prove that every root of the polynomial $$\sum_{k=0}^{n}2^{k(n-k)}x ^{k}$$ is real. If you solve this do not give me the solution, just give me a very very small hint. I don't want to waste a good problem.

Question

anonymous · Answer

This is what I have got so far: I tried a lot of different things, but the thing that I realized is that this sum is recursive(took me a few hours to realize that), so I defined my $$f(a) = 1 + a*x$$
then I defined a series of numbers as $$g_{n}$$ such that $$g _{0} = f(0) = 1$$$$g _{1} = f(1) = 1+ x$$$$g _{2} = f(f(0) + f(1)) = 1+ 2x + x ^{2}$$$$g _{3} = f(3f(0) + f(3f(0) + f(1))) = 1+ 4x + 4x ^{2} + x^3$$$$g _{4} = f(7 + f(15 + f(7 + f(1)))$$$$g_{5} = f(15 + f(63 + f(63 + f(15 + f(1)))))$$ and so on, I think I have got a pattern, it is easy to see as in the last one the number 15 defines the coefficient of the x^4 term. It is easy to write this form, but I am stuck here.

anonymous · Answer

$$\sum_{k=0}^{n}2^{k(n-k)}x^k$$ this is polynomial

anonymous · Answer

this is the polynomial*

empty · Answer

$z=z^*$

mathmate · Answer

Use Descartes rule of signs to rule out any positive root (real or not). Sturm's theorem can help you find the number of negative real roots. See, for example, section 2 of: http://web.cs.iastate.edu/~cs577/handouts/polyroots.pdf