# Continuous Function Proof

• Oct 13th 2007, 03:15 PM
tbyou87
Continuous Function Proof
Hi,
Suppose we have a polynomail of degree n:
p(x) = a0 + a1x + a2x^2 + ... + anx^n
where n is an odd number and "an" (the constant) not equal 0. Show that this polynomail has a real root. You can assume that polynomails are continuous.

So it is pretty intuitive that p(x) as x-> inf is positive and as x-> -inf is negative. I just don't know what facts about polynomials to get
a p(a) = negative number and a p(b) = positive number. Then I know you can use the intermediate value theorem.

Thanks
• Oct 13th 2007, 03:19 PM
Jhevon
Quote:

Originally Posted by tbyou87
Hi,
Suppose we have a polynomail of degree n:
p(x) = a0 + a1x + a2x^2 + ... + anx^n
where n is an odd number and "an" (the constant) not equal 0. Show that this polynomail has a real root. You can assume that polynomails are continuous.

So it is pretty intuitive that p(x) as x-> inf is positive and as x-> -inf is negative. I just don't know what facts about polynomials to get
a p(a) = negative number and a p(b) = positive number. Then I know you can use the intermediate value theorem.

Thanks

first you can assume (without loss of generality) that $\displaystyle a_n >0$. then take the limit as x goes to infinity (this will make the polynomial go to infinity as well), then take the limit as x goes to negative infinity (this causes the polynomial to go to negative infinity as well, since it is an odd polynomial). now you can apply the intermediate value theorem, since clearly $\displaystyle - \infty \le 0 \le \infty$.
• Oct 13th 2007, 04:16 PM
ThePerfectHacker
Say $\displaystyle a_n>0$ without lose of generality. Define the following sequence $\displaystyle x_k = a_n k^n + ... + a_1 k + a_0$. The claim is that $\displaystyle \lim \ x_k = +\infty$. To see that you can write $\displaystyle x_k = k^n \left( a_n + ... + \frac{a_1}{k^{n-1}}+\frac{a_0}{k^n} \right)$ the first term $\displaystyle k^n \to \infty$ but the second term $\displaystyle \to a_n$ since $\displaystyle a_n>0$ it means this limit goes to $\displaystyle +\infty$. Thus there is a $\displaystyle k$ so that $\displaystyle x_n > 1$. Similarly define $\displaystyle y_k = a_n (-k)^n+...+a_1 (-k)+a_0$. And show $\displaystyle \lim y_n = -\infty$ thus there is a $\displaystyle k$ so that $\displaystyle y_k < -1$. Meaning we can choose an interval large enough so that the polynomial changes from -1 to 1. Which completes the proof.