Question

How do I use the intermediate value theorem to prove every polynomial of odd degree has at least one real root?

Answer 1

Given any polynomial `f(x)` of odd degree and positive leading coefficient find `x_1` such that `f(-x_1) < 0` and `f(x_1) > 0`, so `EE x in (-x_1, x_1)` with `f(x) = 0`.

Explanation:

Let `f(x) = a_0x^n+a_1x^(n-1)+...+a_n` with `a_0 != 0`

Note that `f(x)` is a continuous function.

If `x` is sufficiently large and positive, `f(x) > 0`.

To prove that:

Let `x_1 = (1+abs(a_0)+abs(a_1)+abs(a_2)+...+abs(a_n))/abs(a_0)`

Note that `x_1 > 1`, so if `m <= n-1` then `x^m <= x^(n-1)`

So we find:

`f(x_1) >= a_0x_1^n-(abs(a_1)x^(n-1)+abs(a_2)x^(n-2)+...+abs(a_n))`

`>= a_0x_1^n-(abs(a_1)x^(n-1)+abs(a_2)x^(n-1)+...+abs(a_n)x^(x-1))`

`= x^(n-1)(a_0x-(abs(a_1)+abs(a_2)+...+abs(a_n)))`

`= x^(n-1)((1+abs(a_0)+abs(a_1)+abs(a_2)+...+abs(a_n))-(abs(a_1)+abs(a_2)+...+abs(a_n)))`

`= x^(n-1)(1+abs(a_0)) > 0`

If `n` is odd and `x` sufficiently large and negative then `f(x) < 0`.

To prove that, note that if `n` is odd then `a_0(-x)^n = -a_0x^n`, so `-f(-x) = a_0x^n-a_1x^(n-1)+...-a_n`, hence `f(-x_1) < 0`

We have:

• `f(x)` continuous over `[-x_1, x_1]`
• `f(-x_1) < 0 < f(x_1)`

So by the intermediate value theorem:

`EE x in (-x_1, x_1) : f(x) = 0`

If the leading coefficient (`a_0`) of `f(x)` is negative then the same formula for `x_1` yields `f(x_1) < 0 < f(-x_1)`, hence there is a root in `(-x_1, x_1)` again.