Providing Conditions

**frenchguy87** · February 2nd 2010, 11:27 AM

I need to provide a condition on |x-1| that will assure that |x^2 - 1| < 1/n. Any ideas?

**Drexel28** · February 2nd 2010, 12:04 PM

Originally Posted by frenchguy87

I need to provide a condition on |x-1| that will assure that |x^2 - 1| < 1/n. Any ideas?

$\left|x^2-1\right|=\left|x-1\right||x+1|$ ...so... come on.

**HallsofIvy** · February 2nd 2010, 12:04 PM

Originally Posted by frenchguy87

I need to provide a condition on |x-1| that will assure that |x^2 - 1| < 1/n. Any ideas?

$|x^2- 1|= |x-1||x+1|$ to make that less than $\frac{1}{n}$ , you need to have $|x- 1|< \frac{1}{n|x+1|}$ . But we need a constant, not a function of x on the right. That is, we need $|x- 1|< \frac{1}{n(x+1}< \delta$ which means we need an lower bound on $\frac{1}{x+1}$ . If |x-1|< 1 so that -1< x-1< 1, then 2< x+1< 3 (add 2 to each part). As long as |x-1|< 1, |x+1|> 2 so [tex]\frac{1}{|x+1|}< \frac{1}{2}. Then $|x-1|< \frac{1}{n|x+1|}< \frac{1}{2n}$ . In order that both of those be true, you must have |x-1|< the smaller of 1 and $\frac{1}{2n}$ . Assuming that n is a positive integer, of course, $\frac{1}{2n}$ is less than 1 and so is the smaller. The condition you want is $|x-1|< \frac{1}{2n}$ .

**frenchguy87** · February 2nd 2010, 12:21 PM

Thanks a lot! Very helpful

**frenchguy87** · February 3rd 2010, 08:43 PM

You actually kind of lose me here, why do you say

Originally Posted by HallsofIvy

If |x-1|< 1 so that -1< x-1< 1, then 2< x+1< 3 (add 2 to each part). As long as |x-1|< 1, |x+1|> 2 so [tex]\frac{1}{|x+1|}< \frac{1}{2}.

**xalk** · February 4th 2010, 06:32 AM

Originally Posted by HallsofIvy

$|x^2- 1|= |x-1||x+1|$ to make that less than $\frac{1}{n}$ , you need to have $|x- 1|< \frac{1}{n|x+1|}$ . But we need a constant, not a function of x on the right. That is, we need $|x- 1|< \frac{1}{n(x+1}< \delta$ which means we need an lower bound on $\frac{1}{x+1}$ . If |x-1|< 1 so that -1< x-1< 1, then 2< x+1< 3 (add 2 to each part). As long as |x-1|< 1, |x+1|> 2 so [tex]\frac{1}{|x+1|}< \frac{1}{2}. Then $|x-1|< \frac{1}{n|x+1|}< \frac{1}{2n}$ . In order that both of those be true, you must have |x-1|< the smaller of 1 and $\frac{1}{2n}$ . Assuming that n is a positive integer, of course, $\frac{1}{2n}$ is less than 1 and so is the smaller. The condition you want is $|x-1|< \frac{1}{2n}$ .

Shouldn't we choose |x-1|= min{ 1, $\frac{1}{3n}$ ???

,by adding 2 to all sides of -1< x-1<1 give you 1<x+1< 3

**xalk** · February 6th 2010, 01:32 PM

Originally Posted by frenchguy87

I need to provide a condition on |x-1| that will assure that |x^2 - 1| < 1/n. Any ideas?

.

$|x-1|<\frac{1}{3n}$ => |x-1|<1 => |x|-1<1 => |x|+1<3 => |x+1|< 3 => $|x-1||x+1|\leq 3\frac{1}{3n}=\frac{1}{n}$

**CrazyCat87** · February 7th 2010, 03:57 PM

Originally Posted by xalk

.

$|x-1|<\frac{1}{3n}$ => |x-1|<1 => |x|-1<1 => |x|+1<3 => |x+1|< 3 => $|x-1||x+1|\leq 3\frac{1}{3n}=\frac{1}{n}$

Why do you guys all say $|x-1|<1$ , I can't figure out where that comes from...

**HallsofIvy** · February 8th 2010, 04:53 AM

Originally Posted by CrazyCat87

Why do you guys all say $|x-1|<1$ , I can't figure out where that comes from...

The point is to get some kind of bound on x so we can bound x+1. Since we want |x-1| small, |x-1|< 1 is handy. You could use any positive number other than 1 on the right, but 1 is easiest.

**frenchguy87** · February 8th 2010, 07:43 AM

Originally Posted by HallsofIvy

$|x^2- 1|= |x-1||x+1|$ to make that less than $\frac{1}{n}$ , you need to have $|x- 1|< \frac{1}{n|x+1|}$ . But we need a constant, not a function of x on the right. That is, we need $|x- 1|< \frac{1}{n(x+1}< \delta$ which means we need an lower bound on $\frac{1}{x+1}$ . If |x-1|< 1 so that -1< x-1< 1, then 2< x+1< 3 (add 2 to each part). As long as |x-1|< 1, |x+1|> 2 so [tex]\frac{1}{|x+1|}< \frac{1}{2}. Then $|x-1|< \frac{1}{n|x+1|}< \frac{1}{2n}$ . In order that both of those be true, you must have |x-1|< the smaller of 1 and $\frac{1}{2n}$ . Assuming that n is a positive integer, of course, $\frac{1}{2n}$ is less than 1 and so is the smaller. The condition you want is $|x-1|< \frac{1}{2n}$ .

I can almost follow this except for this part,
If |x-1|< 1 so that -1< x-1< 1, then 2< x+1< 3 (add 2 to each part).

Shouldn't we have 1 < x+1 < 3, which would then give

|x+1|> 1 so $\frac{1}{|x+1|}< 1$ . Then $|x-1|< \frac{1}{n|x+1|}< \frac{1}{n}$ ?

**Plato** · February 8th 2010, 08:03 AM

This has gone on too long.

Do you agree that $|x-1|<1$ implies that $|x+1|<3$ ?

If so, then let $|x-1|<\frac{1}{3n}<1$ .

That gives $|x^2-1|={\color{red}|x-1|}{\color{blue}|x+1|}<{\color{red}\left(\frac{1}{ 3n}\right)}{\color{blue}(3)}=\frac{1}{n}$ .

Now you have your condition: $\frac{1}{3n}$ .

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