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Math Help - derivatives

  1. #16
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Plato View Post
    Not sure what you point is? But I can tell you that over a thirty year stint of teaching real analysis, undergraduate and graduate, we never used anything but the epsilon-delta proofs. Limits, continuity, derivatives, and integrals are all defined using epsilon-deltas. We insist, well at least the school of mathematics I represent, a solid proof using epsilon and deltas is the only correct proof.
    Now wait a minute. Say you are trying to prove a theorem. This theorem can be "proven" using a specific technique. This technique has been proven to be correct using an epsilon-delta proof. Why would you then require an epsilon-delta proof of your theorem if you already have a proof using the technique?

    I don't have a problem with doing the epsilon-delta proof anyway, but to actually require one seems a bit, well, fussy to me. As if advanced methods should not be relied upon, even though they rest upon a firm basis.

    Consider, for example, the methods to find a derivative. I am quite happy to accept the "shortcut" formulas without referring directly to the definition of the derivative for each problem because each of the short-cut formulae have been proven. I don't demand that each new derivative I see be done the long way if I have a shorter method that has been proven. Of course, we CAN always do the problem the long way, and it's a good exercise, but surely even the meanest of professors wouldn't demand that I take the derivative of \frac{e^{1/x}}{ln(cos(x))} by using the definition of a derivative.

    -Dan
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  2. #17
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    Quote Originally Posted by Plato View Post
    a solid proof using epsilon and deltas is the only correct proof.
    Not at all.

    Once you proved a theorem through delta-epsilon. You can prove another theorem about limits without relying on on delta-epsilon. In fact, later on delta-epsilon is no longer used because all the important theorem were established by delta-epsilon.

    Here is an example, for n\not =0 prove that,
    \lim_{x\to 0} \frac{\sin nx}{nx} = 1
    If you are going to use delta-epsilon I would love to see how suffer trying to prove it.
    However, you can do this.
    We note that, if
    f=\frac{\sin nx}{nx}
    Then,
    f=g\circ h
    Were,
    h=nx and g=\frac{\sin x}{x}
    Since,
    \lim_{x\to 0}h=0
    And,
    \lim_{x\to 0}\frac{\sin x}{x}=1
    Thus,
    \lim_{x\to 0}\frac{\sin nx}{nx}=1
    I just used the limit composition rule.
    Its proof is based on delta-epsilon.
    But this prove did not have to be.
    See what I am saying.
    ---
    Analogy, most theorems are not proven by set theory. They are proven by other theorems proven by set theory.
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