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Prove a polynomial with opposite signed first and last coefficients has at least one positive zero

by RoRi

Define a polynomial

    \[ f(x) = \sum_{k=0}^n c_k x^k, \]

such that c_0 and c_n have opposite signs. Prove there is some x > 0 such that f(x) = 0.

Proof. Since

    \[ f(0) = \sum_{k=0}^n c_k 0^k = c_0 \]

we see that f(0) has the same sign as c_0.
Next,

    \[ f(x) = c_n x^n + c_{n-1}x^{n-1} + \cdots + c_1 x + c_0 = c_n x^n \left( 1+ \frac{c_{n-1}}{c_n} \cdot \frac{1}{x} + \cdots + \frac{c_0}{c_n} \cdot \frac{1}{x^n} \right). \]

Then we claim that for sufficiently large x,

    \[ 1 + \frac{c_{n-1}}{c_n} \frac{1}{x} + \cdots + \frac{c_0}{c_n} \frac{1}{x^n} > 0; \]

hence, f(x) will have the same sign as c_n for sufficiently large x, since x > 0 and so

    \[ x^n \left( 1 + \frac{c_{n-1}}{c_n} \frac{1}{x} + \cdots + \frac{c_0}{c_n} \frac{1}{x^n} \right) > 0. \]

(So, when we multiply a positive number by c_n the result will have the same sign as c_n.)

So, we need to show that the claimed term is indeed positive. First,

    \[ \left( 1 + \frac{c_{n-1}}{c_n} \cdot \frac{1}{x} + \cdots + \frac{c_0}{c_n} \cdot \frac{1}{x^n} \right) \geq 1 - \left( \left| \frac{c_{n-1}}{c_n} \cdot \frac{1}{x} \right| + \cdots + \left| \frac{c_0}{c_n} \cdot \frac{1}{x^n} \right| \right). \]

This is true since for each term in the sum

    \[ \left| \frac{c_i}{c_n} \cdot \frac{1}{x^{n-i}} \right| \geq \frac{c_i}{c_n} \cdot \frac{1}{x^{n-i}}, \qquad \text{for all } 1 \leq i \leq n-1. \]

Now, since we are showing there is sufficiently large x such that our claim is true, we let

    \[ x > \max \left\{ 1, \left| \frac{c_n}{n \cdot c_a} \right| \right\}, \qquad \text{where} \qquad |c_a| = \max\{ |c_0|, |c_1|, \ldots, |c_n|\}. \]

Since x > 1, we know \frac{1}{x} > \frac{1}{x^n} for all n \in \mathbb{Z}_{>0}, so

    \begin{align*} 1 - \left( \left| \frac{c_{n-1}}{c_n} \frac{1}{x} \right| + \cdots + \left| \frac{c_0}{c_n} \frac{1}{x^n} \right| \right) &> 1 - \left( \left| \frac{c_{n-1}}{c_n} \frac{1}{x} \right| + \cdots + \left| \frac{c_0}{c_n} \frac{1}{x} \right| \right) \\ &= 1 - \left( \left| \frac{c_{n-1}}{c_n} \frac{c_n}{n c_a} \right| + \cdots + \left| \frac{c_0}{c_n} \frac{c_n}{n c_a}\right| \right) \\ &= 1 - \left( \left| \frac{c_{n-1}}{n c_a} \right| + \cdots + \left| \frac{c_0}{nc_a} \right| \right) \\ &> 1 - \left( \left| \frac{c_a}{nc_a} \right| + \cdots + \left| \frac{c_a}{nc_a} \right| \right)\\ &= 1 - \left( \underbrace{\frac{1}{n} + \cdots + \frac{1}{n}}_{n-1\text{ terms}} \right) \\ &= 1 - \left( \frac{n-1}{n} \right) \\ &= \frac{1}{n} > 0. \end{align*}

This proves our claim, and so f(x) has the same sign as c_n for sufficiently large x. Hence, f(0) and f(x) have different signs (since c_0 and c_n had different signs by assumption); thus, there is some c>0 such that f(c) = 0. \qquad \blacksquare

7 comments

  1. Anonymous says:

    Correction:

    You have to prove that, indeed, f(x) > 0 for a LARGE ENOUGH x, since f(x) > 0 doesn’t hold for all x > 0.

  2. Lucian says:

    A related result:

    All polynomials of even degree, with opposite signed first and last coefficients, have at least two real zeroes.

    This follows from writing $P_n(x)=x~P_{n-1}(x)+a_0\iff P_{n-1}(x)=-\dfrac{a_0}x~,~$ and then inspecting the asymptotic behavior of both functions involved for even values of $n=2k.$

  3. pablo says:

    Hello,

    Could you please elaborate on why you make the substitution: 1/x = Cn/NCa when you go from the 6th line from the last line to the 5th line from the last line?

    thanks.

  4. H says:

    another way to prove it, is by Bolzano’s theorem:

    1.we prove that ax^n is continuous for all x>0: (can be proved by epsilon-delta definition or by the limit definition of continuity).[not necessary for the proof, but it helps understanding]

    2.we prove that f(x) is continuous for all x>0 :
    since f(x) is defined at c (for some c>0) and lim f(x)=c as x approaches c (from above and below)
    we conclude that f(x) is continuous for all x>0.

    3. by definition, f at k=0 and f at k=n have opposite signs (since x>0, x^k is always positive ).

    then by Bolzano’s theorem we can find a number k such that f(x)=0.

    • Anonymous says:

      “by definition, f at k=0 and f at k=n have opposite signs”

      False. The fact that x^n > 0 doesn’t imply that f(x) > 0 for all positive x, since the other components of the function (c1)x + (c2)x^2 + … could be negative and be larger (in absolute value) than (cn)x^n, making f(x) < 0.

      You have to prove that, indeed, f(x) for a LARGE ENOUGH x, since there f(x) doesn't hold for all x < 0.

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