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# real analysis - show that all polynomials of odd degree have at least one real root?

show that all polynomials of odd degree have atleast one real root?

Posted in Math, asked by kavana, 6 years ago. 2771 hits.

## 0

What you want to show is this:

If n is odd, then any equation

x^n + a_(n - 1)x^(n - 1) + ... + a_0 = 0

has a root.

Proof:

We obviously want to consider the function

f(x) = x^n + a_(n - 1)x^(n -1) + ... a_0;

we would like to prove that f is sometimes positive and sometimes negative. The intuitive idea is that for large |x|, the function is very much like g(x) = x^n and, since n is odd, this function is positive for large positive x and negative for large negative x. A little algebra is all we need to make this intuitive idea work.

The proper analysis of the function f depends on writing

f(x) = x^n + a_(n - 1)x^(n -1) + ... a_0 = x^n(1 + a_(n-1)/x + ... + a_0/x^n).

Note that

|a_(n - 1)/x + a_(n -2)/x^2 + ... + a_0/x^n| ≤ |a_(n-1)|/|x| + ... + |a_0|/|x^n|

Consequently, if we choose x satisfying

(*) |x| > 1, 2n|a_(n-1)|, ..., 2n|a_0|,

then |x^k| > |x| and

|a_(n - k)|/|x^k| < |a_(n - k)|/|x| < |a_(n-k)|/2n|a_(n-k)| = 1/2n,

so

|a_(n-1)/x + a_(n-2)/x^2 + ... + a_0/x^n| ≤ 1/2n + ... + 1/2n = 1/2

In other words,

-1/2 ≤ a_(n-1)/x + ... + a_0/x^n ≤ 1/2,

which implies that

1/2 ≤ 1 + a_(n-1)/x + ... + a_0/x^n.

Therefore, if we choose an x_1 > 0 which satisfies (*), then

[(x_1)^n]/2 ≤ (x_1)^n(1 + a_(n-1)/x_1 + ... + a_0/(x_1)^n) = f(x_1)

so that f(x_1) > 0. On the other hand, if x_2 < 0 satisfies (*), then (x_2)^n < 0 and

[(x_2)^n]/2 ≥ (x_2)^n(1 + a_(n-1)/x_2 + ... + a_0/(x_2)^n) = f(x_2),

so that f(x_2) < 0.

Now applying the Intermediate Value theorem to the interval [x_1, x_2] we conclude that there is an x in [x_1, x_2] such that f(x) = 0.

Gurudas Kadam - 6 years ago
- Just now

## 0

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