup$
It should be
$$
|x|=frac{pi}{2}-frac{4}{pi}sum_{n=1}^{infty}frac{cos(2n-1)x}{(2n-1)^2}.
$$
Since $|x|$ is even, there are no sine terms, and the coefficient of $cos(n,x)$ is
$$
frac{1}{pi}int_{-pi}^pi|x|cos(n,x),dx=frac{2}{pi}int_0^pi xcos(n,x),dx,quad nge0.
$$
This integral easily computed using integration by parts.
answered Dec 13 '18 at 18:41
Julián AguirreJulián Aguirre
69.1k24096
$endgroup$
It should be
$$
|x|=frac{pi}{2}-frac{4}{pi}sum_{n=1}^{infty}frac{cos(2n-1)x}{(2n-1)^2}.
$$
Since $|x|$ is even, there are no sine terms, and the coefficient of $cos(n,x)$ is
$$
frac{1}{pi}int_{-pi}^pi|x|cos(n,x),dx=frac{2}{pi}int_0^pi xcos(n,x),dx,quad nge0.
$$
This integral easily computed using integration by parts.
answered Dec 13 '18 at 18:41
Julián AguirreJulián Aguirre
69.1k24096
answered Dec 13 '18 at 18:41
Julián AguirreJulián Aguirre
69.1k24096
answered Dec 13 '18 at 18:41
Julián AguirreJulián Aguirre
69.1k24096
answered Dec 13 '18 at 18:41
Julián AguirreJulián Aguirre
69.1k24096
69.1k24096
add a comment |
add a comment |
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$begingroup$
Just a comment. Note that the right hand side is periodic while the left hand side is not (as it is written). What you are doing is in fact the Fourier series of a sawtooth-like function. So the result depends on which interval you pick. Common choices are $[-pi,pi]$ or $[0,2pi]$. Since the right hand side is an even function, your choice is the former. This may be the source of confusion.
$endgroup$
– lcv
Dec 13 '18 at 18:47