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 May 12th, 2010, 08:09 PM #1 Senior Member   Joined: Jul 2008 From: Western Canada Posts: 3,634 Thanks: 39 Looking for a database of constants Reading through old scientific papers, I frequently find formulae containing numerical constants, usually given only to 4 or 5 decimal places, and no explanation of from where they came. Often they are the result of summing a series, and the value will be some simple relationship to a fundamental constant such as pi, or else the log of some integer or simple fraction. At the moment for example, I'm trying to figure out if the value -0.60835 is derived from some simple constant. So, I'm just wondering if there's an online database somewhere where one could search for these constants.
 May 12th, 2010, 09:28 PM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Looking for a database of constants Yes! Neil Sloane's On-Line Encyclopedia of Integer Sequences: http://oeis.org/classic/ If you have any trouble finding a constant there, feel free to post and I'll help you. (I'm involved with that project.) Other possibilities include: RIES http://mrob.com/pub/ries/ Plouffe's Inverter: http://pi.lacim.uqam.ca/ If the value is an obscure infinite product or special constant, essentially your only hope is that it is in the OEIS; the others won't have it. If it's a 'simple' combination of fundamental constants, but not very well-known, RIES may be best at recognizing it. I haven't found much use for Plouffe's Inverter, and his newer Inverse Symbolic Calculator seems to have been lost to the web. (It still wasn't generally as good as the other tools, though...)
 May 13th, 2010, 10:23 PM #3 Senior Member   Joined: Jul 2008 From: Western Canada Posts: 3,634 Thanks: 39 Re: Looking for a database of constants Thank you very much. I looked at Neil Sloane's site, but wasn't able to figure out how it would apply to my current situation. I did try both Plouffe's Inverter, and RIES, both of which came up with possible solutions, which I will now have to work through. Meanwhile, I chopped up my current problem up into four smaller pieces, which I hope will be easier to solve. I now need to find the the following sum: $\sum\limits_{n=1}^{\infty}{\frac{H(an+b)}{n^2}}$ where H(k) is the kth Harmonic number, and a and b are constants, which for the four different parts of my problem, take on the following sets of values: a=1, b=-1 a=2, b=-1 a=1, b=+1 a=2, b=+1 Hence, I need to evaluate these: $\sum\limits_{n=1}^{\infty}{\frac{H(n-1)}{n^2}}$, $\sum\limits_{n=1}^{\infty}{\frac{H(2n-1)}{n^2}}$, $\sum\limits_{n=1}^{\infty}{\frac{H(n+1)}{n^2}}$, $\sum\limits_{n=1}^{\infty}{\frac{H(2n+1)}{n^2}}$ Any idea whether these have been solved previously?
May 14th, 2010, 07:18 AM   #4
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Re: Looking for a database of constants

Quote:
 Originally Posted by Yooklid I looked at Neil Sloane's site, but wasn't able to figure out how it would apply to my current situation.
Enter the constant into the search bar. For pi, type 3.1415926. This will actually search for sequences with 3,1,4,1,5,9,2,6 but as you can see it pulls up the right sequence without any real trouble.

There are a *lot* of constants there, over 5000. Random obscure example: the Khinchin mean $K_{-6}= 1.15655237\ldots$. The others wouldn't be able to pull something like that up.

Quote:
 Originally Posted by Yooklid I did try both Plouffe's Inverter, and RIES, both of which came up with possible solutions, which I will now have to work through. Meanwhile, I chopped up my current problem up into four smaller pieces, which I hope will be easier to solve. I now need to find the the following sum: $\sum\limits_{n=1}^{\infty}{\frac{H(an+b)}{n^2}}$ where H(k) is the kth Harmonic number, and a and b are constants, which for the four different parts of my problem, take on the following sets of values: a=1, b=-1 a=2, b=-1 a=1, b=+1 a=2, b=+1 Hence, I need to evaluate these: $\sum\limits_{n=1}^{\infty}{\frac{H(n-1)}{n^2}}$, $\sum\limits_{n=1}^{\infty}{\frac{H(2n-1)}{n^2}}$, $\sum\limits_{n=1}^{\infty}{\frac{H(n+1)}{n^2}}$, $\sum\limits_{n=1}^{\infty}{\frac{H(2n+1)}{n^2}}$ Any idea whether these have been solved previously?
Stab in the dark, but is the first one zeta(3)? I certainly can't prove it, but numerically it seems plausible. I calculated the number to 11 digits (using an integral to estimate the terms I can't sum to) and it matches:
http://oeis.org/classic/?q=1.20205690315

I calculate the third one as approximately 3.049047873156, but I can't find any simple form for it. I haven't tried the other two yet.

 May 14th, 2010, 07:31 AM #5 Senior Member   Joined: Apr 2010 Posts: 215 Thanks: 0 Re: Looking for a database of constants Wolfram Mathematica says: $\sum\limits_{n=1}^{\infty}{\frac{H(n-1)}{n^2}} = \zeta(3)$ $\sum\limits_{n=1}^{\infty}{\frac{H(2n-1)}{n^2}} = \frac{9\zeta(3)}{4}$ $\sum\limits_{n=1}^{\infty}{\frac{H(n+1)}{n^2}} = \zeta(2) + 2 \zeta(3) - 1$ $\sum\limits_{n=1}^{\infty}{\frac{H(2n+1)}{n^2}} = \zeta(2) + ln(16) + \frac{11 \zeta(3)}{4} - 4$
 May 14th, 2010, 07:47 AM #6 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Looking for a database of constants Impressive. I think that these might be provable by hand by expanding the H(n) and rewriting in terms of what is divided by each n^2, but I haven't tried yet. (I was about to, but then you posted with the answers...) Code: S(lim)=s=0;sum(n=1,lim,s+=1./n;s/(n+1)^2)+(1+Euler)/lim+log(lim)/lim
 May 14th, 2010, 05:21 PM #7 Senior Member   Joined: Jul 2008 From: Western Canada Posts: 3,634 Thanks: 39 Re: Looking for a database of constants CRGreathouse, Thank you for the explanation on looking up constants on OEIS. This will be very useful. Brangelito, Thank you for finding these solutions. Now, I'm embarassed to confess that I made a transcription error on the second and forth summations. They should really have been: $\sum\limits_{n=1}^{\infty}{\frac{H(2n-2)}{n^2}}$ and $\sum\limits_{n=1}^{\infty}{\frac{H(2n+2)}{n^2}}$ Could I possibly twist your arm, and have you look these up too? After my post last night, I wrote a short program to calculate these, and let it run overnight. It's calculating to 100 digits, and isn't terribly fast. It is still running, and is now up to about 21,000,000 terms. The sums, so far are only accurate to about 6 figures. I was hoping that it might generate enough terms to for OEIS to recognize it. Anyway that appears to be moot now. Out of interest, I also included the following sum which isn't part of my problem, but seems to be the simplest case: $\sum\limits_{n=1}^{\infty}{\frac{H(n)}{n^2}}$ and appears (with only 6 significant figures so far) that it may be equal to $2\zeta(3)$ I thought that this last one might be simple enough for me to attempt a hand solution, but when I started expanding the terms, I could find no discernible pattern. I'd be fascinated to learn what process Mathematica used to find these results.
May 15th, 2010, 04:48 AM   #8
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Re: Looking for a database of constants

Mathematica had some problems this time around, it seems. Anyway, here's a picture of the output (I'm not sure how to decipher it):

Quote:
 Out of interest, I also included the following sum which isn't part of my problem, but seems to be the simplest case: $\sum\limits_{n=1}^{\infty}{\frac{H(n)}{n^2}}$ and appears (with only 6 significant figures so far) that it may be equal to $2\zeta(3)$
This is correct!:)

 May 16th, 2010, 02:34 AM #9 Senior Member   Joined: Jul 2008 From: Western Canada Posts: 3,634 Thanks: 39 Re: Looking for a database of constants Thanks Brangelito. I'm afraid that I don't know what the H(2n-2) case means either. However, it gives me an idea how to proceed. Since we now know the value of the H(2n-1) sum, and the H(2n-2) case would differ only by a (1/n) in each term, the overall difference would be the following series: $\sum\limits_{n=1}^{\infty}{\frac{1}{(2n-1)n^2}}$ Then using partial fractions it should be simple to split this into: <> It's too late for me to think clearly right now. I'll look at this again in the morning when I'm actually awake.
 May 16th, 2010, 03:38 PM #10 Senior Member   Joined: Jul 2008 From: Western Canada Posts: 3,634 Thanks: 39 Re: Looking for a database of constants The partial fraction expansion wasn't quite as simple as what I thought in my last post, but still reasonable. So, I worked though the simple series and found: $\sum\limits_{n=1}^{\infty}{\frac{1}{(2n-1)n^2}}=4Ln(2)-\zeta(2)$ So combining that with the H(2n-1) series gives: $\sum\limits_{n=1}^{\infty}{\frac{H_{2n-2}}{n^2}} = \sum\limits_{n=1}^{\infty}{\frac{H_{2n-1}}{n^2}} - \sum\limits_{n=1}^{\infty}{\frac{1}{(2n-1)n^2}} = \frac{9}{4}\zeta(3)-4Ln(2)+\zeta(2)$ or approx. 1.576973 This result agrees with the value my program has now calculated up to 7 figures so far (with 59,000,000 terms calculated).

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