Polynomial roots & Mathematical induction

AI Thread Summary
The discussion revolves around proving the relationship S(n) = -p(S(n-2)) - q(S(n-3)) for n > 3, where S(n) represents the sum of the nth powers of the roots of the polynomial x^3 + px + q = 0. The initial attempts included calculating specific values for S(1), S(2), and S(3) and using the polynomial's rearrangement to derive the recursive formula. Participants noted that the proof for n > 3 does not require validation for n ≤ 3, although caution was advised for cases where p and q are both zero. The conversation emphasizes the importance of understanding the derivation of the recursion rather than just calculating initial values. Overall, the thread provides insights into polynomial roots and the application of mathematical induction.
ben9703
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hi i have this homework question and I am not sure if my thought process is valid.

The Question:

let a, b and c be roots of the polynomial equation: x^3+px+q=0 and S(n)=(a^n)+(b^n)+(c^n)

now prove: that for S(n)= -p(S(n-2))-q(S(n-3)) for n>3my attempt:
-------------

first off i just played around with the equation. S(1)=0 , S(2)=-2p , S(3)=-3q after this all following values can be found by rearranging the polynomial to... x^3=-px-q and multiplying through x for each successive value of n. from this its clear how the above function ( S(n)= -p(S(n-2))-q(S(n-3)) ) arises. what I am not sure about is the pure mathematical proof to proving this (n>3).

if anybody could shine some light on the question or just share ideas i would be very grateful. Thanks :)
 
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ben9703 said:
hi i have this homework question and I am not sure if my thought process is valid.

The Question:

let a, b and c be roots of the polynomial equation: x^3+px+q=0 and S(n)=(a^n)+(b^n)+(c^n)

now prove: that for S(n)= -p(S(n-2))-q(S(n-3)) for n>3


my attempt:
-------------

first off i just played around with the equation. S(1)=0 , S(2)=-2p , S(3)=-3q after this all following values can be found by rearranging the polynomial to... x^3=-px-q and multiplying through x for each successive value of n. from this its clear how the above function ( S(n)= -p(S(n-2))-q(S(n-3)) ) arises. what I am not sure about is the pure mathematical proof to proving this (n>3).

if anybody could shine some light on the question or just share ideas i would be very grateful. Thanks :)

I think you are already done. ##x^n=-px^{n-2}-qx^{n-3}##, by multiplying your relation through by powers of x. So summing over a,b,c gives you S(n)=-p(S(n-2))-q(S(n-3)). I'm not even sure you needed to find those initial values, unless you actually want to calculate the values of S(n) using the recursion.
 
Yeah that's what I was thinking but I'm still not sure how to prove the domain n>3 . thanks for the reply
 
ben9703 said:
Yeah that's what I was thinking but I'm still not sure how to prove the domain n>3 . thanks for the reply

I don't think you need to worry about that. The problem just says to show it works if n>3. It doesn't say you have to show that it doesn't work for n<=3. Though if p=q=0 you might have a problem with that case.
 
Ah yeah I think your right. Thanks for answering when no one else did :)
 
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