Upper bound the Polylogarithm $sum_{n=1}^infty frac{x^n}{n^2}$
$begingroup$
Let $x in (0,1)$ be some real number, we can then consider the Polylogarithm:
$$operatorname{L}_2(x)=sum_{n=1}^infty frac{x^n}{n^2}$$
It is not hard to see that the following upper bound holds:
$$
operatorname{L}_2(x) = x cdot sum_{n=1}^infty frac{x^{n-1}}{n^2} leq x sum_{n=1}^infty frac{1}{n^2} =frac{xpi^2}{6}.
$$
One could continue to get a finer upper bound using the same strategy:
$$
operatorname{L}_2(x) = sum_{n=1}^m frac{x^n}{n^2} + sum_{n=m+1}^infty frac{x^{n}}{n^2}
$$
where we can then upper bound $$sum_{n=m+1}^infty frac{x^{n}}{n^2} leq x^{m+1} cdot sum_{n=m+1}^infty frac{1}{n^2}$$
I am quite pleased with this, but was wondering if there are alternative, better well known bounds.
real-analysis calculus upper-lower-bounds polylogarithm
edited Dec 21 '18 at 11:56
mrtaurho
6,15271641
asked Dec 21 '18 at 11:50
HolyMonkHolyMonk
622416
$endgroup$
add a comment |
$begingroup$
Let $x in (0,1)$ be some real number, we can then consider the Polylogarithm:
$$operatorname{L}_2(x)=sum_{n=1}^infty frac{x^n}{n^2}$$
It is not hard to see that the following upper bound holds:
$$
operatorname{L}_2(x) = x cdot sum_{n=1}^infty frac{x^{n-1}}{n^2} leq x sum_{n=1}^infty frac{1}{n^2} =frac{xpi^2}{6}.
$$
One could continue to get a finer upper bound using the same strategy:
$$
operatorname{L}_2(x) = sum_{n=1}^m frac{x^n}{n^2} + sum_{n=m+1}^infty frac{x^{n}}{n^2}
$$
where we can then upper bound $$sum_{n=m+1}^infty frac{x^{n}}{n^2} leq x^{m+1} cdot sum_{n=m+1}^infty frac{1}{n^2}$$
I am quite pleased with this, but was wondering if there are alternative, better well known bounds.
real-analysis calculus upper-lower-bounds polylogarithm
edited Dec 21 '18 at 11:56
mrtaurho
6,15271641
asked Dec 21 '18 at 11:50
HolyMonkHolyMonk
622416
$endgroup$
1
$begingroup$
Just to be precise it is more common to call this special Polylogarithm "Dilogarithm" $($or Spence's Function$)$ denoted as $operatorname{Li}_2(z)$.
$endgroup$
– mrtaurho
Dec 21 '18 at 11:55
$begingroup$
You can use this formula : $sum_{n=1}^{infty}frac{x^{n+1}}{n(n+1)} = (1-x) log(1-x) + x$. But I don't know if it will be better
$endgroup$
– Damien
Dec 21 '18 at 14:14
$begingroup$
Awesome, this is exactly the type of thing I was looking for! You can post it as an answer if you want to. Just a small thing: your last $x$ should be a $1$ I believe, i.e. $(1-x) log(1-x) + 1$.
$endgroup$
– HolyMonk
Dec 21 '18 at 14:43
add a comment |
$begingroup$
Let $x in (0,1)$ be some real number, we can then consider the Polylogarithm:
$$operatorname{L}_2(x)=sum_{n=1}^infty frac{x^n}{n^2}$$
It is not hard to see that the following upper bound holds:
$$
operatorname{L}_2(x) = x cdot sum_{n=1}^infty frac{x^{n-1}}{n^2} leq x sum_{n=1}^infty frac{1}{n^2} =frac{xpi^2}{6}.
$$
One could continue to get a finer upper bound using the same strategy:
$$
operatorname{L}_2(x) = sum_{n=1}^m frac{x^n}{n^2} + sum_{n=m+1}^infty frac{x^{n}}{n^2}
$$
where we can then upper bound $$sum_{n=m+1}^infty frac{x^{n}}{n^2} leq x^{m+1} cdot sum_{n=m+1}^infty frac{1}{n^2}$$
I am quite pleased with this, but was wondering if there are alternative, better well known bounds.
real-analysis calculus upper-lower-bounds polylogarithm
edited Dec 21 '18 at 11:56
mrtaurho
6,15271641
asked Dec 21 '18 at 11:50
HolyMonkHolyMonk
622416
$endgroup$
Let $x in (0,1)$ be some real number, we can then consider the Polylogarithm:
$$operatorname{L}_2(x)=sum_{n=1}^infty frac{x^n}{n^2}$$
It is not hard to see that the following upper bound holds:
$$
operatorname{L}_2(x) = x cdot sum_{n=1}^infty frac{x^{n-1}}{n^2} leq x sum_{n=1}^infty frac{1}{n^2} =frac{xpi^2}{6}.
$$
One could continue to get a finer upper bound using the same strategy:
$$
operatorname{L}_2(x) = sum_{n=1}^m frac{x^n}{n^2} + sum_{n=m+1}^infty frac{x^{n}}{n^2}
$$
where we can then upper bound $$sum_{n=m+1}^infty frac{x^{n}}{n^2} leq x^{m+1} cdot sum_{n=m+1}^infty frac{1}{n^2}$$
I am quite pleased with this, but was wondering if there are alternative, better well known bounds.
real-analysis calculus upper-lower-bounds polylogarithm
real-analysis calculus upper-lower-bounds polylogarithm
edited Dec 21 '18 at 11:56
mrtaurho
6,15271641
asked Dec 21 '18 at 11:50
HolyMonkHolyMonk
622416
edited Dec 21 '18 at 11:56
mrtaurho
6,15271641
asked Dec 21 '18 at 11:50
HolyMonkHolyMonk
622416
edited Dec 21 '18 at 11:56
mrtaurho
6,15271641
edited Dec 21 '18 at 11:56
mrtaurho
6,15271641
edited Dec 21 '18 at 11:56
mrtaurho
6,15271641
6,15271641
asked Dec 21 '18 at 11:50
HolyMonkHolyMonk
622416
asked Dec 21 '18 at 11:50
HolyMonkHolyMonk
622416
asked Dec 21 '18 at 11:50
HolyMonkHolyMonk
622416
622416
1
$begingroup$
Just to be precise it is more common to call this special Polylogarithm "Dilogarithm" $($or Spence's Function$)$ denoted as $operatorname{Li}_2(z)$.
$endgroup$
– mrtaurho
Dec 21 '18 at 11:55
$begingroup$
You can use this formula : $sum_{n=1}^{infty}frac{x^{n+1}}{n(n+1)} = (1-x) log(1-x) + x$. But I don't know if it will be better
$endgroup$
– Damien
Dec 21 '18 at 14:14
$begingroup$
Awesome, this is exactly the type of thing I was looking for! You can post it as an answer if you want to. Just a small thing: your last $x$ should be a $1$ I believe, i.e. $(1-x) log(1-x) + 1$.
$endgroup$
– HolyMonk
Dec 21 '18 at 14:43
add a comment |
1
$begingroup$
Just to be precise it is more common to call this special Polylogarithm "Dilogarithm" $($or Spence's Function$)$ denoted as $operatorname{Li}_2(z)$.
$endgroup$
– mrtaurho
Dec 21 '18 at 11:55
$begingroup$
You can use this formula : $sum_{n=1}^{infty}frac{x^{n+1}}{n(n+1)} = (1-x) log(1-x) + x$. But I don't know if it will be better
$endgroup$
– Damien
Dec 21 '18 at 14:14
$begingroup$
Awesome, this is exactly the type of thing I was looking for! You can post it as an answer if you want to. Just a small thing: your last $x$ should be a $1$ I believe, i.e. $(1-x) log(1-x) + 1$.
$endgroup$
– HolyMonk
Dec 21 '18 at 14:43
1
1
$begingroup$
Just to be precise it is more common to call this special Polylogarithm "Dilogarithm" $($or Spence's Function$)$ denoted as $operatorname{Li}_2(z)$.
$endgroup$
– mrtaurho
Dec 21 '18 at 11:55
$begingroup$
Just to be precise it is more common to call this special Polylogarithm "Dilogarithm" $($or Spence's Function$)$ denoted as $operatorname{Li}_2(z)$.
$endgroup$
– mrtaurho
Dec 21 '18 at 11:55
$begingroup$
You can use this formula : $sum_{n=1}^{infty}frac{x^{n+1}}{n(n+1)} = (1-x) log(1-x) + x$. But I don't know if it will be better
$endgroup$
– Damien
Dec 21 '18 at 14:14
$begingroup$
You can use this formula : $sum_{n=1}^{infty}frac{x^{n+1}}{n(n+1)} = (1-x) log(1-x) + x$. But I don't know if it will be better
$endgroup$
– Damien
Dec 21 '18 at 14:14
$begingroup$
Awesome, this is exactly the type of thing I was looking for! You can post it as an answer if you want to. Just a small thing: your last $x$ should be a $1$ I believe, i.e. $(1-x) log(1-x) + 1$.
$endgroup$
– HolyMonk
Dec 21 '18 at 14:43
$begingroup$
Awesome, this is exactly the type of thing I was looking for! You can post it as an answer if you want to. Just a small thing: your last $x$ should be a $1$ I believe, i.e. $(1-x) log(1-x) + 1$.
$endgroup$
– HolyMonk
Dec 21 '18 at 14:43
add a comment |
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$begingroup$
Just to be precise it is more common to call this special Polylogarithm "Dilogarithm" $($or Spence's Function$)$ denoted as $operatorname{Li}_2(z)$.
$endgroup$
– mrtaurho
Dec 21 '18 at 11:55
$begingroup$
You can use this formula : $sum_{n=1}^{infty}frac{x^{n+1}}{n(n+1)} = (1-x) log(1-x) + x$. But I don't know if it will be better
$endgroup$
– Damien
Dec 21 '18 at 14:14
$begingroup$
Awesome, this is exactly the type of thing I was looking for! You can post it as an answer if you want to. Just a small thing: your last $x$ should be a $1$ I believe, i.e. $(1-x) log(1-x) + 1$.
$endgroup$
– HolyMonk
Dec 21 '18 at 14:43