User talk:Yahia Kahloune

    How to solve equations of degree "n" is an unknown with sequences "O.E.I.S".

Let F(x)=(x-a)*(x-b)*(x-c).......*(x-p). If "n" is the degree to te equation, we get after development : F(x) = X^n - S1*X^(n-1) + S2*X^(n-2)......+(or)-..........Sp. We have : S1 = a+b+c+..........+p. there are :c(n,1) monomials. S2 =a*b +b*c +c*a+.....a*p there are: c(n,2) monomials. S3 = a*b*c+ b*c*d +.....abp there are: c(n,3) monomials. .............................. Sp =a*b*c.....*p there are: c(p,p)=1 monomial If: a, b, c, .....p are the root of F(x) , expansion of 1/((1-ax)*(1-bx)*(1-cx)......*(1-px) is calculated as follows : a(o)=1 a(1)=S1 a(2)= (S1)^2-S2 a(3)=(S1)^3-2*(S2)*(S1)+(S3) a(4)=(S1)^4-3*(S2)*(S1)^2+2*(S3)*(S1)- (S4)+ (S2)^2. a(5)=(S1)^5-4*(S2)*(S1)^3+3*(S3)*(S1)^2-2*(S4)*(S1)+3*((S2)^2)*(S1)+(S5)- 2*(S3)*(S2).

   Example :1

F(x)= X^4-25*X^3+217*X^2-783*X+990 F(x)=X^4-(S1)*X^3+(S2)*X^2-(S3)*X+ (S4). we have (S1)=25 (S2)=217 (S3)=783 (S4)=990 To calculate the expansion of F(x) we have a(o)=1 a(1)=(S1)=25 a(2)=(S1)^2-(S2)=25^2-217=408 a(3)=(S1)^3-2*(S2)*(s1)+(S3)=25^3-2*217*25+783=5558 a(4)=(S1)^4-3*(S2)*(S1)^2+2*(S3)*(S1)-(S4)+(S2)^2=25^4-3*217*(25^2)+2*783*25-990+(217^2)=68999. By introducing the following :1, 25, 408, 5558, 68999, in the "O.E.I.S" we get sequence A028057, which is the expansion 0f 1/((1-3x)(1-5x)(1-6x)(1-11x). We can conclude that F(X)=X^4-25*X^3+217*X^2-783*X+990=(X-3)(X-5)(X-6)((X-11). The roots of the equation F(x) are 3,5,6,11. Obviously this is only valid for existing sequences "O.E.I.S"

  Example: 2 

F(x)=X^5 - 26*X^4 + 259*X^3 -1226*X^2 + 2728*X - 2240 We have (S1)=26; (S2)=259; (S3)=1226; (S4)=2728; (S5)=2240. To calculate the expansion of F(x) we have: a(o)=1; a(1)=26; a(2)=417; a(3)=5334; a(4)=59824. By introducing the following: 1; 26; 417; 5334; 59829; in the"O.E.I.S" we get sequence A229025 which the expansion of 1/((1-2x)(1-4x)(1-5x)(1-7x)(1-8x)). We can conclude that F(x)=(x-2)(x-4)(x-5)(x-7)(x-8). The roots of the equation F(x) are:2; 4; 5; 7; 8.

 Example:3

F(x)=x^5 - 8*x^4 + 22*x^3 - 28*x^2 + 17*x - 4 we have the expansion of F(x) : if (S1)=8; (S2)=22; (S3)=28; (S4)=17; (S5)=4; a(0)=1 a(1)=(S1)=8 a(2)=(S1)^2-(S2)=42 a(3)=(S1)^3-2*(S2)*(S1)- (S3)=188 a(4)=(S1)^4 - 3*(S2)*(S1)^2 + 2*(S3)*(S1) - (S4) + (S2)^2=788. By introducing the following : 1, 8, 42, 188, 788, in the "O.E.I.S" we get sequence A097788 which is the expansion of 1/((1-4x)(1-x)^4 we can conclude that F(x)= x^5 - 8*x^4 + 22*x^3 -28*x^2 + 17*x - 4= F(x)= (x-4)((x-1)^4. The roots of equation are: 4,1,1,1,1.