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Series acceleration formulae for common transcendental constants

DreamWeaver

Well-known member
Sep 16, 2013
337
OK, so for starters, I'm not asking anyone to re-post results found in other books, papers, etc. That'd be entirely contrary to the basic idea of this 'ere thread.

What I'm really interested in is this: have you - yes YOU, personally - found any new or presumably unique series acceleration formulae for such constants as \(\displaystyle G\), \(\displaystyle \pi\), \(\displaystyle \gamma\), \(\displaystyle \zeta(3)\), \(\displaystyle \log 2\), etc...?


If so, why not share...???


Nearly time for my bed, so I'll add a few tomorrow. (Bug)
 

chisigma

Well-known member
Feb 13, 2012
1,704
OK, so for starters, I'm not asking anyone to re-post results found in other books, papers, etc. That'd be entirely contrary to the basic idea of this 'ere thread.

What I'm really interested in is this: have you - yes YOU, personally - found any new or presumably unique series acceleration formulae for such constants as \(\displaystyle G\), \(\displaystyle \pi\), \(\displaystyle \gamma\), \(\displaystyle \zeta(3)\), \(\displaystyle \log 2\), etc...?


If so, why not share...???


Nearly time for my bed, so I'll add a few tomorrow. (Bug)
Some years ago I arrived to the following Taylor expansion of the function $z\ \ln z$ around z=1...

$\displaystyle z\ \ln z = (z-1) + \sum_{n=2}^{\infty} (-1)^{n}\ \frac{(z-1)^{n}}{n\ (n-1)},\ |z - 1| \le 1\ (1)$

The main task of my work was to demonstrate that for z=0 it is unequivocally $z\ \ln z = 0$ and for consequence for z=0 it is unequivocally $z^{z} = e^{z\ \ln z}=1$. In fact if You set in (1) z=0 You obtain...

$\displaystyle 0\ \ln 0 = - 1 + \sum_{n = 2}^{\infty} \frac{1}{n\ (n-1)} = -1 + 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ... = 0\ (2)$

In addition in (1) the general term goes to 0 as $\frac{1}{n^{2}}$ and that means that quicher convergence respect to series with the general term going to 0 as $\frac{1}{n}$ is guaranted. Setting $z=2$ the series supplies $2\ \ln 2$ and setting $z = 1 + i$ the series supplies $\displaystyle \frac{\ln 2}{2} - \frac{\pi}{4} + i\ (\frac{\ln 2}{2} + \frac{\pi}{4})$. I personally promoted 'investigations' and nobody found the (1) in the existing literature... I wonder why!...

Kind regards

$\chi$ $\sigma$
 
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mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
Er... Actually, I am not quite fond of the rate of convergence of that. If my calculations aren't wrong, I'd say the series, at $z = 2$ converges to $\log(4)$ more like $\sim 10^{-\log n}$ which isn't very fast at all.
 

chisigma

Well-known member
Feb 13, 2012
1,704
Er... Actually, I am not quite fond of the rate of convergence of that. If my calculations aren't wrong, I'd say the series, at $z = 2$ converges to $\log(4)$ more like $\sim 10^{-\log n}$ which isn't very fast at all.
I'm not sure that Your answer concerns to my post... if yes You have to compare the expansions...

$\displaystyle \ln 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + ...\ (1)$

$\displaystyle \ln 2 = \frac{1}{2} + \frac{1}{4} - \frac{1}{12} + \frac{1}{24} - \frac{1}{40} + ...\ (2)$


Which is faster?...

Kind regards

$\chi$ $\sigma$
 
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mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
The series is convergent at $z = 2$, as far as my understanding goes. I am speaking that in general, the series isn't fast at all, which I demonstrate by showing how slowly it converges to $\log(4)$ at $z = 2$.
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
chisigma said:
$\displaystyle \ln 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + ...\ (1)$

$\displaystyle \ln 2 = \frac{1}{2} + \frac{1}{4} - \frac{1}{12} + \frac{1}{24} - \frac{1}{40} + ...\ (2)$


Which is faster?...
Okay, I see you have edited that in. Of course the latter is faster, but that doesn't say that your series is in general fast. You see, according to what I have calculated down, after $n$-steps, you are near to $\log(2)$ only by $10^{-\! \log \! n}$. Do you consider that fast enough?
 

chisigma

Well-known member
Feb 13, 2012
1,704
Okay, I see you have edited that in. Of course the latter is faster, but that doesn't say that your series is in general fast. You see, according to what I have calculated down, after $n$-steps, you are near to $\log(2)$ only by $10^{-\! \log \! n}$. Do you consider that fast enough?
If fast convergence is the main task, then there are two excellent series expansions for $\ln 2$, i.e. the Pierce expansion...


$\displaystyle \ln 2 = 1 - \frac{1}{3} + \frac{1}{36} - ... (1)$


A091846 - OEIS

... and the Engel expansion...


$\displaystyle \ln 2 = \frac{1}{2} + \frac{1}{6} + \frac{1}{42} + \frac{1}{378} + ...\ (2)$


A059180 - OEIS

In fact the OP said '... have you - yes YOU, personally - found any new or presumably unique series acceleration formulae...' and the series I found obeyes to the requirement... as I said before the main task of 'my' series was to demonstrate that is $0^{0}=1$ and I'm very lucky to have been successful in that (Yes)...


Kind regards


$\chi$ $\sigma$
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
chisigma said:
and the series I found obeys to the requirement
I never doubted that anywhere. I merely pointed out the series wasn't quite "fast" in that sense.

PS : Oh, and about fast expansions of $\log(2)$ via acceleration, yes, those series are quite quick, although the best I'd think is CVZ applied to the usual slowly convergent series for $\log$.

Carry on...
 

DreamWeaver

Well-known member
Sep 16, 2013
337
Some years ago I arrived to the following Taylor expansion of the function $z\ \ln z$ around z=1...

$\displaystyle z\ \ln z = (z-1) + \sum_{n=2}^{\infty} (-1)^{n}\ \frac{(z-1)^{n}}{n\ (n-1)},\ |z - 1| \le 1\ (1)$



Kind regards

$\chi$ $\sigma$


Very nice indeed, Chisigma! Thanks for sharing... (Yes)


It strikes me that your series above could be used to find a series acceleration formula for the Dilogarithm. For example, replacing the \(\displaystyle z\) in your series with the dummy variable \(\displaystyle x\), dividing both sides of the expansion by \(\displaystyle (1-x)\), and then integrating from \(\displaystyle 1\) to \(\displaystyle z\) gives


\(\displaystyle \int_1^z \frac{x\log x}{1-x}\,dx= - \int_1^z \,dx + \sum_{k=2}^{\infty} \frac{ (-1)^{k-1} }{k(k-1)} \, \int_1^z (x-1)^{k-1} \,dx=\)


\(\displaystyle 1-z + \sum_{k=2}^{\infty} \frac{1}{k(k-1)} \, \int_1^z (1-x)^{k-1} \,dx\)


Substitute \(\displaystyle y=1-x\) in that last integral to get:


\(\displaystyle 1-z - \sum_{k=2}^{\infty} \frac{1}{k(k-1)} \, \int_0^{1-z} y^{k-1} \,dy=\)


\(\displaystyle (01) \quad 1-z - \sum_{k=2}^{\infty} \frac{(1-z)^k}{k^2(k-1)} \)


On the other hand,


\(\displaystyle \int_1^z \frac{x\log x}{1-x}\,dx= \int_1^z \frac{ [1-(1-x)] }{1-x} \log x \,dx=\)


\(\displaystyle \int_1^z\frac{ \log x}{(1-x)}\,dx - \int_1^z \log x \,dx=\)


\(\displaystyle -\log x\log(1-x)\, \Bigg|_1^z+\int_1^z\frac{\log(1-x)}{x}\, dx - \Bigg[ z\log z - z +1 \Bigg]=\)


\(\displaystyle -\log z \log(1-z) - z\log z +z - 1 + \Bigg[ \int_1^z\frac{\log(1-x)}{x}\, dx \Bigg]=\)


\(\displaystyle -\log z \log(1-z) - z\log z +z - 1 + \Bigg[ \text{Li}_2(1) - \text{Li}_2(z) \Bigg]=\)


\(\displaystyle (02) \quad -\log z \log(1-z) - z\log z +z - 1 + \Bigg[ \zeta(2) - \text{Li}_2(z) \Bigg] \)



Finally, equating (01) and (02) then gives the final result:



\(\displaystyle \text{Li}_2(z)= -\log z \log(1-z) - z\log z +2z - 2 + \zeta(2) + \sum_{k=2}^{\infty} \frac{(1-z)^k}{k^2(k-1)}\)



A few observations:


Firstly, although faster series acceleration formulae exist for the Dilogarithm - search for BBP-type series online - and some modest restrictions would need to be put on \(\displaystyle z\), this series is self-evidently a good deal faster than the standard Dilogarithmic series:


\(\displaystyle \text{Li}_2(z) = \sum_{k=1}^{\infty} \frac{z^k}{k^2}\)


Secondly, in many years of exploring Polylogarithms, and reading many related papers, I'm pretty certain I've not seen this series before.

And finally, repeating the same steps as above on this series will give an acceleration formula for the Trilogarithm. Similarly, repeated iterated integration in this way will lead to series acceleration formulae for additional higher order Polylogarithms.


As I said, Chisigma, very nice work indeed... :D