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    As ever, analysis is turning my brain into an algebraic mush and preventing me from seeing sense.

    I need to show that the function f(x)=e^{-x^4} is uniformly continuous (or rather, I'm asked whether it is, but it clearly is so I just need to show it).

    My attempts were too pathetic and embarrassing to warrant posting here, so I'd appreciate a poke in the right direction to get me started (nothing in the way of full answers though please).

    On a side-note, I don't know what it is about uniform continuity that makes me instantly panic and run for the hills, but getting stumped on bog-standard easy questions like this certainly isn't helping! If anyone knows of any good books or websites on the matter, or indeed for analysis in general, then please let me know -- pure maths is brilliant and the only thing really keeping me from doing as well as I want to do in it is my fear of analysis.
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    (Original post by nuodai)
    As ever, analysis is turning my brain into an algebraic mush and preventing me from seeing sense.

    I need to show that the function f(x)=e^{-x^4} is uniformly continuous (or rather, I'm asked whether it is, but it clearly is so I just need to show it).

    My attempts were too pathetic and embarrassing to warrant posting here, so I'd appreciate a poke in the right direction to get me started (nothing in the way of full answers though please).

    On a side-note, I don't know what it is about uniform continuity that makes me instantly panic and run for the hills, but getting stumped on bog-standard easy questions like this certainly isn't helping! If anyone knows of any good books or websites on the matter, or indeed for analysis in general, then please let me know -- pure maths is brilliant and the only thing really keeping me from doing as well as I want to do in it is my fear of analysis.
    And here was I thinking you were infallible. :P I'll have a think, but first I'm told that Apostol's book on Analysis is v. good, probably worth a look.
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    (From a thread that I recycled, and then decided against recycling):
    (Original post by IrrationalNumber)
    Just a thought, could you use the MVT?
    Bingo!

    I can bound |f'(x)|=|-4x^3e^{-x^4}| by 3, and so given x,y \in \mathbb{R} there is some z between x and y such that |e^{-x^4}-e^{-y^4}| = |4z^3e^{-z^4}||x-y| \le 3|x-y|, so given \varepsilon > 0, if |x-y| < \frac{\varepsilon}{3} then |f(x)-f(y)| < \varepsilon.

    Thanks for that

    (Original post by Haddock3)
    And here was I thinking you were infallible. :P I'll have a think, but first I'm told that Apostol's book on Analysis is v. good, probably worth a look.
    Thanks, I'll have a look into that! And I'm far from infallible; people here just think I'm infallible because I answer questions I know I can answer :p:

    EDIT: That book looks quite good, but it doesn't have much in the way of uniform continuity and uniform convergence in it. It looks like it might be useful for the other parts of my analysis course though (e.g. vector derivatives) so I'll see if I can find a copy. Thanks!

    If anyone has any further book/website suggestions I'd really appreciate it!
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    (Original post by nuodai)
    (From a thread that I recycled, and then decided against recycling):

    Bingo!

    I can bound |f'(x)|=|-4x^3e^{-x^4}| by 3, and so given x,y \in \mathbb{R} there is some z between x and y such that |e^{-x^4}-e^{-y^4}| = |4z^3e^{-z^4}||x-y| \le 3|x-y|, so given \varepsilon > 0, if |x-y| < \frac{\varepsilon}{3} then |f(x)-f(y)| < \varepsilon.

    Thanks for that


    Thanks, I'll have a look into that! And I'm far from infallible; people here just think I'm infallible because I answer questions I know I can answer :p:
    Damn you, MVT! What chance does a 1st year who's only just learnt about it have of thinking to apply it in situations like these... :P

    At the risk of exposing my ignorance here, doesn't the MVT only work on closed bounded intervals? In which case surely the proof doesn't extend to a proof over \Re? While we're on that topic, if we only want a proof working on closed bounded intervals, surely we have it anyway, since exp and -x^4 are both u. continuous on any closed bounded interval...
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    (Original post by Haddock3)
    Damn you, MVT! What chance does a 1st year who's only just learnt about it have of thinking to apply it in situations like these... :P

    At the risk of exposing my ignorance here, doesn't the MVT only work on closed bounded intervals? In which case surely the proof doesn't extend to a proof over \Re? While we're on that topic, if we only want a proof working on closed bounded intervals, surely we have it anyway, since exp and -x^4 are both u. continuous on any closed bounded interval...
    We're talking about uniform continuity on the whole of \mathbb{R}, but in fact it's fine to use the MVT here. If we restrict f to any interval [x,y] \subset \mathbb{R} then we can apply the MVT on that interval to deduce that there is some z \in [x,y] s.t. f(y)-f(x) = f'(z)(y-x). The fact that I can then bound f' means that this I can write it in a form which is x,y (provided |x-y|<\delta, which allows me to deduce uniform convergence.

    Unfortunately it's not quite as simple as composing functions, since x^4 isn't a uniformly continuous function on \mathbb{R} (though it is continuous), and neither are e^x or e^{-x}.
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    (Original post by nuodai)
    Unfortunately it's not quite as simple as composing functions, since x^4 isn't a uniformly continuous function on \mathbb{R} (though it is continuous), and neither are e^x or e^{-x}.
    Hehe I realised THAT, but mentioned it just in case we WERE talking about closed bounded intervals, in which case composition would be fine. :P It was the first bit I didn't get, but do now, cheers. (Love how this thread has turned around entirely, btw )
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    (Original post by nuodai)
    (From a thread that I recycled, and then decided against recycling):
    I much prefer it if people don't delete threads, if only because it provides a large set of questions for other people to practice with
    Bingo!

    I can bound |f'(x)|=|-4x^3e^{-x^4}| by 3, and so given x,y \in \mathbb{R} there is some z between x and y such that |e^{-x^4}-e^{-y^4}| = |4z^3e^{-z^4}||x-y| \le 3|x-y|, so given \varepsilon > 0, if |x-y| < \frac{\varepsilon}{3} then |f(x)-f(y)| < \varepsilon.

    Thanks for that
    I think the following is true and would immediately imply the result as well

    Suppose f is continuous on the real line and f vanishes at +-infinity. Then f is uniformly continuous on the real line.
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    (Original post by nuodai)
    I can bound |f'(x)|=|-4x^3e^{-x^4}| by 3,
    I am wondering how you got 3 as a bound. The easiest bound that I could see was 4.
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    I also think this is true (re post #7).
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    (Original post by IrrationalNumber)
    I much prefer it if people don't delete threads, if only because it provides a large set of questions for other people to practice with
    I think for the number of people who would trawl through threads to find questions to practice with, it's not worth risking someone going to a huge effort to answer a question if I've already answered it myself. [I mean, sometimes threads I made when I was in Y12 get dug up by people responding to them.] Maybe I'm wrong.

    (Original post by IrrationalNumber)
    I think the following is true and would immediately imply the result as well

    Suppose f is continuous on the real line and f vanishes at +-infinity. Then f is uniformly continuous on the real line.
    Yes, that's true; you can pick a closed interval I, on which it must be uniformly continuous, such that |f(x)| < \varepsilon for all x \not \in I, and then the result follows pretty much immediately. Thanks for your help

    (Original post by IrrationalNumber)
    I am wondering how you got 3 as a bound. The easiest bound that I could see was 4.
    A somewhat contrived method involving finding where the maximum of f'(x) is (i.e. where x=\frac{3}{4}).
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    (Original post by DFranklin)
    I also think this is true (re post #7).
    Infact, I think we can generalize the result to f(x) having limits at +-infinity. But now we're getting away from the original thread.
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    (Original post by IrrationalNumber)
    Infact, I think we can generalize the result to f(x) having limits at +-infinity. But now we're getting away from the original thread.
    My problems with analysis go much further than this question; any results like this are helpful.
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    (Original post by nuodai)

    Yes, that's true; you can pick a closed interval I, on which it must be uniformly continuous, such that |f(x)| < \varepsilon for all x \not \in I, and then the result follows pretty much immediately. Thanks for your help
    I think the trickiest part of the proof is showing that if |x-y|<delta and x is in I and y is not in I then |f(x)-f(y)|<eps

    A somewhat contrived method involving finding where the maximum of f'(x) is (i.e. where x=\frac{3}{4}).
    I don't know if I'm talking out my **** here but I generally find analysis easier if I try to do things like estimating rather than trying to find maxima/minima by differentiating. What I did was |4x^3e^{-x^4}|\leq \frac{4|x|^3}{1+x^4} and then splitting into cases of |x|>1 and |x| less than or equal to 1.
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    Again, I'm also all for estimating rather than finding the actual maximum. (Excepting in those cases like contraction mapping where you may need the actual maximum).
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    (Original post by IrrationalNumber)
    I don't know if I'm talking out my **** here but I generally find analysis easier if I try to do things like estimating rather than trying to find maxima/minima by differentiating. What I did was |4x^3e^{-x^4}|\leq \frac{4|x|^3}{1+x^4} and then splitting into cases of |x|>1 and |x| less than or equal to 1.
    (Original post by DFranklin)
    Again, I'm also all for estimating rather than finding the actual maximum. (Excepting in those cases like contraction mapping where you may need the actual maximum).
    Estimation's great if you can spot it; and I often can, but if it gets much more sophisticated than the Cauchy-Schwarz inequality or saying that |\sin x| \le 1 then I just panic. I'm particularly weary of Taylor series, because they bring with them so much baggage (e.g. are all the terms positive? Does it converge there?). I think I've just not had enough practice with it, so I chicken out and just differentiate, forsaking an elegant method for one which just 'gets it done' (which obviously isn't much good if the function isn't differentiable, or if I'm trying to show it's differentiable!).

    So it's good that you mention estimation; it seems an obvious thing to use but I haven't given it much of my time to date, so I'll try and get good at it somehow.
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    Yeah, just gets it done is also fine.

    In the case of x^3 e^{-x^4}, I think (in this context) it would also be fine to say it tends to 0 as x->+/- infty, so is bounded, by M, say.
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    (Original post by nuodai)
    So it's good that you mention estimation; it seems an obvious thing to use but I haven't given it much of my time to date, so I'll try and get good at it somehow.
    Pretty much every proof in analysis involves estimating in some sense so that's possibly the main reason why you're finding it hard.

    Incidentally, another trick that is useful when you're talking about functions is to hop from point to point using the triangle inequality. I'll show you what I mean by that by giving a bit of the proof of that statement I thought was true above.

    Fix eps>0. We know there are x and y such that w\geq x or w\leq y implies |f(w)|\leq \frac{\epsilon}{3} .
    The interval I:=[x,y] is closed and f is continuous on this interval, so there is a delta such that for all u,v in [x,y] with  |u-v|&lt;\delta then  |f(u)-f(v)|\leq \frac{\epsilon}{3} .
    Now if u>y>v and |u-v|&lt;\delta then  |f(u)-f(v)|\leq|f(u)-f(y)|+|f(y)-f(v)|\leq \frac{\epsilon}{3} + |f(y)|+|f(v)| \leq \epsilon
    That's the bit I mean when I'm talking about point hopping.
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    It might be worth noting here that you've used the general result that

    differentiable with bounded derivative => Lipschitz => uniformly continuous

    (the first implication comes from MVT, the second from setting \delta = \epsilon/K in the definition of uniform continuity, where K is the Lipschitz constant)
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    (Original post by .matt)
    It might be worth noting here that you've used the general result that

    differentiable with bounded derivative => Lipschitz => uniformly continuous
    Well, derived, rather than used. You don't need to know anything about Lipschitz conditions (or even how to spell the word!) to prove the result.
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    (Original post by nuodai)
    On a side-note, I don't know what it is about uniform continuity that makes me instantly panic and run for the hills, but getting stumped on bog-standard easy questions like this certainly isn't helping!
    Ha! Ditto and add metric spaces in general to my list.
 
 
 
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