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

**Dark Sun** · November 6th 2012, 03:41 PM

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 $x^n=\sum _{i=0}^{n-1}a_ix^i$ has at least one root in the interval (0,1)?

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

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

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

**chiro** · November 6th 2012, 09:59 PM

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.

**Dark Sun** · November 7th 2012, 05:39 PM

Thanks for your reply Chiro,

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.

**Deveno** · November 7th 2012, 06:51 PM

consider the polynomial:

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

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

if $a_0 < 0$ and $\sum_{i = 0}^{n-1} a_i > 1$ , 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).

this is answer III.

**Dark Sun** · November 17th 2012, 01:49 PM

Thanks so much, this is an excellent answer. Thank you ^_^

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

LinkBack

Thread Tools

Display

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

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

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

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

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

Similar Math Help Forum Discussions

roots of i, polynomial

Sum of Roots of a polynomial

Find the characteristic polynomial of an arbitrary-sized matrix

polynomial roots

Roots of a polynomial.

Search Tags