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**Stangerzv****Member**- Registered: 2012-01-30
- Posts: 209

This is part of my works. I have formulated a new formulation for sums of power and it works for any numbers (i.e. real & complex arithmetic progression). The generalize equation can generate any power p (it works fine also with complex p). Below are the equation for p=2&3,

The value s is the common difference of successive terms in arithmetic progression and

is the sum of arithmetic terms. The paper is pending for publication but you can read on vixra. The beauty of this equation is that when you set n=2, it describes the Fermat's Last Theorem in a polynomial forms and if you set p to be negative, you can get new form of Riemmann's Zeta Function. I would be happy if anyone could point me the references if someone else has found it before me.Here, you can see how the coefficients are repetitive:

*Last edited by Stangerzv (2012-02-03 11:52:38)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Hi Stangerzv;

What is the name of the article on vixra? How do I access it?

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**Stangerzv****Member**- Registered: 2012-01-30
- Posts: 209

Hi Bobbym

You go to vixra dot org/numth/ and type sums of power and all five papers are mine. Select "A Treaty of Symmetric Function Part 1 Sums of Power" and read version 2 (v2).

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Hi;

Okay, thanks for the links.

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**If it ain't broke, fix it until it is.**

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Hi;

I got there but when I typed "sums of power" I got a blank page.

Oh, I have them now.

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

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**wintersolstice****Real Member**- Registered: 2009-06-06
- Posts: 115

I wouldn't mind knowing how you figured this out/found it because I was working on a formula using a completly different method:

two pyramids of numbers generated using Pascals Triangle, then the use of simaltaneous equations

this worked out the indivdual coffiecents of the terms of the formula all you have to do is factorise.

or was it the same method?

Why did the chicken cross the Mobius Band?

To get to the other ...um...!!!

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**Stangerzv****Member**- Registered: 2012-01-30
- Posts: 209

Hi Wintersolstice

It was started more than 17 years ago when I was a teenager, my brother introduced me Fermat's Last Theorem. While thinking about solving Fermat's Last Theorem I got an Idea, why not expressing

into function. That was how it started, not knowing that Jakob Bernoulli and few other guys already developed sums of power for integers few hundred years back I managed to formulated the formulation. The good things about this generalize equation is that it uses the most elementary symmetric function and making it possible to describe almost all series either real or complex numbers. I think this is the first ever formulation that could do sums of power for complex numbers, example as follows:Let's say we got 1000 series, the first term is 3+2i and the common difference between successive terms is s=i+1, find the 3rd power of this series

Using the formula for arithmetic sums and sums of power for 3rd power:

Substituting the arithmetic sums into the 3rd power equation yields the result as follow:

*Last edited by Stangerzv (2012-02-04 12:36:45)*

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**bobbym****Administrator**- From: Bumpkinland
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Hi;

Finished 1012.0032v2. From a referees point of view I

only found one typo in it. Other than that it looks pretty good. Right or wrong, it is a nice piece of work. Interesting historical info too.

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**If it ain't broke, fix it until it is.**

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**Stangerzv****Member**- Registered: 2012-01-30
- Posts: 209

Hi Bobbym

Thanks for your comment. For your information, I have emailed the link to John Coates, Andrew Wiles advisor when he was in Cambridge. He said it could be new since it could do sums of power for arithmetic progression but he hasn't got time to look into the 17-19th century papers and p-adic articles. I need to change the format of the paper to latex before I could submit the paper to a mathematical journal. Thanks again with the latex info, I thought it was difficult to use it.

*Last edited by Stangerzv (2012-02-05 02:49:42)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Hi;

Yes, I know who he is.

I will help you with the latex if you need it. Just post the pieces you are having problems with. Good luck with publication!

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**If it ain't broke, fix it until it is.**

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**Stangerzv****Member**- Registered: 2012-01-30
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Sum of Power of Integers using Generalized Equation for Sums of Power for the Arithmetic Progression.

Where:

and

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**cmowla****Member**- Registered: 2012-06-14
- Posts: 58

Stangerzv wrote:

You go to vixra dot org/numth/ and type sums of power and all five papers are mine. Select "A Treaty of Symmetric Function Part 1 Sums of Power" and read version 2 (v2).

I am trying to understand what you are trying to accomplish (I remember reading something about symmetry), and it's very interesting! I have recently just begun to study the sums of power generalized formula (well, I actually made one from scratch before I saw any existing ones), and one thing from "Part I" caught my attention (among other things, but...)

"The Derivation of Bernoulli Number for the Generalized Equation for Sums of Power", which begins on page 38 and concludes on page 40, inspired me to create my own Bernoulli Number generating function (I think that's what it is) which only relies on the preceding non-zero values from my own general sum of power formula (but it can generate the values for all *m*).

This was the result.

So the first Bernoulli number we can make with this is B[2] = 1/6 (which I believe is what yours starts off with too!)

I was also just adjusted this to make a polynomial (I guess it's considered a Bernoulli Polynomial, although my polynomial is not the same as the Bernoulli Polynomials described on Wikipedia):

, if all a "Bernoulli Polynomial" means is that one will get a Bernoulli number when 1 is substituted in for the variable of the polynomial.

Anyway,

By showing a Bernoulli number function from your generalized sum of power polynomial in your paper, are you saying that your Bernoulli function may be more useful in actual applications to more advanced concepts in number theory than the rest of the Bernoulli generating functions which existed already (and mine which at least exists now, if someone already didn't find it before)? If so, what is that exactly? (I'm not being critical at all. I am just curious because it appears that your work has been created for a "higher purpose" than just another sums of power equation).

Now, I have just mainly skimmed through Part I, so if you already explained this in any of those 5 or so papers, then I would appreciate direction to the paper and page number which answers my question (just in case you don't want to say the same thing twice).

*Last edited by cmowla (2012-07-03 21:41:46)*

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**Stangerzv****Member**- Registered: 2012-01-30
- Posts: 209

Hi cmowla

Basically, there are many Bernoulli's formulations and the finding of new Bernoulli's formulation not that significant. In my paper, the development of new bernoulli's formulation is only a small portion and without this formulation I can get others formulation to get the numbers. Since, Sums of power got bernoulli's numbers in it, I managed to manipulate it to get new forms of bernoulli's formulation but the main purpose is to develop sums of power for arithmetic progression. I do believe finding new sums of power for arithmetic progression is a big thing before people get to know it. It can be used for numerical analysis, Riemman's zeta function, Fermat's Last Theorem, generating function for finding prime numbers etc. I had demonstrated few examples of the use of this formulation.

Fulhaber is known as one of the greatest mathematicians because he developed sums of power for integers. This encourage me to work on bringing this formulation to the world as it is the umbrella for all sums of power because it can do integers, non-integers, integer power, complex power and many more.

Here the bernoulli's formulation:

*Last edited by Stangerzv (2012-07-05 12:31:43)*

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**Stangerzv****Member**- Registered: 2012-01-30
- Posts: 209

I have also formulated the formulation for alternating sums of power for arithmetic progression.

For odd power:

For even power:

*Last edited by Stangerzv (2012-10-12 07:37:39)*

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**Stangerzv****Member**- Registered: 2012-01-30
- Posts: 209

Anybody got an idea how to submit this code "O_{m,j}=n^{2m}+\sum_{j=1}^{m}\left \left [ \left ( -1 \right )^{j}\binom{m}{j} \left ( 2j+1 \right )E_{j}n^{2(m-j)}\frac{\prod_{k=0}^{j-1}(1+2(m-k))}{\prod_{k=0}^{j}(1+2(j-k))}\right]" I couldn't display it.

*Last edited by Stangerzv (2012-10-12 07:41:59)*

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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You had an extra \left in there.

Btw, I would imagine it to be O_{m,k}, because the value of j in the expression isn't really constant...

*Last edited by anonimnystefy (2012-10-12 07:51:22)*

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**Stangerzv****Member**- Registered: 2012-01-30
- Posts: 209

Some of the results:

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**anonimnystefy****Real Member**- From: The Foundation
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Have you seen post #16?

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

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**Stangerzv****Member**- Registered: 2012-01-30
- Posts: 209

Thanks anonimystefy, yes you are right, basically, I need to rewrite the equation but got no time to edit the whole paper. Anyway, Ej is the Euler number. There are two coefficients, Oj and Qj and both are using Euler or Zig/Secant number. The sums of power for arithmetic progression is using Bernoulli's number but the alternating sums of power is using Euler's number instead.

*Last edited by Stangerzv (2012-10-12 08:05:06)*

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**Stangerzv****Member**- Registered: 2012-01-30
- Posts: 209

Other way to get sums of power for smaller p.

Let the expansion of (x+y) as follows:

Now let the T-th term of arithmetic progression as (a+bi).

Thus,

Summing the terms above yields:

Example:

=>

=>

As the p is getting larger, the calculation would be becoming tedious.

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**Stangerzv****Member**- Registered: 2012-01-30
- Posts: 209

Alternative proof for Fermat's Last Theorem Using Sums of Power Formulation for p=3.

Now, let consider n=3,

When n=3, the sums of power for p=3 reduces into:

Let

Then

Or

Now consider this equation

Where

and

Assuming w is an even, thus

and

Therefore

Solving the equation yields

Let s=1, then

w=12

Solving the equation yields,

, andand z=2a=6

Therefore, there is a solution for this equation, which is given as follows

Now consider when n=2 and using the same procedure.

When n=2, the sums of power for p=3 reduces into:

Let

Then

Assuming w is an even, thus

and

Solving the equation yields:

since w=2a=2(s/2)=s,

This is a trivial solution,

or

Consider

and

Solving the equations yields:

Imaginary Solution.*Last edited by Stangerzv (2013-02-22 06:35:26)*

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