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


by mesa
Tags: comprehensive, list, power, summations
mesa
mesa is offline
#1
Jan23-14, 12:50 PM
P: 495
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 10th power in exact form? (even of course)
Phys.Org News Partner Mathematics news on Phys.org
Math modeling handbook now available
Hyperbolic homogeneous polynomials, oh my!
Researchers help Boston Marathon organizers plan for 2014 race
Office_Shredder
Office_Shredder is offline
#2
Jan23-14, 01:11 PM
Mentor
P: 4,499
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/The-first-4...i-Numbers.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?
mesa
mesa is offline
#3
Jan23-14, 03:17 PM
P: 495
Quote Quote by Office_Shredder View Post
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?
The derived function will compute exact values of 'some' even powered summations. At the moment it will produce correct answers for,
$$\sum_{n=1}^{\infty} \frac{1}{n^4}=\frac{Pi^4}{90}$$
$$\sum_{n=1}^{\infty} \frac{1}{n^6}=\frac{Pi^6}{945}$$
$$\sum_{n=1}^{\infty} \frac{1}{n^8}=\frac{Pi^8}{9450}$$
$$\sum_{n=1}^{\infty} \frac{1}{n^{10}}=\frac{Pi^{10}}{93555}$$

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!

mesa
mesa is offline
#4
Jan24-14, 07:37 AM
P: 495

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


Quote Quote by Office_Shredder View Post
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/The-first-4...i-Numbers.html

which says it has a list of the Bernoulli numbers but you might want to check that it is correct.
Hey Office_Shredder, I had a chance to run through these links although I am not clear on what this has to do with the question I posed, am I missing something or was my post just a little too hazy and that led you off track?
Office_Shredder
Office_Shredder is offline
#5
Jan24-14, 08:17 AM
Mentor
P: 4,499
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 B36 = -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
mesa
mesa is offline
#6
Jan24-14, 10:34 AM
P: 495
Quote Quote by Office_Shredder View Post
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 B36 = -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
Okay, it all makes sense now. I didn't notice in that first link to wikipedia that it was a summation for the far right term. Sometimes being so focused it is hard to spot even the obvious...

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...)
mesa
mesa is offline
#7
Jan24-14, 09:00 PM
P: 495
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.


Register to reply

Related Discussions
comprehensive list of inference rules used in physics General Physics 0
comprehensive list of cosmological models and their properties? Special & General Relativity 1
I need a link to a list of properties of real numbers General Math 0
a question in searching a link in a circular list Programming & Computer Science 9
Rice Tops The Power List Current Events 4