...
$begingroup$
I am looking for numerical evidence that
$$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^3}sinleft(frac{n}{m^2}right)=frac{pi^6}{11340}-frac{pi^4}{72}$$
I have proven it, but I just want to be extra sure. Desmos only gives me accuracy to the third decimal place, but I know that some of you (@Claude Leibovici) are able to give me extremely high decimal accuracy.
Proof:
We know that for $|t|leqpi$,
$$t^2=frac{pi^2}3+4sum_{ngeq1}frac{(-1)^n}{n^2}cos nt$$
Solving for the sum then integrating both sides from $0$ to $x$,
$$sum_{ngeq1}frac{(-1)^n}{n^3}sin(nx)=frac{x^3}{12}-frac{pi^2x}{12}$$
Then plugging in $x=frac1{m^2}$ for integer $mgeq1$,
$$sum_{ngeq1}frac{(-1)^n}{n^3}sinbigg(frac{n}{m^2}bigg)=frac{1}{12m^6}-frac{pi^2}{12m^2}$$
then applying $sum_{mgeq1}$ on both sides,
$$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^3}sinbigg(frac{n}{m^2}bigg)=frac1{12}zeta(6)-frac{pi^2}{12}zeta(2)$$
We simplify to reach our conclusion
$$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^3}sinbigg(frac{n}{m^2}bigg)=frac{pi^6}{11340}-frac{pi^4}{72}$$
sequences-and-series proof-verification fourier-series decimal-expansion
edited Dec 25 '18 at 9:35
Did
249k23229468
asked Dec 25 '18 at 1:00
clathratusclathratus
5,3141440
$endgroup$
|
show 1 more comment
$begingroup$
I am looking for numerical evidence that
$$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^3}sinleft(frac{n}{m^2}right)=frac{pi^6}{11340}-frac{pi^4}{72}$$
I have proven it, but I just want to be extra sure. Desmos only gives me accuracy to the third decimal place, but I know that some of you (@Claude Leibovici) are able to give me extremely high decimal accuracy.
Proof:
We know that for $|t|leqpi$,
$$t^2=frac{pi^2}3+4sum_{ngeq1}frac{(-1)^n}{n^2}cos nt$$
Solving for the sum then integrating both sides from $0$ to $x$,
$$sum_{ngeq1}frac{(-1)^n}{n^3}sin(nx)=frac{x^3}{12}-frac{pi^2x}{12}$$
Then plugging in $x=frac1{m^2}$ for integer $mgeq1$,
$$sum_{ngeq1}frac{(-1)^n}{n^3}sinbigg(frac{n}{m^2}bigg)=frac{1}{12m^6}-frac{pi^2}{12m^2}$$
then applying $sum_{mgeq1}$ on both sides,
$$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^3}sinbigg(frac{n}{m^2}bigg)=frac1{12}zeta(6)-frac{pi^2}{12}zeta(2)$$
We simplify to reach our conclusion
$$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^3}sinbigg(frac{n}{m^2}bigg)=frac{pi^6}{11340}-frac{pi^4}{72}$$
sequences-and-series proof-verification fourier-series decimal-expansion
edited Dec 25 '18 at 9:35
Did
249k23229468
asked Dec 25 '18 at 1:00
clathratusclathratus
5,3141440
$endgroup$
1
$begingroup$
This is a great way to obtain some nice problems for math contests or exams :D
$endgroup$
– Zacky
Dec 25 '18 at 1:35
1
$begingroup$
Mathematica is able to find the value of the sum up to $5$ decimal places: $-1.26813$. It agrees with the numerical value of $frac{pi^6}{11340}-frac{pi^4}{72}$.
$endgroup$
– Kemono Chen
Dec 25 '18 at 1:50
2
$begingroup$
@Zacky using a similar method, one can show that $$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^4}cos(n/m^2)=frac{pi^6}{2160}-frac{pi^8}{453600}-frac{7pi^4}{720}$$
$endgroup$
– clathratus
Dec 25 '18 at 1:53
1
$begingroup$
Nicely done, for sure ! $to +1$
$endgroup$
– Claude Leibovici
Dec 25 '18 at 3:25
2
$begingroup$
Consider abandoningbigg(andbigg)and replacing them byleft(andright).
$endgroup$
– Did
Dec 25 '18 at 9:36
|
show 1 more comment
$begingroup$
I am looking for numerical evidence that
$$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^3}sinleft(frac{n}{m^2}right)=frac{pi^6}{11340}-frac{pi^4}{72}$$
I have proven it, but I just want to be extra sure. Desmos only gives me accuracy to the third decimal place, but I know that some of you (@Claude Leibovici) are able to give me extremely high decimal accuracy.
Proof:
We know that for $|t|leqpi$,
$$t^2=frac{pi^2}3+4sum_{ngeq1}frac{(-1)^n}{n^2}cos nt$$
Solving for the sum then integrating both sides from $0$ to $x$,
$$sum_{ngeq1}frac{(-1)^n}{n^3}sin(nx)=frac{x^3}{12}-frac{pi^2x}{12}$$
Then plugging in $x=frac1{m^2}$ for integer $mgeq1$,
$$sum_{ngeq1}frac{(-1)^n}{n^3}sinbigg(frac{n}{m^2}bigg)=frac{1}{12m^6}-frac{pi^2}{12m^2}$$
then applying $sum_{mgeq1}$ on both sides,
$$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^3}sinbigg(frac{n}{m^2}bigg)=frac1{12}zeta(6)-frac{pi^2}{12}zeta(2)$$
We simplify to reach our conclusion
$$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^3}sinbigg(frac{n}{m^2}bigg)=frac{pi^6}{11340}-frac{pi^4}{72}$$
sequences-and-series proof-verification fourier-series decimal-expansion
edited Dec 25 '18 at 9:35
Did
249k23229468
asked Dec 25 '18 at 1:00
clathratusclathratus
5,3141440
$endgroup$
I am looking for numerical evidence that
$$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^3}sinleft(frac{n}{m^2}right)=frac{pi^6}{11340}-frac{pi^4}{72}$$
I have proven it, but I just want to be extra sure. Desmos only gives me accuracy to the third decimal place, but I know that some of you (@Claude Leibovici) are able to give me extremely high decimal accuracy.
Proof:
We know that for $|t|leqpi$,
$$t^2=frac{pi^2}3+4sum_{ngeq1}frac{(-1)^n}{n^2}cos nt$$
Solving for the sum then integrating both sides from $0$ to $x$,
$$sum_{ngeq1}frac{(-1)^n}{n^3}sin(nx)=frac{x^3}{12}-frac{pi^2x}{12}$$
Then plugging in $x=frac1{m^2}$ for integer $mgeq1$,
$$sum_{ngeq1}frac{(-1)^n}{n^3}sinbigg(frac{n}{m^2}bigg)=frac{1}{12m^6}-frac{pi^2}{12m^2}$$
then applying $sum_{mgeq1}$ on both sides,
$$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^3}sinbigg(frac{n}{m^2}bigg)=frac1{12}zeta(6)-frac{pi^2}{12}zeta(2)$$
We simplify to reach our conclusion
$$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^3}sinbigg(frac{n}{m^2}bigg)=frac{pi^6}{11340}-frac{pi^4}{72}$$
sequences-and-series proof-verification fourier-series decimal-expansion
sequences-and-series proof-verification fourier-series decimal-expansion
edited Dec 25 '18 at 9:35
Did
249k23229468
asked Dec 25 '18 at 1:00
clathratusclathratus
5,3141440
edited Dec 25 '18 at 9:35
Did
249k23229468
asked Dec 25 '18 at 1:00
clathratusclathratus
5,3141440
edited Dec 25 '18 at 9:35
Did
249k23229468
edited Dec 25 '18 at 9:35
Did
249k23229468
edited Dec 25 '18 at 9:35
Did
249k23229468
249k23229468
asked Dec 25 '18 at 1:00
clathratusclathratus
5,3141440
asked Dec 25 '18 at 1:00
clathratusclathratus
5,3141440
asked Dec 25 '18 at 1:00
clathratusclathratus
5,3141440
5,3141440
1
$begingroup$
This is a great way to obtain some nice problems for math contests or exams :D
$endgroup$
– Zacky
Dec 25 '18 at 1:35
1
$begingroup$
Mathematica is able to find the value of the sum up to $5$ decimal places: $-1.26813$. It agrees with the numerical value of $frac{pi^6}{11340}-frac{pi^4}{72}$.
$endgroup$
– Kemono Chen
Dec 25 '18 at 1:50
2
$begingroup$
@Zacky using a similar method, one can show that $$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^4}cos(n/m^2)=frac{pi^6}{2160}-frac{pi^8}{453600}-frac{7pi^4}{720}$$
$endgroup$
– clathratus
Dec 25 '18 at 1:53
1
$begingroup$
Nicely done, for sure ! $to +1$
$endgroup$
– Claude Leibovici
Dec 25 '18 at 3:25
2
$begingroup$
Consider abandoningbigg(andbigg)and replacing them byleft(andright).
$endgroup$
– Did
Dec 25 '18 at 9:36
|
show 1 more comment
1
$begingroup$
This is a great way to obtain some nice problems for math contests or exams :D
$endgroup$
– Zacky
Dec 25 '18 at 1:35
1
$begingroup$
Mathematica is able to find the value of the sum up to $5$ decimal places: $-1.26813$. It agrees with the numerical value of $frac{pi^6}{11340}-frac{pi^4}{72}$.
$endgroup$
– Kemono Chen
Dec 25 '18 at 1:50
2
$begingroup$
@Zacky using a similar method, one can show that $$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^4}cos(n/m^2)=frac{pi^6}{2160}-frac{pi^8}{453600}-frac{7pi^4}{720}$$
$endgroup$
– clathratus
Dec 25 '18 at 1:53
1
$begingroup$
Nicely done, for sure ! $to +1$
$endgroup$
– Claude Leibovici
Dec 25 '18 at 3:25
2
$begingroup$
Consider abandoningbigg(andbigg)and replacing them byleft(andright).
$endgroup$
– Did
Dec 25 '18 at 9:36
1
1
$begingroup$
This is a great way to obtain some nice problems for math contests or exams :D
$endgroup$
– Zacky
Dec 25 '18 at 1:35
$begingroup$
This is a great way to obtain some nice problems for math contests or exams :D
$endgroup$
– Zacky
Dec 25 '18 at 1:35
1
1
$begingroup$
Mathematica is able to find the value of the sum up to $5$ decimal places: $-1.26813$. It agrees with the numerical value of $frac{pi^6}{11340}-frac{pi^4}{72}$.
$endgroup$
– Kemono Chen
Dec 25 '18 at 1:50
$begingroup$
Mathematica is able to find the value of the sum up to $5$ decimal places: $-1.26813$. It agrees with the numerical value of $frac{pi^6}{11340}-frac{pi^4}{72}$.
$endgroup$
– Kemono Chen
Dec 25 '18 at 1:50
2
2
$begingroup$
@Zacky using a similar method, one can show that $$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^4}cos(n/m^2)=frac{pi^6}{2160}-frac{pi^8}{453600}-frac{7pi^4}{720}$$
$endgroup$
– clathratus
Dec 25 '18 at 1:53
$begingroup$
@Zacky using a similar method, one can show that $$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^4}cos(n/m^2)=frac{pi^6}{2160}-frac{pi^8}{453600}-frac{7pi^4}{720}$$
$endgroup$
– clathratus
Dec 25 '18 at 1:53
1
1
$begingroup$
Nicely done, for sure ! $to +1$
$endgroup$
– Claude Leibovici
Dec 25 '18 at 3:25
$begingroup$
Nicely done, for sure ! $to +1$
$endgroup$
– Claude Leibovici
Dec 25 '18 at 3:25
2
2
$begingroup$
Consider abandoning
bigg( and bigg) and replacing them by left( and right).$endgroup$
– Did
Dec 25 '18 at 9:36
$begingroup$
Consider abandoning
bigg( and bigg) and replacing them by left( and right).$endgroup$
– Did
Dec 25 '18 at 9:36
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
After you so elegant solution, I am quite ashamed to have done it the following way.
$$sum_{n=1}^inftyfrac{(-1)^n}{n^3}sin(nx)=frac{i}{2} left(text{Li}_3left(-e^{-i x}right)-text{Li}_3left(-e^{i x}right)right)$$
Replace $x=frac 1 {m^2}$ and numerically compute
$$frac{i}{2}sum_{m=1}^infty left(text{Li}_3left(-e^{-frac i {m^2} }right)-text{Li}_3left(-e^{frac i {m^2}}right)right)$$
It takes a long time to arrive to $-1.268125453640218644419$.
Update
I watched the news, I read the newspaper, I had a heavy breakfast and came back to my computer
$$-1.26812545364021864441879478985797582963162172461540$$
Edit
I tried to understand why this computation was taking so much computing time.
Let us consider
$$a_m=frac{i}{2} left(text{Li}_3left(-e^{-frac i {m^2} }right)-text{Li}_3left(-e^{frac i {m^2}}right)right)$$ For large values of $m$, we have
$$a_m=-frac{pi ^2}{12 m^2}+frac{1}{12 m^6}+Oleft(frac{1}{m^{14}}right)$$ So
$$sum_{m=1}^infty a_msimeqsum_{m=1}^p a_m+sum_{m=p+1}^inftyleft(-frac{pi ^2}{12 m^2}+frac{1}{12 m^6} right)=sum_{m=1}^p a_m+frac{psi ^{(5)}(p+1)-120 pi ^2 psi ^{(1)}(p+1)}{1440}$$ and the last term is almost
$$frac{psi ^{(5)}(p+1)-120 pi ^2 psi ^{(1)}(p+1)}{1440}simeq -frac{pi^2}{12p}left(1-frac{1}{2 p}+frac{1}{6 p^2}+Oleft(frac{1}{p^4}right) right)$$ In other words, for $k$ significant figures, we need to add a lot terms.
This explains that.
answered Dec 25 '18 at 4:03
Claude LeiboviciClaude Leibovici
127k1158135
$endgroup$
$begingroup$
Elegant or not, I really appreciate the answer. Thanks Claude :)
$endgroup$
– clathratus
Dec 25 '18 at 4:07
1
$begingroup$
@clathratus. It is not only elegant; it is beautiful ! Beauty is part of mathematics.
$endgroup$
– Claude Leibovici
Dec 25 '18 at 4:12
$begingroup$
Nice! Mind if I ask you what programm do you use for such computations?
$endgroup$
– Zacky
Dec 25 '18 at 15:04
$begingroup$
@Zacky. I have an old CAS that, in my group over more than twenty years, we adapted to our specific needs. I suppose that this could easily be done using Mathematica.
$endgroup$
– Claude Leibovici
Dec 25 '18 at 15:09
$begingroup$
I see, pretty nice, I only have the online version for Mathematica. If you have some free time, can you please check it's power against this? $$int_0^infty frac{1-cos x}{8-4xsin x +x^2(1-cos x)}dx=frac{pi}{4}$$ My wolfram can't even spit $1$ decimal out... So I would be glad if I know that the result it's correct. It is $I_5$ that I put here: math.stackexchange.com/a/3027759/515527
$endgroup$
– Zacky
Dec 25 '18 at 15:11
|
show 2 more comments
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1 Answer
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$begingroup$
After you so elegant solution, I am quite ashamed to have done it the following way.
$$sum_{n=1}^inftyfrac{(-1)^n}{n^3}sin(nx)=frac{i}{2} left(text{Li}_3left(-e^{-i x}right)-text{Li}_3left(-e^{i x}right)right)$$
Replace $x=frac 1 {m^2}$ and numerically compute
$$frac{i}{2}sum_{m=1}^infty left(text{Li}_3left(-e^{-frac i {m^2} }right)-text{Li}_3left(-e^{frac i {m^2}}right)right)$$
It takes a long time to arrive to $-1.268125453640218644419$.
Update
I watched the news, I read the newspaper, I had a heavy breakfast and came back to my computer
$$-1.26812545364021864441879478985797582963162172461540$$
Edit
I tried to understand why this computation was taking so much computing time.
Let us consider
$$a_m=frac{i}{2} left(text{Li}_3left(-e^{-frac i {m^2} }right)-text{Li}_3left(-e^{frac i {m^2}}right)right)$$ For large values of $m$, we have
$$a_m=-frac{pi ^2}{12 m^2}+frac{1}{12 m^6}+Oleft(frac{1}{m^{14}}right)$$ So
$$sum_{m=1}^infty a_msimeqsum_{m=1}^p a_m+sum_{m=p+1}^inftyleft(-frac{pi ^2}{12 m^2}+frac{1}{12 m^6} right)=sum_{m=1}^p a_m+frac{psi ^{(5)}(p+1)-120 pi ^2 psi ^{(1)}(p+1)}{1440}$$ and the last term is almost
$$frac{psi ^{(5)}(p+1)-120 pi ^2 psi ^{(1)}(p+1)}{1440}simeq -frac{pi^2}{12p}left(1-frac{1}{2 p}+frac{1}{6 p^2}+Oleft(frac{1}{p^4}right) right)$$ In other words, for $k$ significant figures, we need to add a lot terms.
This explains that.
answered Dec 25 '18 at 4:03
Claude LeiboviciClaude Leibovici
127k1158135
$endgroup$
$begingroup$
Elegant or not, I really appreciate the answer. Thanks Claude :)
$endgroup$
– clathratus
Dec 25 '18 at 4:07
1
$begingroup$
@clathratus. It is not only elegant; it is beautiful ! Beauty is part of mathematics.
$endgroup$
– Claude Leibovici
Dec 25 '18 at 4:12
$begingroup$
Nice! Mind if I ask you what programm do you use for such computations?
$endgroup$
– Zacky
Dec 25 '18 at 15:04
$begingroup$
@Zacky. I have an old CAS that, in my group over more than twenty years, we adapted to our specific needs. I suppose that this could easily be done using Mathematica.
$endgroup$
– Claude Leibovici
Dec 25 '18 at 15:09
$begingroup$
I see, pretty nice, I only have the online version for Mathematica. If you have some free time, can you please check it's power against this? $$int_0^infty frac{1-cos x}{8-4xsin x +x^2(1-cos x)}dx=frac{pi}{4}$$ My wolfram can't even spit $1$ decimal out... So I would be glad if I know that the result it's correct. It is $I_5$ that I put here: math.stackexchange.com/a/3027759/515527
$endgroup$
– Zacky
Dec 25 '18 at 15:11
|
show 2 more comments
$begingroup$
After you so elegant solution, I am quite ashamed to have done it the following way.
$$sum_{n=1}^inftyfrac{(-1)^n}{n^3}sin(nx)=frac{i}{2} left(text{Li}_3left(-e^{-i x}right)-text{Li}_3left(-e^{i x}right)right)$$
Replace $x=frac 1 {m^2}$ and numerically compute
$$frac{i}{2}sum_{m=1}^infty left(text{Li}_3left(-e^{-frac i {m^2} }right)-text{Li}_3left(-e^{frac i {m^2}}right)right)$$
It takes a long time to arrive to $-1.268125453640218644419$.
Update
I watched the news, I read the newspaper, I had a heavy breakfast and came back to my computer
$$-1.26812545364021864441879478985797582963162172461540$$
Edit
I tried to understand why this computation was taking so much computing time.
Let us consider
$$a_m=frac{i}{2} left(text{Li}_3left(-e^{-frac i {m^2} }right)-text{Li}_3left(-e^{frac i {m^2}}right)right)$$ For large values of $m$, we have
$$a_m=-frac{pi ^2}{12 m^2}+frac{1}{12 m^6}+Oleft(frac{1}{m^{14}}right)$$ So
$$sum_{m=1}^infty a_msimeqsum_{m=1}^p a_m+sum_{m=p+1}^inftyleft(-frac{pi ^2}{12 m^2}+frac{1}{12 m^6} right)=sum_{m=1}^p a_m+frac{psi ^{(5)}(p+1)-120 pi ^2 psi ^{(1)}(p+1)}{1440}$$ and the last term is almost
$$frac{psi ^{(5)}(p+1)-120 pi ^2 psi ^{(1)}(p+1)}{1440}simeq -frac{pi^2}{12p}left(1-frac{1}{2 p}+frac{1}{6 p^2}+Oleft(frac{1}{p^4}right) right)$$ In other words, for $k$ significant figures, we need to add a lot terms.
This explains that.
answered Dec 25 '18 at 4:03
Claude LeiboviciClaude Leibovici
127k1158135
$endgroup$
$begingroup$
Elegant or not, I really appreciate the answer. Thanks Claude :)
$endgroup$
– clathratus
Dec 25 '18 at 4:07
1
$begingroup$
@clathratus. It is not only elegant; it is beautiful ! Beauty is part of mathematics.
$endgroup$
– Claude Leibovici
Dec 25 '18 at 4:12
$begingroup$
Nice! Mind if I ask you what programm do you use for such computations?
$endgroup$
– Zacky
Dec 25 '18 at 15:04
$begingroup$
@Zacky. I have an old CAS that, in my group over more than twenty years, we adapted to our specific needs. I suppose that this could easily be done using Mathematica.
$endgroup$
– Claude Leibovici
Dec 25 '18 at 15:09
$begingroup$
I see, pretty nice, I only have the online version for Mathematica. If you have some free time, can you please check it's power against this? $$int_0^infty frac{1-cos x}{8-4xsin x +x^2(1-cos x)}dx=frac{pi}{4}$$ My wolfram can't even spit $1$ decimal out... So I would be glad if I know that the result it's correct. It is $I_5$ that I put here: math.stackexchange.com/a/3027759/515527
$endgroup$
– Zacky
Dec 25 '18 at 15:11
|
show 2 more comments
$begingroup$
After you so elegant solution, I am quite ashamed to have done it the following way.
$$sum_{n=1}^inftyfrac{(-1)^n}{n^3}sin(nx)=frac{i}{2} left(text{Li}_3left(-e^{-i x}right)-text{Li}_3left(-e^{i x}right)right)$$
Replace $x=frac 1 {m^2}$ and numerically compute
$$frac{i}{2}sum_{m=1}^infty left(text{Li}_3left(-e^{-frac i {m^2} }right)-text{Li}_3left(-e^{frac i {m^2}}right)right)$$
It takes a long time to arrive to $-1.268125453640218644419$.
Update
I watched the news, I read the newspaper, I had a heavy breakfast and came back to my computer
$$-1.26812545364021864441879478985797582963162172461540$$
Edit
I tried to understand why this computation was taking so much computing time.
Let us consider
$$a_m=frac{i}{2} left(text{Li}_3left(-e^{-frac i {m^2} }right)-text{Li}_3left(-e^{frac i {m^2}}right)right)$$ For large values of $m$, we have
$$a_m=-frac{pi ^2}{12 m^2}+frac{1}{12 m^6}+Oleft(frac{1}{m^{14}}right)$$ So
$$sum_{m=1}^infty a_msimeqsum_{m=1}^p a_m+sum_{m=p+1}^inftyleft(-frac{pi ^2}{12 m^2}+frac{1}{12 m^6} right)=sum_{m=1}^p a_m+frac{psi ^{(5)}(p+1)-120 pi ^2 psi ^{(1)}(p+1)}{1440}$$ and the last term is almost
$$frac{psi ^{(5)}(p+1)-120 pi ^2 psi ^{(1)}(p+1)}{1440}simeq -frac{pi^2}{12p}left(1-frac{1}{2 p}+frac{1}{6 p^2}+Oleft(frac{1}{p^4}right) right)$$ In other words, for $k$ significant figures, we need to add a lot terms.
This explains that.
answered Dec 25 '18 at 4:03
Claude LeiboviciClaude Leibovici
127k1158135
$endgroup$
After you so elegant solution, I am quite ashamed to have done it the following way.
$$sum_{n=1}^inftyfrac{(-1)^n}{n^3}sin(nx)=frac{i}{2} left(text{Li}_3left(-e^{-i x}right)-text{Li}_3left(-e^{i x}right)right)$$
Replace $x=frac 1 {m^2}$ and numerically compute
$$frac{i}{2}sum_{m=1}^infty left(text{Li}_3left(-e^{-frac i {m^2} }right)-text{Li}_3left(-e^{frac i {m^2}}right)right)$$
It takes a long time to arrive to $-1.268125453640218644419$.
Update
I watched the news, I read the newspaper, I had a heavy breakfast and came back to my computer
$$-1.26812545364021864441879478985797582963162172461540$$
Edit
I tried to understand why this computation was taking so much computing time.
Let us consider
$$a_m=frac{i}{2} left(text{Li}_3left(-e^{-frac i {m^2} }right)-text{Li}_3left(-e^{frac i {m^2}}right)right)$$ For large values of $m$, we have
$$a_m=-frac{pi ^2}{12 m^2}+frac{1}{12 m^6}+Oleft(frac{1}{m^{14}}right)$$ So
$$sum_{m=1}^infty a_msimeqsum_{m=1}^p a_m+sum_{m=p+1}^inftyleft(-frac{pi ^2}{12 m^2}+frac{1}{12 m^6} right)=sum_{m=1}^p a_m+frac{psi ^{(5)}(p+1)-120 pi ^2 psi ^{(1)}(p+1)}{1440}$$ and the last term is almost
$$frac{psi ^{(5)}(p+1)-120 pi ^2 psi ^{(1)}(p+1)}{1440}simeq -frac{pi^2}{12p}left(1-frac{1}{2 p}+frac{1}{6 p^2}+Oleft(frac{1}{p^4}right) right)$$ In other words, for $k$ significant figures, we need to add a lot terms.
This explains that.
answered Dec 25 '18 at 4:03
Claude LeiboviciClaude Leibovici
127k1158135
edited Dec 25 '18 at 14:39
answered Dec 25 '18 at 4:03
Claude LeiboviciClaude Leibovici
127k1158135
answered Dec 25 '18 at 4:03
Claude LeiboviciClaude Leibovici
127k1158135
answered Dec 25 '18 at 4:03
Claude LeiboviciClaude Leibovici
127k1158135
127k1158135
$begingroup$
Elegant or not, I really appreciate the answer. Thanks Claude :)
$endgroup$
– clathratus
Dec 25 '18 at 4:07
1
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@clathratus. It is not only elegant; it is beautiful ! Beauty is part of mathematics.
$endgroup$
– Claude Leibovici
Dec 25 '18 at 4:12
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Nice! Mind if I ask you what programm do you use for such computations?
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– Zacky
Dec 25 '18 at 15:04
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@Zacky. I have an old CAS that, in my group over more than twenty years, we adapted to our specific needs. I suppose that this could easily be done using Mathematica.
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– Claude Leibovici
Dec 25 '18 at 15:09
$begingroup$
I see, pretty nice, I only have the online version for Mathematica. If you have some free time, can you please check it's power against this? $$int_0^infty frac{1-cos x}{8-4xsin x +x^2(1-cos x)}dx=frac{pi}{4}$$ My wolfram can't even spit $1$ decimal out... So I would be glad if I know that the result it's correct. It is $I_5$ that I put here: math.stackexchange.com/a/3027759/515527
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– Zacky
Dec 25 '18 at 15:11
|
show 2 more comments
$begingroup$
Elegant or not, I really appreciate the answer. Thanks Claude :)
$endgroup$
– clathratus
Dec 25 '18 at 4:07
1
$begingroup$
@clathratus. It is not only elegant; it is beautiful ! Beauty is part of mathematics.
$endgroup$
– Claude Leibovici
Dec 25 '18 at 4:12
$begingroup$
Nice! Mind if I ask you what programm do you use for such computations?
$endgroup$
– Zacky
Dec 25 '18 at 15:04
$begingroup$
@Zacky. I have an old CAS that, in my group over more than twenty years, we adapted to our specific needs. I suppose that this could easily be done using Mathematica.
$endgroup$
– Claude Leibovici
Dec 25 '18 at 15:09
$begingroup$
I see, pretty nice, I only have the online version for Mathematica. If you have some free time, can you please check it's power against this? $$int_0^infty frac{1-cos x}{8-4xsin x +x^2(1-cos x)}dx=frac{pi}{4}$$ My wolfram can't even spit $1$ decimal out... So I would be glad if I know that the result it's correct. It is $I_5$ that I put here: math.stackexchange.com/a/3027759/515527
$endgroup$
– Zacky
Dec 25 '18 at 15:11
$begingroup$
Elegant or not, I really appreciate the answer. Thanks Claude :)
$endgroup$
– clathratus
Dec 25 '18 at 4:07
$begingroup$
Elegant or not, I really appreciate the answer. Thanks Claude :)
$endgroup$
– clathratus
Dec 25 '18 at 4:07
1
1
$begingroup$
@clathratus. It is not only elegant; it is beautiful ! Beauty is part of mathematics.
$endgroup$
– Claude Leibovici
Dec 25 '18 at 4:12
$begingroup$
@clathratus. It is not only elegant; it is beautiful ! Beauty is part of mathematics.
$endgroup$
– Claude Leibovici
Dec 25 '18 at 4:12
$begingroup$
Nice! Mind if I ask you what programm do you use for such computations?
$endgroup$
– Zacky
Dec 25 '18 at 15:04
$begingroup$
Nice! Mind if I ask you what programm do you use for such computations?
$endgroup$
– Zacky
Dec 25 '18 at 15:04
$begingroup$
@Zacky. I have an old CAS that, in my group over more than twenty years, we adapted to our specific needs. I suppose that this could easily be done using Mathematica.
$endgroup$
– Claude Leibovici
Dec 25 '18 at 15:09
$begingroup$
@Zacky. I have an old CAS that, in my group over more than twenty years, we adapted to our specific needs. I suppose that this could easily be done using Mathematica.
$endgroup$
– Claude Leibovici
Dec 25 '18 at 15:09
$begingroup$
I see, pretty nice, I only have the online version for Mathematica. If you have some free time, can you please check it's power against this? $$int_0^infty frac{1-cos x}{8-4xsin x +x^2(1-cos x)}dx=frac{pi}{4}$$ My wolfram can't even spit $1$ decimal out... So I would be glad if I know that the result it's correct. It is $I_5$ that I put here: math.stackexchange.com/a/3027759/515527
$endgroup$
– Zacky
Dec 25 '18 at 15:11
$begingroup$
I see, pretty nice, I only have the online version for Mathematica. If you have some free time, can you please check it's power against this? $$int_0^infty frac{1-cos x}{8-4xsin x +x^2(1-cos x)}dx=frac{pi}{4}$$ My wolfram can't even spit $1$ decimal out... So I would be glad if I know that the result it's correct. It is $I_5$ that I put here: math.stackexchange.com/a/3027759/515527
$endgroup$
– Zacky
Dec 25 '18 at 15:11
|
show 2 more comments
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This is a great way to obtain some nice problems for math contests or exams :D
$endgroup$
– Zacky
Dec 25 '18 at 1:35
1
$begingroup$
Mathematica is able to find the value of the sum up to $5$ decimal places: $-1.26813$. It agrees with the numerical value of $frac{pi^6}{11340}-frac{pi^4}{72}$.
$endgroup$
– Kemono Chen
Dec 25 '18 at 1:50
2
$begingroup$
@Zacky using a similar method, one can show that $$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^4}cos(n/m^2)=frac{pi^6}{2160}-frac{pi^8}{453600}-frac{7pi^4}{720}$$
$endgroup$
– clathratus
Dec 25 '18 at 1:53
1
$begingroup$
Nicely done, for sure ! $to +1$
$endgroup$
– Claude Leibovici
Dec 25 '18 at 3:25
2
$begingroup$
Consider abandoning
bigg(andbigg)and replacing them byleft(andright).$endgroup$
– Did
Dec 25 '18 at 9:36