For a cubic polynomial a3x3+a2x2+a1x+a0, the roots r1, r2, and r3 come in three symmetric combinations: σ1=r1+r2+r3, σ2=r1r2+r1r3+r2r3, and σ3=r1r2r3.

The sum of the kth powers of the roots is defined as Sk=r^k1+r^k2+r^k3.

a) Prove that a3*S1+a2=0.
b) Prove that a3*S2+a2*s1+2*a1=0.
c) Generalize to something that starts with a3^Sk. Generalize further to polynomials of degree n.

Question

For a cubic polynomial a3x3+a2x2+a1x+a0, the roots r1, r2, and r3 come in three symmetric combinations: σ1=r1+r2+r3, σ2=r1r2+r1r3+r2r3, and σ3=r1r2r3.

The sum of the kth powers of the roots is defined as Sk=r^k1+r^k2+r^k3.

a) Prove that a3*S1+a2=0.
b) Prove that a3*S2+a2*s1+2*a1=0.
c) Generalize to something that starts with a3^Sk. Generalize further to polynomials of degree n.

caozeyuan · Answer

@experimentX @whpalmer4 help me please！

caozeyuan · Answer

@mukushla  look at this please

experimentx · Answer

$$ a_3x^3+a_2x^2+a_1x+a_0 = 0  \implies x = r_1, r_2, r_3$$

experimentx · Answer

what are k's?

caozeyuan · Answer

any real number can be k

caozeyuan · Answer

usually integers to make life easier, though

experimentx · Answer

no ,, there are k1,k2,k3 <-- they are not defined

caozeyuan · Answer

I think it's related with Newton's Sum in some way

caozeyuan · Answer

it's actually r1^k+r2^k+r3^k. I made a typo earlier

experimentx · Answer

nvm  got that.

experimentx · Answer

Do you know Vieta's fomula?

experimentx · Answer

$$ a_3x^3+a_2x^2+a_1x+a_0 = a_3 (x - r_1)(x-r_2)(x-r_3)  $$
compare the coefficients of x ... it should work out smoothly.

caozeyuan · Answer

great， thx！ but how about generalize it to  polynomials with degree n?

experimentx · Answer

as you stated this is called newton's sum generalization of proof is here http://digitalcommons.unl.edu/cgi/viewcontent.cgi?article=1049&context=mathfacpub

experimentx · Answer

interesting proof ... cleverly manipulated

caozeyuan · Answer

too complicated to  understand.......

experimentx · Answer

which part you don't understand?

caozeyuan · Answer

All of them! I'm just a high school stu, not a univ mathematician！

experimentx · Answer

do you know basic stuff like differentiation? fundamental theorem of calc?

caozeyuan · Answer

Yeap, I've done AP cal BC,so I know all the stuff about it

experimentx · Answer

Let 
$$ x^n + a_n x^{n-1} + .. +a_1$$
be your polynomial. and $ r_1, r_2, ... r_n$ be it's zeros.
from fundamental theorem of calc you have
$$ x^n + a_n x^{n-1} + .. +a_1 = (x-r_1) (x-r_2) ... (x-r_n)$$

caozeyuan · Answer

is that fund. theor. of cal or of alg?

experimentx · Answer

typo ..algebra sorry.

experimentx · Answer

differentiate it one ... on LHS, one term is missing.
$$ Q(x) = (x-r_1) (x-r_2) ... (x-r_n)$$

experimentx · Answer

Q'(x) = (x-r2)...(x-rn)+(x-r1)(x-r3) ... (x-rn)+ ....+(x-r1)...(x-r_n-1)
we are using product rule.

experimentx · Answer

divide Q'(x)/Q(x) and see what you get

caozeyuan · Answer

my head just exploded!