Define a polynomial

such that and have opposite signs. Prove there is some such that .

* Proof. * Since

we see that has the same sign as .

Next,

Then we claim that for sufficiently large ,

hence, will have the same sign as for sufficiently large , since and so

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

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

This is true since for each term in the sum

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

Since , we know for all , so

This proves our claim, and so has the same sign as for sufficiently large . Hence, and have different signs (since and had different signs by assumption); thus, there is some such that

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.

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 $n=2k.$

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.

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.

correction, in step 2. lim f(x)=B (for some constant B) as x approaches c from above and below.

“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.