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Math Help - Epsilon Delta proof, need help

  1. #1
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    Epsilon Delta proof, need help

    Hey,
    I was reading a document about epsilon-delta proof when I reached a point I did not understand.
    The document is http://www.ocf.berkeley.edu/~yosenl/...ilon-delta.pdf and my question is number 4, it starts on page 3.
    On the beginning of page 4 they write :
    "Notice that the right hand side is at the minimum when x + 3 is at itsmaximum. Since the maximum of x + 3 is 6, we know that
    |x-2| < e/|x+3| < e/6"

    How could that be true ? |x+3| is smaller than 6, therefor e/|x+3| should be bigger than e/6, Am I wrong ?
    Thanks a lot
    Michael Engstler.
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  2. #2
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    Re: Epsilon Delta proof, need help

    I believe you are right. The phrase "the right hand side is at the minimum when x + 3 is at its maximum" shows that they found the minimum of e / |x + 3|. But in order to bound this expression from above, they should have found its maximum, which is achieved when |x + 3| is at its minimum, i.e., 4. Therefore, e / |x + 3| < e / 4, not e / 6. Good catch.
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    Re: Epsilon Delta proof, need help

    Quote Originally Posted by MichaelEngstler View Post
    How could that be true ? |x+3| is smaller than 6, therefor e/|x+3| should be bigger than e/6, Am I wrong ?
    Actually the paper is correct.
    It may be poorly worded but it is correct.
    Here is idea.
    |x+3||x-2|<6|x-2| but we want it to be |x+3||x-2|<\epsilon.

    Therefore we need |x-2|<\frac{\epsilon}{6} so \delta=\text{min}\left\{\frac{\epsilon}{6},1\right  \}
    Thanks from emakarov
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  4. #4
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    Re: Epsilon Delta proof, need help

    Thanks, you are right. We want |x - 2| < e / |x + 3| for all 1 < x < 3, so we find the minimum of the right-hand side on that interval. That is, contrary to what I said before, we bound e / |x + 3| from below, not from above. The paper is indeed poorly worded when it says, "Since the maximum of x + 3 is 6, we know that |x-2| < e/|x+3| < e/6," giving an impression that e / |x + 3| < e / 6 holds for all 1 < x < 3.
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    Re: Epsilon Delta proof, need help

    Thanks a lot for your replies I really appreciate it, But I did not fully understand both of your explanations.
    My quest starts by limiting delta <= 1.
    (*)This limits x it the following way : 4< x+3 < 6.
    My quest continues by searching for a second limitation to delta.
    As I understand from both of your explanations, you decided that |x-2|< e/|x+3| can become with the help of (*) to |x-2| < e/6.
    Is that legal ? We know that x+3 is smaller than 6, maybe when x+3 = 6 than |x-2| is bigger or equal to e/6 ?

    So my question is :
    When given 4< x+3 < 6
    and |x-2|< e/|x+3|
    I do not understand how to get to |x-2| < e/6 in a legal way

    P.S
    Plato, How do you write mathematics expressions as done in your reply ?
    Thanks.
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  6. #6
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    Re: Epsilon Delta proof, need help

    There's a really good thread about epsilon-delta proofs right here on the MHF Calculus forum - it's one of the four "sticky" threads.

    When you're trying to figure out what the limits on delta should be, it's a different way of thinking - you're working backwards. Then when you write the proof, it works in the forward direction. In this case, we want |x-3| to be as large as necessary so that in our proof we give a bound for delta that's tight enough to work.

    - Hollywood
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    Re: Epsilon Delta proof, need help

    Quote Originally Posted by MichaelEngstler View Post
    Thanks a lot for your replies I really appreciate it, But I did not fully understand both of your explanations.
    My quest starts by limiting delta <= 1.
    (*)This limits x it the following way : 4< x+3 < 6.
    My quest continues by searching for a second limitation to delta.
    As I understand from both of your explanations, you decided that |x-2|< e/|x+3| can become with the help of (*) to |x-2| < e/6.
    My advice to is to drop the very idea \left| {x - 2} \right| < \frac{\varepsilon }{{\left| {x + 3} \right|}}.
    That has nothing to do with the proof. It should not even be there.

    Here is what we want |x+3||x-2|<\varepsilon.
    We have control over the size of |x-2|.

    So we start with |x-2|<1 giving us 1<x<3.
    Looking at the product we build the other factor: 4<x+3<6.
    At this point we know that |x+3|<6.
    Because we control |x-2| make it less than \min \left\{ {1,\frac{\varepsilon }{6}} \right\}

    Having done that, we can now say:
    |x+3||x-2|<6|x-2|<(6)\left(\frac{\varepsilon }{6}\right)=\varepsilon

    Using LaTex [tex]|x+3||x-2|<6|x-2|<(6)\left(\frac{\varepsilon }{6}\right)=\varepsilon [/tex] gives us |x+3||x-2|<6|x-2|<(6)\left(\frac{\varepsilon }{6}\right)=\varepsilon. The tab \boxed{\Sigma} on the toolbar inserts the [tex]..[/tex] wrap.
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  8. #8
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    Re: Epsilon Delta proof, need help

    Quote Originally Posted by MichaelEngstler View Post
    So my question is :
    When given 4< x+3 < 6
    and |x-2|< e/|x+3|
    I do not understand how to get to |x-2| < e/6 in a legal way
    If by getting from |x - 2| < e / |x + 3| to |x - 2| < e / 6 "in a legal way" you mean that the first statement implies the second one (for all 1 < x < 3), then this is indeed impossible. However, as Hollywood pointed out, in this first stage of problem solving we are working backwards. In the PDF file, right in the beginning of problem 4 on p. 3, there is a multi-line formula with arrows ==>. In fact, we are interested in the opposite arrows. We would like to find such an upper bound on |x - 2| that it would imply that |f(x) - L| < e, and not the other way around. Now, all arrows in that formula are reversible, i.e., the statements on each line are equivalent for all 1 < x < 3. Therefore, what is left is to find an upper bound d (which depends on e but not on x) such that |x - 2| < d would imply |x - 2| < e / |x + 3|. For this it is sufficient to take d to be the minimum of e / |x + 3| on the segment [1, 3], which is e / 6.

    To summarize: if 1 < x < 3 and |x - 2| < e / 6, then, since e / 6 <= e / |x + 3|, we have |x - 2| < e / |x + 3|, and by reversing the arrows, we arrive at |f(x) - 4| < e, as required. On the other hand, if 1 < x < 3, then |x - 2| < e / |x + 3| implies that |x - 2| < e / 4, but this is not used in the proof.
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  9. #9
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    Re: Epsilon Delta proof, need help

    Thanks a lot ! For your replies, I am starting to understand.
    I read the pdf and I saw the following statement that is simillar to the statement presented in the topic.
    I would like to understand if the statement is legal: (It appears in the pdf, page 3, example 3)

    "|x^2 − 25| < e <==> |x − 5||x + 5| < e < ==> |x − 5| < e/|x + 5|. (1)
    This situation differs from Examples 1 and 2 in that we want to define delta = e/|x + 5|, but we cannot since delta
    is supposed to be a number depending only on e, not a function of x. Here is one way to get around this
    difficulty: we will replace |x + 5| in (1) by a number M which satisfies |x + 5| <= M. In so doing, we rewrite
    (1) as
    |x^2 − 25| < e <==> |x − 5||x + 5| < e <==> |x − 5|M < e <==> |x − 5| < e/M"

    How is the bold part legal ? Is states that M is bigger or equal to |x+5|, if it is bigger, who can promiss me |x-5|M is still smaller than e ?
    I apolagize for all my questions, I just get it hard to understand how this statement (That is near identical to the statement in the topic) are legal.
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  10. #10
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    Re: Epsilon Delta proof, need help

    Yes, it does seem that there is one extra arrow in the second statement.

    The first statement,

    |x^2-25| <  \epsilon \Longleftrightarrow |x-5||x + 5| <  \epsilon \Longleftrightarrow |x-5| <  \epsilon / |x + 5|

    is correct. They are only different by basic algebraic steps. But once M \ge |x+5| is introduced, some of the inequalities become stronger than others, so we no longer have the implication in both directions.

    Since this is the "scratch work" section, we are working backwards of normal. So here is the thinking:

    We want |x^2-25| <  \epsilon .
    This is equivalent to |x-5| |x+5| <  \epsilon .
    If we can prove |x-5| M <  \epsilon we will be able to replace M by something smaller, |x+5|, to prove the preceding statement.
    This is equivalent to |x-5|  <  \epsilon / M.

    So the arrow between |x-5| |x+5| <  \epsilon and |x-5| M <  \epsilon should point only to the left.

    - Hollywood
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  11. #11
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    Re: Epsilon Delta proof, need help

    I do not have the original .tex to edit it again, so I don't think I could edit the problem.
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    Re: Epsilon Delta proof, need help

    Thanks a lot for your replies !
    I really appreciate it
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