Find the closed form of $sum_{k=1}^{n}cos^{2m+1}left(frac{(2k-1)pi}{2n+1}right)$












1














Add it: Let $m,n$ be postive integers. Find the closed form of



$$f=sum_{k=1}^{n} cos^{2m+1}{left(dfrac{(2k-1)pi}{2n+1}right)}$$



for $m, n in mathbb{N}^{+}$.



Maybe use Euler



begin{align}
2cos{x} &=e^{ix}+e^{-ix} \
dfrac{pi}{2n+1} &=x
end{align}



then



begin{align}
f &= dfrac{1}{2^{2m+1}}sum_{k=1}^{n}(w^{(2k-1)}+w^{-(2k-1)})^{2m+1} \
&=dfrac{1}{2^{2m+1}}sum_{k=1}^{n}sum_{i=0}^{2m+1}binom{2m+1}{i}w^{-(2k-1)i}w^{((2m+1)-i)(2k-1)}\
&=dfrac{1}{(2w)^{2m+1}}sum_{i=0}^{2m+1}binom{2m+1}{i}w^{2i}sum_{k=1}^{n}w^{(4m-4i+2)k}
end{align}



where $w=e^{ix}$.






trigonometry







share|cite|improve this question




















  • 3




    What have you tried? You seem to have a demand rather than a question.
    – JimB
    Dec 26 '18 at 4:13










  • Last sum is geometric, isn't it?
    – saulspatz
    Dec 26 '18 at 8:00






  • 3




    It is misleading to use $i$ both as imaginary unit and a run index.
    – user
    Dec 26 '18 at 8:13












  • One simplification is to note that f=1/2 whenever $n>m$. Also, one could argue that what you present is already a "closed-form" solution
    – JimB
    Dec 27 '18 at 5:55












  • One of my previous questions is about a similar sum, and has a nice answer math.stackexchange.com/a/2395003/309055 which might help
    – MeMyselfI
    Jan 1 at 12:24
















1














Add it: Let $m,n$ be postive integers. Find the closed form of



$$f=sum_{k=1}^{n} cos^{2m+1}{left(dfrac{(2k-1)pi}{2n+1}right)}$$



for $m, n in mathbb{N}^{+}$.



Maybe use Euler



begin{align}
2cos{x} &=e^{ix}+e^{-ix} \
dfrac{pi}{2n+1} &=x
end{align}



then



begin{align}
f &= dfrac{1}{2^{2m+1}}sum_{k=1}^{n}(w^{(2k-1)}+w^{-(2k-1)})^{2m+1} \
&=dfrac{1}{2^{2m+1}}sum_{k=1}^{n}sum_{i=0}^{2m+1}binom{2m+1}{i}w^{-(2k-1)i}w^{((2m+1)-i)(2k-1)}\
&=dfrac{1}{(2w)^{2m+1}}sum_{i=0}^{2m+1}binom{2m+1}{i}w^{2i}sum_{k=1}^{n}w^{(4m-4i+2)k}
end{align}



where $w=e^{ix}$.






trigonometry







share|cite|improve this question




















  • 3




    What have you tried? You seem to have a demand rather than a question.
    – JimB
    Dec 26 '18 at 4:13










  • Last sum is geometric, isn't it?
    – saulspatz
    Dec 26 '18 at 8:00






  • 3




    It is misleading to use $i$ both as imaginary unit and a run index.
    – user
    Dec 26 '18 at 8:13












  • One simplification is to note that f=1/2 whenever $n>m$. Also, one could argue that what you present is already a "closed-form" solution
    – JimB
    Dec 27 '18 at 5:55












  • One of my previous questions is about a similar sum, and has a nice answer math.stackexchange.com/a/2395003/309055 which might help
    – MeMyselfI
    Jan 1 at 12:24














1












1








1


4





Add it: Let $m,n$ be postive integers. Find the closed form of



$$f=sum_{k=1}^{n} cos^{2m+1}{left(dfrac{(2k-1)pi}{2n+1}right)}$$



for $m, n in mathbb{N}^{+}$.



Maybe use Euler



begin{align}
2cos{x} &=e^{ix}+e^{-ix} \
dfrac{pi}{2n+1} &=x
end{align}



then



begin{align}
f &= dfrac{1}{2^{2m+1}}sum_{k=1}^{n}(w^{(2k-1)}+w^{-(2k-1)})^{2m+1} \
&=dfrac{1}{2^{2m+1}}sum_{k=1}^{n}sum_{i=0}^{2m+1}binom{2m+1}{i}w^{-(2k-1)i}w^{((2m+1)-i)(2k-1)}\
&=dfrac{1}{(2w)^{2m+1}}sum_{i=0}^{2m+1}binom{2m+1}{i}w^{2i}sum_{k=1}^{n}w^{(4m-4i+2)k}
end{align}



where $w=e^{ix}$.






trigonometry







share|cite|improve this question















Add it: Let $m,n$ be postive integers. Find the closed form of



$$f=sum_{k=1}^{n} cos^{2m+1}{left(dfrac{(2k-1)pi}{2n+1}right)}$$



for $m, n in mathbb{N}^{+}$.



Maybe use Euler



begin{align}
2cos{x} &=e^{ix}+e^{-ix} \
dfrac{pi}{2n+1} &=x
end{align}



then



begin{align}
f &= dfrac{1}{2^{2m+1}}sum_{k=1}^{n}(w^{(2k-1)}+w^{-(2k-1)})^{2m+1} \
&=dfrac{1}{2^{2m+1}}sum_{k=1}^{n}sum_{i=0}^{2m+1}binom{2m+1}{i}w^{-(2k-1)i}w^{((2m+1)-i)(2k-1)}\
&=dfrac{1}{(2w)^{2m+1}}sum_{i=0}^{2m+1}binom{2m+1}{i}w^{2i}sum_{k=1}^{n}w^{(4m-4i+2)k}
end{align}



where $w=e^{ix}$.






trigonometry




trigonometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 26 '18 at 7:53







inequality

















asked Dec 26 '18 at 4:04









inequalityinequality

701520




701520








  • 3




    What have you tried? You seem to have a demand rather than a question.
    – JimB
    Dec 26 '18 at 4:13










  • Last sum is geometric, isn't it?
    – saulspatz
    Dec 26 '18 at 8:00






  • 3




    It is misleading to use $i$ both as imaginary unit and a run index.
    – user
    Dec 26 '18 at 8:13












  • One simplification is to note that f=1/2 whenever $n>m$. Also, one could argue that what you present is already a "closed-form" solution
    – JimB
    Dec 27 '18 at 5:55












  • One of my previous questions is about a similar sum, and has a nice answer math.stackexchange.com/a/2395003/309055 which might help
    – MeMyselfI
    Jan 1 at 12:24














  • 3




    What have you tried? You seem to have a demand rather than a question.
    – JimB
    Dec 26 '18 at 4:13










  • Last sum is geometric, isn't it?
    – saulspatz
    Dec 26 '18 at 8:00






  • 3




    It is misleading to use $i$ both as imaginary unit and a run index.
    – user
    Dec 26 '18 at 8:13












  • One simplification is to note that f=1/2 whenever $n>m$. Also, one could argue that what you present is already a "closed-form" solution
    – JimB
    Dec 27 '18 at 5:55












  • One of my previous questions is about a similar sum, and has a nice answer math.stackexchange.com/a/2395003/309055 which might help
    – MeMyselfI
    Jan 1 at 12:24








3




3




What have you tried? You seem to have a demand rather than a question.
– JimB
Dec 26 '18 at 4:13




What have you tried? You seem to have a demand rather than a question.
– JimB
Dec 26 '18 at 4:13












Last sum is geometric, isn't it?
– saulspatz
Dec 26 '18 at 8:00




Last sum is geometric, isn't it?
– saulspatz
Dec 26 '18 at 8:00




3




3




It is misleading to use $i$ both as imaginary unit and a run index.
– user
Dec 26 '18 at 8:13






It is misleading to use $i$ both as imaginary unit and a run index.
– user
Dec 26 '18 at 8:13














One simplification is to note that f=1/2 whenever $n>m$. Also, one could argue that what you present is already a "closed-form" solution
– JimB
Dec 27 '18 at 5:55






One simplification is to note that f=1/2 whenever $n>m$. Also, one could argue that what you present is already a "closed-form" solution
– JimB
Dec 27 '18 at 5:55














One of my previous questions is about a similar sum, and has a nice answer math.stackexchange.com/a/2395003/309055 which might help
– MeMyselfI
Jan 1 at 12:24




One of my previous questions is about a similar sum, and has a nice answer math.stackexchange.com/a/2395003/309055 which might help
– MeMyselfI
Jan 1 at 12:24










1 Answer
1






active

oldest

votes


















1














The function $frac{(2n+1)/z}{z^{2n+1}-1}$ has residue $1$ at each $2n+1^text{st}$ root of unity. So we need to account for the residues at $0$ and $infty$.



$$newcommand{Res}{operatorname*{Res}}
begin{align}
sum_{k=1}^ncos^{2m+1}left(pi,frac{2k-1}{2n+1}right)
&=frac12-frac12sum_{k=0}^{2n}cos^{2m+1}left(2pi,frac{k}{2n+1}right)\
&=frac12+frac12Res_{z=0}left(frac1{2^{2m+1}}frac{(2n+1)/z}{z^{2n+1}-1}left(z+frac1zright)^{2m+1}right)\
&-frac12Res_{z=infty}left(frac1{2^{2m+1}}frac{(2n+1)/z}{z^{2n+1}-1}left(z+frac1zright)^{2m+1}right)\
&=frac12-frac{2n+1}{2^{2m+2}}left[z^0right]left(sum_{k=0}^infty z^{(2n+1)k}left(z+frac1zright)^{2m+1}right)\
&-frac{2n+1}{2^{2m+2}}left[z^0right]left(sum_{k=1}^infty z^{-(2n+1)k}left(z+frac1zright)^{2m+1}right)\
&=frac12-frac{2n+1}{2^{2m+1}}sum_{k=0}^mbinom{2m+1}{k}[,2n+1,|,2m+1-2k,]
end{align}
$$
where $[dots]$ are Iverson Brackets.



Note that for $mlt n$, the sum is $frac12$ because $2n+1$ cannot divide $2m+1-2k$.






share|cite|improve this answer























  • The original sum is finite, but contains cosines that are non-rational. This answer shows that the sum is rational and does not require the evaluation of any cosines.
    – robjohn♦
    yesterday











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














The function $frac{(2n+1)/z}{z^{2n+1}-1}$ has residue $1$ at each $2n+1^text{st}$ root of unity. So we need to account for the residues at $0$ and $infty$.



$$newcommand{Res}{operatorname*{Res}}
begin{align}
sum_{k=1}^ncos^{2m+1}left(pi,frac{2k-1}{2n+1}right)
&=frac12-frac12sum_{k=0}^{2n}cos^{2m+1}left(2pi,frac{k}{2n+1}right)\
&=frac12+frac12Res_{z=0}left(frac1{2^{2m+1}}frac{(2n+1)/z}{z^{2n+1}-1}left(z+frac1zright)^{2m+1}right)\
&-frac12Res_{z=infty}left(frac1{2^{2m+1}}frac{(2n+1)/z}{z^{2n+1}-1}left(z+frac1zright)^{2m+1}right)\
&=frac12-frac{2n+1}{2^{2m+2}}left[z^0right]left(sum_{k=0}^infty z^{(2n+1)k}left(z+frac1zright)^{2m+1}right)\
&-frac{2n+1}{2^{2m+2}}left[z^0right]left(sum_{k=1}^infty z^{-(2n+1)k}left(z+frac1zright)^{2m+1}right)\
&=frac12-frac{2n+1}{2^{2m+1}}sum_{k=0}^mbinom{2m+1}{k}[,2n+1,|,2m+1-2k,]
end{align}
$$
where $[dots]$ are Iverson Brackets.



Note that for $mlt n$, the sum is $frac12$ because $2n+1$ cannot divide $2m+1-2k$.






share|cite|improve this answer























  • The original sum is finite, but contains cosines that are non-rational. This answer shows that the sum is rational and does not require the evaluation of any cosines.
    – robjohn♦
    yesterday
















1














The function $frac{(2n+1)/z}{z^{2n+1}-1}$ has residue $1$ at each $2n+1^text{st}$ root of unity. So we need to account for the residues at $0$ and $infty$.



$$newcommand{Res}{operatorname*{Res}}
begin{align}
sum_{k=1}^ncos^{2m+1}left(pi,frac{2k-1}{2n+1}right)
&=frac12-frac12sum_{k=0}^{2n}cos^{2m+1}left(2pi,frac{k}{2n+1}right)\
&=frac12+frac12Res_{z=0}left(frac1{2^{2m+1}}frac{(2n+1)/z}{z^{2n+1}-1}left(z+frac1zright)^{2m+1}right)\
&-frac12Res_{z=infty}left(frac1{2^{2m+1}}frac{(2n+1)/z}{z^{2n+1}-1}left(z+frac1zright)^{2m+1}right)\
&=frac12-frac{2n+1}{2^{2m+2}}left[z^0right]left(sum_{k=0}^infty z^{(2n+1)k}left(z+frac1zright)^{2m+1}right)\
&-frac{2n+1}{2^{2m+2}}left[z^0right]left(sum_{k=1}^infty z^{-(2n+1)k}left(z+frac1zright)^{2m+1}right)\
&=frac12-frac{2n+1}{2^{2m+1}}sum_{k=0}^mbinom{2m+1}{k}[,2n+1,|,2m+1-2k,]
end{align}
$$
where $[dots]$ are Iverson Brackets.



Note that for $mlt n$, the sum is $frac12$ because $2n+1$ cannot divide $2m+1-2k$.






share|cite|improve this answer























  • The original sum is finite, but contains cosines that are non-rational. This answer shows that the sum is rational and does not require the evaluation of any cosines.
    – robjohn♦
    yesterday














1












1








1






The function $frac{(2n+1)/z}{z^{2n+1}-1}$ has residue $1$ at each $2n+1^text{st}$ root of unity. So we need to account for the residues at $0$ and $infty$.



$$newcommand{Res}{operatorname*{Res}}
begin{align}
sum_{k=1}^ncos^{2m+1}left(pi,frac{2k-1}{2n+1}right)
&=frac12-frac12sum_{k=0}^{2n}cos^{2m+1}left(2pi,frac{k}{2n+1}right)\
&=frac12+frac12Res_{z=0}left(frac1{2^{2m+1}}frac{(2n+1)/z}{z^{2n+1}-1}left(z+frac1zright)^{2m+1}right)\
&-frac12Res_{z=infty}left(frac1{2^{2m+1}}frac{(2n+1)/z}{z^{2n+1}-1}left(z+frac1zright)^{2m+1}right)\
&=frac12-frac{2n+1}{2^{2m+2}}left[z^0right]left(sum_{k=0}^infty z^{(2n+1)k}left(z+frac1zright)^{2m+1}right)\
&-frac{2n+1}{2^{2m+2}}left[z^0right]left(sum_{k=1}^infty z^{-(2n+1)k}left(z+frac1zright)^{2m+1}right)\
&=frac12-frac{2n+1}{2^{2m+1}}sum_{k=0}^mbinom{2m+1}{k}[,2n+1,|,2m+1-2k,]
end{align}
$$
where $[dots]$ are Iverson Brackets.



Note that for $mlt n$, the sum is $frac12$ because $2n+1$ cannot divide $2m+1-2k$.






share|cite|improve this answer














The function $frac{(2n+1)/z}{z^{2n+1}-1}$ has residue $1$ at each $2n+1^text{st}$ root of unity. So we need to account for the residues at $0$ and $infty$.



$$newcommand{Res}{operatorname*{Res}}
begin{align}
sum_{k=1}^ncos^{2m+1}left(pi,frac{2k-1}{2n+1}right)
&=frac12-frac12sum_{k=0}^{2n}cos^{2m+1}left(2pi,frac{k}{2n+1}right)\
&=frac12+frac12Res_{z=0}left(frac1{2^{2m+1}}frac{(2n+1)/z}{z^{2n+1}-1}left(z+frac1zright)^{2m+1}right)\
&-frac12Res_{z=infty}left(frac1{2^{2m+1}}frac{(2n+1)/z}{z^{2n+1}-1}left(z+frac1zright)^{2m+1}right)\
&=frac12-frac{2n+1}{2^{2m+2}}left[z^0right]left(sum_{k=0}^infty z^{(2n+1)k}left(z+frac1zright)^{2m+1}right)\
&-frac{2n+1}{2^{2m+2}}left[z^0right]left(sum_{k=1}^infty z^{-(2n+1)k}left(z+frac1zright)^{2m+1}right)\
&=frac12-frac{2n+1}{2^{2m+1}}sum_{k=0}^mbinom{2m+1}{k}[,2n+1,|,2m+1-2k,]
end{align}
$$
where $[dots]$ are Iverson Brackets.



Note that for $mlt n$, the sum is $frac12$ because $2n+1$ cannot divide $2m+1-2k$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited yesterday

























answered yesterday









robjohn♦robjohn

265k27303625




265k27303625












  • The original sum is finite, but contains cosines that are non-rational. This answer shows that the sum is rational and does not require the evaluation of any cosines.
    – robjohn♦
    yesterday


















  • The original sum is finite, but contains cosines that are non-rational. This answer shows that the sum is rational and does not require the evaluation of any cosines.
    – robjohn♦
    yesterday
















The original sum is finite, but contains cosines that are non-rational. This answer shows that the sum is rational and does not require the evaluation of any cosines.
– robjohn♦
yesterday




The original sum is finite, but contains cosines that are non-rational. This answer shows that the sum is rational and does not require the evaluation of any cosines.
– robjohn♦
yesterday


















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