Mixing
Prove $text{Li}_2(e^{-2 i x})+text{Li}_2(e^{2 i x})=frac{1}{3} (6 x^2-6 pi x+pi ^2)$ when $0<x<pi$
$begingroup$
This is an identity I deduced when playing with the initial-boundary value problem of heat conduction equation asked here. It's easy to verify numerically with Mathematica:
Plot[{PolyLog[2, E^(-2 I x)] + PolyLog[2, E^(2 I x)],
1/3 (π^2 - 6 π x + 6 x^2)}, {x, -1, 4},
PlotStyle -> {Automatic, {Thick, Dashed}}]
However, Mathematica doesn't know how to simplify $text{Li}_2(e^{-2 i x})+text{Li}_2(e^{2 i x})$ to $frac{1}{3} (6 x^2-6 pi x+pi ^2)$ symbolically, so I'm wondering, how can I prove it by hand?
special-functions polylogarithm
asked Jan 19 at 11:32
xzczdxzczd
192112
$endgroup$
add a comment |
$begingroup$
This is an identity I deduced when playing with the initial-boundary value problem of heat conduction equation asked here. It's easy to verify numerically with Mathematica:
Plot[{PolyLog[2, E^(-2 I x)] + PolyLog[2, E^(2 I x)],
1/3 (π^2 - 6 π x + 6 x^2)}, {x, -1, 4},
PlotStyle -> {Automatic, {Thick, Dashed}}]
However, Mathematica doesn't know how to simplify $text{Li}_2(e^{-2 i x})+text{Li}_2(e^{2 i x})$ to $frac{1}{3} (6 x^2-6 pi x+pi ^2)$ symbolically, so I'm wondering, how can I prove it by hand?
special-functions polylogarithm
asked Jan 19 at 11:32
xzczdxzczd
192112
$endgroup$
add a comment |
$begingroup$
This is an identity I deduced when playing with the initial-boundary value problem of heat conduction equation asked here. It's easy to verify numerically with Mathematica:
Plot[{PolyLog[2, E^(-2 I x)] + PolyLog[2, E^(2 I x)],
1/3 (π^2 - 6 π x + 6 x^2)}, {x, -1, 4},
PlotStyle -> {Automatic, {Thick, Dashed}}]
However, Mathematica doesn't know how to simplify $text{Li}_2(e^{-2 i x})+text{Li}_2(e^{2 i x})$ to $frac{1}{3} (6 x^2-6 pi x+pi ^2)$ symbolically, so I'm wondering, how can I prove it by hand?
special-functions polylogarithm
asked Jan 19 at 11:32
xzczdxzczd
192112
$endgroup$
This is an identity I deduced when playing with the initial-boundary value problem of heat conduction equation asked here. It's easy to verify numerically with Mathematica:
Plot[{PolyLog[2, E^(-2 I x)] + PolyLog[2, E^(2 I x)],
1/3 (π^2 - 6 π x + 6 x^2)}, {x, -1, 4},
PlotStyle -> {Automatic, {Thick, Dashed}}]
However, Mathematica doesn't know how to simplify $text{Li}_2(e^{-2 i x})+text{Li}_2(e^{2 i x})$ to $frac{1}{3} (6 x^2-6 pi x+pi ^2)$ symbolically, so I'm wondering, how can I prove it by hand?
special-functions polylogarithm
special-functions polylogarithm
asked Jan 19 at 11:32
xzczdxzczd
192112
asked Jan 19 at 11:32
xzczdxzczd
192112
asked Jan 19 at 11:32
xzczdxzczd
192112
asked Jan 19 at 11:32
xzczdxzczd
192112
asked Jan 19 at 11:32
xzczdxzczd
192112
192112
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
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newcommand{ds}[1]{displaystyle{#1}}
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newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
&bbox[10px,#ffd]{left.vphantom{LARGE A}%
mrm{Li}_{2}pars{expo{-2ic x}} +
mrm{Li}_{2}pars{expo{2ic x}},rightvert_{ 0 < x < pi}} =
mrm{Li}_{2}pars{exppars{2piic,{x over pi}}} +
mrm{Li}_{2}pars{exppars{-2piic,{x over pi}}}
\[5mm] = &
-,{pars{2piic}^{2} over 2!},mathrm{B}_{2}pars{x over pi}
end{align}
which is Jonqui$grave{mrm{e}}$re Inversion Formula. $ds{mrm{B}_{n}}$ is a Bernoulli Polynomial.
Note that $ds{mrm{B}_{2}pars{z} = z^{2} - z + {1 over 6}}$ such that
begin{align}
&bbox[10px,#ffd]{left.vphantom{LARGE A}%
mrm{Li}_{2}pars{expo{-2ic x}} +
mrm{Li}_{2}pars{expo{2ic x}},rightvert_{ 0 < x < pi}} =
2pi^{2}bracks{pars{x over pi}^{2} - {x over pi} + {1 over 6}}
\[5mm] = &
bbx{2x^{2} - 2pi x + {pi^{2} over 3}}
end{align}
answered Jan 19 at 18:24
Felix MarinFelix Marin
68.2k7109143
$endgroup$
add a comment |
$begingroup$
$$operatorname{Li}_2(e^{-2 i x})+operatorname{Li}_2(e^{2 i x})\
=2Reoperatorname{Li}_2(e^{2i x})\
=2operatorname{Sl}_2(2x)\
=2(frac{pi^2}{6}+pi x+x^2)$$
Where $operatorname{Sl}$ is the SL-type Clausen function.
answered Jan 19 at 11:38
Kemono ChenKemono Chen
3,1741844
$endgroup$
1
$begingroup$
Most of this is just definitions, the only property that's used is here. Pretty simple. +1
$endgroup$
– Jakobian
Jan 19 at 11:45
add a comment |
$begingroup$
Try a series expansion around $x=0$ and get
$$text{Li}_2left(e^{-2 i x}right)+text{Li}_2left(e^{2 i x}right)=frac{pi ^2}{3}-2 pi x+2 x^2+Oleft(x^{99}right)$$
answered Jan 19 at 13:23
Claude LeiboviciClaude Leibovici
123k1157134
$endgroup$
$begingroup$
Er… how to calculate the series expansion of LHS?
$endgroup$
– xzczd
Jan 19 at 13:37
$begingroup$
@xzczd. Ask Mathematica to do it ... and let me know if it works ! Cheers
$endgroup$
– Claude Leibovici
Jan 19 at 13:39
$begingroup$
OK… Mathematica can't find the general form of the series coefficient,SeriesCoefficient[PolyLog[2, E^(-2 I x)] + PolyLog[2, E^(2 I x)], {x, 0, n}]returns the input, but e.g.Simplify[Series[PolyLog[2, E^(-2 I x)] + PolyLog[2, E^(2 I x)], {x, 0, 10}] // Normal, 0 < x < Pi]returns1/3 (Pi^2 - 6 Pi x + 6 x^2), so the identity is verified in another way. Thx.
$endgroup$
– xzczd
Jan 19 at 13:49
$begingroup$
@xzczd. Coulod you try the first instruction not with $n$ but with $6$ for example ? Thanks to let me know. I suppose that you need to assume $x>0$.
$endgroup$
– Claude Leibovici
Jan 19 at 14:00
$begingroup$
OK ! Try Series instead of SeriesCoefficient still with $6$ and not $n$. I think that I had things like that years ago when I had Mma at university.
$endgroup$
– Claude Leibovici
Jan 19 at 14:06
|
show 3 more comments
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
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oldest
votes
$begingroup$
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begin{align}
&bbox[10px,#ffd]{left.vphantom{LARGE A}%
mrm{Li}_{2}pars{expo{-2ic x}} +
mrm{Li}_{2}pars{expo{2ic x}},rightvert_{ 0 < x < pi}} =
mrm{Li}_{2}pars{exppars{2piic,{x over pi}}} +
mrm{Li}_{2}pars{exppars{-2piic,{x over pi}}}
\[5mm] = &
-,{pars{2piic}^{2} over 2!},mathrm{B}_{2}pars{x over pi}
end{align}
which is Jonqui$grave{mrm{e}}$re Inversion Formula. $ds{mrm{B}_{n}}$ is a Bernoulli Polynomial.
Note that $ds{mrm{B}_{2}pars{z} = z^{2} - z + {1 over 6}}$ such that
begin{align}
&bbox[10px,#ffd]{left.vphantom{LARGE A}%
mrm{Li}_{2}pars{expo{-2ic x}} +
mrm{Li}_{2}pars{expo{2ic x}},rightvert_{ 0 < x < pi}} =
2pi^{2}bracks{pars{x over pi}^{2} - {x over pi} + {1 over 6}}
\[5mm] = &
bbx{2x^{2} - 2pi x + {pi^{2} over 3}}
end{align}
answered Jan 19 at 18:24
Felix MarinFelix Marin
68.2k7109143
$endgroup$
add a comment |
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
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newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
&bbox[10px,#ffd]{left.vphantom{LARGE A}%
mrm{Li}_{2}pars{expo{-2ic x}} +
mrm{Li}_{2}pars{expo{2ic x}},rightvert_{ 0 < x < pi}} =
mrm{Li}_{2}pars{exppars{2piic,{x over pi}}} +
mrm{Li}_{2}pars{exppars{-2piic,{x over pi}}}
\[5mm] = &
-,{pars{2piic}^{2} over 2!},mathrm{B}_{2}pars{x over pi}
end{align}
which is Jonqui$grave{mrm{e}}$re Inversion Formula. $ds{mrm{B}_{n}}$ is a Bernoulli Polynomial.
Note that $ds{mrm{B}_{2}pars{z} = z^{2} - z + {1 over 6}}$ such that
begin{align}
&bbox[10px,#ffd]{left.vphantom{LARGE A}%
mrm{Li}_{2}pars{expo{-2ic x}} +
mrm{Li}_{2}pars{expo{2ic x}},rightvert_{ 0 < x < pi}} =
2pi^{2}bracks{pars{x over pi}^{2} - {x over pi} + {1 over 6}}
\[5mm] = &
bbx{2x^{2} - 2pi x + {pi^{2} over 3}}
end{align}
answered Jan 19 at 18:24
Felix MarinFelix Marin
68.2k7109143
$endgroup$
add a comment |
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
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begin{align}
&bbox[10px,#ffd]{left.vphantom{LARGE A}%
mrm{Li}_{2}pars{expo{-2ic x}} +
mrm{Li}_{2}pars{expo{2ic x}},rightvert_{ 0 < x < pi}} =
mrm{Li}_{2}pars{exppars{2piic,{x over pi}}} +
mrm{Li}_{2}pars{exppars{-2piic,{x over pi}}}
\[5mm] = &
-,{pars{2piic}^{2} over 2!},mathrm{B}_{2}pars{x over pi}
end{align}
which is Jonqui$grave{mrm{e}}$re Inversion Formula. $ds{mrm{B}_{n}}$ is a Bernoulli Polynomial.
Note that $ds{mrm{B}_{2}pars{z} = z^{2} - z + {1 over 6}}$ such that
begin{align}
&bbox[10px,#ffd]{left.vphantom{LARGE A}%
mrm{Li}_{2}pars{expo{-2ic x}} +
mrm{Li}_{2}pars{expo{2ic x}},rightvert_{ 0 < x < pi}} =
2pi^{2}bracks{pars{x over pi}^{2} - {x over pi} + {1 over 6}}
\[5mm] = &
bbx{2x^{2} - 2pi x + {pi^{2} over 3}}
end{align}
answered Jan 19 at 18:24
Felix MarinFelix Marin
68.2k7109143
$endgroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
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newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
&bbox[10px,#ffd]{left.vphantom{LARGE A}%
mrm{Li}_{2}pars{expo{-2ic x}} +
mrm{Li}_{2}pars{expo{2ic x}},rightvert_{ 0 < x < pi}} =
mrm{Li}_{2}pars{exppars{2piic,{x over pi}}} +
mrm{Li}_{2}pars{exppars{-2piic,{x over pi}}}
\[5mm] = &
-,{pars{2piic}^{2} over 2!},mathrm{B}_{2}pars{x over pi}
end{align}
which is Jonqui$grave{mrm{e}}$re Inversion Formula. $ds{mrm{B}_{n}}$ is a Bernoulli Polynomial.
Note that $ds{mrm{B}_{2}pars{z} = z^{2} - z + {1 over 6}}$ such that
begin{align}
&bbox[10px,#ffd]{left.vphantom{LARGE A}%
mrm{Li}_{2}pars{expo{-2ic x}} +
mrm{Li}_{2}pars{expo{2ic x}},rightvert_{ 0 < x < pi}} =
2pi^{2}bracks{pars{x over pi}^{2} - {x over pi} + {1 over 6}}
\[5mm] = &
bbx{2x^{2} - 2pi x + {pi^{2} over 3}}
end{align}
answered Jan 19 at 18:24
Felix MarinFelix Marin
68.2k7109143
answered Jan 19 at 18:24
Felix MarinFelix Marin
68.2k7109143
answered Jan 19 at 18:24
Felix MarinFelix Marin
68.2k7109143
answered Jan 19 at 18:24
Felix MarinFelix Marin
68.2k7109143
68.2k7109143
add a comment |
add a comment |
$begingroup$
$$operatorname{Li}_2(e^{-2 i x})+operatorname{Li}_2(e^{2 i x})\
=2Reoperatorname{Li}_2(e^{2i x})\
=2operatorname{Sl}_2(2x)\
=2(frac{pi^2}{6}+pi x+x^2)$$
Where $operatorname{Sl}$ is the SL-type Clausen function.
answered Jan 19 at 11:38
Kemono ChenKemono Chen
3,1741844
$endgroup$
1
$begingroup$
Most of this is just definitions, the only property that's used is here. Pretty simple. +1
$endgroup$
– Jakobian
Jan 19 at 11:45
add a comment |
$begingroup$
$$operatorname{Li}_2(e^{-2 i x})+operatorname{Li}_2(e^{2 i x})\
=2Reoperatorname{Li}_2(e^{2i x})\
=2operatorname{Sl}_2(2x)\
=2(frac{pi^2}{6}+pi x+x^2)$$
Where $operatorname{Sl}$ is the SL-type Clausen function.
answered Jan 19 at 11:38
Kemono ChenKemono Chen
3,1741844
$endgroup$
1
$begingroup$
Most of this is just definitions, the only property that's used is here. Pretty simple. +1
$endgroup$
– Jakobian
Jan 19 at 11:45
add a comment |
$begingroup$
$$operatorname{Li}_2(e^{-2 i x})+operatorname{Li}_2(e^{2 i x})\
=2Reoperatorname{Li}_2(e^{2i x})\
=2operatorname{Sl}_2(2x)\
=2(frac{pi^2}{6}+pi x+x^2)$$
Where $operatorname{Sl}$ is the SL-type Clausen function.
answered Jan 19 at 11:38
Kemono ChenKemono Chen
3,1741844
$endgroup$
$$operatorname{Li}_2(e^{-2 i x})+operatorname{Li}_2(e^{2 i x})\
=2Reoperatorname{Li}_2(e^{2i x})\
=2operatorname{Sl}_2(2x)\
=2(frac{pi^2}{6}+pi x+x^2)$$
Where $operatorname{Sl}$ is the SL-type Clausen function.
answered Jan 19 at 11:38
Kemono ChenKemono Chen
3,1741844
answered Jan 19 at 11:38
Kemono ChenKemono Chen
3,1741844
answered Jan 19 at 11:38
Kemono ChenKemono Chen
3,1741844
answered Jan 19 at 11:38
Kemono ChenKemono Chen
3,1741844
3,1741844
1
$begingroup$
Most of this is just definitions, the only property that's used is here. Pretty simple. +1
$endgroup$
– Jakobian
Jan 19 at 11:45
add a comment |
1
$begingroup$
Most of this is just definitions, the only property that's used is here. Pretty simple. +1
$endgroup$
– Jakobian
Jan 19 at 11:45
1
1
$begingroup$
Most of this is just definitions, the only property that's used is here. Pretty simple. +1
$endgroup$
– Jakobian
Jan 19 at 11:45
$begingroup$
Most of this is just definitions, the only property that's used is here. Pretty simple. +1
$endgroup$
– Jakobian
Jan 19 at 11:45
add a comment |
$begingroup$
Try a series expansion around $x=0$ and get
$$text{Li}_2left(e^{-2 i x}right)+text{Li}_2left(e^{2 i x}right)=frac{pi ^2}{3}-2 pi x+2 x^2+Oleft(x^{99}right)$$
answered Jan 19 at 13:23
Claude LeiboviciClaude Leibovici
123k1157134
$endgroup$
$begingroup$
Er… how to calculate the series expansion of LHS?
$endgroup$
– xzczd
Jan 19 at 13:37
$begingroup$
@xzczd. Ask Mathematica to do it ... and let me know if it works ! Cheers
$endgroup$
– Claude Leibovici
Jan 19 at 13:39
$begingroup$
OK… Mathematica can't find the general form of the series coefficient,SeriesCoefficient[PolyLog[2, E^(-2 I x)] + PolyLog[2, E^(2 I x)], {x, 0, n}]returns the input, but e.g.Simplify[Series[PolyLog[2, E^(-2 I x)] + PolyLog[2, E^(2 I x)], {x, 0, 10}] // Normal, 0 < x < Pi]returns1/3 (Pi^2 - 6 Pi x + 6 x^2), so the identity is verified in another way. Thx.
$endgroup$
– xzczd
Jan 19 at 13:49
$begingroup$
@xzczd. Coulod you try the first instruction not with $n$ but with $6$ for example ? Thanks to let me know. I suppose that you need to assume $x>0$.
$endgroup$
– Claude Leibovici
Jan 19 at 14:00
$begingroup$
OK ! Try Series instead of SeriesCoefficient still with $6$ and not $n$. I think that I had things like that years ago when I had Mma at university.
$endgroup$
– Claude Leibovici
Jan 19 at 14:06
|
show 3 more comments
$begingroup$
Try a series expansion around $x=0$ and get
$$text{Li}_2left(e^{-2 i x}right)+text{Li}_2left(e^{2 i x}right)=frac{pi ^2}{3}-2 pi x+2 x^2+Oleft(x^{99}right)$$
answered Jan 19 at 13:23
Claude LeiboviciClaude Leibovici
123k1157134
$endgroup$
$begingroup$
Er… how to calculate the series expansion of LHS?
$endgroup$
– xzczd
Jan 19 at 13:37
$begingroup$
@xzczd. Ask Mathematica to do it ... and let me know if it works ! Cheers
$endgroup$
– Claude Leibovici
Jan 19 at 13:39
$begingroup$
OK… Mathematica can't find the general form of the series coefficient,SeriesCoefficient[PolyLog[2, E^(-2 I x)] + PolyLog[2, E^(2 I x)], {x, 0, n}]returns the input, but e.g.Simplify[Series[PolyLog[2, E^(-2 I x)] + PolyLog[2, E^(2 I x)], {x, 0, 10}] // Normal, 0 < x < Pi]returns1/3 (Pi^2 - 6 Pi x + 6 x^2), so the identity is verified in another way. Thx.
$endgroup$
– xzczd
Jan 19 at 13:49
$begingroup$
@xzczd. Coulod you try the first instruction not with $n$ but with $6$ for example ? Thanks to let me know. I suppose that you need to assume $x>0$.
$endgroup$
– Claude Leibovici
Jan 19 at 14:00
$begingroup$
OK ! Try Series instead of SeriesCoefficient still with $6$ and not $n$. I think that I had things like that years ago when I had Mma at university.
$endgroup$
– Claude Leibovici
Jan 19 at 14:06
|
show 3 more comments
$begingroup$
Try a series expansion around $x=0$ and get
$$text{Li}_2left(e^{-2 i x}right)+text{Li}_2left(e^{2 i x}right)=frac{pi ^2}{3}-2 pi x+2 x^2+Oleft(x^{99}right)$$
answered Jan 19 at 13:23
Claude LeiboviciClaude Leibovici
123k1157134
$endgroup$
Try a series expansion around $x=0$ and get
$$text{Li}_2left(e^{-2 i x}right)+text{Li}_2left(e^{2 i x}right)=frac{pi ^2}{3}-2 pi x+2 x^2+Oleft(x^{99}right)$$
answered Jan 19 at 13:23
Claude LeiboviciClaude Leibovici
123k1157134
answered Jan 19 at 13:23
Claude LeiboviciClaude Leibovici
123k1157134
answered Jan 19 at 13:23
Claude LeiboviciClaude Leibovici
123k1157134
answered Jan 19 at 13:23
Claude LeiboviciClaude Leibovici
123k1157134
123k1157134
$begingroup$
Er… how to calculate the series expansion of LHS?
$endgroup$
– xzczd
Jan 19 at 13:37
$begingroup$
@xzczd. Ask Mathematica to do it ... and let me know if it works ! Cheers
$endgroup$
– Claude Leibovici
Jan 19 at 13:39
$begingroup$
OK… Mathematica can't find the general form of the series coefficient,SeriesCoefficient[PolyLog[2, E^(-2 I x)] + PolyLog[2, E^(2 I x)], {x, 0, n}]returns the input, but e.g.Simplify[Series[PolyLog[2, E^(-2 I x)] + PolyLog[2, E^(2 I x)], {x, 0, 10}] // Normal, 0 < x < Pi]returns1/3 (Pi^2 - 6 Pi x + 6 x^2), so the identity is verified in another way. Thx.
$endgroup$
– xzczd
Jan 19 at 13:49
$begingroup$
@xzczd. Coulod you try the first instruction not with $n$ but with $6$ for example ? Thanks to let me know. I suppose that you need to assume $x>0$.
$endgroup$
– Claude Leibovici
Jan 19 at 14:00
$begingroup$
OK ! Try Series instead of SeriesCoefficient still with $6$ and not $n$. I think that I had things like that years ago when I had Mma at university.
$endgroup$
– Claude Leibovici
Jan 19 at 14:06
|
show 3 more comments
$begingroup$
Er… how to calculate the series expansion of LHS?
$endgroup$
– xzczd
Jan 19 at 13:37
$begingroup$
@xzczd. Ask Mathematica to do it ... and let me know if it works ! Cheers
$endgroup$
– Claude Leibovici
Jan 19 at 13:39
$begingroup$
OK… Mathematica can't find the general form of the series coefficient,SeriesCoefficient[PolyLog[2, E^(-2 I x)] + PolyLog[2, E^(2 I x)], {x, 0, n}]returns the input, but e.g.Simplify[Series[PolyLog[2, E^(-2 I x)] + PolyLog[2, E^(2 I x)], {x, 0, 10}] // Normal, 0 < x < Pi]returns1/3 (Pi^2 - 6 Pi x + 6 x^2), so the identity is verified in another way. Thx.
$endgroup$
– xzczd
Jan 19 at 13:49
$begingroup$
@xzczd. Coulod you try the first instruction not with $n$ but with $6$ for example ? Thanks to let me know. I suppose that you need to assume $x>0$.
$endgroup$
– Claude Leibovici
Jan 19 at 14:00
$begingroup$
OK ! Try Series instead of SeriesCoefficient still with $6$ and not $n$. I think that I had things like that years ago when I had Mma at university.
$endgroup$
– Claude Leibovici
Jan 19 at 14:06
$begingroup$
Er… how to calculate the series expansion of LHS?
$endgroup$
– xzczd
Jan 19 at 13:37
$begingroup$
Er… how to calculate the series expansion of LHS?
$endgroup$
– xzczd
Jan 19 at 13:37
$begingroup$
@xzczd. Ask Mathematica to do it ... and let me know if it works ! Cheers
$endgroup$
– Claude Leibovici
Jan 19 at 13:39
$begingroup$
@xzczd. Ask Mathematica to do it ... and let me know if it works ! Cheers
$endgroup$
– Claude Leibovici
Jan 19 at 13:39
$begingroup$
OK… Mathematica can't find the general form of the series coefficient,
SeriesCoefficient[PolyLog[2, E^(-2 I x)] + PolyLog[2, E^(2 I x)], {x, 0, n}] returns the input, but e.g. Simplify[Series[PolyLog[2, E^(-2 I x)] + PolyLog[2, E^(2 I x)], {x, 0, 10}] // Normal, 0 < x < Pi] returns 1/3 (Pi^2 - 6 Pi x + 6 x^2), so the identity is verified in another way. Thx.$endgroup$
– xzczd
Jan 19 at 13:49
$begingroup$
OK… Mathematica can't find the general form of the series coefficient,
SeriesCoefficient[PolyLog[2, E^(-2 I x)] + PolyLog[2, E^(2 I x)], {x, 0, n}] returns the input, but e.g. Simplify[Series[PolyLog[2, E^(-2 I x)] + PolyLog[2, E^(2 I x)], {x, 0, 10}] // Normal, 0 < x < Pi] returns 1/3 (Pi^2 - 6 Pi x + 6 x^2), so the identity is verified in another way. Thx.$endgroup$
– xzczd
Jan 19 at 13:49
$begingroup$
@xzczd. Coulod you try the first instruction not with $n$ but with $6$ for example ? Thanks to let me know. I suppose that you need to assume $x>0$.
$endgroup$
– Claude Leibovici
Jan 19 at 14:00
$begingroup$
@xzczd. Coulod you try the first instruction not with $n$ but with $6$ for example ? Thanks to let me know. I suppose that you need to assume $x>0$.
$endgroup$
– Claude Leibovici
Jan 19 at 14:00
$begingroup$
OK ! Try Series instead of SeriesCoefficient still with $6$ and not $n$. I think that I had things like that years ago when I had Mma at university.
$endgroup$
– Claude Leibovici
Jan 19 at 14:06
$begingroup$
OK ! Try Series instead of SeriesCoefficient still with $6$ and not $n$. I think that I had things like that years ago when I had Mma at university.
$endgroup$
– Claude Leibovici
Jan 19 at 14:06
|
show 3 more comments
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