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    \displaystyle\int_0^{\infty} \delta x\; dx

    I can't figure out if it would be zero or infinite? Or maybe 1?
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    (Original post by Plato's Trousers)
    \displaystyle\int_0^{\infty} \delta x\; dx

    I can't figure out if it would be zero or infinite? Or maybe 1?
    You meant \displaystyle\int_0^{\infty} \delta y\; dx
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    (Original post by Get me off the £\?%!^@ computer)
    You meant \displaystyle\int_0^{\infty} \delta y\; dx
    apologies, I did indeed.

    What's the answer
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    (Original post by Plato's Trousers)
    apologies, I did indeed.

    What's the answer
    I don't know. Infinity.
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    Unless I very much miss my guess, you have no sensible definition of what "infinitesimal" means here. So the question has no meaning.
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    (Original post by DFranklin)
    Unless I very much miss my guess, you have no sensible definition of what "infinitesimal" means here. So the question has no meaning.
    it is the usual meaning of \delta (as used in working out derivatives from first principles, eg x+\delta x).
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    Not from the attachment you've posted it isn't. (In fact a proper proof of derivatives from first principles almost certainly won't use the word infinitesimal except in an illustrative sense).

    And if it is, the answer is infinity.
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    (Original post by DFranklin)
    Not from the attachment you've posted it isn't.
    What is the difference? The graph shows a line which is an infinitesimal (or if you prefer, arbitrarily small) distance from the y=0 line. Thus the width of the band is \delta y

    How is that different to the \delta used in the derivation of the derivative?
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    When you derive the derivative, you don't use the word infinitesimal. Therefore, "the infinitesimal distance \delta y" is a different object from the \delta y you might see in a derivative proof. One of them is infinitesimal (whatever you think that means), and one of them is not.
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    I was assuming your delta y to have some constant positive value, no matter how small. With this assumption infinity is the answer.
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    (Original post by DFranklin)
    When you derive the derivative, you don't use the word infinitesimal. Therefore, "the infinitesimal distance \delta y" is a different object from the \delta y you might see in a derivative proof. One of them is infinitesimal (whatever you think that means), and one of them is not.
    I think I see what you mean. When we derive a derivative, we say that \delta x is very small and then we let it tend to zero. So I guess it's not infinitesimal.

    So are you saying infinitesimal has no meaning? Does it not mean "in the limit \delta x \rightarrow 0
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    It might do. But doing so doesn't really make sense here.

    Formally, you have two main ways of doing that here:

    \int_0^\infty \lim_{\delta y \to 0^+} \delta y \, dx = \int_0^\infty 0 \,dx = 0

    or

    \lim_{\delta y \to 0^+} \int_0^\infty \delta y \, dx = \lim_{\delta y \to 0^+} \infty = \infty

    neither of which seems very useful.

    (or you could not worry about ensuring the limit is from above, in which case the 2nd approach gives an undefined integral. Again not useful).
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    (Original post by DFranklin)
    It might do. But doing so doesn't really make sense here.

    Formally, you have two main ways of doing that here:

    \int_0^\infty \lim_{\delta y \to 0^+} \delta y \, dx = \int_0^\infty 0 \,dx = 0

    or

    \lim_{\delta y \to 0^+} \int_0^\infty \delta y \, dx = \lim_{\delta y \to 0^+} \infty = \infty

    neither of which seems very useful.

    (or you could not worry about ensuring the limit is from above, in which case the 2nd approach gives an undefined integral. Again not useful).
    Ok, thanks for this.

    So a bit of a cloth-eared question then
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    Well, I suspect there's a fair bit of context you haven't given...
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    (Original post by DFranklin)
    Well, I suspect there's a fair bit of context you haven't given...
    :nope: that was all there was (the attachment). It was billed as being this big, difficult problem that even the finest minds couldn't solve etc

    I think you have shown it for what it is...
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    Well, I'm guessing they really *do* mean infinitesimal (so not the same as \delta y in \lim_{\delta y \to 0}, where \delta y is just a number).
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    (Original post by DFranklin)
    Well, I'm guessing they really *do* mean infinitesimal (so not the same as \delta y in \lim_{\delta y \to 0}, where \delta y is just a number).
    but hang on, I thought you were saying "infinitesimal" is fairly meaningless?

    Or do you mean that \delta y really is infinitely small (ie more akin to dx) and so the vale of the integral is zero as you explained?
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    (Original post by Plato's Trousers)
    but hang on, I thought you were saying "infinitesimal" is fairly meaningless?
    If you just say the word, then yeah - because what does it actually mean?

    But you can do it properly: http://en.wikipedia.org/wiki/Non-standard_analysis

    I've not really done any NSA, but from analogy with measure theory I'd guess you can't actually make this integral make sense even under that framework.

    On the gripping hand, I suspect all they meant was some vague woolly idea of infinitesimal either. Conceptually the problem is "either it's 0, and the integral's 0, or it isn't 0, and the integral's infinite". Hard to get something inbetween...
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    (Original post by DFranklin)
    If you just say the word, then yeah - because what does it actually mean?

    But you can do it properly: http://en.wikipedia.org/wiki/Non-standard_analysis

    I've not really done any NSA, but from analogy with measure theory I'd guess you can't actually make this integral make sense even under that framework.

    On the gripping hand, I suspect all they meant was some vague woolly idea of infinitesimal either. Conceptually the problem is "either it's 0, and the integral's 0, or it isn't 0, and the integral's infinite". Hard to get something inbetween...
    interesting. Thanks

    BTW what does "on the gripping hand" mean?
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    http://www.catb.org/jargon/html/O/on...ping-hand.html
 
 
 
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