Recent content by Jack Jenkins

Hi everyone, sorry for the long delay I had the flu! All better now. I actually managed to get a directed study at my college approved that looks into this problem. What I'm concentrating on are the fact that certain numbers, such as 1, sqrt(2), sqrt(3), and (1+sqrt(5))/2 (the golden ratio)...

There is a function \pi(x) that takes the natural numbers as its input and outputs the number of primes less than or equal to x. There are some nice properties of this function. For instance, the growth rate as x approaches infinity approaches x/lnx

Oops indeed! Math and the flu don't mix well. Once I'm feeling better I'll run through some more possibilities using the "family" of equalities above; I bet that there are other ways to represent P(n) as some easily tractable formula of "smaller" P(n)'s. Unfortunately, it seems that we're left...

to extend further, \begin{align*} P(n) &= xP(n-1) - P(n-2) \\ &= x\left( xP(n-2) - P(n-3) \right) - P(n-2) \\ &= (x^2-1)P(n-2) - xP(n-3) &= (x^3-x-1)P(n-3)-(x^2-1)P(n-2) \end{align*} and we could continue to extend this procedure to obtain as many repeated roots as we like.

Thank you all for your constructive comments! @coolul007, you're right, there do appear to be roots that occur over and over again, not only 0 and ±1 for n odd or even respectively, but for the square roots of certain integer numbers. For example, P(n), n=4,8,12,16 is zero at x=±\sqrt{2}...

Hello all, I have a series of polynomials P(n), given by the recursive formula P(n)=xP(n-1)-P(n-2) with initial values P(0)=1 and P(1)=x. P(2)=xx-1=x2-1 P(3)=x(x2-1)-(x)=x3-2x ... I am confident that all of the roots of P(n) lie on the real line. So for P(n), I hope to find these roots. I...