Complex roots question (1 Viewer)

[khorne]

The question is, given a^5 = 1, a=/=1

Prove sum of a^(rs)*x^r from r = 0 to 4 is equal to (1-x^(5))/(1-a^(s)x)

I am not sure if my method is valid but:

I said, let w = a^(s)x

Thus the LHS becomes 1 + w + w^2 + w^3 + w^4

Which is equal to (w^5 - 1)/(w-1) = (a^(5s)x^5 - 1)/(a^(s)x - 1)
= (x^5 - 1)/(a^(s)x -1) = RHS

 

[Hermes1]

I dont see anything wrong with ur method, just wait for drongoski, he'll be able to give you a more definitive answer

 

[khorne]

What about for this:

Part two of the above Q:

(1+x+x^2 + x^3 +x^4)(1+ax+a^2x^2 +a^3x^3+a^4x^4)(1+a^2x+a^4x^2 +a x^3 +a^3x^4)(1+a^3x+ax^2 + a^4x^3 +a^2x^4)(1+a^4x+a^3x^2 + a^2x^3 +ax^4) = (1-x^5)^4

For this, I subbed: w = x for the first, w = ax for the second, w = a^2x for the third, w = a^3 x and a^4 x for the last two, which gives (x^5 - 1)^4/((x-1)(ax-1)(a^x-1)(a^3x-1)(a^4x-1))

upon expanding the denominator, it cancels down to x^5 -1, giving the required value for the question.

However, is there an easier way? expanding ~26 terms is time consuming

 

[deterministic]

Clearly are the 5 distinct roots of (sub them in to see) so the denominator is just the product of all factors of

 

[khorne]

Thanks for that, totally missed it. I've been having a few sloppy maths days, maybe time to give it a rest.