Mixing
How to calculate the sum of the series $sum_{n=1}^{infty}frac{cos^3(nx)}{n^2}$?
As the title says, I need to calculate the sum of the function series $$sum_{n=1}^{infty}frac{cos^3(nx)}{n^2}$$ so that I can find out if the series does simple or uniform converge to something.
Usually, for problems like this I need to write the sum like a difference of sums so when we write the sums, some terms will go away, but I can't think of a way to write this one.
Can you help me?
UPDATE
I managed to show that the series is uniform convergent even if I do not know to who it converges. Still, I wonder if there is any way to calculate the sum of the series.
sequences-and-series
edited Nov 21 '18 at 21:05
amWhy
192k28225439
asked Nov 21 '18 at 20:57
GhostGhost
553210
add a comment |
As the title says, I need to calculate the sum of the function series $$sum_{n=1}^{infty}frac{cos^3(nx)}{n^2}$$ so that I can find out if the series does simple or uniform converge to something.
Usually, for problems like this I need to write the sum like a difference of sums so when we write the sums, some terms will go away, but I can't think of a way to write this one.
Can you help me?
UPDATE
I managed to show that the series is uniform convergent even if I do not know to who it converges. Still, I wonder if there is any way to calculate the sum of the series.
sequences-and-series
edited Nov 21 '18 at 21:05
amWhy
192k28225439
asked Nov 21 '18 at 20:57
GhostGhost
553210
1
The answer will involve complex dilogarithms.
– Lord Shark the Unknown
Nov 21 '18 at 21:03
@LordSharktheUnknown Ok... it's a bit too much for me and I probably won't understand since I am not soo advanced in math knowledge. Thanks for the info anyway.
– Ghost
Nov 21 '18 at 21:15
add a comment |
As the title says, I need to calculate the sum of the function series $$sum_{n=1}^{infty}frac{cos^3(nx)}{n^2}$$ so that I can find out if the series does simple or uniform converge to something.
Usually, for problems like this I need to write the sum like a difference of sums so when we write the sums, some terms will go away, but I can't think of a way to write this one.
Can you help me?
UPDATE
I managed to show that the series is uniform convergent even if I do not know to who it converges. Still, I wonder if there is any way to calculate the sum of the series.
sequences-and-series
edited Nov 21 '18 at 21:05
amWhy
192k28225439
asked Nov 21 '18 at 20:57
GhostGhost
553210
As the title says, I need to calculate the sum of the function series $$sum_{n=1}^{infty}frac{cos^3(nx)}{n^2}$$ so that I can find out if the series does simple or uniform converge to something.
Usually, for problems like this I need to write the sum like a difference of sums so when we write the sums, some terms will go away, but I can't think of a way to write this one.
Can you help me?
UPDATE
I managed to show that the series is uniform convergent even if I do not know to who it converges. Still, I wonder if there is any way to calculate the sum of the series.
sequences-and-series
sequences-and-series
edited Nov 21 '18 at 21:05
amWhy
192k28225439
asked Nov 21 '18 at 20:57
GhostGhost
553210
edited Nov 21 '18 at 21:05
amWhy
192k28225439
asked Nov 21 '18 at 20:57
GhostGhost
553210
edited Nov 21 '18 at 21:05
amWhy
192k28225439
edited Nov 21 '18 at 21:05
amWhy
192k28225439
edited Nov 21 '18 at 21:05
amWhy
192k28225439
192k28225439
asked Nov 21 '18 at 20:57
GhostGhost
553210
asked Nov 21 '18 at 20:57
GhostGhost
553210
asked Nov 21 '18 at 20:57
GhostGhost
553210
553210
1
The answer will involve complex dilogarithms.
– Lord Shark the Unknown
Nov 21 '18 at 21:03
@LordSharktheUnknown Ok... it's a bit too much for me and I probably won't understand since I am not soo advanced in math knowledge. Thanks for the info anyway.
– Ghost
Nov 21 '18 at 21:15
add a comment |
1
The answer will involve complex dilogarithms.
– Lord Shark the Unknown
Nov 21 '18 at 21:03
@LordSharktheUnknown Ok... it's a bit too much for me and I probably won't understand since I am not soo advanced in math knowledge. Thanks for the info anyway.
– Ghost
Nov 21 '18 at 21:15
1
1
The answer will involve complex dilogarithms.
– Lord Shark the Unknown
Nov 21 '18 at 21:03
The answer will involve complex dilogarithms.
– Lord Shark the Unknown
Nov 21 '18 at 21:03
@LordSharktheUnknown Ok... it's a bit too much for me and I probably won't understand since I am not soo advanced in math knowledge. Thanks for the info anyway.
– Ghost
Nov 21 '18 at 21:15
@LordSharktheUnknown Ok... it's a bit too much for me and I probably won't understand since I am not soo advanced in math knowledge. Thanks for the info anyway.
– Ghost
Nov 21 '18 at 21:15
add a comment |
2 Answers
2
active
oldest
votes
$$sum_{ngeq 1}frac{cos(nx)}{n^2} = text{Re},text{Li}_2(e^{ix})$$
is a periodic and piecewise-parabolic function, as the primitive of the sawtooth wave $sum_{ngeq 1}frac{sin(nx)}{n}$.
Since $cos^3(nx)=frac{1}{4}cos(3nx)+frac{3}{4}cos(nx)$ your function is a piecewise-parabolic function too.
In explicit terms it is a $2pi$-periodic, even function which equals $frac{3}{4}left(x-frac{pi}{3}right)left(x-frac{2pi}{3}right)$ over $[0,2pi/3]$ and $frac{3}{4}left(x-frac{2pi}{3}right)left(x-frac{4pi}{3}right)$ over $[2pi/3,pi]$.
This is very simple to derive by interpolation once the original series is evaluated at $xinleft{0,frac{pi}{3},frac{2pi}{3},pi,frac{4pi}{3},frac{5pi}{3}right}$.
answered Nov 21 '18 at 21:53
Jack D'AurizioJack D'Aurizio
287k33280658
add a comment |
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
"As the title says, I need to calculate the sum of the function series..."
begin{equation}
bbx{mbox{It's clear that}
sum_{n=1}^{infty}{cos^{3}pars{nx} over n^{2}} = {pi^{2} over 6phantom{^{2}}} mbox{when}
braces{verts{x} over 2pi} = 0}
end{equation}
Hereafter, I'll consider the case
$ds{braces{verts{x} over 2pi} not= 0}$.
Lets $ds{r equiv 2pibraces{verts{x} over 2pi}}$ with $ds{r in left[0,2piright)}$. Then
begin{align}
&bbox[10px,#ffd]{sum_{n=1}^{infty}{cos^{3}pars{nx} over n^{2}}} =
sum_{n=1}^{infty}{cos^{3}pars{nverts{x}} over n^{2}} =
sum_{n=1}^{infty}{cos^{3}pars{nr} over n^{2}} \[5mm] = &
sum_{n=1}^{infty}{bracks{3cospars{nr} + cospars{3nr}}/4 over n^{2}}
\[5mm] = &
{3 over 4},Resum_{n = 1}^{infty}
{pars{expo{ic r}}^{n} over n^{2}} +
{1 over 4},Resum_{n = 1}^{infty}
{pars{expo{3ic r}}^{n} over n^{2}}
\[5mm] = &
{3 over 4},
Remrm{Li}_{2}pars{exppars{ic r}} +
{1 over 4},
Remrm{Li}_{2}pars{exppars{ictilde{r}}}
\[2mm] &
mbox{where}
tilde{r} equiv 2pibraces{3r over 2pi} =
2pibraces{3braces{verts{x} over 2pi}}
end{align}
$ds{mrm{Li}_{s}}$ is the
Polylogarithm Function. Note that
$ds{3r in left[0,6piright)}$.
With
Jonqui$grave{mathrm{e}}$re Inversion Formula, it's found
begin{align}
&left.bbox[10px,#ffd]{sum_{n=1}^{infty}{cos^{3}pars{nx} over n^{2}}},rightvert
_{ verts{x}/pars{2pi} not= 0}
\[5mm] = &
bbx{{3 over 4},bracks{pi^{2},mrm{B}_{2}pars{braces{verts{x} over 2pi}}} +
{1 over 4},bracks{pi^{2},
mrm{B}_{2}pars{braces{3braces{verts{x} over 2pi}}}}}
end{align}
$ds{mrm{B}_{n}}$ is a
Bernoulli Polynomial.
Note that
$ds{mrm{B}_{2}pars{z} = z^{2} - z + {1 over 6}}$.
answered Nov 24 '18 at 19:57
Felix MarinFelix Marin
67.3k7107141
add a comment |
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3008350%2fhow-to-calculate-the-sum-of-the-series-sum-n-1-infty-frac-cos3nxn2%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$$sum_{ngeq 1}frac{cos(nx)}{n^2} = text{Re},text{Li}_2(e^{ix})$$
is a periodic and piecewise-parabolic function, as the primitive of the sawtooth wave $sum_{ngeq 1}frac{sin(nx)}{n}$.
Since $cos^3(nx)=frac{1}{4}cos(3nx)+frac{3}{4}cos(nx)$ your function is a piecewise-parabolic function too.
In explicit terms it is a $2pi$-periodic, even function which equals $frac{3}{4}left(x-frac{pi}{3}right)left(x-frac{2pi}{3}right)$ over $[0,2pi/3]$ and $frac{3}{4}left(x-frac{2pi}{3}right)left(x-frac{4pi}{3}right)$ over $[2pi/3,pi]$.
This is very simple to derive by interpolation once the original series is evaluated at $xinleft{0,frac{pi}{3},frac{2pi}{3},pi,frac{4pi}{3},frac{5pi}{3}right}$.
answered Nov 21 '18 at 21:53
Jack D'AurizioJack D'Aurizio
287k33280658
add a comment |
$$sum_{ngeq 1}frac{cos(nx)}{n^2} = text{Re},text{Li}_2(e^{ix})$$
is a periodic and piecewise-parabolic function, as the primitive of the sawtooth wave $sum_{ngeq 1}frac{sin(nx)}{n}$.
Since $cos^3(nx)=frac{1}{4}cos(3nx)+frac{3}{4}cos(nx)$ your function is a piecewise-parabolic function too.
In explicit terms it is a $2pi$-periodic, even function which equals $frac{3}{4}left(x-frac{pi}{3}right)left(x-frac{2pi}{3}right)$ over $[0,2pi/3]$ and $frac{3}{4}left(x-frac{2pi}{3}right)left(x-frac{4pi}{3}right)$ over $[2pi/3,pi]$.
This is very simple to derive by interpolation once the original series is evaluated at $xinleft{0,frac{pi}{3},frac{2pi}{3},pi,frac{4pi}{3},frac{5pi}{3}right}$.
answered Nov 21 '18 at 21:53
Jack D'AurizioJack D'Aurizio
287k33280658
add a comment |
$$sum_{ngeq 1}frac{cos(nx)}{n^2} = text{Re},text{Li}_2(e^{ix})$$
is a periodic and piecewise-parabolic function, as the primitive of the sawtooth wave $sum_{ngeq 1}frac{sin(nx)}{n}$.
Since $cos^3(nx)=frac{1}{4}cos(3nx)+frac{3}{4}cos(nx)$ your function is a piecewise-parabolic function too.
In explicit terms it is a $2pi$-periodic, even function which equals $frac{3}{4}left(x-frac{pi}{3}right)left(x-frac{2pi}{3}right)$ over $[0,2pi/3]$ and $frac{3}{4}left(x-frac{2pi}{3}right)left(x-frac{4pi}{3}right)$ over $[2pi/3,pi]$.
This is very simple to derive by interpolation once the original series is evaluated at $xinleft{0,frac{pi}{3},frac{2pi}{3},pi,frac{4pi}{3},frac{5pi}{3}right}$.
answered Nov 21 '18 at 21:53
Jack D'AurizioJack D'Aurizio
287k33280658
$$sum_{ngeq 1}frac{cos(nx)}{n^2} = text{Re},text{Li}_2(e^{ix})$$
is a periodic and piecewise-parabolic function, as the primitive of the sawtooth wave $sum_{ngeq 1}frac{sin(nx)}{n}$.
Since $cos^3(nx)=frac{1}{4}cos(3nx)+frac{3}{4}cos(nx)$ your function is a piecewise-parabolic function too.
In explicit terms it is a $2pi$-periodic, even function which equals $frac{3}{4}left(x-frac{pi}{3}right)left(x-frac{2pi}{3}right)$ over $[0,2pi/3]$ and $frac{3}{4}left(x-frac{2pi}{3}right)left(x-frac{4pi}{3}right)$ over $[2pi/3,pi]$.
This is very simple to derive by interpolation once the original series is evaluated at $xinleft{0,frac{pi}{3},frac{2pi}{3},pi,frac{4pi}{3},frac{5pi}{3}right}$.
answered Nov 21 '18 at 21:53
Jack D'AurizioJack D'Aurizio
287k33280658
edited Nov 21 '18 at 22:03
answered Nov 21 '18 at 21:53
Jack D'AurizioJack D'Aurizio
287k33280658
answered Nov 21 '18 at 21:53
Jack D'AurizioJack D'Aurizio
287k33280658
answered Nov 21 '18 at 21:53
Jack D'AurizioJack D'Aurizio
287k33280658
287k33280658
add a comment |
add a comment |
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
"As the title says, I need to calculate the sum of the function series..."
begin{equation}
bbx{mbox{It's clear that}
sum_{n=1}^{infty}{cos^{3}pars{nx} over n^{2}} = {pi^{2} over 6phantom{^{2}}} mbox{when}
braces{verts{x} over 2pi} = 0}
end{equation}
Hereafter, I'll consider the case
$ds{braces{verts{x} over 2pi} not= 0}$.
Lets $ds{r equiv 2pibraces{verts{x} over 2pi}}$ with $ds{r in left[0,2piright)}$. Then
begin{align}
&bbox[10px,#ffd]{sum_{n=1}^{infty}{cos^{3}pars{nx} over n^{2}}} =
sum_{n=1}^{infty}{cos^{3}pars{nverts{x}} over n^{2}} =
sum_{n=1}^{infty}{cos^{3}pars{nr} over n^{2}} \[5mm] = &
sum_{n=1}^{infty}{bracks{3cospars{nr} + cospars{3nr}}/4 over n^{2}}
\[5mm] = &
{3 over 4},Resum_{n = 1}^{infty}
{pars{expo{ic r}}^{n} over n^{2}} +
{1 over 4},Resum_{n = 1}^{infty}
{pars{expo{3ic r}}^{n} over n^{2}}
\[5mm] = &
{3 over 4},
Remrm{Li}_{2}pars{exppars{ic r}} +
{1 over 4},
Remrm{Li}_{2}pars{exppars{ictilde{r}}}
\[2mm] &
mbox{where}
tilde{r} equiv 2pibraces{3r over 2pi} =
2pibraces{3braces{verts{x} over 2pi}}
end{align}
$ds{mrm{Li}_{s}}$ is the
Polylogarithm Function. Note that
$ds{3r in left[0,6piright)}$.
With
Jonqui$grave{mathrm{e}}$re Inversion Formula, it's found
begin{align}
&left.bbox[10px,#ffd]{sum_{n=1}^{infty}{cos^{3}pars{nx} over n^{2}}},rightvert
_{ verts{x}/pars{2pi} not= 0}
\[5mm] = &
bbx{{3 over 4},bracks{pi^{2},mrm{B}_{2}pars{braces{verts{x} over 2pi}}} +
{1 over 4},bracks{pi^{2},
mrm{B}_{2}pars{braces{3braces{verts{x} over 2pi}}}}}
end{align}
$ds{mrm{B}_{n}}$ is a
Bernoulli Polynomial.
Note that
$ds{mrm{B}_{2}pars{z} = z^{2} - z + {1 over 6}}$.
answered Nov 24 '18 at 19:57
Felix MarinFelix Marin
67.3k7107141
add a comment |
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
"As the title says, I need to calculate the sum of the function series..."
begin{equation}
bbx{mbox{It's clear that}
sum_{n=1}^{infty}{cos^{3}pars{nx} over n^{2}} = {pi^{2} over 6phantom{^{2}}} mbox{when}
braces{verts{x} over 2pi} = 0}
end{equation}
Hereafter, I'll consider the case
$ds{braces{verts{x} over 2pi} not= 0}$.
Lets $ds{r equiv 2pibraces{verts{x} over 2pi}}$ with $ds{r in left[0,2piright)}$. Then
begin{align}
&bbox[10px,#ffd]{sum_{n=1}^{infty}{cos^{3}pars{nx} over n^{2}}} =
sum_{n=1}^{infty}{cos^{3}pars{nverts{x}} over n^{2}} =
sum_{n=1}^{infty}{cos^{3}pars{nr} over n^{2}} \[5mm] = &
sum_{n=1}^{infty}{bracks{3cospars{nr} + cospars{3nr}}/4 over n^{2}}
\[5mm] = &
{3 over 4},Resum_{n = 1}^{infty}
{pars{expo{ic r}}^{n} over n^{2}} +
{1 over 4},Resum_{n = 1}^{infty}
{pars{expo{3ic r}}^{n} over n^{2}}
\[5mm] = &
{3 over 4},
Remrm{Li}_{2}pars{exppars{ic r}} +
{1 over 4},
Remrm{Li}_{2}pars{exppars{ictilde{r}}}
\[2mm] &
mbox{where}
tilde{r} equiv 2pibraces{3r over 2pi} =
2pibraces{3braces{verts{x} over 2pi}}
end{align}
$ds{mrm{Li}_{s}}$ is the
Polylogarithm Function. Note that
$ds{3r in left[0,6piright)}$.
With
Jonqui$grave{mathrm{e}}$re Inversion Formula, it's found
begin{align}
&left.bbox[10px,#ffd]{sum_{n=1}^{infty}{cos^{3}pars{nx} over n^{2}}},rightvert
_{ verts{x}/pars{2pi} not= 0}
\[5mm] = &
bbx{{3 over 4},bracks{pi^{2},mrm{B}_{2}pars{braces{verts{x} over 2pi}}} +
{1 over 4},bracks{pi^{2},
mrm{B}_{2}pars{braces{3braces{verts{x} over 2pi}}}}}
end{align}
$ds{mrm{B}_{n}}$ is a
Bernoulli Polynomial.
Note that
$ds{mrm{B}_{2}pars{z} = z^{2} - z + {1 over 6}}$.
answered Nov 24 '18 at 19:57
Felix MarinFelix Marin
67.3k7107141
add a comment |
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
"As the title says, I need to calculate the sum of the function series..."
begin{equation}
bbx{mbox{It's clear that}
sum_{n=1}^{infty}{cos^{3}pars{nx} over n^{2}} = {pi^{2} over 6phantom{^{2}}} mbox{when}
braces{verts{x} over 2pi} = 0}
end{equation}
Hereafter, I'll consider the case
$ds{braces{verts{x} over 2pi} not= 0}$.
Lets $ds{r equiv 2pibraces{verts{x} over 2pi}}$ with $ds{r in left[0,2piright)}$. Then
begin{align}
&bbox[10px,#ffd]{sum_{n=1}^{infty}{cos^{3}pars{nx} over n^{2}}} =
sum_{n=1}^{infty}{cos^{3}pars{nverts{x}} over n^{2}} =
sum_{n=1}^{infty}{cos^{3}pars{nr} over n^{2}} \[5mm] = &
sum_{n=1}^{infty}{bracks{3cospars{nr} + cospars{3nr}}/4 over n^{2}}
\[5mm] = &
{3 over 4},Resum_{n = 1}^{infty}
{pars{expo{ic r}}^{n} over n^{2}} +
{1 over 4},Resum_{n = 1}^{infty}
{pars{expo{3ic r}}^{n} over n^{2}}
\[5mm] = &
{3 over 4},
Remrm{Li}_{2}pars{exppars{ic r}} +
{1 over 4},
Remrm{Li}_{2}pars{exppars{ictilde{r}}}
\[2mm] &
mbox{where}
tilde{r} equiv 2pibraces{3r over 2pi} =
2pibraces{3braces{verts{x} over 2pi}}
end{align}
$ds{mrm{Li}_{s}}$ is the
Polylogarithm Function. Note that
$ds{3r in left[0,6piright)}$.
With
Jonqui$grave{mathrm{e}}$re Inversion Formula, it's found
begin{align}
&left.bbox[10px,#ffd]{sum_{n=1}^{infty}{cos^{3}pars{nx} over n^{2}}},rightvert
_{ verts{x}/pars{2pi} not= 0}
\[5mm] = &
bbx{{3 over 4},bracks{pi^{2},mrm{B}_{2}pars{braces{verts{x} over 2pi}}} +
{1 over 4},bracks{pi^{2},
mrm{B}_{2}pars{braces{3braces{verts{x} over 2pi}}}}}
end{align}
$ds{mrm{B}_{n}}$ is a
Bernoulli Polynomial.
Note that
$ds{mrm{B}_{2}pars{z} = z^{2} - z + {1 over 6}}$.
answered Nov 24 '18 at 19:57
Felix MarinFelix Marin
67.3k7107141
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
"As the title says, I need to calculate the sum of the function series..."
begin{equation}
bbx{mbox{It's clear that}
sum_{n=1}^{infty}{cos^{3}pars{nx} over n^{2}} = {pi^{2} over 6phantom{^{2}}} mbox{when}
braces{verts{x} over 2pi} = 0}
end{equation}
Hereafter, I'll consider the case
$ds{braces{verts{x} over 2pi} not= 0}$.
Lets $ds{r equiv 2pibraces{verts{x} over 2pi}}$ with $ds{r in left[0,2piright)}$. Then
begin{align}
&bbox[10px,#ffd]{sum_{n=1}^{infty}{cos^{3}pars{nx} over n^{2}}} =
sum_{n=1}^{infty}{cos^{3}pars{nverts{x}} over n^{2}} =
sum_{n=1}^{infty}{cos^{3}pars{nr} over n^{2}} \[5mm] = &
sum_{n=1}^{infty}{bracks{3cospars{nr} + cospars{3nr}}/4 over n^{2}}
\[5mm] = &
{3 over 4},Resum_{n = 1}^{infty}
{pars{expo{ic r}}^{n} over n^{2}} +
{1 over 4},Resum_{n = 1}^{infty}
{pars{expo{3ic r}}^{n} over n^{2}}
\[5mm] = &
{3 over 4},
Remrm{Li}_{2}pars{exppars{ic r}} +
{1 over 4},
Remrm{Li}_{2}pars{exppars{ictilde{r}}}
\[2mm] &
mbox{where}
tilde{r} equiv 2pibraces{3r over 2pi} =
2pibraces{3braces{verts{x} over 2pi}}
end{align}
$ds{mrm{Li}_{s}}$ is the
Polylogarithm Function. Note that
$ds{3r in left[0,6piright)}$.
With
Jonqui$grave{mathrm{e}}$re Inversion Formula, it's found
begin{align}
&left.bbox[10px,#ffd]{sum_{n=1}^{infty}{cos^{3}pars{nx} over n^{2}}},rightvert
_{ verts{x}/pars{2pi} not= 0}
\[5mm] = &
bbx{{3 over 4},bracks{pi^{2},mrm{B}_{2}pars{braces{verts{x} over 2pi}}} +
{1 over 4},bracks{pi^{2},
mrm{B}_{2}pars{braces{3braces{verts{x} over 2pi}}}}}
end{align}
$ds{mrm{B}_{n}}$ is a
Bernoulli Polynomial.
Note that
$ds{mrm{B}_{2}pars{z} = z^{2} - z + {1 over 6}}$.
answered Nov 24 '18 at 19:57
Felix MarinFelix Marin
67.3k7107141
edited Nov 25 '18 at 2:00
answered Nov 24 '18 at 19:57
Felix MarinFelix Marin
67.3k7107141
answered Nov 24 '18 at 19:57
Felix MarinFelix Marin
67.3k7107141
answered Nov 24 '18 at 19:57
Felix MarinFelix Marin
67.3k7107141
67.3k7107141
add a comment |
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3008350%2fhow-to-calculate-the-sum-of-the-series-sum-n-1-infty-frac-cos3nxn2%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
1
The answer will involve complex dilogarithms.
– Lord Shark the Unknown
Nov 21 '18 at 21:03
@LordSharktheUnknown Ok... it's a bit too much for me and I probably won't understand since I am not soo advanced in math knowledge. Thanks for the info anyway.
– Ghost
Nov 21 '18 at 21:15