Sum of a speciale series
$begingroup$
I would like to know if what i have found is correct or not
what is the value of this series
$$sum_{n=2,n;text{even} }^{infty}dfrac{sin^{2}(nt)}{n^2(n^2-1)}$$
where $t in [0 , 2pi]$
Thanks in adavance
sequences-and-series fourier-series
$endgroup$
add a comment |
$begingroup$
I would like to know if what i have found is correct or not
what is the value of this series
$$sum_{n=2,n;text{even} }^{infty}dfrac{sin^{2}(nt)}{n^2(n^2-1)}$$
where $t in [0 , 2pi]$
Thanks in adavance
sequences-and-series fourier-series
$endgroup$
$begingroup$
Why don't you report what you found ?
$endgroup$
– Claude Leibovici
Jan 28 at 7:15
add a comment |
$begingroup$
I would like to know if what i have found is correct or not
what is the value of this series
$$sum_{n=2,n;text{even} }^{infty}dfrac{sin^{2}(nt)}{n^2(n^2-1)}$$
where $t in [0 , 2pi]$
Thanks in adavance
sequences-and-series fourier-series
$endgroup$
I would like to know if what i have found is correct or not
what is the value of this series
$$sum_{n=2,n;text{even} }^{infty}dfrac{sin^{2}(nt)}{n^2(n^2-1)}$$
where $t in [0 , 2pi]$
Thanks in adavance
sequences-and-series fourier-series
sequences-and-series fourier-series
edited Jan 28 at 0:31
user
5,94011031
5,94011031
asked Jan 27 at 22:34
BernsteinBernstein
12012
12012
$begingroup$
Why don't you report what you found ?
$endgroup$
– Claude Leibovici
Jan 28 at 7:15
add a comment |
$begingroup$
Why don't you report what you found ?
$endgroup$
– Claude Leibovici
Jan 28 at 7:15
$begingroup$
Why don't you report what you found ?
$endgroup$
– Claude Leibovici
Jan 28 at 7:15
$begingroup$
Why don't you report what you found ?
$endgroup$
– Claude Leibovici
Jan 28 at 7:15
add a comment |
1 Answer
1
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oldest
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$begingroup$
As $nge 2$ even integer we can write $n=2k$ where $kge1$:
$sumlimits_{k=1 }^{infty}dfrac{sin^{2}(2kt)}{(2k)^2((2k)^2-1)}=dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{1-cos(4kt)}{(2k)^2-1}-dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{1-cos(4kt)}{(2k)^2}tag1$
Regrouping (1):
$dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{1}{(2k)^2-1}-dfrac{1}{8}sumlimits_{k=1 }^{infty}dfrac{1}{k^2}-dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{cos(4kt)}{(2k)^2-1}+dfrac{1}{8}sumlimits_{k=1 }^{infty}dfrac{cos(4kt)}{k^2}tag2$
Let's calculate the sums of (2) one after the other:
First one is telescoping $S_1=dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{1}{(2k)^2-1}=dfrac{1}{4}sumlimits_{k=1 }^{infty}Big(dfrac{1}{2k-1}-dfrac{1}{2k+1}big)=dfrac{1}{4}tag{2.1}$
The second one is Riemann zeta function:
$S_2=-dfrac{1}{8}sumlimits_{k=1 }^{infty}dfrac{1}{k^2}=dfrac{-zeta(2)}{8}=-dfrac{pi^2}{48}tag{2.2}$
- At the third one we have to realize that the sum is proportional to the Fourier series of
$| sin(2t)|$ can be found on the link //hu.wikipedia.org/wiki/Fourier-transzform%C3%A1ci%C3%B3#Fourier-sorok;
$|sin(2t)|=dfrac{2}{pi}-dfrac{4}{pi}sumlimits_{k=1 }^{infty}dfrac{cos(4kt)}{(2k)^2-1}$
So $S_3=-dfrac{1}{4}+dfrac{pi}{8}|sin(2t)|tag{2.3}$
- The last one can be expressed with dilogarithm function:
As $cos(4kt)=dfrac{e^{4ikt}+e^{-4ikt}}{2}$
$S_4=-dfrac{1}{16}Big(Li_2(e^{4it})+Li_2(e^{-4it})Big)=-dfrac{1}{8}Ci_2(4t)tag{2.4}$
where $Ci_2$ is the real part of Clausen function of order 2.
Finally the result:
$sum_{n=2,n;text{even} }^{infty}dfrac{sin^{2}(nt)}{n^2(n^2-1)}=dfrac{pi}{8}|sin(2t)|-dfrac{pi^2}{48}-dfrac{1}{8}Ci_2(4t)tag3$
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As $nge 2$ even integer we can write $n=2k$ where $kge1$:
$sumlimits_{k=1 }^{infty}dfrac{sin^{2}(2kt)}{(2k)^2((2k)^2-1)}=dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{1-cos(4kt)}{(2k)^2-1}-dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{1-cos(4kt)}{(2k)^2}tag1$
Regrouping (1):
$dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{1}{(2k)^2-1}-dfrac{1}{8}sumlimits_{k=1 }^{infty}dfrac{1}{k^2}-dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{cos(4kt)}{(2k)^2-1}+dfrac{1}{8}sumlimits_{k=1 }^{infty}dfrac{cos(4kt)}{k^2}tag2$
Let's calculate the sums of (2) one after the other:
First one is telescoping $S_1=dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{1}{(2k)^2-1}=dfrac{1}{4}sumlimits_{k=1 }^{infty}Big(dfrac{1}{2k-1}-dfrac{1}{2k+1}big)=dfrac{1}{4}tag{2.1}$
The second one is Riemann zeta function:
$S_2=-dfrac{1}{8}sumlimits_{k=1 }^{infty}dfrac{1}{k^2}=dfrac{-zeta(2)}{8}=-dfrac{pi^2}{48}tag{2.2}$
- At the third one we have to realize that the sum is proportional to the Fourier series of
$| sin(2t)|$ can be found on the link //hu.wikipedia.org/wiki/Fourier-transzform%C3%A1ci%C3%B3#Fourier-sorok;
$|sin(2t)|=dfrac{2}{pi}-dfrac{4}{pi}sumlimits_{k=1 }^{infty}dfrac{cos(4kt)}{(2k)^2-1}$
So $S_3=-dfrac{1}{4}+dfrac{pi}{8}|sin(2t)|tag{2.3}$
- The last one can be expressed with dilogarithm function:
As $cos(4kt)=dfrac{e^{4ikt}+e^{-4ikt}}{2}$
$S_4=-dfrac{1}{16}Big(Li_2(e^{4it})+Li_2(e^{-4it})Big)=-dfrac{1}{8}Ci_2(4t)tag{2.4}$
where $Ci_2$ is the real part of Clausen function of order 2.
Finally the result:
$sum_{n=2,n;text{even} }^{infty}dfrac{sin^{2}(nt)}{n^2(n^2-1)}=dfrac{pi}{8}|sin(2t)|-dfrac{pi^2}{48}-dfrac{1}{8}Ci_2(4t)tag3$
$endgroup$
add a comment |
$begingroup$
As $nge 2$ even integer we can write $n=2k$ where $kge1$:
$sumlimits_{k=1 }^{infty}dfrac{sin^{2}(2kt)}{(2k)^2((2k)^2-1)}=dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{1-cos(4kt)}{(2k)^2-1}-dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{1-cos(4kt)}{(2k)^2}tag1$
Regrouping (1):
$dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{1}{(2k)^2-1}-dfrac{1}{8}sumlimits_{k=1 }^{infty}dfrac{1}{k^2}-dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{cos(4kt)}{(2k)^2-1}+dfrac{1}{8}sumlimits_{k=1 }^{infty}dfrac{cos(4kt)}{k^2}tag2$
Let's calculate the sums of (2) one after the other:
First one is telescoping $S_1=dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{1}{(2k)^2-1}=dfrac{1}{4}sumlimits_{k=1 }^{infty}Big(dfrac{1}{2k-1}-dfrac{1}{2k+1}big)=dfrac{1}{4}tag{2.1}$
The second one is Riemann zeta function:
$S_2=-dfrac{1}{8}sumlimits_{k=1 }^{infty}dfrac{1}{k^2}=dfrac{-zeta(2)}{8}=-dfrac{pi^2}{48}tag{2.2}$
- At the third one we have to realize that the sum is proportional to the Fourier series of
$| sin(2t)|$ can be found on the link //hu.wikipedia.org/wiki/Fourier-transzform%C3%A1ci%C3%B3#Fourier-sorok;
$|sin(2t)|=dfrac{2}{pi}-dfrac{4}{pi}sumlimits_{k=1 }^{infty}dfrac{cos(4kt)}{(2k)^2-1}$
So $S_3=-dfrac{1}{4}+dfrac{pi}{8}|sin(2t)|tag{2.3}$
- The last one can be expressed with dilogarithm function:
As $cos(4kt)=dfrac{e^{4ikt}+e^{-4ikt}}{2}$
$S_4=-dfrac{1}{16}Big(Li_2(e^{4it})+Li_2(e^{-4it})Big)=-dfrac{1}{8}Ci_2(4t)tag{2.4}$
where $Ci_2$ is the real part of Clausen function of order 2.
Finally the result:
$sum_{n=2,n;text{even} }^{infty}dfrac{sin^{2}(nt)}{n^2(n^2-1)}=dfrac{pi}{8}|sin(2t)|-dfrac{pi^2}{48}-dfrac{1}{8}Ci_2(4t)tag3$
$endgroup$
add a comment |
$begingroup$
As $nge 2$ even integer we can write $n=2k$ where $kge1$:
$sumlimits_{k=1 }^{infty}dfrac{sin^{2}(2kt)}{(2k)^2((2k)^2-1)}=dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{1-cos(4kt)}{(2k)^2-1}-dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{1-cos(4kt)}{(2k)^2}tag1$
Regrouping (1):
$dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{1}{(2k)^2-1}-dfrac{1}{8}sumlimits_{k=1 }^{infty}dfrac{1}{k^2}-dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{cos(4kt)}{(2k)^2-1}+dfrac{1}{8}sumlimits_{k=1 }^{infty}dfrac{cos(4kt)}{k^2}tag2$
Let's calculate the sums of (2) one after the other:
First one is telescoping $S_1=dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{1}{(2k)^2-1}=dfrac{1}{4}sumlimits_{k=1 }^{infty}Big(dfrac{1}{2k-1}-dfrac{1}{2k+1}big)=dfrac{1}{4}tag{2.1}$
The second one is Riemann zeta function:
$S_2=-dfrac{1}{8}sumlimits_{k=1 }^{infty}dfrac{1}{k^2}=dfrac{-zeta(2)}{8}=-dfrac{pi^2}{48}tag{2.2}$
- At the third one we have to realize that the sum is proportional to the Fourier series of
$| sin(2t)|$ can be found on the link //hu.wikipedia.org/wiki/Fourier-transzform%C3%A1ci%C3%B3#Fourier-sorok;
$|sin(2t)|=dfrac{2}{pi}-dfrac{4}{pi}sumlimits_{k=1 }^{infty}dfrac{cos(4kt)}{(2k)^2-1}$
So $S_3=-dfrac{1}{4}+dfrac{pi}{8}|sin(2t)|tag{2.3}$
- The last one can be expressed with dilogarithm function:
As $cos(4kt)=dfrac{e^{4ikt}+e^{-4ikt}}{2}$
$S_4=-dfrac{1}{16}Big(Li_2(e^{4it})+Li_2(e^{-4it})Big)=-dfrac{1}{8}Ci_2(4t)tag{2.4}$
where $Ci_2$ is the real part of Clausen function of order 2.
Finally the result:
$sum_{n=2,n;text{even} }^{infty}dfrac{sin^{2}(nt)}{n^2(n^2-1)}=dfrac{pi}{8}|sin(2t)|-dfrac{pi^2}{48}-dfrac{1}{8}Ci_2(4t)tag3$
$endgroup$
As $nge 2$ even integer we can write $n=2k$ where $kge1$:
$sumlimits_{k=1 }^{infty}dfrac{sin^{2}(2kt)}{(2k)^2((2k)^2-1)}=dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{1-cos(4kt)}{(2k)^2-1}-dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{1-cos(4kt)}{(2k)^2}tag1$
Regrouping (1):
$dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{1}{(2k)^2-1}-dfrac{1}{8}sumlimits_{k=1 }^{infty}dfrac{1}{k^2}-dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{cos(4kt)}{(2k)^2-1}+dfrac{1}{8}sumlimits_{k=1 }^{infty}dfrac{cos(4kt)}{k^2}tag2$
Let's calculate the sums of (2) one after the other:
First one is telescoping $S_1=dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{1}{(2k)^2-1}=dfrac{1}{4}sumlimits_{k=1 }^{infty}Big(dfrac{1}{2k-1}-dfrac{1}{2k+1}big)=dfrac{1}{4}tag{2.1}$
The second one is Riemann zeta function:
$S_2=-dfrac{1}{8}sumlimits_{k=1 }^{infty}dfrac{1}{k^2}=dfrac{-zeta(2)}{8}=-dfrac{pi^2}{48}tag{2.2}$
- At the third one we have to realize that the sum is proportional to the Fourier series of
$| sin(2t)|$ can be found on the link //hu.wikipedia.org/wiki/Fourier-transzform%C3%A1ci%C3%B3#Fourier-sorok;
$|sin(2t)|=dfrac{2}{pi}-dfrac{4}{pi}sumlimits_{k=1 }^{infty}dfrac{cos(4kt)}{(2k)^2-1}$
So $S_3=-dfrac{1}{4}+dfrac{pi}{8}|sin(2t)|tag{2.3}$
- The last one can be expressed with dilogarithm function:
As $cos(4kt)=dfrac{e^{4ikt}+e^{-4ikt}}{2}$
$S_4=-dfrac{1}{16}Big(Li_2(e^{4it})+Li_2(e^{-4it})Big)=-dfrac{1}{8}Ci_2(4t)tag{2.4}$
where $Ci_2$ is the real part of Clausen function of order 2.
Finally the result:
$sum_{n=2,n;text{even} }^{infty}dfrac{sin^{2}(nt)}{n^2(n^2-1)}=dfrac{pi}{8}|sin(2t)|-dfrac{pi^2}{48}-dfrac{1}{8}Ci_2(4t)tag3$
answered Jan 28 at 18:54
JV.StalkerJV.Stalker
93349
93349
add a comment |
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$begingroup$
Why don't you report what you found ?
$endgroup$
– Claude Leibovici
Jan 28 at 7:15