1 Cubic Power Algorithm II Using polynomials Author and researcher Zeolla Gabriel Martín The discovery of a new algorithm, which went unnoticed for centuries, now comes to light to show its characteristics and its contribution to the use of polynomials. Registrado en La ciudad de La Plata, Buenos Aires Argentina. 2 Title: Cubic Power Algorithm, using polynomials. Sub title: Cube of a binomial, trinomial, tetranomial and pentanomial. Author: Zeolla, Gabriel Martín Comments: 29 pages [email protected] Abstract: This document develops and demonstrates the discovery of a new cube potentiation algorithm that works absolutely with all the numbers using the formula of the square of a binomial, trinomial, tetranomial and pentanomial. This presents the expansion of terms to the cube, the ideal order of the coefficients to obtain a sum that generates the results of the power. Chapter 1: Square of a binomial, trinomial, tetranomial and pentanomial. Example nº1 Binomial (a+b)3 = (a+b)*(a+b)*(a+b) Right distribution of terms a3 +3a2b + 3ab2 + b3 Coefficient of terms 1331 We can get the coefficients of Pascal's triangle 1 1 1 1 2 1 1 3 3 1 3 Demonstration of cubic power Example (25)3= 15.625 a=2 b=5 Right distribution of terms a3 +3a2b + 3ab2 + b3 253 = 23+3 ∗ 22 ∗ 5 + 3 ∗ 2 ∗ 52 ∗ 53 253 = 8 + 60 + 150 + 125 8 *1.000 6 0 *100 1 5 0 *10 + 1 2 5 *1 1 5 6 2 5 Result The figure is a pattern that will be present in all the numbers of two digits squared. We multiply the first term by 1000, the second term by 100, the third term by 10, and the four term by 1. In all cases when we use the cube of a whole number. 4 Example nº2 Trinomial (a+b+c)3= (a+b+c)*(a+b+c)*(a+b+c) Right distribution of terms a3 +3a2b + 3a2c + 3ab2+ 6abc + 3ac2+ b3+ 3b2c+3bc2+ c3 Coefficient of terms 1333631331 We can obtain the coefficients by multiplying Pascal's triangle (a+b)0 1 x1 (a+b)1 1 1 x3 (a+b)2 1 2 1 x3 (a+b)3 1 3 3 1 x1 (a+b+c) 3 1 3 3 3 6 3 1 3 3 1 Demonstration of cubic power Example (132)3= 2.299.968 a=1 b=3 c=2 Right distribution of terms a3 +3a2b + 3a2c + 3ab2+ 6abc + 3ac2+ b3+ 3b2c+3bc2+ c3 ퟏퟑ+ ퟑ ∗ ퟏퟐ ∗ ퟑ + ퟑ ∗ ퟏퟐ ∗ ퟐ + ퟑ ∗ ퟏ ∗ ퟑퟐ + ퟔ ∗ ퟏ ∗ ퟑ ∗ ퟐ + ퟑ ∗ ퟏ ∗ ퟐퟐ + ퟑퟑ+ ퟑ ∗ ퟑퟐ ∗ ퟐ + ퟑ ∗ ퟑퟐ ∗ ퟐ + ퟐퟑ 1 + 6 + 9 + 27 + 36 + 12 + 27 + 108 + 54 + 8 5 1 *1.000.000 0 9 *100.000 0 6 *10.000 2 7 *10.000 3 6 *1.000 1 2 *100 2 7 *1.000 5 4 *100 3 6 *10 0 8 *1 2 2 9 9 9 6 8 Result The shape that is formed here is a pattern that will always be formed when we have three cube digits. We can see that the geometric figure contains the figure of example 1 (cube of a binomial). We add following this model, ordering the numbers from left to right. 6 Example nº3 Tetranomial (a+b+c+d)3 = (a+b+c+d)*(a+b+c+d)*(a+b+c+d) Right distribution of terms a3+3a2b +3a2c + 3a2d+3ab2+ 6abc + 6abd +3ac2+6acd+3ad2+b3+3b2c + 3b2d + 3bc2+ 6bcd + 3bd2+c3+ 3c2d+ 3cd2+ d3 Coefficient of terms 13333663631333631331 We can obtain the coefficients by multiplying Pyramid of three terms 0 (a + b + c) 1 x1 1 (a + b + c) 1 1 1 x3 2 (a + b + c) 1 2 2 1 2 1 x3 3 (a + b + c) 1 3 3 3 6 3 1 3 3 1 x1 (a + b + c + d)3 1 3 3 3 3 6 6 3 6 3 1 3 3 3 6 3 1 3 3 1 Demonstration of cubic power Example (1.234)3= 1.879.080.904 a =1 b =2 c =3 d =4 a3+3a2b +3a2c + 3a2d+3ab2+ 6abc + 6abd +3ac2+6acd+3ad2+b3+3b2c + 3b2d + 3bc2+ 6bcd + 3bd2+c3+ 3c2d+ 3cd2+ d3 13+3*12*2 +3*12*3 + 3*12*4+3*1*22+ 6*1*2*3 + 6*1*2*4 +3*1*32+6*1*3*4+3*1*42+23+3*22*3 + 3*22*4 + 3*2*32+ 6*2*3*4 + 3*2*42+33+ 3*32*4+ 3*3*42+ 43 1+6+9+12+12+36+48+27+72+48+8+36+48+54+144+96+27+108+144+64 7 1 *1.000.000.000 0 6 *100.000.000 0 9 *10.000.000 1 2 *1.000.000 1 2 *10.000.000 3 6 *1.000.000 4 8 *100.000 2 7 *100.000 7 2 *10.000 4 8 *1.000 0 8 *1.000.000 3 6 *100.000 4 8 *10.000 5 4 *10.000 1 4 4 *1.000 9 6 *100 2 7 *1.000 1 0 8 *100 1 4 4 *10 6 4 *1 1 8 7 9 0 8 0 9 0 4 Result The figure is a pattern that will be formed with all the numbers with a maximum of 4 digits. To add we use this model, ordering the numbers from left to right. This pattern contains the patterns of examples 1 and 2 within itself. 8 Example nº4 Pentanomial (a+b+c+d+e)3 =(a+b+c+d+e)* (a+b+c+d+e)* (a+b+c+d+e) Right distribution of terms a3 + 3a2b + 3a2c + 3a2d + 3a2e+ 3ab2 + 6abc + 6abd + 6abe + 3ac2 + 6acd +6ace + 3ad2+ 6ade +3ae2+ b3 + 3cb2 + 3b2d +3b2e + 3c2b + 6bcd + 6bce + 3bd2 + 6bde+ 3be2+ c3+ 3dc2 + 3c2e+ 3d2c + 6cde +3ce2 + d3+ 3d2e + 3de2 + e3 Coefficient of terms 13333366636636313333663631333631331 We can obtain the coefficients by multiplying Pyramid of four terms 0 (a + b + c + d) 1 x1 1 (a + b + c + d) 1 1 1 1 x3 2 (a + b + c + d) 1 2 2 2 1 2 2 1 2 1 x3 3 (a + b + c + d) 1 3 3 3 3 6 6 3 6 3 1 3 3 3 6 3 1 3 3 1 x1 (a+b+c+d+e) 3 1 3 3 3 3 3 6 6 6 3 6 6 3 6 3 1 3 3 3 3 6 6 3 6 3 1 3 3 3 6 3 1 3 3 1 Demonstration of cubic power Example (12.345)3= 1.881.365.963.625 a=1 b=2 c=3 d=4 e=5 13 + 3*12*2 + 3*12*3 + 3*12*4 + 3*12*5+ 3*1*22 + 6*1*2*3 + 6*1*2*4 + 6*1*2*5 + 3*1*32 + 6*1*3*4 +6*1*3*5 + 3*1*42+ 6*1*4*5 + 3*1*52 + 23 + 3*3*22 + 3*22*4 + 3*22*5 + 3*32*2 + 6*2*3*4 + 6*2*3*5 + 3*2*42 + 6*2*4*5+ 3*2*52+ 33+ 3*4*32 + 3*32*5+ 3*42*3 + 6*3*4*5 +3*3*52 + 43+ 3*42*5 + 3*4*52 + 53 9 1 + 6 + 9 + 12 + 15+ 12 + 36 + 48 + 60 + 27 + 72 +90 + 48+ 120 + 75 + 8 + 36 + 48 + 60 + 54 + 144 + 180 + 96 + 240+ 150+ 27+ 108 + 135+ 144 + 360 +225 + 64+ 240 + 300+ 125 1 *1.000.000.000.000 0 6 *100.000.000.000 0 9 *10.000.000.000 1 2 *1.000.000.000 1 5 *100.000.000 1 2 *10.000.000.000 3 6 *1.000.000.000 4 8 *100.000.000 6 0 *10.000.000 2 7 *100.000.000 7 2 *10.000.000 9 0 *1.000.000 4 8 *1.000.000 1 2 0 100.000 7 5 10.000 0 8 *1.000.000.000 3 6 *100.000.000 4 8 *10.000.000 6 0 *1.000.000 5 4 10.000.000 1 4 4 1.000.000 1 8 0 100.000 9 6 100.000 2 4 0 10.000 1 5 0 1.000 2 7 *1.000.000 1 0 8 *100.000 1 3 5 *10.000 1 4 4 *10.000 3 6 0 *1.000 2 2 5 *100 6 4 *1.000 2 4 0 *100 + 3 0 0 *10 1 2 5 *1 1 8 8 1 3 6 5 9 6 3 6 2 5 Result 10 The figure is a pattern that will be formed with all the numbers with a maximum of 5 digits. To add we use this model, ordering the numbers from left to right. This pattern contains the patterns of examples 1, 2 and 3 within itself. The red numbers are the values that were cube in the formula. These are ordered multiplying each other by 1000. Example: 1; 1000; 1.000.000; 1.000.000.000; 1.000.000.000.000 The multinomial theorem The multinomial theorem, which gives us a simple formula for any coefficient we might want. It is possible to "read off" the multinomial coefficients from the terms by using the multinomial coefficient formula. 푛 (푥1 + 푥2 + 푥3 + ⋯ 푥푚 ) 풏! 풂풌ퟏ풃풌ퟐ풄풌ퟑ … . &풌풎 = 풌ퟏ! 풌ퟐ! 풌ퟑ! … … 푲풎! 푬풙풂풎풑풍풆: 풂 + 풃 + 풄 ퟑ = a3 +3a2b + 3a2c + 3ab2+ 6abc + 3ac2+ b3+ 3b2c+3bc2+ c3 푛! 3! 6 푎2푏1푐0 = = = = ퟑ 푘1! 푘2! 푘3! 2! 1! 0! 2 The coefficient for this term will be number 3.