# derivatives

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• Sep 10th 2006, 05:54 PM
topsquark
Quote:

Originally Posted by Plato
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
• Sep 10th 2006, 06:40 PM
ThePerfectHacker
Quote:

Originally Posted by Plato
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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