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Looking for comprehensive list (or link) of even power summations 
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#1
Jan2314, 12:50 PM

P: 559

I have a function that results in 'exact' values for even powered summation series but it gives odd results for powers of '2' and '12+', how exciting! Unfortunately this also means the function is a far cry from a 'general solution'...
Does anyone have a comprehensive list of power summations in the form of, $$\sum_{n=1}^{\infty} 1/n^m$$ where 'm' is greater than the 10^{th} power in exact form? (even of course) 


#2
Jan2314, 01:11 PM

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PF Gold
P: 4,500

That's just [itex] \zeta(m)[/itex]. The Bernoulli numbers can be calculated explicitly and can be written in terms of zeta values, see
http://en.wikipedia.org/wiki/Bernoul..._approximation for the relationship between Bernoulli numbers and zeta values, and http://en.wikipedia.org/wiki/Bernoul...cit_definition for a formula for calculating them. I also found this page http://www.fullbooks.com/Thefirst4...iNumbers.html which says it has a list of the Bernoulli numbers but you might want to check that it is correct. It's not clear what your post means when you say you have a function that results in exact values, but gives results that you think are wrong for almost all powers  in what manner is it supposed to be exact? 


#3
Jan2314, 03:17 PM

P: 559

$$\sum_{n=1}^{\infty} \frac{1}{n^4}=\frac{Pi^4}{90}$$ $$\sum_{n=1}^{\infty} \frac{1}{n^6}=\frac{Pi^6}{945}$$ $$\sum_{n=1}^{\infty} \frac{1}{n^8}=\frac{Pi^8}{9450}$$ $$\sum_{n=1}^{\infty} \frac{1}{n^{10}}=\frac{Pi^{10}}{93555}$$ But for powers of '12' and higher the results don't look right and I know '2' is also incorrect (It is giving Pi^2/20, not Pi^2/6). I can easily fix the '2' but I need to see what is going on with the other end since it is best to work both sides of the problem. I could post the function although it is incomplete as far as being a 'general solution'. These discrepancies can likely be fixed but in order to do so I need a more comprehensive list of exact values for these power summations. On another note, thanks for the links! 


#4
Jan2414, 07:37 AM

P: 559

Looking for comprehensive list (or link) of even power summations



#5
Jan2414, 08:17 AM

Emeritus
Sci Advisor
PF Gold
P: 4,500

The first link gives the formula
[tex] B_{2m} = (1)^{m+1} \frac{2 (2m)!}{(2\pi)^{2m}} \left( \sum_{n=1}^{\infty} 1/n^{2m} \right)[/tex] which can be solved to give [tex] \sum_{n=1}^{\infty} 1/n^{2m} = (1)^{m+1} B_{2m} \frac{ (2\pi)^{2m}}{2(2m)!} [/tex] So to calculate your sum on the left you need to evaluate the thing on the right, which is easy except for the B[sub]2m[/sum] part. I gave a link that will let you explicitly calculate it as a formula (if you wanted to compare your formula to the formula for the B's) and also a link which gives numerical evaluations of the first couple hundred. For example, from the last link B_{36} = 26315271553053477373/1919190, so letting m=18 above, [tex] \sum_{n=1}^{\infty} 1/n^{36} = 26315271553053477373/1919190 * \frac{ (2\pi)^{36}}{2(36)!} [/tex] We can confirm these are in fact the same number. http://www.wolframalpha.com/input/?i=zeta%2836%29 http://www.wolframalpha.com/input/?i...%2836%29%21%7D 


#6
Jan2414, 10:34 AM

P: 559

I am pleased to see that, [tex] \sum_{n=1}^{\infty} 1/n^{12} = \frac{ 691Pi^{12}}{638512875} [/tex] I 'threw out' my solution as being anomalous because it had a 'non 1' numerated fraction for an answer (just like this does). I unfortunately left my notebook at home (probably crumpled up buried under a pile of sheets and pillows :P) although I will check this once I get home. So if this is in fact a 'general solution' then it is also a unique version of this Riemann Zeta function that doesn't require an input of your Bernoulli numbers, neat! (That is of course assuming the solutions for powers of 12+ are correct...) 


#7
Jan2414, 09:00 PM

P: 559

Office_Shredder, okay I checked it. The correct answer for n=12 is,
$$\sum_{n=1}^{\infty} 1/n^{12} = \frac{ 691Pi^{12}}{638512875}$$ My function gives, $$\sum_{n=1}^{\infty} 1/n^{12} = \frac{ 733Pi^{12}}{638512875}$$ Close but not correct. I will work this from both ends and post when I have a solution. Thanks for the help. 


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