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You are not logged in. #1 20120203 13:15:42
New Formulation For Sums of PowerThis 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 (20120204 10:52:38) #2 20120203 17:56:14
Re: New Formulation For Sums of PowerHi Stangerzv; In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #3 20120203 22:21:59
Re: New Formulation For Sums of PowerHi Bobbym #4 20120203 22:26:31
Re: New Formulation For Sums of PowerHi; In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #5 20120204 05:24:34
Re: New Formulation For Sums of PowerHi; In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #6 20120205 07:00:55
Re: New Formulation For Sums of PowerI wouldn't mind knowing how you figured this out/found it because I was working on a formula using a completly different method: Why did the chicken cross the Mobius Band? To get to the other ...um...!!! #7 20120205 11:18:33
Re: New Formulation For Sums of PowerHi Wintersolstice 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 (20120205 11:36:45) #8 20120205 22:21:54
Re: New Formulation For Sums of PowerHi; In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #9 20120205 23:32:44
Re: New Formulation For Sums of PowerHi Bobbym Last edited by Stangerzv (20120206 01:49:42) #10 20120206 03:17:02
Re: New Formulation For Sums of PowerHi; In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #11 20120423 03:41:16
Re: New Formulation For Sums of PowerSum of Power of Integers using Generalized Equation for Sums of Power for the Arithmetic Progression. Where: and #12 20120704 19:38:27
Re: New Formulation For Sums of Power
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...) 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 (20120704 19:41:46) #13 20120705 15:19:31
Re: New Formulation For Sums of PowerHi cmowla Last edited by Stangerzv (20120706 10:31:43) #14 20121013 06:03:03
Re: New Formulation For Sums of PowerI have also formulated the formulation for alternating sums of power for arithmetic progression. For even power: Last edited by Stangerzv (20121013 06:37:39) #15 20121013 06:38:27
Re: New Formulation For Sums of PowerAnybody 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(mj)}\frac{\prod_{k=0}^{j1}(1+2(mk))}{\prod_{k=0}^{j}(1+2(jk))}\right]" I couldn't display it. Last edited by Stangerzv (20121013 06:41:59) #16 20121013 06:49:57
Re: New Formulation For Sums of PowerYou 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 (20121013 06:51:22) The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #18 20121013 06:54:58
Re: New Formulation For Sums of PowerHave you seen post #16? The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #19 20121013 07:03:55
Re: New Formulation For Sums of PowerThanks 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 (20121013 07:05:06) #20 20121111 05:55:37
Re: New Formulation For Sums of PowerOther way to get sums of power for smaller p. Now let the Tth 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. #21 20130223 03:50:13
Re: New Formulation For Sums of PowerAlternative proof for Fermat's Last Theorem Using Sums of Power Formulation for p=3. 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, , and and 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 (20130223 05:35:26) 