Polynomials Question. (1 Viewer)

[seanieg89]

If x,y,z are complex numbers such that: , prove that:

 

[Carrotsticks]

Let x,y,z be the roots of a cubic polynomial:



We define:



It's not *too* difficult to show that S_n satisfies the following recursive relationship for n > 3. I can put the proof in if you like.



Where the base cases are:



So we want to find S_11, and we will do so using the recursive relationship we have:



And so working recursively and with a bit of algebra, we acquire:



Which is the desired result.

 

[Carrotsticks]

I just realised, we don't really need to prove the recursive relationship (do we?). We can see by inspection that its Characteristic Polynomial is just P(x) !

 

[deswa1]

How long do you reckon it would take to multiply out the whole of the RHS and simplify?

 

[Carrotsticks]

How long do you reckon it would take to multiply out the whole of the RHS and simplify?

I can't imagine it being too long if you can see what to do...

x + y + z = 0

x = -(y+z)

Raise both sides to the power of 11:

x^11 = -(y+z)^11

Expand using Binomial expansions but GROUP the first term with the last term, second with second last etc etc and use symmetry of Pascals triangle (or the C(n,k) = C(n,n-k) argument, same thing) to group terms. You will get a pretty nice expression.

Since we have 11 in the factorial (in the numerator), we can factorise it out and combine with the LHS once we've moved the y^11 and z^11 terms over, yielding the LHS of the required expression.

So our job now is to prove that what we have left (in the RHS) is the same as seanieg's RHS.

 

[seanieg89]

I just realised, we don't really need to prove the recursive relationship (do we?). We can see by inspection that its Characteristic Polynomial is just P(x) !

Nice way of looking at it! Of course the direct proof of the recursive relation is just summing the equation (x^k)P(x)=0 over the roots of P, where k is a natural number. Would rep but need to share it apparently...