Can induction prove that P(x) has a unique positive zero?

[ehrenfest]

[SOLVED] what is wrong with this

What is wrong with this.

I want to show this: Let a_1,...,a_n be positive real numbers. Prove that the polynomial
P(x) = x^n-a_1 x^{n-1}-...-a_n has a unique positive zero.

Q(x) = x^n+a_1 x^{n-1} + ...+ a_n has n complex nonzero zeroes. For each of them, we have that

0 = |x^n+a_1 x^{n-1} + ...+ a_n | \geq ||x|^n-a_1 |x|^{n-1} - ...- a_n |

which implies that |x| is a zero y^n-a_1 y^{n-1} - ...- a_n. But that implies that there could be more than one unless it is somehow true that all of the zeros of Q(x) lie on a circle in the complex plane!

 

[ehrenfest]

never mind, I figured it out

 

[uart]

Hi ehrenfest. I wonder if you thought of using induction on this one?

Note that if the proposition is true for n-1 then the derivative P'(x) also has one unique positive zero which in turn implies the desired property for P(n). (as it's trivial to show that for P(x) to have more than one positive zero that it must also have more than one +ive turning point).

So it's dead easy to show that the proposition for n-1 implies the proposition for n.