Help with Fourier Series $sum_j=1^infty frac1j^2ksin2pi j x$
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
I've found the limit of a similar expression in the literature (see bottom), which uses Bernoulli polynomials:
$sum_j=1^infty frac1j^2k+1sin2pi j x=frac-(-1)^k(2pi)^2k+12(2k+1)B_2k+1(x)$
My question is, how about the below, does it also have a known limit? (The aforementioned reference doesn't have it.)
$sum_j=1^infty frac1j^2ksin2pi j x$
I would guess not, but unlikely. Reason is cause this is similar to the non-existence of closed-forms for say $zeta(2k+1)$ as opposed to $zeta(2k)$, but need to confirm. (Yes, searching the literature for the novelty of things is one of the most challenging tasks.)
Reference:
M. Abramowitz, I. Stegun (Eds.), Handbook of Mathematical Functions with Formulas, Graphs and Mathematical Tables, Dover Publications, New York, 1972:
fourier-series bernoulli-polynomials
asked Aug 21 at 0:34
JR S.
177
add a comment |Â
up vote
0
down vote
favorite
I've found the limit of a similar expression in the literature (see bottom), which uses Bernoulli polynomials:
$sum_j=1^infty frac1j^2k+1sin2pi j x=frac-(-1)^k(2pi)^2k+12(2k+1)B_2k+1(x)$
My question is, how about the below, does it also have a known limit? (The aforementioned reference doesn't have it.)
$sum_j=1^infty frac1j^2ksin2pi j x$
I would guess not, but unlikely. Reason is cause this is similar to the non-existence of closed-forms for say $zeta(2k+1)$ as opposed to $zeta(2k)$, but need to confirm. (Yes, searching the literature for the novelty of things is one of the most challenging tasks.)
Reference:
M. Abramowitz, I. Stegun (Eds.), Handbook of Mathematical Functions with Formulas, Graphs and Mathematical Tables, Dover Publications, New York, 1972:
fourier-series bernoulli-polynomials
asked Aug 21 at 0:34
JR S.
177
1
You are just dealing with the real/imaginary part of $textLi_m(e^2pi i x)$.
– Jack D'Aurizio♦
Aug 21 at 1:07
Is there an integral representation for it?
– JR S.
Aug 21 at 1:09
Of course, but it is kind of tautological. You may compute the $(n+1)$-th Bernoulli polynomial by integrating the $n$-th Bernoulli polynomial and choosing the suitable constant. The same happens here.
– Jack D'Aurizio♦
Aug 21 at 1:13
@JackD'Aurizio Would you mind demonstrating how that's done in an answer? ;)
– JR S.
Aug 21 at 1:34
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I've found the limit of a similar expression in the literature (see bottom), which uses Bernoulli polynomials:
$sum_j=1^infty frac1j^2k+1sin2pi j x=frac-(-1)^k(2pi)^2k+12(2k+1)B_2k+1(x)$
My question is, how about the below, does it also have a known limit? (The aforementioned reference doesn't have it.)
$sum_j=1^infty frac1j^2ksin2pi j x$
I would guess not, but unlikely. Reason is cause this is similar to the non-existence of closed-forms for say $zeta(2k+1)$ as opposed to $zeta(2k)$, but need to confirm. (Yes, searching the literature for the novelty of things is one of the most challenging tasks.)
Reference:
M. Abramowitz, I. Stegun (Eds.), Handbook of Mathematical Functions with Formulas, Graphs and Mathematical Tables, Dover Publications, New York, 1972:
fourier-series bernoulli-polynomials
asked Aug 21 at 0:34
JR S.
177
I've found the limit of a similar expression in the literature (see bottom), which uses Bernoulli polynomials:
$sum_j=1^infty frac1j^2k+1sin2pi j x=frac-(-1)^k(2pi)^2k+12(2k+1)B_2k+1(x)$
My question is, how about the below, does it also have a known limit? (The aforementioned reference doesn't have it.)
$sum_j=1^infty frac1j^2ksin2pi j x$
I would guess not, but unlikely. Reason is cause this is similar to the non-existence of closed-forms for say $zeta(2k+1)$ as opposed to $zeta(2k)$, but need to confirm. (Yes, searching the literature for the novelty of things is one of the most challenging tasks.)
Reference:
M. Abramowitz, I. Stegun (Eds.), Handbook of Mathematical Functions with Formulas, Graphs and Mathematical Tables, Dover Publications, New York, 1972:
fourier-series bernoulli-polynomials
asked Aug 21 at 0:34
JR S.
177
edited Aug 21 at 1:07
asked Aug 21 at 0:34
JR S.
177
asked Aug 21 at 0:34
JR S.
177
asked Aug 21 at 0:34
JR S.
177
177
1
You are just dealing with the real/imaginary part of $textLi_m(e^2pi i x)$.
– Jack D'Aurizio♦
Aug 21 at 1:07
Is there an integral representation for it?
– JR S.
Aug 21 at 1:09
Of course, but it is kind of tautological. You may compute the $(n+1)$-th Bernoulli polynomial by integrating the $n$-th Bernoulli polynomial and choosing the suitable constant. The same happens here.
– Jack D'Aurizio♦
Aug 21 at 1:13
@JackD'Aurizio Would you mind demonstrating how that's done in an answer? ;)
– JR S.
Aug 21 at 1:34
add a comment |Â
1
You are just dealing with the real/imaginary part of $textLi_m(e^2pi i x)$.
– Jack D'Aurizio♦
Aug 21 at 1:07
Is there an integral representation for it?
– JR S.
Aug 21 at 1:09
Of course, but it is kind of tautological. You may compute the $(n+1)$-th Bernoulli polynomial by integrating the $n$-th Bernoulli polynomial and choosing the suitable constant. The same happens here.
– Jack D'Aurizio♦
Aug 21 at 1:13
@JackD'Aurizio Would you mind demonstrating how that's done in an answer? ;)
– JR S.
Aug 21 at 1:34
1
1
You are just dealing with the real/imaginary part of $textLi_m(e^2pi i x)$.
– Jack D'Aurizio♦
Aug 21 at 1:07
You are just dealing with the real/imaginary part of $textLi_m(e^2pi i x)$.
– Jack D'Aurizio♦
Aug 21 at 1:07
Is there an integral representation for it?
– JR S.
Aug 21 at 1:09
Is there an integral representation for it?
– JR S.
Aug 21 at 1:09
Of course, but it is kind of tautological. You may compute the $(n+1)$-th Bernoulli polynomial by integrating the $n$-th Bernoulli polynomial and choosing the suitable constant. The same happens here.
– Jack D'Aurizio♦
Aug 21 at 1:13
Of course, but it is kind of tautological. You may compute the $(n+1)$-th Bernoulli polynomial by integrating the $n$-th Bernoulli polynomial and choosing the suitable constant. The same happens here.
– Jack D'Aurizio♦
Aug 21 at 1:13
@JackD'Aurizio Would you mind demonstrating how that's done in an answer? ;)
– JR S.
Aug 21 at 1:34
@JackD'Aurizio Would you mind demonstrating how that's done in an answer? ;)
– JR S.
Aug 21 at 1:34
add a comment |Â
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f2889355%2fhelp-with-fourier-series-sum-j-1-infty-frac1j2k-sin2-pi-j-x%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
1
You are just dealing with the real/imaginary part of $textLi_m(e^2pi i x)$.
– Jack D'Aurizio♦
Aug 21 at 1:07
Is there an integral representation for it?
– JR S.
Aug 21 at 1:09
Of course, but it is kind of tautological. You may compute the $(n+1)$-th Bernoulli polynomial by integrating the $n$-th Bernoulli polynomial and choosing the suitable constant. The same happens here.
– Jack D'Aurizio♦
Aug 21 at 1:13
@JackD'Aurizio Would you mind demonstrating how that's done in an answer? ;)
– JR S.
Aug 21 at 1:34