# Math Help - Epsilon-delta problem

1. ## Epsilon-delta problem

Information:
Given $f: [-\sqrt\pi, \sqrt\pi] \rightarrow [-1,1]$ defined by $f(x)= sin(x^2)$

The problem:

Find a $\delta$ so that $|x - y| \leq \delta$ implicates that $|f(x) - f(y)| \leq 0.1$ for all $x$ and $y$ in $[-\sqrt\pi, \sqrt\pi]$

I was told that the mean value theorem can be used to find our $\delta$. Don't know how though. Any ideas on how to proceed?

2. You are correct about using the mean value theorem.
From that we know that $\left| {f(b) - f(a)} \right| = \left| {b - a} \right|\left| {2c\cos (c^2 )} \right| \leqslant 5$ for $- \sqrt \pi \leqslant a < c < b \leqslant \sqrt \pi$.

So what is $\delta?$

3. Originally Posted by Plato
From that we know that $\left| {f(b) - f(a)} \right| = \left| {b - a} \right|\left| {2c\cos (c^2 )} \right| \leqslant 5$
Where do you get the "5" from? $|{f(b) - f(a)}| = |sin(b^2) - sin(a^2)|$ which is less or equal to 2 (?)

4. Sorry it should have been $\le 5|b-a|$.

5. So that:

$\left| {f(b) - f(a)} \right| = \left| {b - a} \right|\left| {2c\cos (c^2 )} \right| \leqslant 5 |b-a|?$

That would have to mean that: $|2c\cos(c^2)| \leqslant 5$

$cos(c^2)$ is always $\leqslant 1$. And c is a value between $-\sqrt\pi$ and $\sqrt\pi$. But how is all of this less than, or equal to 5?

6. Graph $\left| {2x\cos (x^2 )} \right|$ on $\left[ { - \sqrt \pi ,\sqrt \pi } \right]$.

You will see that 5 is good bound.

7. This is what I did. Please correct any mistakes:

Given $f: [-\sqrt\pi, \sqrt\pi] \rightarrow [-1,1]$ defined by $f(x)= sin(x^2)$.

We want to find a $\delta$ so that $|x-y| \leq \delta$ implicates that $|f(x) - f(y)| \leq 0.1$ for all $x$ and $y$ in $[-\sqrt\pi, \sqrt\pi]$

The mean value theorem says that since f is countinous for $[-\sqrt\pi, \sqrt\pi]$, we have:

$\left| {f(x) - f(y)} \right| = \left| {x - y} \right|\left| f'(c) \right|$
where $|f'(c)| = |2c cos(c^2)|$

By graphing $|2c cos(c^2)|$ we see that this (absolute) value is never greater than 4 (or 5, or 6...). We now have that:

$| {f(x) - f(y)} | < 4| {x - y} |$

Let's say that $\delta = \epsilon / 4$, where we have been told that $\epsilon = 0.1$. Since $|x-y| \leq \delta$, then $|x-y| \leq \epsilon / 4 = 0.1/4$.

So finally:
$| {f(x) - f(y)} | < 4| {x - y} | \leq 4 * (\0.1/4) = 0.1 = \epsilon$

Therefore $\delta = 0.1/4 = 1/40$ is a value for $\delta$ that makes $|x-y| \leq \delta$

8. Originally Posted by jenkki
So that:

$\left| {f(b) - f(a)} \right| = \left| {b - a} \right|\left| {2c\cos (c^2 )} \right| \leqslant 5 |b-a|?$

That would have to mean that: $|2c\cos(c^2)| \leqslant 5$

$cos(c^2)$ is always $\leqslant 1$. And c is a value between $-\sqrt\pi$ and $\sqrt\pi$. But how is all of this less than, or equal to 5?
Yes, $-1\le cos(c^2)\le 1$ so $-2c\le 2c cos(c^2)\le 2c$. Since $c\le \sqrt{\pi}$, $2c\le 2\sqrt{\pi}$ which is about $2\sqrt{3.14}$ or 3.5. 4 could have been used but any number that is less than 3.5... is certainly less than 5. I'm not sure why Plato used "5" instead of "4" but it is valid.

9. Originally Posted by HallsofIvy
I'm not sure why Plato used "5" instead of "4" but it is valid.
Because in the original post it required $|f(x)-f(y)|<0.1$.
The 5 just makes for nice numbers.

10. Originally Posted by jenkki
This is what I did. Please correct any mistakes:
No one? I'd really like to know if I've got the right answer