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Math Help - Providing Conditions

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    Providing Conditions

    I need to provide a condition on |x-1| that will assure that |x^2 - 1| < 1/n. Any ideas?
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    Quote Originally Posted by frenchguy87 View Post
    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.
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    Quote Originally Posted by frenchguy87 View Post
    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}.
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    Thanks a lot! Very helpful
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    You actually kind of lose me here, why do you say
    Quote Originally Posted by HallsofIvy View Post
    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}.
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    Quote Originally Posted by HallsofIvy View Post
    |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
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    Quote Originally Posted by frenchguy87 View Post
    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}
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  8. #8
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    Quote Originally Posted by xalk View Post
    .

    |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...
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    Quote Originally Posted by CrazyCat87 View Post
    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.
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  10. #10
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    Quote Originally Posted by HallsofIvy View Post
    |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} ?
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  11. #11
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    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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