How to find sum of roots of equatons of different degree?
Answers
p(x)=xn+p1xn−1+⋯+pn−1x+p0=(x−r1)(x−r2)⋯(x−rn).
When you multiply out (x−r1)⋯(x−rn), and then put together the powers of x, you get a polynomial expression in which the coefficients are written in terms of the roots. For example, to get the coefficient of x, note that you will get an x when you expand and multiply the x in the first binomial by each of the constants in the rest, giving you (−1)n−1(r2⋯rn)x. Another when you multiply the x in the second binomial by each of the constants in the rest; another when you multiply the x in the third binomial by the constants in the rest; etc.
If you work this out, you'll find that when you expand (x−r1)⋯(x−rn) and then group together the powers of x, you will have:
The coefficient of xn is 1.
The coefficient of xn−1 is −(r1+⋯+rn).
The coefficient of xn−2 is
r1r2+r1r3+⋯+r1rn+r2r3+⋯+rn−1rn.
The coefficient of xn−3 is
−(r1r2r3+r1r2r4+⋯+r1r2rn+r1r3r4+⋯+rn−2rn−1rn).
⋮
The coefficient of x is
(−1)n−1(r1r2⋯rn−1+r1r2⋯rn−2rn+⋯+r2⋯rn).
The constant coefficient is
(−1)n(r1⋯rn).
But for two polynomials to be equal they have to be equal coefficient by coefficient. So that means that:
p1 =(−1)1(r1+⋯+rn) p2 =(−1)2(r1r2+r1r3+⋯+r1rn+r2r3+⋯+rn−1rn) p3 =(−1)3(r1r2r3+r1r2r4+⋯+r1r2rn+r1r3r4+⋯+rn−2rn−1rn) ⋮ pn =(−1)n(r1⋯rn).
That is:
The sum of the roots is −p1;
The sum of all products of two roots is (−1)2p2;
The sum of all products of three roots is (−1)3p3;
The sum of all products of four roots is (−1)4p4;
⋮
The sum of all products of n−1 roots is (−1)n−1pn−1;
The product of all roots is (−1)npn.
Which is what the text you cite says.
How about a polynomial that is not monic,
p(x)=p0xn+p1xn−1+⋯+pn−1x+pn,p0≠0 ?
Note that r is a root of p(x) if and only if it is a root of
P(x)=
1
p0
p(x)=xn+
p1
p0
xn−1+⋯+
pn−1
p0
x+
pn
p0
.
So the argument above applies to P(x).