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    http://img34.imageshack.us/img34/198...s123523456.jpg


    I'm trying to work through some examples, but im not sure where the following comes from:



    1. circled in black -- how do i get the ?<1?



    2. circled in red -- how do I get 0<x<2, i.e. x?(0,2)?



    3. cirlced in blue -- how do i get |x^2+x+3|<9



    Thanks, most appreciated.
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    1. In the second line it says that you set \delta &lt; \text{min} \{ 1, \frac{\varepsilon}{9} \}, so it's definitely less than 1 no matter what its value is.

    2. Well you know that \delta &lt; 1, so |x-1| &lt; \delta &lt; 1, so ... (you do this bit :p:)

    3. Where does the maximum value of x^2+x+9 occur in [0,2]? So what can you bound it by in (0,2)?

    EDIT: This is a really strange proof of the continuity of a polynomial, and I think it's quite mean. It doesn't give much insight, since you have to guess from the start that \frac{\varepsilon}{9} might be a useful bound for \delta, and it uses all sorts of weird tricks. I prefer to show continuity of polynomials first by using the binomial theorem to prove that, for all n, x^n is continuous at every point, and then using the fact that linear combinations of continuous functions are continuous.
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    (Original post by rpan161)
    http://img34.imageshack.us/img34/198...s123523456.jpg


    I'm trying to work through some examples, but im not sure where the following comes from:



    1. circled in black -- how do i get the ?<1?
    THere is a restriction. Let
     0&lt;\delta&lt;min\left [1,\frac{\epsilon}{9}\right ]
    this means that \delta&lt;1

    2. circled in red -- how do I get 0<x<2, i.e. x?(0,2)?
    |x-1|&lt;\delta \rightarrow 1-\delta&lt;x&lt;1+\delta
    As max(delta)<1 sub delta=1 so 0<x<2

    3. cirlced in blue -- how do i get |x^2+x+3|<9
    Thanks, most appreciated.
    As max(x)<2 sub x=2
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    (Original post by ztibor)
    |x-1|&lt;\delta \rightarrow 1-\delta&lt;x&lt;1+\delta
    As max(delta)<1 sub delta=1 so 0<x<2
    Just a couple of nitpicks (which are fairly insignificant, but when it comes to analysis, everything matters):

    \text{max}(\delta) doesn't really make much sense; you've already picked \delta, and so its maximum value is simply \delta, in the same way that max(3)=3 and so on. However, we know that \delta &lt; 1, so that's enough. Secondly, we can't substitute \delta=1, because we've already said that \delta &lt; 1. However, this does allow us to replace "\delta" by "1" in the inequalities, since 1+\delta &lt; 1+1 and 1-\delta &gt; 1-1.


    (Original post by ztibor)
    As max(x)<2 sub x=2
    This requires more work. For example, |x^2-6x+9|&lt;9 in (0,2), but putting x=2 gives 1; however it's certainly not true that |x^2-6x+9|&lt;1 in (0,2)! You need to show that the function in the mod brackets is increasing on the interval (0,2) (and, in particular, that it's strictly increasing near 2).
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    (Original post by nuodai)
    Just a couple of nitpicks (which are fairly insignificant, but when it comes to analysis, everything matters):

    \text{max}(\delta) doesn't really make much sense; you've already picked \delta, and so its maximum value is simply \delta, in the same way that max(3)=3 and so on. However, we know that \delta &lt; 1, so that's enough. Secondly, we can't substitute \delta=1, because we've already said that \delta &lt; 1.
    I do not think so. The upper bound for \delta is 1. I substituted this to get bounds for x.
    However, this does allow us to replace "\delta" by "1" in the inequalities, since 1+\delta &lt; 1+1 and 1-\delta &gt; 1-1.



    This requires more work. For example, |x^2-6x+9|&lt;9 in (0,2), but putting x=2 gives 1; however it's certainly not true that |x^2-6x+9|&lt;1 in (0,2)! You need to show that the function in the mod brackets is increasing on the interval (0,2) (and, in particular, that it's strictly increasing near 2).
    THanks. Sorry, I do not now why I thought this to be increasing at the given interval.
    Without any showing but minimal knowledge it is clear this is the (x-3 )^2
    parabola which in this case always nonnegative, and the pole of it is at (3,0)
    pont, so it is decreasing in the (0,2) interval.
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    (Original post by nuodai)
    EDIT: This is a really strange proof of the continuity of a polynomial, and I think it's quite mean. It doesn't give much insight, since you have to guess from the start that \frac{\varepsilon}{9} might be a useful bound for \delta, and it uses all sorts of weird tricks. I prefer to show continuity of polynomials first by using the binomial theorem to prove that, for all n, x^n is continuous at every point, and then using the fact that linear combinations of continuous functions are continuous.
    I have to say, it's probably the way I'd prove a particular poly was cts from first principles. Although there's not a lot of explanation, and the value for epsilon is obviously "written in" once the question is complete - as you say, it's impossible to guess it at the start.

    But loosely:

    Given a poly p(x), to show p is cts at x=a, we look at p(x)-p(a) to see (x-a) | (p(x)-p(a)). Divide out the RHS to get p(x)-p(a) = (x-a)q(x) for some poly q, then bound q near a; deduce result.
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    thanks everyone.

    (Original post by nuodai)
    1. In the second line it says that you set \delta &lt; \text{min} \{ 1, \frac{\varepsilon}{9} \}, so it's definitely less than 1 no matter what its value is.

    2. Well you know that \delta &lt; 1, so |x-1| &lt; \delta &lt; 1, so ... (you do this bit :p:)

    3. Where does the maximum value of x^2+x+9 occur in [0,2]? So what can you bound it by in (0,2)?

    EDIT: This is a really strange proof of the continuity of a polynomial, and I think it's quite mean. It doesn't give much insight, since you have to guess from the start that \frac{\varepsilon}{9} might be a useful bound for \delta, and it uses all sorts of weird tricks. I prefer to show continuity of polynomials first by using the binomial theorem to prove that, for all n, x^n is continuous at every point, and then using the fact that linear combinations of continuous functions are continuous.

    Thanks, do you know a website/somewhere that teaches a better way to show proof of continuity?
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    (Original post by rpan161)
    Thanks, do you know a website/somewhere that teaches a better way to show proof of continuity?
    Build it up in steps. Show that the sum of continuous functions is continuous, and the product of continuous functions is continuous, and that constant functions and the identity function are continuous.
 
 
 
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