Use Fourier coefficients of $f(t)=t$ to show $sum_{n=1}^{infty} frac{1}{n^2}= frac{pi^2}{6}$ [duplicate]
This question already has an answer here:
Fourier serie property for $sum_{n=1}^infty frac{1}{n^2}=frac{pi^2}{6}$
1 answer
Using Fourier coefficients of $f in L_2(mathbb{T})$ given by $f(t)=t$ for almost all $t in ,,]-pi, pi[$ show that
$$sum_{n=1}^{infty} frac{1}{n^2}= frac{pi^2}{6}$$
I think it’s an easy exercise but I didn’t understand the Fourier analysis part I had in the functional analysis course I attented and I don’t know how to solve this problem
sequences-and-series functional-analysis fourier-analysis fourier-series lp-spaces
asked Jan 4 at 18:48
Maggie94Maggie94
1026
marked as duplicate by José Carlos Santos, Dietrich Burde, mrtaurho, RRL, jgon Jan 4 at 21:13
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
This question already has an answer here:
Fourier serie property for $sum_{n=1}^infty frac{1}{n^2}=frac{pi^2}{6}$
1 answer
Using Fourier coefficients of $f in L_2(mathbb{T})$ given by $f(t)=t$ for almost all $t in ,,]-pi, pi[$ show that
$$sum_{n=1}^{infty} frac{1}{n^2}= frac{pi^2}{6}$$
I think it’s an easy exercise but I didn’t understand the Fourier analysis part I had in the functional analysis course I attented and I don’t know how to solve this problem
sequences-and-series functional-analysis fourier-analysis fourier-series lp-spaces
asked Jan 4 at 18:48
Maggie94Maggie94
1026
marked as duplicate by José Carlos Santos, Dietrich Burde, mrtaurho, RRL, jgon Jan 4 at 21:13
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Do you know how to compute the Fourier coefficients?
– Wojowu
Jan 4 at 18:51
I retract my close vote because I have realized the linked problem finds the value of the sum differently
– Wojowu
Jan 4 at 18:58
add a comment |
This question already has an answer here:
Fourier serie property for $sum_{n=1}^infty frac{1}{n^2}=frac{pi^2}{6}$
1 answer
Using Fourier coefficients of $f in L_2(mathbb{T})$ given by $f(t)=t$ for almost all $t in ,,]-pi, pi[$ show that
$$sum_{n=1}^{infty} frac{1}{n^2}= frac{pi^2}{6}$$
I think it’s an easy exercise but I didn’t understand the Fourier analysis part I had in the functional analysis course I attented and I don’t know how to solve this problem
sequences-and-series functional-analysis fourier-analysis fourier-series lp-spaces
asked Jan 4 at 18:48
Maggie94Maggie94
1026
This question already has an answer here:
Fourier serie property for $sum_{n=1}^infty frac{1}{n^2}=frac{pi^2}{6}$
1 answer
Using Fourier coefficients of $f in L_2(mathbb{T})$ given by $f(t)=t$ for almost all $t in ,,]-pi, pi[$ show that
$$sum_{n=1}^{infty} frac{1}{n^2}= frac{pi^2}{6}$$
I think it’s an easy exercise but I didn’t understand the Fourier analysis part I had in the functional analysis course I attented and I don’t know how to solve this problem
This question already has an answer here:
Fourier serie property for $sum_{n=1}^infty frac{1}{n^2}=frac{pi^2}{6}$
1 answer
sequences-and-series functional-analysis fourier-analysis fourier-series lp-spaces
sequences-and-series functional-analysis fourier-analysis fourier-series lp-spaces
asked Jan 4 at 18:48
Maggie94Maggie94
1026
asked Jan 4 at 18:48
Maggie94Maggie94
1026
asked Jan 4 at 18:48
Maggie94Maggie94
1026
asked Jan 4 at 18:48
Maggie94Maggie94
1026
asked Jan 4 at 18:48
Maggie94Maggie94
1026
1026
marked as duplicate by José Carlos Santos, Dietrich Burde, mrtaurho, RRL, jgon Jan 4 at 21:13
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by José Carlos Santos, Dietrich Burde, mrtaurho, RRL, jgon Jan 4 at 21:13
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Do you know how to compute the Fourier coefficients?
– Wojowu
Jan 4 at 18:51
I retract my close vote because I have realized the linked problem finds the value of the sum differently
– Wojowu
Jan 4 at 18:58
add a comment |
Do you know how to compute the Fourier coefficients?
– Wojowu
Jan 4 at 18:51
I retract my close vote because I have realized the linked problem finds the value of the sum differently
– Wojowu
Jan 4 at 18:58
Do you know how to compute the Fourier coefficients?
– Wojowu
Jan 4 at 18:51
Do you know how to compute the Fourier coefficients?
– Wojowu
Jan 4 at 18:51
I retract my close vote because I have realized the linked problem finds the value of the sum differently
– Wojowu
Jan 4 at 18:58
I retract my close vote because I have realized the linked problem finds the value of the sum differently
– Wojowu
Jan 4 at 18:58
add a comment |
1 Answer
1
active
oldest
votes
If $f$ is a continuous function on $[-pi,pi]$ then
$$int_{-pi}^pi |f(x)|^2,dx=2pisum_{n=-infty}^infty |c_n|^2$$
where
$$c_n=frac1{2pi}int_{-pi}^pi e^{-inx}f(x),dx$$
is the $n$-th Fourier coefficient of $f$.
In your case, you need to compute
$$int_{-pi}^pi x^2,dx$$
and
$$int_{-pi}^pi xe^{-inx},dx$$
for each integer $n$.
answered Jan 4 at 18:55
Lord Shark the UnknownLord Shark the Unknown
102k959132
Yes I did it and in the first integral I get $frac{2 pi^3}{3}$, while in the second, integrating by parts, I get $-pi e^{-inpi} - pi e^{inpi} + e^ {inpi} - e^{-inpi}$, right? And from here?
– Maggie94
yesterday
The are all definite integrals. There should be no $t$s in them.
– Lord Shark the Unknown
yesterday
Yes the t were all $ pi$, sorry
– Maggie94
yesterday
You should know what $e^{ipi}$ is.
– Lord Shark the Unknown
yesterday
Yes, it is 1. But $e^{in pi}$ could be 1 or -1 depending on $n$ (if it is odd it is -1), right?
– Maggie94
yesterday
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
If $f$ is a continuous function on $[-pi,pi]$ then
$$int_{-pi}^pi |f(x)|^2,dx=2pisum_{n=-infty}^infty |c_n|^2$$
where
$$c_n=frac1{2pi}int_{-pi}^pi e^{-inx}f(x),dx$$
is the $n$-th Fourier coefficient of $f$.
In your case, you need to compute
$$int_{-pi}^pi x^2,dx$$
and
$$int_{-pi}^pi xe^{-inx},dx$$
for each integer $n$.
answered Jan 4 at 18:55
Lord Shark the UnknownLord Shark the Unknown
102k959132
Yes I did it and in the first integral I get $frac{2 pi^3}{3}$, while in the second, integrating by parts, I get $-pi e^{-inpi} - pi e^{inpi} + e^ {inpi} - e^{-inpi}$, right? And from here?
– Maggie94
yesterday
The are all definite integrals. There should be no $t$s in them.
– Lord Shark the Unknown
yesterday
Yes the t were all $ pi$, sorry
– Maggie94
yesterday
You should know what $e^{ipi}$ is.
– Lord Shark the Unknown
yesterday
Yes, it is 1. But $e^{in pi}$ could be 1 or -1 depending on $n$ (if it is odd it is -1), right?
– Maggie94
yesterday
add a comment |
If $f$ is a continuous function on $[-pi,pi]$ then
$$int_{-pi}^pi |f(x)|^2,dx=2pisum_{n=-infty}^infty |c_n|^2$$
where
$$c_n=frac1{2pi}int_{-pi}^pi e^{-inx}f(x),dx$$
is the $n$-th Fourier coefficient of $f$.
In your case, you need to compute
$$int_{-pi}^pi x^2,dx$$
and
$$int_{-pi}^pi xe^{-inx},dx$$
for each integer $n$.
answered Jan 4 at 18:55
Lord Shark the UnknownLord Shark the Unknown
102k959132
Yes I did it and in the first integral I get $frac{2 pi^3}{3}$, while in the second, integrating by parts, I get $-pi e^{-inpi} - pi e^{inpi} + e^ {inpi} - e^{-inpi}$, right? And from here?
– Maggie94
yesterday
The are all definite integrals. There should be no $t$s in them.
– Lord Shark the Unknown
yesterday
Yes the t were all $ pi$, sorry
– Maggie94
yesterday
You should know what $e^{ipi}$ is.
– Lord Shark the Unknown
yesterday
Yes, it is 1. But $e^{in pi}$ could be 1 or -1 depending on $n$ (if it is odd it is -1), right?
– Maggie94
yesterday
add a comment |
If $f$ is a continuous function on $[-pi,pi]$ then
$$int_{-pi}^pi |f(x)|^2,dx=2pisum_{n=-infty}^infty |c_n|^2$$
where
$$c_n=frac1{2pi}int_{-pi}^pi e^{-inx}f(x),dx$$
is the $n$-th Fourier coefficient of $f$.
In your case, you need to compute
$$int_{-pi}^pi x^2,dx$$
and
$$int_{-pi}^pi xe^{-inx},dx$$
for each integer $n$.
answered Jan 4 at 18:55
Lord Shark the UnknownLord Shark the Unknown
102k959132
If $f$ is a continuous function on $[-pi,pi]$ then
$$int_{-pi}^pi |f(x)|^2,dx=2pisum_{n=-infty}^infty |c_n|^2$$
where
$$c_n=frac1{2pi}int_{-pi}^pi e^{-inx}f(x),dx$$
is the $n$-th Fourier coefficient of $f$.
In your case, you need to compute
$$int_{-pi}^pi x^2,dx$$
and
$$int_{-pi}^pi xe^{-inx},dx$$
for each integer $n$.
answered Jan 4 at 18:55
Lord Shark the UnknownLord Shark the Unknown
102k959132
answered Jan 4 at 18:55
Lord Shark the UnknownLord Shark the Unknown
102k959132
answered Jan 4 at 18:55
Lord Shark the UnknownLord Shark the Unknown
102k959132
answered Jan 4 at 18:55
Lord Shark the UnknownLord Shark the Unknown
102k959132
102k959132
Yes I did it and in the first integral I get $frac{2 pi^3}{3}$, while in the second, integrating by parts, I get $-pi e^{-inpi} - pi e^{inpi} + e^ {inpi} - e^{-inpi}$, right? And from here?
– Maggie94
yesterday
The are all definite integrals. There should be no $t$s in them.
– Lord Shark the Unknown
yesterday
Yes the t were all $ pi$, sorry
– Maggie94
yesterday
You should know what $e^{ipi}$ is.
– Lord Shark the Unknown
yesterday
Yes, it is 1. But $e^{in pi}$ could be 1 or -1 depending on $n$ (if it is odd it is -1), right?
– Maggie94
yesterday
add a comment |
Yes I did it and in the first integral I get $frac{2 pi^3}{3}$, while in the second, integrating by parts, I get $-pi e^{-inpi} - pi e^{inpi} + e^ {inpi} - e^{-inpi}$, right? And from here?
– Maggie94
yesterday
The are all definite integrals. There should be no $t$s in them.
– Lord Shark the Unknown
yesterday
Yes the t were all $ pi$, sorry
– Maggie94
yesterday
You should know what $e^{ipi}$ is.
– Lord Shark the Unknown
yesterday
Yes, it is 1. But $e^{in pi}$ could be 1 or -1 depending on $n$ (if it is odd it is -1), right?
– Maggie94
yesterday
Yes I did it and in the first integral I get $frac{2 pi^3}{3}$, while in the second, integrating by parts, I get $-pi e^{-inpi} - pi e^{inpi} + e^ {inpi} - e^{-inpi}$, right? And from here?
– Maggie94
yesterday
Yes I did it and in the first integral I get $frac{2 pi^3}{3}$, while in the second, integrating by parts, I get $-pi e^{-inpi} - pi e^{inpi} + e^ {inpi} - e^{-inpi}$, right? And from here?
– Maggie94
yesterday
The are all definite integrals. There should be no $t$s in them.
– Lord Shark the Unknown
yesterday
The are all definite integrals. There should be no $t$s in them.
– Lord Shark the Unknown
yesterday
Yes the t were all $ pi$, sorry
– Maggie94
yesterday
Yes the t were all $ pi$, sorry
– Maggie94
yesterday
You should know what $e^{ipi}$ is.
– Lord Shark the Unknown
yesterday
You should know what $e^{ipi}$ is.
– Lord Shark the Unknown
yesterday
Yes, it is 1. But $e^{in pi}$ could be 1 or -1 depending on $n$ (if it is odd it is -1), right?
– Maggie94
yesterday
Yes, it is 1. But $e^{in pi}$ could be 1 or -1 depending on $n$ (if it is odd it is -1), right?
– Maggie94
yesterday
add a comment |
Do you know how to compute the Fourier coefficients?
– Wojowu
Jan 4 at 18:51
I retract my close vote because I have realized the linked problem finds the value of the sum differently
– Wojowu
Jan 4 at 18:58