Polynomial roots & Mathematical induction

In summary, the conversation involves a discussion about proving the equation S(n)= -p(S(n-2))-q(S(n-3)) for n>3, given that a, b, and c are roots of a polynomial equation. The person asking for help has played around with the equation and found initial values, but is unsure about how to prove the domain n>3. Another person points out that the proof may not be necessary and suggests that the person only needs to show that the equation works for n>3. The conversation ends with the person thanking the other for answering their question.
  • #1
ben9703
3
0
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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  • #2
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.
 
  • #3
Yeah that's what I was thinking but I'm still not sure how to prove the domain n>3 . thanks for the reply
 
  • #4
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.
 
  • #5
Ah yeah I think your right. Thanks for answering when no one else did :)
 

1. What are polynomial roots?

Polynomial roots are the values of the variable that make the polynomial equation equal to zero. These values are also known as solutions or zeros of the polynomial.

2. How do you find polynomial roots?

There are various methods to find polynomial roots, such as the rational root theorem, synthetic division, factoring, and using the quadratic formula for higher degree polynomials. The most efficient method depends on the type and complexity of the polynomial.

3. What is mathematical induction?

Mathematical induction is a proof technique used to show that a statement is true for all natural numbers. It involves proving a base case and then showing that if the statement is true for a given number, it is also true for the next consecutive number.

4. How is mathematical induction used?

Mathematical induction is used to prove statements about natural numbers, such as formulas, equations, and inequalities. It is also commonly used in computer science to prove the correctness of algorithms and programs.

5. Can mathematical induction be used to prove any statement?

No, mathematical induction can only be used to prove statements about natural numbers. It cannot be used for statements involving real or complex numbers, as they do not follow the same pattern as natural numbers.

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