stangerzv Posted January 9, 2020 Share Posted January 9, 2020 (edited) 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). Let the p-th power of an arithmetic series as follows [math]\sum_{i=1}^n x_i^p = x_1^p + x_2^p + x_3^p + \cdots + x_n^p[/math] The general equation for the sum is given as follows [math]\sum^{n}_{i=1} {x_i}^p=\sum^{u}_{m=0}\phi_m s^{2m}\frac{[\sum^{n}_{i=1} x_i] ^{p-2m}}{n^{p-(2m+1)}}[/math] where: [math]p-(2m+1)\ge 1[/math] if p is even and [math]p-(2m+1)\le 1[/math] if p is odd, [math]\phi_m[/math] is a coefficient,[math]\sum^{n}_{i=1} {x_i}[/math] sum of n-th term, [math]u=\frac {p-1}{2}[/math] for odd p and [math]u=\frac {p}{2}[/math] for even p and s is the difference between terms (i.e [math]s=x_{i+1}-x_{i}[/math]). Below are the equations for p=2-7 [math]\\\sum_{i=0}^{n}x_{i}^{2}=\frac{\left [ \sum_{i=0}^{n}x_{i} \right ]^2}{n}+\frac{n(n^2-1)s^2}{12}\\\\\\\sum_{i=0}^{n}x_i^3=\frac{\left [ \sum_{i=0}^{n}x_i \right ]^3}{n^2}+\frac{(n^2-1)s^2\left [ \sum_{i=0}^{n} x_i\right ]}{4}[/math] The value s is the common difference of successive terms in arithmetic progression and [math]\sum_{i=0}^{n}x_i[/math] is the sum of arithmetic terms. 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. Here, you can see how the coefficients are repetitive: [math]\\\sum_{i=0}^{n}x_i^4=\frac{\left [ \sum_{i=0}^{n}x_i \right ]^4}{n^3}+\frac{(n^2-1)s^2\left [ \sum_{i=0}^{n}x_i \right ]^2}{2n}+\frac{n(3n^2-7)(n^2-1)s^4}{240}\\\\\\\sum_{i=0}^{n}x_i^5=\frac{\left [ \sum_{i=0}^{n}x_i \right ]^5}{n^4}+\frac{5(n^2-1)s^2\left [ \sum_{i=0}^{n} x_i\right ]^3}{6n^2}+\frac{(3n^2-7)(n^2-1)s^4\left [ \sum_{i=0}^{n}x_i \right ]}{48}[/math] [math]\\\sum_{i=0}^{n}x_i^6=\frac{\left [ \sum_{i=0}^{n}x_i \right ]^6}{n^5}+\frac{5(n^2-1)s^2\left [ \sum_{i=0}^{n}x_i \right ]^4}{4n^3}+\frac{(3n^2-7)(n^2-1)s^4\left [ \sum_{i=0}^{n}x_i \right ]^2}{16n}+\frac{n(3n^4-18n^2+31) (n^2-1)s^6}{1344}\\\\\\\sum_{i=0}^{n}x_i^7=\frac{\left [ \sum_{i=0}^{n}x_i \right ]^7}{n^6}+\frac{7(n^2-1)s^2\left [ \sum_{i=0}^{n}x_i \right ]^5}{4n^4}+\frac{7(3n^2-7)(n^2-1)s^4\left [ \sum_{i=0}^{n}x_i \right ]^3}{48n^2}+\frac{(3n^4-18n^2+31) (n^2-1)s^6\left [ \sum_{i=0}^{n}x_i \right ]}{192}[/math] Perhaps, by looking at this new formulation, someone could work out an alternative shorter proof for Fermat's Last Theorem. Edited January 9, 2020 by stangerzv Link to comment Share on other sites More sharing options...
mathematic Posted January 9, 2020 Share Posted January 9, 2020 (edited) How are x_i related? Edited January 9, 2020 by mathematic Link to comment Share on other sites More sharing options...
stangerzv Posted January 10, 2020 Author Share Posted January 10, 2020 x_i is an arithmetic series (e.g x_1, x_2, ...x_n=2,3,..,1000) Link to comment Share on other sites More sharing options...
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