Questions about proof of upper and lower bound theorem for polynomials

[RChristenk]

Homework Statement

Proof of upper and lower bound theorem for polynomials

Relevant Equations

None

To prove the upper bound: Let $c>0$, divide it into $f(x)$ and the coefficients in the final line of the synthetic division tableau are all non-negative. Thus $f(x)=(x-c)q(x)+r$, where $r \geq 0$ (since the coefficients are given as all non-negative) and is a constant because it's degree must be less than $(x-c)$.
Now let $b>c>0$, then $f(b)=(b-c)q(b)+r$. $b-c >0$, $q(b)>0$ (since all coefficients of $q(x)$ are non-negative) and $r \geq 0$. Hence $f(b) > 0$, which means it never touches the $x$-axis and therefore proves $c>0$ is a upper bound.

If all the coefficients are non-positive, and $b>c>0$, then in $f(b)=(b-c)q(b)+r$, $b-c>0$, $q(b)<0$ since all coefficients in $q(x)$ are negative, and $r \leq 0$ because it is given all the coefficients are non-positive. Therefore $f(b)=$ negative $+$ negative/zero, or $f(b)<0$. I have a question here: I don't understand why $f(b)<0$ is considered an upper bound? Doesn't this contradict $f(b)>0$ shown previously?

As for proving the lower bound, I'm told the lower bound for the negative real zeros of $f(x)$ is equal to the upper bound of the positive real zeros of $f(-x)$. I'm completely lost here and don't understand what it means.

 

[andrewkirk]

It means you first make a new polynomial $g(x)$ by substituting $-x$ into the formula for $f(x)$. So for instance if $f(x)=3x^2+2x+1$ then $g(x)=3(-x)^2+2(-x)+1=3x^2-2x+1$.

Then you use the Upper Bound Theorem to try to find an upper bound for roots of $g(x)$. If you find $c$ as such an upper bound, then $-c$ is a lower bound for roots of $f(x)$.

It turns out that's the same condition as the one described in the second bullet of your post.

 

[RChristenk]

I understand $f(-x)$ leads to a polynomial with alternating signs. But if I use the upper bound theory on $f(-x)$, isn't there a contradiction? The upper bound theory says the coefficients of the quotient and remainder are all non-negative for a $c>0$ divided synthetically into $f(x)$. But $f(-x)$ has alternating signs, so how can I be sure that in $f(-x)=(-x-c)q(-x)+r$, the signs of the coefficients of $q(-x)$ and $r$ are all the same?

 

[Office_Shredder]

If f(x) has alternating signs, then f(-x) has only one sign. Does that help?

 

[RChristenk]

Ah so the upper bound assumes $f(x)$ starts with all non-negative coefficients, and the lower bound assumes $f(x)$ starts with alternating signs. I kept thinking $f(x)$ was strictly positive for both cases. So if $f(x)$ alternate signs, then $f(-x)$ is positive for odd degrees and negative for even degrees. But then I can't use the upper bound theory for polynomials with even degrees because its negative. I'm stuck here.

 

[Office_Shredder]

Ah so the upper bound assumes $f(x)$ starts with all non-negative coefficients, and the lower bound assumes $f(x)$ starts with alternating signs. I kept thinking $f(x)$ was strictly positive for both cases. So if $f(x)$ alternate signs, then $f(-x)$ is positive for odd degrees and negative for even degrees. But then I can't use the upper bound theory for polynomials with even degrees because its negative. I'm stuck here.

If all the coefficients are negative I think you already observed earlier that $f(b)<0$ when $b>c$. But then it's not zero. It doesn't matter if f is positive or negative, the important part is it's not zero.

Alternatively, apply the upper bound theorem to $-f$ and notice $f$ and $-f$ have the same set of zeroes.

 

[RChristenk]

If all the coefficients are negative, after applying the upper bound theorem I get $f(b)<0$. But I just realized the upper bound theorem requires the final line in the synthetic division to be strictly non-negative. So doesn't that make $f(b)<0$ meaningless?

 

[fresh_42]

If all the coefficients are negative, after applying the upper bound theorem I get $f(b)<0$. But I just realized the upper bound theorem requires the final line in the synthetic division to be strictly non-negative. So doesn't that make $f(b)<0$ meaningless?

Write your calculations with $g=-f$ instead of using the same $f$ and apply the first part to $g.$ This lowers the risk of a sign error somewhere.

 

[RChristenk]

According to the theorem as written, any $f(x)$ with all-negative coefficients have no bound (because the upper bound requires all non-negative signs and the lower bound requires alternating signs). But this is incorrect? Can someone explain to me please?

Regarding the lower bound, let $c<0$, and let $f(x)=(x+c)q(x)+r$ have alternating signs. I need to prove $b<c<0$ to show there is a lower bound. Change $x$ to $-x$, and now $f(-x)$ is strictly all positive or all negative. Now I can use the upper bound theory, $-b>-c>0$. Then $f(-x)=(-x+c)q(-x)+r$ or $f(-b)=(-b+c)q(-b)+r$. Now $-b+c < 0$, $q(-b)$ is strictly negative or positive, and $r$ is strictly negative or positive. I need $f(-b) \neq 0$ to prove $-b$ is the upper bound (or $b$ is the lower bound). But then $(-b+c)q(-b)$ and $r$ need to be the same sign to rule out the possibility of them adding up to zero. How do I determine their signs?

 

[Office_Shredder]

If all the coefficients of q, and r, are negative when you divide by $x-c$ then what is the sign of $(b-c)q(b)+r$ when $b>c$? In particular can it be zero?

 

[RChristenk]

If all the coefficients of q, and r, are negative when you divide by $x-c$ then what is the sign of $(b-c)q(b)+r$ when $b>c$? In particular can it be zero?

You are right. It cannot be zero because $b-c$ is positive, $q(b)$ is negative and $r$ is negative. $f(b)$ perpetually is less than zero. So this proves if all the coefficients are negative $b$ is the upper bound? Then why does the theorem states all coefficients need to be non-negative? That's so weird.

 

[Office_Shredder]

You are right. It cannot be zero because $b-c$ is positive, $q(b)$ is negative and $r$ is negative. $f(b)$ perpetually is less than zero. So this proves if all the coefficients are negative $b$ is the upper bound? Then why does the theorem states all coefficients need to be non-negative? That's so weird.

It's not that uncommon for theorems to omit some cases or be less general than they could be. Part of math is recognizing how to use the tools you have to solve new similar problems in front of you. This was a pretty good example I think.