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Hello.
I open this 'thread', in number theory, but he also wears "calculation".
I've done a little research, I share with you.
[tex]Let \ r_1, r_2, \cdots, r_n[/tex], roots of the polynomial.
[tex]P(x)=p_0x^n+p_1x^{n-1}+ \cdots+p_{n-1}x+p_n[/tex]
[tex]Let \ Q(x)=q_0 x^n+q_1x^{n-1}+ \cdots +q_n[/tex], such that its roots are:
[tex]r_1-1, r_2-1, \cdots, r_n-1[/tex]
[tex]Let \ T(x)=t_0^n+t_1x^{n-1}+ \cdots +t_n[/tex], such that its roots are:
[tex]r_1+1, r_2+1, \cdots, r_n+1[/tex]
I will assume:
[tex]p_0=q_0=t_0=1[/tex]Therefore:
[tex]\displaystyle\sum_{i=0}^n(p_i)=q_n[/tex]
and
[tex]\displaystyle\sum_{i=0}^n(p_i)(-1)^{i+1}=t_n[/tex], if “n” it's even.
[tex]\displaystyle\sum_{i=0}^n(p_i)(-1)^i=t_n[/tex], if “n” it's odd.
Also I have found how to calculate "complete", recurrently cited polynomials.
Example:
[tex]P(x)=x^9-33x^8+149x^7+4431x^6-45669x^5+9081x^4+1506119x^3-7038363x^2+12556936x-7987980[/tex]
Sum of coefficients=-995328, addition and subtraction of alternate coefficients=-29030400
Roots:-11, -7, 2, 2, 3, 5, 7, 13, 19.[tex]Q(x)=x^9-24x^8-79x^7+4634x^6-17676x^5-149768x^4+1177824x^3-2853504x^2+2833920x-995328[/tex]
Sum of coefficients=0, addition and subtraction of alternate coefficients=-7987980
Roots:-12, -8, 1, 1, 2, 4, 6, 12, 18.[tex]T(x)=x^9-42x^8+449x^7+2380x^6-67152x^5+296240x^4+931632x^3-10983168x^2+30862080x-29030400[/tex]
Sum of coefficients=-7987980, addition and subtraction of alternate coefficients=-71442000
Roots:-10, -6, 3, 3, 4, 6, 8, 14, 20.
This procedure can be useful for the analysis of the possible roots of the polynomial, and its factorization.
Regards.
I open this 'thread', in number theory, but he also wears "calculation".
I've done a little research, I share with you.
[tex]Let \ r_1, r_2, \cdots, r_n[/tex], roots of the polynomial.
[tex]P(x)=p_0x^n+p_1x^{n-1}+ \cdots+p_{n-1}x+p_n[/tex]
[tex]Let \ Q(x)=q_0 x^n+q_1x^{n-1}+ \cdots +q_n[/tex], such that its roots are:
[tex]r_1-1, r_2-1, \cdots, r_n-1[/tex]
[tex]Let \ T(x)=t_0^n+t_1x^{n-1}+ \cdots +t_n[/tex], such that its roots are:
[tex]r_1+1, r_2+1, \cdots, r_n+1[/tex]
I will assume:
[tex]p_0=q_0=t_0=1[/tex]Therefore:
[tex]\displaystyle\sum_{i=0}^n(p_i)=q_n[/tex]
and
[tex]\displaystyle\sum_{i=0}^n(p_i)(-1)^{i+1}=t_n[/tex], if “n” it's even.
[tex]\displaystyle\sum_{i=0}^n(p_i)(-1)^i=t_n[/tex], if “n” it's odd.
Also I have found how to calculate "complete", recurrently cited polynomials.
Example:
[tex]P(x)=x^9-33x^8+149x^7+4431x^6-45669x^5+9081x^4+1506119x^3-7038363x^2+12556936x-7987980[/tex]
Sum of coefficients=-995328, addition and subtraction of alternate coefficients=-29030400
Roots:-11, -7, 2, 2, 3, 5, 7, 13, 19.[tex]Q(x)=x^9-24x^8-79x^7+4634x^6-17676x^5-149768x^4+1177824x^3-2853504x^2+2833920x-995328[/tex]
Sum of coefficients=0, addition and subtraction of alternate coefficients=-7987980
Roots:-12, -8, 1, 1, 2, 4, 6, 12, 18.[tex]T(x)=x^9-42x^8+449x^7+2380x^6-67152x^5+296240x^4+931632x^3-10983168x^2+30862080x-29030400[/tex]
Sum of coefficients=-7987980, addition and subtraction of alternate coefficients=-71442000
Roots:-10, -6, 3, 3, 4, 6, 8, 14, 20.
This procedure can be useful for the analysis of the possible roots of the polynomial, and its factorization.
Regards.
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