# Roots of an arbitrary polynomial: x^n=a0+a1x+a2x^2+...+a(n-1)x^(n-1)

• Nov 6th 2012, 03:41 PM
Dark Sun
Roots of an arbitrary polynomial: x^n=a0+a1x+a2x^2+...+a(n-1)x^(n-1)
I could use some input on this one: Let n be an integer greater than 1.

Which of the following conditions guarantee that the equation $\displaystyle x^n=\sum _{i=0}^{n-1}a_ix^i$ has at least one root in the interval (0,1)?

I. $\displaystyle a_0>0\ \& \sum _{i=0}^{n-1}a_i<1$

II. $\displaystyle a_0>0\ \& \sum _{i=0}^{n-1}a_i>1$

III. $\displaystyle a_0<0\ \& \sum _{i=0}^{n-1}a_i>1$
• Nov 6th 2012, 09:59 PM
chiro
Re: Roots of an arbitrary polynomial: x^n=a0+a1x+a2x^2+...+a(n-1)x^(n-1)
Hey Dark Sun.

In the [0,1] range x^n =< x^m if m < n and only equal if m=n or x = 0 or 1. This rules out II.

For III., the condition is definitely satisfied as the RHS will overtake LHS and result in a root (ie LHS = RHS).

For I. I don't know whether you can say for sure and you would have to look at the specifics.

If you want to look at proofs consider the result above with the m's and n's.
• Nov 7th 2012, 05:39 PM
Dark Sun
Re: Roots of an arbitrary polynomial: x^n=a0+a1x+a2x^2+...+a(n-1)x^(n-1)

I totally see what you said about II and III. I am looking at various graphs of I now, and they seem to indicate that I. is also valid. I still want to develop a little more for I. before I move on. Will report back if I find anything.
• Nov 7th 2012, 06:51 PM
Deveno
Re: Roots of an arbitrary polynomial: x^n=a0+a1x+a2x^2+...+a(n-1)x^(n-1)
consider the polynomial:

$\displaystyle p(x) = x^n - \sum_{i = 0}^{n-1} a_ix^i$.

note that: $\displaystyle p(0) = -a_0$ and $\displaystyle p(1) = 1 - \sum_{i = 0}^{n-1}a_i$.

if $\displaystyle a_0 < 0$ and $\displaystyle \sum_{i = 0}^{n-1} a_i > 1$, we see that p(0) is positive, and p(1) is negative, and by the continuity of polynomials, p must cross the x-axis somewhere on (0,1).