# Roots and the IVT

• Jun 17th 2013, 05:50 PM
phys251
Roots and the IVT
Here is the question:

a_0 + a_1/2 + ... + a_n/(n+1) = 0, all a's are real.
Show that at least one real root betw. 0 and 1 exists for a_0 + a_1*x + ... + a_n*x^n = 0.

I decided to go at this with induction. The n=1 step is easy; just solve a 2X2 system of equations. But I am stuck on the inductive step. I asked a classmate, and he suggested integrating the polynomial and then using the intermediate value theorem. But integrating the polynomial with respect to x doesn't produce anything that seems relevant or usable. What should I do?
• Jun 18th 2013, 01:07 AM
BobP
Re: Roots and the IVT
The integral of the polynomial between 0 and 1 is zero.
That means that either the polynomial is identically zero or the area lying above the x axis is equal to the area below the x axis.
Which means that .... .
• Jun 18th 2013, 11:34 AM
Drexel28
Re: Roots and the IVT
Quote:

Originally Posted by phys251
Here is the question:

a_0 + a_1/2 + ... + a_n/(n+1) = 0, all a's are real.
Show that at least one real root betw. 0 and 1 exists for a_0 + a_1*x + ... + a_n*x^n = 0.

I decided to go at this with induction. The n=1 step is easy; just solve a 2X2 system of equations. But I am stuck on the inductive step. I asked a classmate, and he suggested integrating the polynomial and then using the intermediate value theorem. But integrating the polynomial with respect to x doesn't produce anything that seems relevant or usable. What should I do?

Hint: Let $\displaystyle \displaystyle F(x)=a_0x+\frac{a_1}{2}x^2+\cdots+\frac{a_n}{n+1}x ^{n+1}$. Note that $\displaystyle F(0)=F(1)=0$. What does Rolle's theorem tell you?

Quote:

Originally Posted by BobP
The integral of the polynomial between 0 and 1 is zero.
That means that either the polynomial is identically zero or the area lying above the x axis is equal to the area below the x axis.
Which means that .... .

Unfortunately this does not actually work. Consider, for example, the fact that $\displaystyle \displaystyle \int_{-1}^{1}x^3=0$ yet $\displaystyle x^3$ is not simultaneously zero on that interval (you can use this to cook up an example for $\displaystyle [0,1]$. If you know that $\displaystyle p(x)$ was nonnegative on $\displaystyle [0,1]$ this would work (since $\displaystyle p(x)$ would need to be zero a.e. and thus zero everywhere since it's continuous).

Best,
Alex
• Jun 18th 2013, 03:21 PM
BobP
Re: Roots and the IVT
Hi Drexel28

Could you expand on your objection to the statement I made in my earlier post ?

The example you give, $\displaystyle \int^{1}_{-1}x^{3}dx=0,$ doesn't contradict my statement.

My statement would say that the integral is equal to zero $\displaystyle \bold{either}$ because the integrand is identically zero, (which clearly it isn't),$\displaystyle \bold{or}$ because the part of the area lying above the x-axis is equal to the part of the area lying below the x-axis, (which is the case here).

The implication of that is that $\displaystyle f(x)=x^{3}$ is zero at at least one point in the interval, (in order that the curve crosses the x-axis).
• Jun 18th 2013, 08:28 PM
Drexel28
Re: Roots and the IVT
Quote:

Originally Posted by BobP
Hi Drexel28

Could you expand on your objection to the statement I made in my earlier post ?

The example you give, $\displaystyle \int^{1}_{-1}x^{3}dx=0,$ doesn't contradict my statement.

My statement would say that the integral is equal to zero $\displaystyle \bold{either}$ because the integrand is identically zero, (which clearly it isn't),$\displaystyle \bold{or}$ because the part of the area lying above the x-axis is equal to the part of the area lying below the x-axis, (which is the case here).

The implication of that is that $\displaystyle f(x)=x^{3}$ is zero at at least one point in the interval, (in order that the curve crosses the x-axis).

BobP,

I apologize for my previous statement. I too hastily read your argument. Everything seems correct with that--very clever!

Best,
Alex
• Jun 21st 2013, 04:10 PM
phys251
Re: Roots and the IVT
Quote:

Originally Posted by BobP
The integral of the polynomial between 0 and 1 is zero.
That means that either the polynomial is identically zero or the area lying above the x axis is equal to the area below the x axis.
Which means that .... .

Ohhhh, DEFINITE integration. Yep, that did the trick. Thanks.