# Intermediate Value Theorem Problem

1. Oct 3, 2009

### jgens

1. The problem statement, all variables and given/known data

Let the function $\phi$ be continuous on the reals and suppose that $\lim_{x \to \infty}\phi(x)/x^n = 0 = \lim_{x \to -\infty}\phi(x)/x^n$. Prove that if $n$ is odd, then there exists a number $x$ such that $\phi(x) + x^n = 0$.

2. Relevant equations

N/A

3. The attempt at a solution

This is a challenge problem from my book and I don't know if I've done the problem correctly. Any suggestions are appreciated. Thanks!

The criterion with limits means that for some $\varepsilon > 0$ if $x > M$ then $|\phi(x)/x^n| < \varepsilon$. Choosing $M$ so large that $x > M > 0$ implies that $|\phi(x)/x^n| < 1$ and consequently $-\phi(x) \leq |\phi(x)| < |x^n| = x^n$ since $x > 0$. Therefore, for $x > M > 0$ we have that $x^n + \phi(x) > 0$. Using a similar argument we can prove that as long as $0 < m < x$ we have that $0 > x^n + \phi(x)$.

Now consider the closed interval $[M + 1, m -1]$. Since the sum $\phi(x) + x^n$ is continuous and $\phi(x) + x^n > 0$ for some $x$ in $[M + 1, m -1]$ and $\phi(x) + x^n < 0$ for some $x$ in $[M + 1, m -1]$, by the IVT we have that there exists an $x$ in $[M + 1, m -1]$ such that $\phi(x) + x^n = 0$, completing the proof.

2. Oct 3, 2009

### jbunniii

If I'm understanding you correctly, you are choosing both $m$ and $M$ to be positive. It's not clear to me why that would necessarily work. Where are you using the fact that $n$ is odd? I think if you try to write out the details of the "using a similar argument..." part, you will find that it doesn't work.

Also I think you're making it a little more complicated than you need to by forcing yourself to pick a range that extends beyond $[m, M]$.

Your general idea is right, though.

Here's how I would set it up.

(1) Because of the first limit, I can find $M > 0$ such that $|\phi(M)/M^{n}| < 1$.

(2) Because of the second limit, I can find $m < 0$ such that $|\phi(m)/m^{n}| < 1$.

(3) Since $n$ is odd, it is also true that $m^n < 0$ and $M^n > 0$.

Now use the facts that $m^n < 0$ and $M^n > 0$ in order to conclude that $\phi(m) + m^n < 0$ and $\phi(M) + M^n > 0$. You will then be able to apply the IVT directly to the interval $[m,M]$.

Last edited: Oct 4, 2009
3. Oct 3, 2009

### jgens

Sorry for the confusion, I didn't read through my post after I posted. I'll fix some of the slip ups. Doh! *smacks palm against head*

Edit: Alright, I think I should have fixed up my notational error for the most part (I had it down right on paper). Reading through your post I'll agree with all of your criticisms given what I had posted at the time. Hopefully I should have reduced/eliminated all of that confusion. Thanks for the suggestion about the intervals though, it would make the proof look a lot nicer (as well as simplify a few details)!

Ugh, I need to make sure I'm writing what I mean with LaTeX.

Last edited: Oct 3, 2009
4. Oct 3, 2009

### jgens

The criterion with limits means that for some $\varepsilon > 0$ if $x > M$ then $|\phi(x)/x^n| < \varepsilon$. Choosing $M$ so large that $x > M > 0$ implies that $|\phi(x)/x^n| < 1$ and consequently $-\phi(x) \leq |\phi(x)| < |x^n| = x^n$ since $x > 0$. Therefore, for $x > M > 0$ we have that $x^n + \phi(x) > 0$. Using a similar argument we can prove that as long as $0 > m > x$ we have that $0 > x^n + \phi(x)$.

Now consider the closed interval $[m-1, M+1]$. Since the sum $\phi(x) + x^n$ is continuous and $\phi(x) + x^n > 0$ for some $x$ in $[m-1, M+1]$ and $\phi(x) + x^n < 0$ for some $x$ in $[m-1,M+1]$, by the IVT we have that there exists an $x$ in $[m-1, M+1]$ such that $\phi(x) + x^n = 0$, completing the proof.

5. Oct 3, 2009

### jgens

Okay, here's essentially the same proof with some minor changes . . .

The first limit criterion implies that for some $\varepsilon > 0$ if $x > M$ then $|\phi(x)/x^n| < \varepsilon$. Consequently, we can choose a real number $\beta > 0$ such that $|\phi(\beta)/\beta^n| < 1$ and similarly $-\phi(\beta) \leq |\phi(\beta)| < |\beta^n| = \beta^n$ since $\beta > 0$. Therefore, $\beta^n + \phi(\beta) > 0$.

The second limit criterion implies that for some $\varepsilon > 0$ if $x < m$ then $|\phi(x)/x^n| < \varepsilon$. Consequently, we can choose a real number $\alpha < 0$ such that $|\phi(\alpha)/\alpha^n| < 1$ and similarly $\alpha^n < -\phi(\alpha) \leq |\phi(\alpha)| < |\alpha^n|$ since $\alpha < 0$ and $n$ is odd. Therefore, $\alpha^n + \phi(\alpha) < 0$.

Now, considering the interval $[\alpha,\beta]$, we have that the function $x^n + \phi(x) > 0$ for some $x \in [\alpha,\beta]$ and $x^n + \phi(x) < 0$ for some $x \in [\alpha,\beta]$. Since both $x^n$ and $\phi(x)$ are continuous, their sum must be continuous as well, and by the IVT, there must be some $x \in [\alpha,\beta]$ such that $x^n + \phi(x) = 0$ completing the proof.

6. Oct 4, 2009

### jgens

Can anyone comment on the new proof?

7. Oct 4, 2009

### jbunniii

Your new proof is mostly correct, but I would make a few clarifications, and one correction, as follows:

The first limit criterion implies that given any $\epsilon > 0$, there exists $M \in \mathbb{R}$ such that $|\phi(x)/x^n| < \epsilon$ for all $x > M$. Choosing $\epsilon = 1$, we may therefore select $\beta > 0$ such that $|\phi(\beta)/\beta^n| < 1$, and so $-\phi(\beta) \leq |\phi(\beta)| < |\beta^n| = \beta^n$ since $\beta > 0$ and $n$ is odd. Therefore, $\phi(\beta) + \beta^n > 0$.

The second limit criterion implies that given any $\epsilon > 0$, there exists $m \in \mathbb{R}$ such that $|\phi(x)/x^n| < \epsilon$ for all $x < m$. Choosing $\epsilon = 1$, we may therefore select $\alpha < 0$ such that $|\phi(\alpha)/\alpha^n| < 1$, and so [correction: note difference vs. yours:] $\phi(\alpha) \leq |\phi(\alpha)| < |\alpha^n| = -\alpha^n$ since $\alpha < 0$ and $n$ is odd. Therefore, $\phi(\alpha) + \alpha^n < 0$.

We now have $\phi(\alpha) + \alpha^n < 0$ and $\phi(\beta) + \beta^n > 0$. The functions $\phi(x)$ and $x^n$ are both continuous on $[a,b]$, and thus so is $\phi(x) + x^n$. The IVT therefore implies that $\phi(x) + x^n = 0$ for some $x \in (\alpha, \beta)$.

8. Oct 4, 2009

### jgens

Thanks for the suggestions! I have a clarifying question about your correction though . . .

Since $\alpha < 0$ and $n$ is odd this means that $\alpha^n < 0 < |\alpha^n|$. Additionally, since $|\phi(\alpha)|<|\alpha^n|$, this means that $\alpha^n < \phi(\alpha) < -\alpha^n$ or equivalently $-\alpha^n > -\phi(\alpha) > \alpha^n$. Therefore, we have that $\alpha^n < -\phi(\alpha) \leq |\phi(\alpha)| < |\alpha^n|$ which is the inequality I have in my revised proof. Aren't our two inequalities really saying the same thing?

9. Oct 4, 2009

### jbunniii

OK, I see how you got that second inequality now. I think my inequality is a little easier to follow because it's clear why each step is true:

$\phi(\alpha) \leq |\phi(\alpha)|$ (true for any absolute value)
$|\phi(\alpha)| < |\alpha^n|$ (simply cross multiply $|\phi(\alpha)/\alpha^n|<1$)
$|\alpha^n| = -\alpha^n$ (since $\alpha < 0$ and $n$ is odd)

whereas when I tried to read your inequality step by step I started with

$\alpha^n < -\phi(\alpha)$

and thought, wait... that's what we're trying to prove!

In other words, the inequality you wrote is true, but as you read it from left to right, it's not immediately clear where each step follows from. It's better, when possible, to arrange the inequality in the same order as the sequence of logical steps you are taking. (Especially when someone else is assigning a grade to the proof!)

10. Oct 4, 2009

### jgens

Thanks again. Your inequality is definately easier to follow, I just wanted to make sure that I wasn't making a mistake somewhere.