Is there a closed form of $sum_{k=1}^{infty}frac{sin (2k-1)pi x}{(2k-1)^5pi^5}$ [closed]
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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
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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 |
$begingroup$
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
$endgroup$
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.
$begingroup$
You should tell us what you tried
$endgroup$
– Ankit Kumar
Dec 28 '18 at 19:45
add a comment |
$begingroup$
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
$endgroup$
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.
$begingroup$
You should tell us what you tried
$endgroup$
– Ankit Kumar
Dec 28 '18 at 19:45
add a comment |
$begingroup$
You should tell us what you tried
$endgroup$
– Ankit Kumar
Dec 28 '18 at 19:45
$begingroup$
You should tell us what you tried
$endgroup$
– Ankit Kumar
Dec 28 '18 at 19:45
$begingroup$
You should tell us what you tried
$endgroup$
– Ankit Kumar
Dec 28 '18 at 19:45
add a comment |
2 Answers
2
active
oldest
votes
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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
290k33282662
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$begingroup$
thank you very much
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– omarbaker100
Dec 28 '18 at 23:58
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I think the rectangle wave function is $frac{1}{4}$ not $frac{pi}{4}$
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– E.H.E
Dec 29 '18 at 0:16
add a comment |
$begingroup$
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
11k41432
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but I think your solution is not easy
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– omarbaker100
Dec 28 '18 at 19:51
1
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It is a closed form, but if the solution is "not easy," well... it is "not easy."
$endgroup$
– David G. Stork
Dec 28 '18 at 20:01
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thanks but I will wait if there is another form easier
$endgroup$
– omarbaker100
Dec 28 '18 at 20:05
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How is that possible?
$endgroup$
– David G. Stork
Dec 28 '18 at 20:06
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@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.
$endgroup$
– 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
$begingroup$
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
290k33282662
$endgroup$
$begingroup$
thank you very much
$endgroup$
– omarbaker100
Dec 28 '18 at 23:58
$begingroup$
I think the rectangle wave function is $frac{1}{4}$ not $frac{pi}{4}$
$endgroup$
– E.H.E
Dec 29 '18 at 0:16
add a comment |
$begingroup$
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
290k33282662
$endgroup$
$begingroup$
thank you very much
$endgroup$
– omarbaker100
Dec 28 '18 at 23:58
$begingroup$
I think the rectangle wave function is $frac{1}{4}$ not $frac{pi}{4}$
$endgroup$
– E.H.E
Dec 29 '18 at 0:16
add a comment |
$begingroup$
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
290k33282662
$endgroup$
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
290k33282662
edited Dec 29 '18 at 1:08
answered Dec 28 '18 at 21:09
Jack D'AurizioJack D'Aurizio
290k33282662
answered Dec 28 '18 at 21:09
Jack D'AurizioJack D'Aurizio
290k33282662
answered Dec 28 '18 at 21:09
Jack D'AurizioJack D'Aurizio
290k33282662
290k33282662
$begingroup$
thank you very much
$endgroup$
– omarbaker100
Dec 28 '18 at 23:58
$begingroup$
I think the rectangle wave function is $frac{1}{4}$ not $frac{pi}{4}$
$endgroup$
– E.H.E
Dec 29 '18 at 0:16
add a comment |
$begingroup$
thank you very much
$endgroup$
– omarbaker100
Dec 28 '18 at 23:58
$begingroup$
I think the rectangle wave function is $frac{1}{4}$ not $frac{pi}{4}$
$endgroup$
– E.H.E
Dec 29 '18 at 0:16
$begingroup$
thank you very much
$endgroup$
– omarbaker100
Dec 28 '18 at 23:58
$begingroup$
thank you very much
$endgroup$
– omarbaker100
Dec 28 '18 at 23:58
$begingroup$
I think the rectangle wave function is $frac{1}{4}$ not $frac{pi}{4}$
$endgroup$
– E.H.E
Dec 29 '18 at 0:16
$begingroup$
I think the rectangle wave function is $frac{1}{4}$ not $frac{pi}{4}$
$endgroup$
– E.H.E
Dec 29 '18 at 0:16
add a comment |
$begingroup$
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
11k41432
$endgroup$
$begingroup$
but I think your solution is not easy
$endgroup$
– omarbaker100
Dec 28 '18 at 19:51
1
$begingroup$
It is a closed form, but if the solution is "not easy," well... it is "not easy."
$endgroup$
– David G. Stork
Dec 28 '18 at 20:01
$begingroup$
thanks but I will wait if there is another form easier
$endgroup$
– omarbaker100
Dec 28 '18 at 20:05
$begingroup$
How is that possible?
$endgroup$
– David G. Stork
Dec 28 '18 at 20:06
$begingroup$
@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.
$endgroup$
– Jack D'Aurizio
Dec 28 '18 at 21:10
add a comment |
$begingroup$
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
11k41432
$endgroup$
$begingroup$
but I think your solution is not easy
$endgroup$
– omarbaker100
Dec 28 '18 at 19:51
1
$begingroup$
It is a closed form, but if the solution is "not easy," well... it is "not easy."
$endgroup$
– David G. Stork
Dec 28 '18 at 20:01
$begingroup$
thanks but I will wait if there is another form easier
$endgroup$
– omarbaker100
Dec 28 '18 at 20:05
$begingroup$
How is that possible?
$endgroup$
– David G. Stork
Dec 28 '18 at 20:06
$begingroup$
@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.
$endgroup$
– Jack D'Aurizio
Dec 28 '18 at 21:10
add a comment |
$begingroup$
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
11k41432
$endgroup$
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
11k41432
answered Dec 28 '18 at 19:47
David G. StorkDavid G. Stork
11k41432
answered Dec 28 '18 at 19:47
David G. StorkDavid G. Stork
11k41432
answered Dec 28 '18 at 19:47
David G. StorkDavid G. Stork
11k41432
11k41432
$begingroup$
but I think your solution is not easy
$endgroup$
– omarbaker100
Dec 28 '18 at 19:51
1
$begingroup$
It is a closed form, but if the solution is "not easy," well... it is "not easy."
$endgroup$
– David G. Stork
Dec 28 '18 at 20:01
$begingroup$
thanks but I will wait if there is another form easier
$endgroup$
– omarbaker100
Dec 28 '18 at 20:05
$begingroup$
How is that possible?
$endgroup$
– David G. Stork
Dec 28 '18 at 20:06
$begingroup$
@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.
$endgroup$
– Jack D'Aurizio
Dec 28 '18 at 21:10
add a comment |
$begingroup$
but I think your solution is not easy
$endgroup$
– omarbaker100
Dec 28 '18 at 19:51
1
$begingroup$
It is a closed form, but if the solution is "not easy," well... it is "not easy."
$endgroup$
– David G. Stork
Dec 28 '18 at 20:01
$begingroup$
thanks but I will wait if there is another form easier
$endgroup$
– omarbaker100
Dec 28 '18 at 20:05
$begingroup$
How is that possible?
$endgroup$
– David G. Stork
Dec 28 '18 at 20:06
$begingroup$
@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.
$endgroup$
– Jack D'Aurizio
Dec 28 '18 at 21:10
$begingroup$
but I think your solution is not easy
$endgroup$
– omarbaker100
Dec 28 '18 at 19:51
$begingroup$
but I think your solution is not easy
$endgroup$
– omarbaker100
Dec 28 '18 at 19:51
1
1
$begingroup$
It is a closed form, but if the solution is "not easy," well... it is "not easy."
$endgroup$
– David G. Stork
Dec 28 '18 at 20:01
$begingroup$
It is a closed form, but if the solution is "not easy," well... it is "not easy."
$endgroup$
– David G. Stork
Dec 28 '18 at 20:01
$begingroup$
thanks but I will wait if there is another form easier
$endgroup$
– omarbaker100
Dec 28 '18 at 20:05
$begingroup$
thanks but I will wait if there is another form easier
$endgroup$
– omarbaker100
Dec 28 '18 at 20:05
$begingroup$
How is that possible?
$endgroup$
– David G. Stork
Dec 28 '18 at 20:06
$begingroup$
How is that possible?
$endgroup$
– David G. Stork
Dec 28 '18 at 20:06
$begingroup$
@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.
$endgroup$
– Jack D'Aurizio
Dec 28 '18 at 21:10
$begingroup$
@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.
$endgroup$
– Jack D'Aurizio
Dec 28 '18 at 21:10
add a comment |
$begingroup$
You should tell us what you tried
$endgroup$
– Ankit Kumar
Dec 28 '18 at 19:45