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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3052625%2ffind-the-closed-form-of-sum-k-1n-cos2m1-left-frac2k-1-pi2n1-r%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3052625%2ffind-the-closed-form-of-sum-k-1n-cos2m1-left-frac2k-1-pi2n1-r%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Mario Kart Wii

Image
Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro
Read more

What does “Dominus providebit” mean?

Image
5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1
Read more

Antonio Litta Visconti Arese

Image
Antonio Litta Visconti Arese I duca Litta, VII marchese di Gambolò In carica 1797 – 1820 Predecessore Pompeo Litta Visconti Arese Successore Alfonso Litta Visconti Arese Trattamento Sua grazia Altri titoli Duca Litta Conte del Regno napoleonico d'Italia Patrizio milanese Don Nascita Milano, 2 maggio 1748 Morte Milano, 14 agosto 1820 Dinastia Litta Visconti Arese Padre Pompeo Litta Visconti Arese Madre Maria Elisabetta Visconti Borromeo Consorte Barbara Barbiano di Belgioioso Religione cattolicesimo Antonio Litta Visconti Arese , I duca Litta, VII marchese di Gambolò (Milano, 2 maggio 1748 – Milano, 14 agosto 1820), è stato un nobile e politico italiano. Composizione "Il pastore svizzero" di Pietro Morlacchi, con dedica Indice 1 Biografia 2 Onorificenze 3 Araldica 4 Albero genealogico 5 Note 6 Altri progetti Biografia | Antonio Litta Visco
Read more