Thanks bobbym

here is 5 as far as I know it is fully factorised

1^6 + 2^6 + 3^6 + ... + n^6

1^7 + 2^7 + 3^7 + ... + n^7

1^8 + 2^8 + 3^8 + ... + n^8

I know it's almost been a year since this thread's last post, but I made an "Adjusted Pascal's Triangle" to create these power sum formulas for positive integers, and I finally got around to posting it on this forum.



Which I drew from the following formula I derived "from scratch":

(Putting in the n at the top of the triangle for sum of i^0 worked out perfectly, even though the formula itself cannot compute the correct value for R = 0).

Adjusting this triangle and using the method that knighthawk used in the previous post, I created an "Adjusted Pascal's Triangle" to compute the Bernoulli numbers which can be viewed here (it's too wide to post here, I think).

Basically, I found that

, and I used that to create that "Bernoulli Triangle".

I then wrote the following recursion formula from that Bernoulli triangle.

where

, where you can calculate a Bernoulli number in terms of its predecessor Bernoulli numbers.


Has anyone seen similar images like these "Adjusted Pascal Triangles" to compute the power sum and Bernoulli numbers visually?  I'm curious because I never heard of either, particularly the sum of power one, in high school or college math courses.

Last edited by cmowla (2013-03-09 11:26:05)

Stangerzv wrote:

Hi cmowla

I happened to read the article regarding the sums of power of integer using Bernoulli & Pascal long time ago. Could look similar but you can read it here www.sanjosemathcircle.org/handouts/2008-2009/20081112.pdf

When I was in the process of deriving the formula which I used to create the "adjusted pascal triangle" for the sum of power formulas, I actually went through the same process that Bernoulli did for a portion of my research.

Leaving out all of the ugly trial and error that I did (it was helpful to have a CAS handy through it all), here is my full derivation of my formula

which I used to create that "adjusted Pascal's Triangle" for the power sum polynomial formulas.


[Step 1]:  I derived a formula for sum i^2 from scratch.

We need to show that

[Step 2]:  Extrapolate

By applying the same pattern

to different power sums, I found that a general formula for sum i^R is simply:

, which has been known for a while, but I didn't know that unfortunately until after I found it.

This is the only step which is not fully justified, should one consider this entire derivation to possibly be a single proof, and is why I said I derived it "from scratch" (emphasizing with the quotes that there was one point of intuition).  However, one could simply find a proof for that formula, since it has existed for a while, and combine it with the remaining to have a complete proof that my "Adjusted Pascal's Triangle" is indeed correct to construct the power sum polynomials for positive integers.


[Step 3] Make a recurrence relation.

The above general formula isn't useful unless we can somehow use it to calculate the power sum sequence for the next power using all previous formulas.  The following is the recurrence relation I came up with which proved to be useful.  If the expression in the summation of the formula below is evaluated at any positive integer R>0, then the greatest exponent of i is R-1, i.e., i^(R-1), which is precisely what we need in a recurrence relation.

So we need to show that:

[Step 4]: Transforming into a pattern which can be generated by a diagram (which turned out to be an "adjusted Pascal Triangle")

This was the trickiest part.  I probably spent more time doing this step than anything else because I didn't even know if it was possible to create a formula which could generate an intuitive diagram.

So we show that

Last edited by cmowla (2013-03-30 08:15:03)