Show that $frac4pisum_ngeq 1frac1-(-1)^nn^3sin(nx)=frac8pisum_ngeq 1frac1(2n-1)^3sin(2n-1)x.$ The Next CEO of Stack Overflowalternating series test for $sum_n=1^infty(-1)^nfracsqrtnn+4$Convergence of double series $sum_n=1^infty sum_m=1^infty fracsin(sin(nm))n^2+m^2$Convergence of an alternating series : $ sum_ngeq 1 fracsin nn$Is there a closed-form of $sum_n=1^infty fracsin(n)n^4$When the series $sum_n=2^infty fracsin^2fracpinn^p-1$ convergesA community project: prove (or disprove) that $sum_ngeq 1fracsin(2^n)n$ is convergentWhat is the closed form of $sum_ngeq 1(-1)^n-1psi'(n)^2$?Convergence of $sum_n=1^infty fracsin(n)n$Is the series $sum_n=1^infty fracsin(nx)n^3$ termwise differentiable on an interval $Isubseteq mathbbR$?Is there an intuitive way to understand $fracxspace dy-yspace dxx^2+y^2=d(arctanfrac yx)$
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Show that $frac4pisum_ngeq 1frac1-(-1)^nn^3sin(nx)=frac8pisum_ngeq 1frac1(2n-1)^3sin(2n-1)x.$
The Next CEO of Stack Overflowalternating series test for $sum_n=1^infty(-1)^nfracsqrtnn+4$Convergence of double series $sum_n=1^infty sum_m=1^infty fracsin(sin(nm))n^2+m^2$Convergence of an alternating series : $ sum_ngeq 1 frac(-1)^nn$Is there a closed-form of $sum_n=1^infty fracsin(n)n^4$When the series $sum_n=2^infty fracsin^2fracpinn^p-1$ convergesA community project: prove (or disprove) that $sum_ngeq 1fracsin(2^n)n$ is convergentWhat is the closed form of $sum_ngeq 1(-1)^n-1psi'(n)^2$?Convergence of $sum_n=1^infty fracsin(n)n$Is the series $sum_n=1^infty fracsin(nx)n^3$ termwise differentiable on an interval $Isubseteq mathbbR$?Is there an intuitive way to understand $fracxspace dy-yspace dxx^2+y^2=d(arctanfrac yx)$
$begingroup$
So after solving another problem, I got the answer on the LHS below. The book says it's the RHS. However, both of them are correct but I want to know how to go from my answer to the books. Thus the problem I need help with is
Show that $$frac4pisum_ngeq
1frac1-(-1)^nn^3sin(nx)=frac8pisum_ngeq
1frac1(2n-1)^3sin((2n-1)x).tag 1$$
The original problem was to find the fourier series of the function $f(x)=x(pi-|x|).$
calculus sequences-and-series
asked Mar 19 at 21:08
ParsevalParseval
3,0771719
$endgroup$
add a comment |
$begingroup$
So after solving another problem, I got the answer on the LHS below. The book says it's the RHS. However, both of them are correct but I want to know how to go from my answer to the books. Thus the problem I need help with is
Show that $$frac4pisum_ngeq
1frac1-(-1)^nn^3sin(nx)=frac8pisum_ngeq
1frac1(2n-1)^3sin((2n-1)x).tag 1$$
The original problem was to find the fourier series of the function $f(x)=x(pi-|x|).$
calculus sequences-and-series
asked Mar 19 at 21:08
ParsevalParseval
3,0771719
$endgroup$
add a comment |
$begingroup$
So after solving another problem, I got the answer on the LHS below. The book says it's the RHS. However, both of them are correct but I want to know how to go from my answer to the books. Thus the problem I need help with is
Show that $$frac4pisum_ngeq
1frac1-(-1)^nn^3sin(nx)=frac8pisum_ngeq
1frac1(2n-1)^3sin((2n-1)x).tag 1$$
The original problem was to find the fourier series of the function $f(x)=x(pi-|x|).$
calculus sequences-and-series
asked Mar 19 at 21:08
ParsevalParseval
3,0771719
$endgroup$
So after solving another problem, I got the answer on the LHS below. The book says it's the RHS. However, both of them are correct but I want to know how to go from my answer to the books. Thus the problem I need help with is
Show that $$frac4pisum_ngeq
1frac1-(-1)^nn^3sin(nx)=frac8pisum_ngeq
1frac1(2n-1)^3sin((2n-1)x).tag 1$$
The original problem was to find the fourier series of the function $f(x)=x(pi-|x|).$
calculus sequences-and-series
calculus sequences-and-series
asked Mar 19 at 21:08
ParsevalParseval
3,0771719
asked Mar 19 at 21:08
ParsevalParseval
3,0771719
asked Mar 19 at 21:08
ParsevalParseval
3,0771719
asked Mar 19 at 21:08
ParsevalParseval
3,0771719
asked Mar 19 at 21:08
ParsevalParseval
3,0771719
3,0771719
add a comment |
add a comment |
2 Answers
2
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oldest
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$begingroup$
Not hard to prove.$$frac4pisum_ngeq
1frac1-(-1)^nn^3sin(nx)=frac4pisum_ngeq
1\n text is oddfrac1-(-1)^nn^3sin(nx)+frac4pisum_ngeq
1\n text is evenfrac1-(-1)^nn^3sin(nx)\=frac4pisum_ngeq
1\n text is oddfrac1-(-1)n^3sin(nx)+frac4pisum_ngeq
1\n text is evenfrac1-(1)n^3sin(nx)\=frac4pisum_ngeq
1\n text is oddfrac2n^3sin(nx)\=frac8pisum_ngeq
1\n=2k-1\kge 1frac1n^3sin(nx)\=frac8pisum_kge 1frac1(2k-1)^3sinBig((2k-1)xBig)$$
answered Mar 19 at 23:12
Mostafa AyazMostafa Ayaz
18.3k31040
$endgroup$
add a comment |
$begingroup$
The coefficient $1^n-(-1)^n$ in the LHS is $0$ for even $n$ and $2$ for odd $n$.
Therefore the resulting coefficients:
$$
frac4picdotfrac1^n-(-1)^nn^3=begincases
frac8pifrac1n^3,&n-textodd\
0,&n-texteven\
endcases
$$
are identical with these in RHS, where the change of the index $nmapsto 2n-1$ was performed to ensure summation over odd numbers.
answered Mar 19 at 21:15
useruser
6,13811031
$endgroup$
$begingroup$
Should it not be $frac8pifrac1(2n-1)^3?$
$endgroup$
– Parseval
Mar 19 at 22:09
$begingroup$
Of course it is as every odd positive number can be written as $2k-1$ with $k$ being a positive integer number with arbitrary parity. Important is to use the same representation both in the prefactor and in the argument of sine.
$endgroup$
– user
Mar 19 at 22:16
add a comment |
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2 Answers
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2 Answers
2
active
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$begingroup$
Not hard to prove.$$frac4pisum_ngeq
1frac1-(-1)^nn^3sin(nx)=frac4pisum_ngeq
1\n text is oddfrac1-(-1)^nn^3sin(nx)+frac4pisum_ngeq
1\n text is evenfrac1-(-1)^nn^3sin(nx)\=frac4pisum_ngeq
1\n text is oddfrac1-(-1)n^3sin(nx)+frac4pisum_ngeq
1\n text is evenfrac1-(1)n^3sin(nx)\=frac4pisum_ngeq
1\n text is oddfrac2n^3sin(nx)\=frac8pisum_ngeq
1\n=2k-1\kge 1frac1n^3sin(nx)\=frac8pisum_kge 1frac1(2k-1)^3sinBig((2k-1)xBig)$$
answered Mar 19 at 23:12
Mostafa AyazMostafa Ayaz
18.3k31040
$endgroup$
add a comment |
$begingroup$
Not hard to prove.$$frac4pisum_ngeq
1frac1-(-1)^nn^3sin(nx)=frac4pisum_ngeq
1\n text is oddfrac1-(-1)^nn^3sin(nx)+frac4pisum_ngeq
1\n text is evenfrac1-(-1)^nn^3sin(nx)\=frac4pisum_ngeq
1\n text is oddfrac1-(-1)n^3sin(nx)+frac4pisum_ngeq
1\n text is evenfrac1-(1)n^3sin(nx)\=frac4pisum_ngeq
1\n text is oddfrac2n^3sin(nx)\=frac8pisum_ngeq
1\n=2k-1\kge 1frac1n^3sin(nx)\=frac8pisum_kge 1frac1(2k-1)^3sinBig((2k-1)xBig)$$
answered Mar 19 at 23:12
Mostafa AyazMostafa Ayaz
18.3k31040
$endgroup$
add a comment |
$begingroup$
Not hard to prove.$$frac4pisum_ngeq
1frac1-(-1)^nn^3sin(nx)=frac4pisum_ngeq
1\n text is oddfrac1-(-1)^nn^3sin(nx)+frac4pisum_ngeq
1\n text is evenfrac1-(-1)^nn^3sin(nx)\=frac4pisum_ngeq
1\n text is oddfrac1-(-1)n^3sin(nx)+frac4pisum_ngeq
1\n text is evenfrac1-(1)n^3sin(nx)\=frac4pisum_ngeq
1\n text is oddfrac2n^3sin(nx)\=frac8pisum_ngeq
1\n=2k-1\kge 1frac1n^3sin(nx)\=frac8pisum_kge 1frac1(2k-1)^3sinBig((2k-1)xBig)$$
answered Mar 19 at 23:12
Mostafa AyazMostafa Ayaz
18.3k31040
$endgroup$
Not hard to prove.$$frac4pisum_ngeq
1frac1-(-1)^nn^3sin(nx)=frac4pisum_ngeq
1\n text is oddfrac1-(-1)^nn^3sin(nx)+frac4pisum_ngeq
1\n text is evenfrac1-(-1)^nn^3sin(nx)\=frac4pisum_ngeq
1\n text is oddfrac1-(-1)n^3sin(nx)+frac4pisum_ngeq
1\n text is evenfrac1-(1)n^3sin(nx)\=frac4pisum_ngeq
1\n text is oddfrac2n^3sin(nx)\=frac8pisum_ngeq
1\n=2k-1\kge 1frac1n^3sin(nx)\=frac8pisum_kge 1frac1(2k-1)^3sinBig((2k-1)xBig)$$
answered Mar 19 at 23:12
Mostafa AyazMostafa Ayaz
18.3k31040
answered Mar 19 at 23:12
Mostafa AyazMostafa Ayaz
18.3k31040
answered Mar 19 at 23:12
Mostafa AyazMostafa Ayaz
18.3k31040
answered Mar 19 at 23:12
Mostafa AyazMostafa Ayaz
18.3k31040
18.3k31040
add a comment |
add a comment |
$begingroup$
The coefficient $1^n-(-1)^n$ in the LHS is $0$ for even $n$ and $2$ for odd $n$.
Therefore the resulting coefficients:
$$
frac4picdotfrac1^n-(-1)^nn^3=begincases
frac8pifrac1n^3,&n-textodd\
0,&n-texteven\
endcases
$$
are identical with these in RHS, where the change of the index $nmapsto 2n-1$ was performed to ensure summation over odd numbers.
answered Mar 19 at 21:15
useruser
6,13811031
$endgroup$
$begingroup$
Should it not be $frac8pifrac1(2n-1)^3?$
$endgroup$
– Parseval
Mar 19 at 22:09
$begingroup$
Of course it is as every odd positive number can be written as $2k-1$ with $k$ being a positive integer number with arbitrary parity. Important is to use the same representation both in the prefactor and in the argument of sine.
$endgroup$
– user
Mar 19 at 22:16
add a comment |
$begingroup$
The coefficient $1^n-(-1)^n$ in the LHS is $0$ for even $n$ and $2$ for odd $n$.
Therefore the resulting coefficients:
$$
frac4picdotfrac1^n-(-1)^nn^3=begincases
frac8pifrac1n^3,&n-textodd\
0,&n-texteven\
endcases
$$
are identical with these in RHS, where the change of the index $nmapsto 2n-1$ was performed to ensure summation over odd numbers.
answered Mar 19 at 21:15
useruser
6,13811031
$endgroup$
$begingroup$
Should it not be $frac8pifrac1(2n-1)^3?$
$endgroup$
– Parseval
Mar 19 at 22:09
$begingroup$
Of course it is as every odd positive number can be written as $2k-1$ with $k$ being a positive integer number with arbitrary parity. Important is to use the same representation both in the prefactor and in the argument of sine.
$endgroup$
– user
Mar 19 at 22:16
add a comment |
$begingroup$
The coefficient $1^n-(-1)^n$ in the LHS is $0$ for even $n$ and $2$ for odd $n$.
Therefore the resulting coefficients:
$$
frac4picdotfrac1^n-(-1)^nn^3=begincases
frac8pifrac1n^3,&n-textodd\
0,&n-texteven\
endcases
$$
are identical with these in RHS, where the change of the index $nmapsto 2n-1$ was performed to ensure summation over odd numbers.
answered Mar 19 at 21:15
useruser
6,13811031
$endgroup$
The coefficient $1^n-(-1)^n$ in the LHS is $0$ for even $n$ and $2$ for odd $n$.
Therefore the resulting coefficients:
$$
frac4picdotfrac1^n-(-1)^nn^3=begincases
frac8pifrac1n^3,&n-textodd\
0,&n-texteven\
endcases
$$
are identical with these in RHS, where the change of the index $nmapsto 2n-1$ was performed to ensure summation over odd numbers.
answered Mar 19 at 21:15
useruser
6,13811031
edited Mar 20 at 10:21
answered Mar 19 at 21:15
useruser
6,13811031
answered Mar 19 at 21:15
useruser
6,13811031
answered Mar 19 at 21:15
useruser
6,13811031
6,13811031
$begingroup$
Should it not be $frac8pifrac1(2n-1)^3?$
$endgroup$
– Parseval
Mar 19 at 22:09
$begingroup$
Of course it is as every odd positive number can be written as $2k-1$ with $k$ being a positive integer number with arbitrary parity. Important is to use the same representation both in the prefactor and in the argument of sine.
$endgroup$
– user
Mar 19 at 22:16
add a comment |
$begingroup$
Should it not be $frac8pifrac1(2n-1)^3?$
$endgroup$
– Parseval
Mar 19 at 22:09
$begingroup$
Of course it is as every odd positive number can be written as $2k-1$ with $k$ being a positive integer number with arbitrary parity. Important is to use the same representation both in the prefactor and in the argument of sine.
$endgroup$
– user
Mar 19 at 22:16
$begingroup$
Should it not be $frac8pifrac1(2n-1)^3?$
$endgroup$
– Parseval
Mar 19 at 22:09
$begingroup$
Should it not be $frac8pifrac1(2n-1)^3?$
$endgroup$
– Parseval
Mar 19 at 22:09
$begingroup$
Of course it is as every odd positive number can be written as $2k-1$ with $k$ being a positive integer number with arbitrary parity. Important is to use the same representation both in the prefactor and in the argument of sine.
$endgroup$
– user
Mar 19 at 22:16
$begingroup$
Of course it is as every odd positive number can be written as $2k-1$ with $k$ being a positive integer number with arbitrary parity. Important is to use the same representation both in the prefactor and in the argument of sine.
$endgroup$
– user
Mar 19 at 22:16
add a comment |
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