Is there a closed form of $sum_{k=1}^{infty}frac{sin (2k-1)pi x}{(2k-1)^5pi^5}$ [closed]
Is there a closed form of $$sum_{k=1}^{infty}frac{sin (2k-1)pi x}{(2k-1)^5pi^5}$$
The above sum if a solution of some problem so I need to simplify it to get a clear equation if possible .......thanks
summation
asked Dec 28 '18 at 19:43
omarbaker100omarbaker100
32
closed as off-topic by RRL, amWhy, Zacky, Cesareo, Leucippus Dec 29 '18 at 1:31
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, amWhy, Zacky, Cesareo, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
Is there a closed form of $$sum_{k=1}^{infty}frac{sin (2k-1)pi x}{(2k-1)^5pi^5}$$
The above sum if a solution of some problem so I need to simplify it to get a clear equation if possible .......thanks
summation
asked Dec 28 '18 at 19:43
omarbaker100omarbaker100
32
closed as off-topic by RRL, amWhy, Zacky, Cesareo, Leucippus Dec 29 '18 at 1:31
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, amWhy, Zacky, Cesareo, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
You should tell us what you tried
– Ankit Kumar
Dec 28 '18 at 19:45
add a comment |
Is there a closed form of $$sum_{k=1}^{infty}frac{sin (2k-1)pi x}{(2k-1)^5pi^5}$$
The above sum if a solution of some problem so I need to simplify it to get a clear equation if possible .......thanks
summation
asked Dec 28 '18 at 19:43
omarbaker100omarbaker100
32
Is there a closed form of $$sum_{k=1}^{infty}frac{sin (2k-1)pi x}{(2k-1)^5pi^5}$$
The above sum if a solution of some problem so I need to simplify it to get a clear equation if possible .......thanks
summation
summation
asked Dec 28 '18 at 19:43
omarbaker100omarbaker100
32
asked Dec 28 '18 at 19:43
omarbaker100omarbaker100
32
edited Dec 28 '18 at 19:50
omarbaker100
asked Dec 28 '18 at 19:43
omarbaker100omarbaker100
32
asked Dec 28 '18 at 19:43
omarbaker100omarbaker100
32
asked Dec 28 '18 at 19:43
omarbaker100omarbaker100
32
32
closed as off-topic by RRL, amWhy, Zacky, Cesareo, Leucippus Dec 29 '18 at 1:31
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, amWhy, Zacky, Cesareo, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by RRL, amWhy, Zacky, Cesareo, Leucippus Dec 29 '18 at 1:31
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, amWhy, Zacky, Cesareo, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
You should tell us what you tried
– Ankit Kumar
Dec 28 '18 at 19:45
add a comment |
You should tell us what you tried
– Ankit Kumar
Dec 28 '18 at 19:45
You should tell us what you tried
– Ankit Kumar
Dec 28 '18 at 19:45
You should tell us what you tried
– Ankit Kumar
Dec 28 '18 at 19:45
add a comment |
2 Answers
2
active
oldest
votes
It is pretty well-known (and not difficult to prove) that
$$ sum_{ngeq 0}frac{sin((2n+1)pi x)}{(2n+1)pi} $$
is the Fourier (sine) series of the rectangle wave which equals $frac{1}{4}$ over $(0,1)$ and $-frac{1}{4}$ over $(1,2)$.
By applying termwise integration four times, it follows that
$$ sum_{ngeq 0}frac{sin((2n+1)pi x)}{(2n+1)^5 pi^5} $$
is a piecewise-quartic polynomial which equals $color{red}{frac{x(1-x)(1+x-x^2)}{96}}$ over $[0,1]$ and fulfills $f(x+1)=-f(x)$.
answered Dec 28 '18 at 21:09
Jack D'AurizioJack D'Aurizio
287k33280658
thank you very much
– omarbaker100
Dec 28 '18 at 23:58
I think the rectangle wave function is $frac{1}{4}$ not $frac{pi}{4}$
– E.H.E
Dec 29 '18 at 0:16
add a comment |
Mathematica gives:
$$-frac{i e^{-i text{$pi $x}} left(e^{2 i text{$pi $x}} Phi left(e^{2 i text{$pi
$x}},5,frac{1}{2}right)-Phi left(e^{-2 i text{$pi $x}},5,frac{1}{2}right)right)}{64 pi ^5}$$
answered Dec 28 '18 at 19:47
David G. StorkDavid G. Stork
10k21332
but I think your solution is not easy
– omarbaker100
Dec 28 '18 at 19:51
1
It is a closed form, but if the solution is "not easy," well... it is "not easy."
– David G. Stork
Dec 28 '18 at 20:01
thanks but I will wait if there is another form easier
– omarbaker100
Dec 28 '18 at 20:05
How is that possible?
– David G. Stork
Dec 28 '18 at 20:06
@DavidG.Stork: it is easy. This is just an example of the fact that Mathematica is not able to recognize Bernoulli polynomials from their Fourier series. I do not believe this is really an answer.
– Jack D'Aurizio
Dec 28 '18 at 21:10
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
It is pretty well-known (and not difficult to prove) that
$$ sum_{ngeq 0}frac{sin((2n+1)pi x)}{(2n+1)pi} $$
is the Fourier (sine) series of the rectangle wave which equals $frac{1}{4}$ over $(0,1)$ and $-frac{1}{4}$ over $(1,2)$.
By applying termwise integration four times, it follows that
$$ sum_{ngeq 0}frac{sin((2n+1)pi x)}{(2n+1)^5 pi^5} $$
is a piecewise-quartic polynomial which equals $color{red}{frac{x(1-x)(1+x-x^2)}{96}}$ over $[0,1]$ and fulfills $f(x+1)=-f(x)$.
answered Dec 28 '18 at 21:09
Jack D'AurizioJack D'Aurizio
287k33280658
thank you very much
– omarbaker100
Dec 28 '18 at 23:58
I think the rectangle wave function is $frac{1}{4}$ not $frac{pi}{4}$
– E.H.E
Dec 29 '18 at 0:16
add a comment |
It is pretty well-known (and not difficult to prove) that
$$ sum_{ngeq 0}frac{sin((2n+1)pi x)}{(2n+1)pi} $$
is the Fourier (sine) series of the rectangle wave which equals $frac{1}{4}$ over $(0,1)$ and $-frac{1}{4}$ over $(1,2)$.
By applying termwise integration four times, it follows that
$$ sum_{ngeq 0}frac{sin((2n+1)pi x)}{(2n+1)^5 pi^5} $$
is a piecewise-quartic polynomial which equals $color{red}{frac{x(1-x)(1+x-x^2)}{96}}$ over $[0,1]$ and fulfills $f(x+1)=-f(x)$.
answered Dec 28 '18 at 21:09
Jack D'AurizioJack D'Aurizio
287k33280658
thank you very much
– omarbaker100
Dec 28 '18 at 23:58
I think the rectangle wave function is $frac{1}{4}$ not $frac{pi}{4}$
– E.H.E
Dec 29 '18 at 0:16
add a comment |
It is pretty well-known (and not difficult to prove) that
$$ sum_{ngeq 0}frac{sin((2n+1)pi x)}{(2n+1)pi} $$
is the Fourier (sine) series of the rectangle wave which equals $frac{1}{4}$ over $(0,1)$ and $-frac{1}{4}$ over $(1,2)$.
By applying termwise integration four times, it follows that
$$ sum_{ngeq 0}frac{sin((2n+1)pi x)}{(2n+1)^5 pi^5} $$
is a piecewise-quartic polynomial which equals $color{red}{frac{x(1-x)(1+x-x^2)}{96}}$ over $[0,1]$ and fulfills $f(x+1)=-f(x)$.
answered Dec 28 '18 at 21:09
Jack D'AurizioJack D'Aurizio
287k33280658
It is pretty well-known (and not difficult to prove) that
$$ sum_{ngeq 0}frac{sin((2n+1)pi x)}{(2n+1)pi} $$
is the Fourier (sine) series of the rectangle wave which equals $frac{1}{4}$ over $(0,1)$ and $-frac{1}{4}$ over $(1,2)$.
By applying termwise integration four times, it follows that
$$ sum_{ngeq 0}frac{sin((2n+1)pi x)}{(2n+1)^5 pi^5} $$
is a piecewise-quartic polynomial which equals $color{red}{frac{x(1-x)(1+x-x^2)}{96}}$ over $[0,1]$ and fulfills $f(x+1)=-f(x)$.
answered Dec 28 '18 at 21:09
Jack D'AurizioJack D'Aurizio
287k33280658
edited Dec 29 '18 at 1:08
answered Dec 28 '18 at 21:09
Jack D'AurizioJack D'Aurizio
287k33280658
answered Dec 28 '18 at 21:09
Jack D'AurizioJack D'Aurizio
287k33280658
answered Dec 28 '18 at 21:09
Jack D'AurizioJack D'Aurizio
287k33280658
287k33280658
thank you very much
– omarbaker100
Dec 28 '18 at 23:58
I think the rectangle wave function is $frac{1}{4}$ not $frac{pi}{4}$
– E.H.E
Dec 29 '18 at 0:16
add a comment |
thank you very much
– omarbaker100
Dec 28 '18 at 23:58
I think the rectangle wave function is $frac{1}{4}$ not $frac{pi}{4}$
– E.H.E
Dec 29 '18 at 0:16
thank you very much
– omarbaker100
Dec 28 '18 at 23:58
thank you very much
– omarbaker100
Dec 28 '18 at 23:58
I think the rectangle wave function is $frac{1}{4}$ not $frac{pi}{4}$
– E.H.E
Dec 29 '18 at 0:16
I think the rectangle wave function is $frac{1}{4}$ not $frac{pi}{4}$
– E.H.E
Dec 29 '18 at 0:16
add a comment |
Mathematica gives:
$$-frac{i e^{-i text{$pi $x}} left(e^{2 i text{$pi $x}} Phi left(e^{2 i text{$pi
$x}},5,frac{1}{2}right)-Phi left(e^{-2 i text{$pi $x}},5,frac{1}{2}right)right)}{64 pi ^5}$$
answered Dec 28 '18 at 19:47
David G. StorkDavid G. Stork
10k21332
but I think your solution is not easy
– omarbaker100
Dec 28 '18 at 19:51
1
It is a closed form, but if the solution is "not easy," well... it is "not easy."
– David G. Stork
Dec 28 '18 at 20:01
thanks but I will wait if there is another form easier
– omarbaker100
Dec 28 '18 at 20:05
How is that possible?
– David G. Stork
Dec 28 '18 at 20:06
@DavidG.Stork: it is easy. This is just an example of the fact that Mathematica is not able to recognize Bernoulli polynomials from their Fourier series. I do not believe this is really an answer.
– Jack D'Aurizio
Dec 28 '18 at 21:10
add a comment |
Mathematica gives:
$$-frac{i e^{-i text{$pi $x}} left(e^{2 i text{$pi $x}} Phi left(e^{2 i text{$pi
$x}},5,frac{1}{2}right)-Phi left(e^{-2 i text{$pi $x}},5,frac{1}{2}right)right)}{64 pi ^5}$$
answered Dec 28 '18 at 19:47
David G. StorkDavid G. Stork
10k21332
but I think your solution is not easy
– omarbaker100
Dec 28 '18 at 19:51
1
It is a closed form, but if the solution is "not easy," well... it is "not easy."
– David G. Stork
Dec 28 '18 at 20:01
thanks but I will wait if there is another form easier
– omarbaker100
Dec 28 '18 at 20:05
How is that possible?
– David G. Stork
Dec 28 '18 at 20:06
@DavidG.Stork: it is easy. This is just an example of the fact that Mathematica is not able to recognize Bernoulli polynomials from their Fourier series. I do not believe this is really an answer.
– Jack D'Aurizio
Dec 28 '18 at 21:10
add a comment |
Mathematica gives:
$$-frac{i e^{-i text{$pi $x}} left(e^{2 i text{$pi $x}} Phi left(e^{2 i text{$pi
$x}},5,frac{1}{2}right)-Phi left(e^{-2 i text{$pi $x}},5,frac{1}{2}right)right)}{64 pi ^5}$$
answered Dec 28 '18 at 19:47
David G. StorkDavid G. Stork
10k21332
Mathematica gives:
$$-frac{i e^{-i text{$pi $x}} left(e^{2 i text{$pi $x}} Phi left(e^{2 i text{$pi
$x}},5,frac{1}{2}right)-Phi left(e^{-2 i text{$pi $x}},5,frac{1}{2}right)right)}{64 pi ^5}$$
answered Dec 28 '18 at 19:47
David G. StorkDavid G. Stork
10k21332
answered Dec 28 '18 at 19:47
David G. StorkDavid G. Stork
10k21332
answered Dec 28 '18 at 19:47
David G. StorkDavid G. Stork
10k21332
answered Dec 28 '18 at 19:47
David G. StorkDavid G. Stork
10k21332
10k21332
but I think your solution is not easy
– omarbaker100
Dec 28 '18 at 19:51
1
It is a closed form, but if the solution is "not easy," well... it is "not easy."
– David G. Stork
Dec 28 '18 at 20:01
thanks but I will wait if there is another form easier
– omarbaker100
Dec 28 '18 at 20:05
How is that possible?
– David G. Stork
Dec 28 '18 at 20:06
@DavidG.Stork: it is easy. This is just an example of the fact that Mathematica is not able to recognize Bernoulli polynomials from their Fourier series. I do not believe this is really an answer.
– Jack D'Aurizio
Dec 28 '18 at 21:10
add a comment |
but I think your solution is not easy
– omarbaker100
Dec 28 '18 at 19:51
1
It is a closed form, but if the solution is "not easy," well... it is "not easy."
– David G. Stork
Dec 28 '18 at 20:01
thanks but I will wait if there is another form easier
– omarbaker100
Dec 28 '18 at 20:05
How is that possible?
– David G. Stork
Dec 28 '18 at 20:06
@DavidG.Stork: it is easy. This is just an example of the fact that Mathematica is not able to recognize Bernoulli polynomials from their Fourier series. I do not believe this is really an answer.
– Jack D'Aurizio
Dec 28 '18 at 21:10
but I think your solution is not easy
– omarbaker100
Dec 28 '18 at 19:51
but I think your solution is not easy
– omarbaker100
Dec 28 '18 at 19:51
1
1
It is a closed form, but if the solution is "not easy," well... it is "not easy."
– David G. Stork
Dec 28 '18 at 20:01
It is a closed form, but if the solution is "not easy," well... it is "not easy."
– David G. Stork
Dec 28 '18 at 20:01
thanks but I will wait if there is another form easier
– omarbaker100
Dec 28 '18 at 20:05
thanks but I will wait if there is another form easier
– omarbaker100
Dec 28 '18 at 20:05
How is that possible?
– David G. Stork
Dec 28 '18 at 20:06
How is that possible?
– David G. Stork
Dec 28 '18 at 20:06
@DavidG.Stork: it is easy. This is just an example of the fact that Mathematica is not able to recognize Bernoulli polynomials from their Fourier series. I do not believe this is really an answer.
– Jack D'Aurizio
Dec 28 '18 at 21:10
@DavidG.Stork: it is easy. This is just an example of the fact that Mathematica is not able to recognize Bernoulli polynomials from their Fourier series. I do not believe this is really an answer.
– Jack D'Aurizio
Dec 28 '18 at 21:10
add a comment |
You should tell us what you tried
– Ankit Kumar
Dec 28 '18 at 19:45