2Thanks
• 1 Post By chiro
• 1 Post By Deveno

# Thread:

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

2. ## 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.

3. ## 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.

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

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.

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

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