Mixing
$sum_n=1^inftyfrac1n^6=fracpi^6945$ by Fourier series of $x^2$
Clash Royale CLAN TAG#URR8PPP
up vote
4
down vote
favorite
Prove that
$$sum_n=1^inftyfrac1n^6=fracpi^6945$$ by the Fourier series of $x^2$.
By Parseval's identity, I can only show $sum_n=1^inftyfrac1n^4=fracpi^490$. Could you please give me some hints?
sequences-and-series fourier-series
edited Oct 1 '16 at 5:21
Parcly Taxel
33.5k136588
asked Sep 30 '16 at 16:14
Longitude
429310
add a comment |Â
up vote
4
down vote
favorite
Prove that
$$sum_n=1^inftyfrac1n^6=fracpi^6945$$ by the Fourier series of $x^2$.
By Parseval's identity, I can only show $sum_n=1^inftyfrac1n^4=fracpi^490$. Could you please give me some hints?
sequences-and-series fourier-series
edited Oct 1 '16 at 5:21
Parcly Taxel
33.5k136588
asked Sep 30 '16 at 16:14
Longitude
429310
2
I think you need to use the fourier series of x^3 to directly use Parseval's identity.
– BenLaurense
Sep 30 '16 at 16:20
... or integrate the Fourier series of $x^2$, then apply Parseval's theorem. However, in order to implement this idea it is better to take a different quadratic polynomial, namely $B_2left(fracx2piright)$.
– Jack D'Aurizio♦
Sep 30 '16 at 16:57
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Prove that
$$sum_n=1^inftyfrac1n^6=fracpi^6945$$ by the Fourier series of $x^2$.
By Parseval's identity, I can only show $sum_n=1^inftyfrac1n^4=fracpi^490$. Could you please give me some hints?
sequences-and-series fourier-series
edited Oct 1 '16 at 5:21
Parcly Taxel
33.5k136588
asked Sep 30 '16 at 16:14
Longitude
429310
Prove that
$$sum_n=1^inftyfrac1n^6=fracpi^6945$$ by the Fourier series of $x^2$.
By Parseval's identity, I can only show $sum_n=1^inftyfrac1n^4=fracpi^490$. Could you please give me some hints?
sequences-and-series fourier-series
edited Oct 1 '16 at 5:21
Parcly Taxel
33.5k136588
asked Sep 30 '16 at 16:14
Longitude
429310
edited Oct 1 '16 at 5:21
Parcly Taxel
33.5k136588
edited Oct 1 '16 at 5:21
Parcly Taxel
33.5k136588
edited Oct 1 '16 at 5:21
Parcly Taxel
33.5k136588
33.5k136588
asked Sep 30 '16 at 16:14
Longitude
429310
asked Sep 30 '16 at 16:14
Longitude
429310
asked Sep 30 '16 at 16:14
Longitude
429310
429310
2
I think you need to use the fourier series of x^3 to directly use Parseval's identity.
– BenLaurense
Sep 30 '16 at 16:20
... or integrate the Fourier series of $x^2$, then apply Parseval's theorem. However, in order to implement this idea it is better to take a different quadratic polynomial, namely $B_2left(fracx2piright)$.
– Jack D'Aurizio♦
Sep 30 '16 at 16:57
add a comment |Â
2
I think you need to use the fourier series of x^3 to directly use Parseval's identity.
– BenLaurense
Sep 30 '16 at 16:20
... or integrate the Fourier series of $x^2$, then apply Parseval's theorem. However, in order to implement this idea it is better to take a different quadratic polynomial, namely $B_2left(fracx2piright)$.
– Jack D'Aurizio♦
Sep 30 '16 at 16:57
2
2
I think you need to use the fourier series of x^3 to directly use Parseval's identity.
– BenLaurense
Sep 30 '16 at 16:20
I think you need to use the fourier series of x^3 to directly use Parseval's identity.
– BenLaurense
Sep 30 '16 at 16:20
... or integrate the Fourier series of $x^2$, then apply Parseval's theorem. However, in order to implement this idea it is better to take a different quadratic polynomial, namely $B_2left(fracx2piright)$.
– Jack D'Aurizio♦
Sep 30 '16 at 16:57
... or integrate the Fourier series of $x^2$, then apply Parseval's theorem. However, in order to implement this idea it is better to take a different quadratic polynomial, namely $B_2left(fracx2piright)$.
– Jack D'Aurizio♦
Sep 30 '16 at 16:57
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
We may start from
$$ f_1(x) = sum_ngeq 1fracsin(nx)n tag1$$
that is the Fourier series of a sawtooth wave, equal to $fracpi-x2$ on the interval $I=(0,2pi)$.
By termwise integration, we get that
$$ forall xin I,quad sum_ngeq 1frac1-cos(nx)n^2 = frac2pi x-x^24$$
hence:
$$ forall xin I,quad f_2(x) = sum_ngeq 1fraccos(nx)n^2=fracpi^26-fracpi x2+fracx^24tag2 $$
$$ forall xin I,quad f_3(x) = sum_ngeq 1fracsin(nx)n^3=fracpi^2 x6-fracpi x^24+fracx^312tag3 $$
(the integration constants are computed from the fact that $f_j(x)$ has to have mean zero over $I$)
and by Parseval's theorem
$$ zeta(6) = frac1piint_0^2pif_3(x)^2,dx = frac2pi^69int_0^1left[x(x-1)(2x-1)right]^2,dx=colorredfracpi^6945.tag4$$
With the same approach it is not difficult to prove that for any $ngeq 1$, $zeta(2n)$ is a rational multiple of $pi^2nint_0^1B_n(x)^2,dx$, where $B_n(x)$ is a Bernoulli polynomial.
answered Sep 30 '16 at 16:32
Jack D'Aurizio♦
269k30261625
it is not 100% clear that you substract the mean such that $f_j(x)$ has zero mean
– reuns
Sep 30 '16 at 16:40
@user1952009: the expression for $f_2(x)$ is derived from $f_2(x)=zeta(2)-sum_ngeq 1frac1-cos(nx)n^2$ and the expression for $f_3(x)$ is derived from $f_3(pi)=0$. Anyway, answer improved. Thanks for the suggestion.
– Jack D'Aurizio♦
Sep 30 '16 at 16:43
But where is the Fourier series of $x^2$?
– Longitude
Sep 30 '16 at 16:56
@Longitude: it is part of $(2)$. See my comment under your question.
– Jack D'Aurizio♦
Sep 30 '16 at 16:57
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
We may start from
$$ f_1(x) = sum_ngeq 1fracsin(nx)n tag1$$
that is the Fourier series of a sawtooth wave, equal to $fracpi-x2$ on the interval $I=(0,2pi)$.
By termwise integration, we get that
$$ forall xin I,quad sum_ngeq 1frac1-cos(nx)n^2 = frac2pi x-x^24$$
hence:
$$ forall xin I,quad f_2(x) = sum_ngeq 1fraccos(nx)n^2=fracpi^26-fracpi x2+fracx^24tag2 $$
$$ forall xin I,quad f_3(x) = sum_ngeq 1fracsin(nx)n^3=fracpi^2 x6-fracpi x^24+fracx^312tag3 $$
(the integration constants are computed from the fact that $f_j(x)$ has to have mean zero over $I$)
and by Parseval's theorem
$$ zeta(6) = frac1piint_0^2pif_3(x)^2,dx = frac2pi^69int_0^1left[x(x-1)(2x-1)right]^2,dx=colorredfracpi^6945.tag4$$
With the same approach it is not difficult to prove that for any $ngeq 1$, $zeta(2n)$ is a rational multiple of $pi^2nint_0^1B_n(x)^2,dx$, where $B_n(x)$ is a Bernoulli polynomial.
answered Sep 30 '16 at 16:32
Jack D'Aurizio♦
269k30261625
it is not 100% clear that you substract the mean such that $f_j(x)$ has zero mean
– reuns
Sep 30 '16 at 16:40
@user1952009: the expression for $f_2(x)$ is derived from $f_2(x)=zeta(2)-sum_ngeq 1frac1-cos(nx)n^2$ and the expression for $f_3(x)$ is derived from $f_3(pi)=0$. Anyway, answer improved. Thanks for the suggestion.
– Jack D'Aurizio♦
Sep 30 '16 at 16:43
But where is the Fourier series of $x^2$?
– Longitude
Sep 30 '16 at 16:56
@Longitude: it is part of $(2)$. See my comment under your question.
– Jack D'Aurizio♦
Sep 30 '16 at 16:57
add a comment |Â
up vote
3
down vote
We may start from
$$ f_1(x) = sum_ngeq 1fracsin(nx)n tag1$$
that is the Fourier series of a sawtooth wave, equal to $fracpi-x2$ on the interval $I=(0,2pi)$.
By termwise integration, we get that
$$ forall xin I,quad sum_ngeq 1frac1-cos(nx)n^2 = frac2pi x-x^24$$
hence:
$$ forall xin I,quad f_2(x) = sum_ngeq 1fraccos(nx)n^2=fracpi^26-fracpi x2+fracx^24tag2 $$
$$ forall xin I,quad f_3(x) = sum_ngeq 1fracsin(nx)n^3=fracpi^2 x6-fracpi x^24+fracx^312tag3 $$
(the integration constants are computed from the fact that $f_j(x)$ has to have mean zero over $I$)
and by Parseval's theorem
$$ zeta(6) = frac1piint_0^2pif_3(x)^2,dx = frac2pi^69int_0^1left[x(x-1)(2x-1)right]^2,dx=colorredfracpi^6945.tag4$$
With the same approach it is not difficult to prove that for any $ngeq 1$, $zeta(2n)$ is a rational multiple of $pi^2nint_0^1B_n(x)^2,dx$, where $B_n(x)$ is a Bernoulli polynomial.
answered Sep 30 '16 at 16:32
Jack D'Aurizio♦
269k30261625
it is not 100% clear that you substract the mean such that $f_j(x)$ has zero mean
– reuns
Sep 30 '16 at 16:40
@user1952009: the expression for $f_2(x)$ is derived from $f_2(x)=zeta(2)-sum_ngeq 1frac1-cos(nx)n^2$ and the expression for $f_3(x)$ is derived from $f_3(pi)=0$. Anyway, answer improved. Thanks for the suggestion.
– Jack D'Aurizio♦
Sep 30 '16 at 16:43
But where is the Fourier series of $x^2$?
– Longitude
Sep 30 '16 at 16:56
@Longitude: it is part of $(2)$. See my comment under your question.
– Jack D'Aurizio♦
Sep 30 '16 at 16:57
add a comment |Â
up vote
3
down vote
up vote
3
down vote
We may start from
$$ f_1(x) = sum_ngeq 1fracsin(nx)n tag1$$
that is the Fourier series of a sawtooth wave, equal to $fracpi-x2$ on the interval $I=(0,2pi)$.
By termwise integration, we get that
$$ forall xin I,quad sum_ngeq 1frac1-cos(nx)n^2 = frac2pi x-x^24$$
hence:
$$ forall xin I,quad f_2(x) = sum_ngeq 1fraccos(nx)n^2=fracpi^26-fracpi x2+fracx^24tag2 $$
$$ forall xin I,quad f_3(x) = sum_ngeq 1fracsin(nx)n^3=fracpi^2 x6-fracpi x^24+fracx^312tag3 $$
(the integration constants are computed from the fact that $f_j(x)$ has to have mean zero over $I$)
and by Parseval's theorem
$$ zeta(6) = frac1piint_0^2pif_3(x)^2,dx = frac2pi^69int_0^1left[x(x-1)(2x-1)right]^2,dx=colorredfracpi^6945.tag4$$
With the same approach it is not difficult to prove that for any $ngeq 1$, $zeta(2n)$ is a rational multiple of $pi^2nint_0^1B_n(x)^2,dx$, where $B_n(x)$ is a Bernoulli polynomial.
answered Sep 30 '16 at 16:32
Jack D'Aurizio♦
269k30261625
We may start from
$$ f_1(x) = sum_ngeq 1fracsin(nx)n tag1$$
that is the Fourier series of a sawtooth wave, equal to $fracpi-x2$ on the interval $I=(0,2pi)$.
By termwise integration, we get that
$$ forall xin I,quad sum_ngeq 1frac1-cos(nx)n^2 = frac2pi x-x^24$$
hence:
$$ forall xin I,quad f_2(x) = sum_ngeq 1fraccos(nx)n^2=fracpi^26-fracpi x2+fracx^24tag2 $$
$$ forall xin I,quad f_3(x) = sum_ngeq 1fracsin(nx)n^3=fracpi^2 x6-fracpi x^24+fracx^312tag3 $$
(the integration constants are computed from the fact that $f_j(x)$ has to have mean zero over $I$)
and by Parseval's theorem
$$ zeta(6) = frac1piint_0^2pif_3(x)^2,dx = frac2pi^69int_0^1left[x(x-1)(2x-1)right]^2,dx=colorredfracpi^6945.tag4$$
With the same approach it is not difficult to prove that for any $ngeq 1$, $zeta(2n)$ is a rational multiple of $pi^2nint_0^1B_n(x)^2,dx$, where $B_n(x)$ is a Bernoulli polynomial.
answered Sep 30 '16 at 16:32
Jack D'Aurizio♦
269k30261625
edited Sep 30 '16 at 16:45
answered Sep 30 '16 at 16:32
Jack D'Aurizio♦
269k30261625
answered Sep 30 '16 at 16:32
Jack D'Aurizio♦
269k30261625
answered Sep 30 '16 at 16:32
Jack D'Aurizio♦
269k30261625
269k30261625
it is not 100% clear that you substract the mean such that $f_j(x)$ has zero mean
– reuns
Sep 30 '16 at 16:40
@user1952009: the expression for $f_2(x)$ is derived from $f_2(x)=zeta(2)-sum_ngeq 1frac1-cos(nx)n^2$ and the expression for $f_3(x)$ is derived from $f_3(pi)=0$. Anyway, answer improved. Thanks for the suggestion.
– Jack D'Aurizio♦
Sep 30 '16 at 16:43
But where is the Fourier series of $x^2$?
– Longitude
Sep 30 '16 at 16:56
@Longitude: it is part of $(2)$. See my comment under your question.
– Jack D'Aurizio♦
Sep 30 '16 at 16:57
add a comment |Â
it is not 100% clear that you substract the mean such that $f_j(x)$ has zero mean
– reuns
Sep 30 '16 at 16:40
@user1952009: the expression for $f_2(x)$ is derived from $f_2(x)=zeta(2)-sum_ngeq 1frac1-cos(nx)n^2$ and the expression for $f_3(x)$ is derived from $f_3(pi)=0$. Anyway, answer improved. Thanks for the suggestion.
– Jack D'Aurizio♦
Sep 30 '16 at 16:43
But where is the Fourier series of $x^2$?
– Longitude
Sep 30 '16 at 16:56
@Longitude: it is part of $(2)$. See my comment under your question.
– Jack D'Aurizio♦
Sep 30 '16 at 16:57
it is not 100% clear that you substract the mean such that $f_j(x)$ has zero mean
– reuns
Sep 30 '16 at 16:40
it is not 100% clear that you substract the mean such that $f_j(x)$ has zero mean
– reuns
Sep 30 '16 at 16:40
@user1952009: the expression for $f_2(x)$ is derived from $f_2(x)=zeta(2)-sum_ngeq 1frac1-cos(nx)n^2$ and the expression for $f_3(x)$ is derived from $f_3(pi)=0$. Anyway, answer improved. Thanks for the suggestion.
– Jack D'Aurizio♦
Sep 30 '16 at 16:43
@user1952009: the expression for $f_2(x)$ is derived from $f_2(x)=zeta(2)-sum_ngeq 1frac1-cos(nx)n^2$ and the expression for $f_3(x)$ is derived from $f_3(pi)=0$. Anyway, answer improved. Thanks for the suggestion.
– Jack D'Aurizio♦
Sep 30 '16 at 16:43
But where is the Fourier series of $x^2$?
– Longitude
Sep 30 '16 at 16:56
But where is the Fourier series of $x^2$?
– Longitude
Sep 30 '16 at 16:56
@Longitude: it is part of $(2)$. See my comment under your question.
– Jack D'Aurizio♦
Sep 30 '16 at 16:57
@Longitude: it is part of $(2)$. See my comment under your question.
– Jack D'Aurizio♦
Sep 30 '16 at 16:57
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1948206%2fsum-n-1-infty-frac1n6-frac-pi6945-by-fourier-series-of-x2%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
2
I think you need to use the fourier series of x^3 to directly use Parseval's identity.
– BenLaurense
Sep 30 '16 at 16:20
... or integrate the Fourier series of $x^2$, then apply Parseval's theorem. However, in order to implement this idea it is better to take a different quadratic polynomial, namely $B_2left(fracx2piright)$.
– Jack D'Aurizio♦
Sep 30 '16 at 16:57