# Find a positive number $\delta$?

Find a positive number $\delta$ such that when $|x-1|<\delta$, then $|x^2-1| < 0.45$

Part 2:

Find the LARGEST number $\lambda$ with the property that when $|x-1|< \lambda$, then $|x^2-1| < 0.45$

I don't understand this at all. Help?

-
What part(s) of the question you don't understand? – M Turgeon Sep 4 '12 at 20:59
how to find the numbers – Taylor Sep 4 '12 at 21:01

Note that

$$|x+1|=|x-1+2|\le |x-1|+2<2+\delta$$

by the triangle inequality. Then

$$|x^2-1|=|x-1||x+1|<\delta(2+\delta)$$

and we require that $\delta(2+\delta)<0.45$. There are many values of $\delta$ that work, for instance $\delta=0.1$ has $\delta(2+\delta)=0.21<0.45$.

You find the maximum $\delta$ that works (called $\lambda$ in the original question) by solving the quadratic equation $\lambda(2+\lambda)=0.45$. The result is $\lambda\approx 0.204$.

-
Got it. so, i have another question. How would i find a positive number δ of this: |x - 3| < δ then 1/ (|x - 3|^6) > 10030? – Taylor Sep 4 '12 at 22:47
In that case you have $1/|x-3|^6>1/\delta^6$. – Jonathan Sep 4 '12 at 22:51

For example $\delta = 1/8$ solve. in fact, \begin{eqnarray} |x^2-1| &=& |x-1||x+1|\\ &\le& |x-1|(|x|+1)\\ &\le & 1/8 \cdot 2 = 1/4 <0,45. \end{eqnarray} Because $$|x| \le |x-1| +|1| \le 2.$$

Part 2. If $x>1, |x^2-1| = x^2-1$ and $x^2-1<0,45 \Rightarrow x< \sqrt{1,45}$. If $x<1, |x^2-1| = 1-x^2$ and $1-x^2<0,45 \Rightarrow x> \sqrt{0,55}$.Then $\delta = \max (|1-\sqrt{0,55}|,|1-\sqrt{1,45}|)$.

-
How did you come up with that number? – Mariano Suárez-Alvarez Sep 4 '12 at 21:34
Only a small conveniente number, because you any small enouth number musty satisfy. – user29999 Sep 4 '12 at 21:38
You shoud probably add an explanation to the answer: notice that the question is how do I find the number? not how do I check that a number works? :=) – Mariano Suárez-Alvarez Sep 4 '12 at 21:38

For a given $x$, either $x^2-1$ is negative, or it is not. Consider the two separate cases. This and the inequality will give you different restrictions on $x$.

Here is an example: for what $x$ do we have $|x-2|<3$? If $x\geq 2$, then $|x-2|=x-2$, and so $$|x-2|<3 \iff x-2<3 \iff x<5.$$ If $x< 2$, then $|x-2|=2-x$, and so $$|x-2|<3 \iff 2-x<3 \iff x>-1.$$

All in all, we see that $|x-2|<3$ if and only if $x\in (-1,2)\cup [2,5)=(-1,5)$.

To bring this back to your problem, you first have to find for what values of $x$ is the inequality $|x^2-1|<0.45$ satisfied. Then, you have to find how close such an $x$ has to be to 1 in order to ensure these conditions are satisfied. For example, going back to the example above, we see that if $|x-1|<2$, which is equivalent to $x\in(-1,3)$, then by the above computation, the inequality $|x-2|<3$ is satisfied. In short, in this situation, a choice for $\delta$ would be 2, which incindently also gives you the $\lambda$.

-
how do i find values then? – Taylor Sep 4 '12 at 21:54
@Taylor Here is a short summary: find under what condition the inequality $|x^2-1|<0.45$ is satisfied; this will give you an interval. All values in this interval satisfy the inequality. Then the $\lambda$ you are looking for is the largest positive number such that $(1-\lambda,1+\lambda)$ is contained in this interval. This maximal $\lambda$ is an instance of the $\delta$ you are also looking for. – M Turgeon Sep 4 '12 at 21:58
ok so how would i find a positive number δ such that when |x - 3| < δ then 1/ (|x - 3|^6) > 10030? – Taylor Sep 4 '12 at 22:26
@Taylor Where did you pick those numbers? I think you are very confused. – M Turgeon Sep 4 '12 at 22:32
its a different problem – Taylor Sep 4 '12 at 22:38

If you find $\lambda$, it or any smaller positive number can be $\delta$. Start by taking apart the absolute value. If $x \gt 1, |x^2-1| \lt 0.45 \implies 1 \lt x^2 \lt 1.45$ The maximum allowable $x$ is $\sqrt {1.45}$, so the candidate $\lambda$ is ???? If $x \lt 1, |x^2-1| \lt 0.45 \implies 0.55 \lt x^2 \lt 1$, so $x \gt \sqrt {0.55}$. This gives you another candidate $\lambda$ and you need to take the smaller.

-

Suppose $0 < \epsilon < 1$. Consider the case $x^2 - 1 \ge 0$ first, i.e., $x \in (-\infty, -1] \cup [1, \infty)$. In this case, the inequality $|x^2 - 1| < \epsilon$ becomes $$x^2 - 1 - \epsilon < 0.$$ Factor this as $$(x - \sqrt{1 + \epsilon})(x + \sqrt{1 + \epsilon}) < 0.$$ This is true when $x \in (-\sqrt{1 + \epsilon}, \sqrt{1 + \epsilon})$. So we get $x \in (-\sqrt{1 + \epsilon}, -1] \cup [1, \sqrt{1 + \epsilon})$ for this case.

The other case is when $x^2 - 1 \le 0$, i.e., $x \in [-1, 1]$, the inequality $|x^2 - 1| < \epsilon$ becomes $$x^2 - 1 + \epsilon > 0.$$ Factor this as $$(x - \sqrt{1 - \epsilon})(x + \sqrt{1 - \epsilon}) > 0.$$ This is true when $x \in (-\infty, -\sqrt{1 - \epsilon}) \cup (\sqrt{1 - \epsilon}, \infty)$, so we get $x \in [-1, -\sqrt{1 - \epsilon}) \cup (\sqrt{1 - \epsilon}, 1]$ for this case. Combining them both, we get

$$x \in (-\sqrt{1 + \epsilon}, -\sqrt{1 - \epsilon}) \cup (\sqrt{1 - \epsilon}, \sqrt{1 + \epsilon}).$$

So to pick $\lambda$, make $\lambda = 1 - \sqrt{1 - \epsilon}$. This is because $1 - \sqrt{1 - \epsilon} > \sqrt{1 + \epsilon} - 1$.

-