Timeline for Why is differentiating mechanics and integration art?
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| May 30, 2011 at 23:33 | comment | added | Peter Luthy | @Ryan: yeah, good point. As I was writing the previous comment, I realized it wasn't so tough to just get the right leading coefficient; I usually prove it just using the product rule ($x^n=x\times...\times x) and do the first few examples to avoid explicitly teaching induction because a lot of students really don't get the finite sum stuff. Although, I have done some stuff with the Riemann sums they seem to like if you phrase it correctly and avoid anything resembling induction :D | |
| May 30, 2011 at 19:21 | comment | added | Ryan Reich | @Peter: Knowing that $\Delta n^k = kn^{k - 1} + O(n^{k - 2})$ gives you $\sum n^{k - 1} = n^k/k + O(n^{k - 1})$ by induction. Agreed that this is more subtle than most calculus classes, but it's the same as the proof of the power rule for derivatives (and thus integrals). | |
| May 30, 2011 at 15:58 | comment | added | Peter Luthy | That should be $n^{318}$. How shameful! | |
| May 30, 2011 at 15:57 | comment | added | Peter Luthy | @Ryan: ok, I see the distinction you're making. There is an argument one must make to be able to do constant step sizes (the function need be continuous). And I agree that in this case it comes down to computing those sums. I never remember the formulas for those sums aside from the first couple. Do you remember them for $n^318$, for example, or have any way to easily figure out what it is in such a case? :). I believe that business is rather involved (although all you really need is the coefficient of the highest power). | |
| May 30, 2011 at 14:08 | comment | added | Ryan Reich | @Peter Luthy: I said "formal calculus" for that reason :) I've been interpreting the question to refer to symbolic computation, since that's the way in which integration is usually said to be an art. However, if you get down to it, polynomials are among the easiest functions to compute Riemann sums of because we know how to do discrete integration (that is, $\sum_{n = 1}^N n^2$, etc.). This is as long as you use a constant step size, anyway, which everyone does in a calculus class. | |
| May 30, 2011 at 6:12 | comment | added | vonjd | @Michael: I think this question 'why this way around' hits the nail on its head! | |
| May 30, 2011 at 4:42 | comment | added | Peter Luthy | Ryan --- I think all of my calculus students would disagree that Riemann sums are "easy" for polynomials. Indeed, I would have to agree with them! The computation only really becomes simple when you apply the fundamental theorem of calculus. | |
| May 30, 2011 at 3:50 | comment | added | Thierry Zell | @Michael: You have a fair point. As I mentioned, this whole business of direct/inverse is stuff I heard in talks and conversations, but when I tried t pin down a specific reference while composing my answer, I found that it was like looking for a needle in a haystack because the terminology is used more loosely in scientific contexts (basically, to denote modeling, as far as I can tell). I'm pretty sure that you can make a good argument why integration is inverse, but right now the only thing I can think of is the non-uniqueness. | |
| May 30, 2011 at 3:11 | comment | added | Ryan Reich | @Michael Lugo: Perhaps it depends how you define the problem. Formal calculus on polynomials is certainly equally easy for either operation. Likewise finite Fourier series. In both cases, we can apply linearity and consult a finite table of derivatives or antiderivatives. However, composition usually produces too many functions to admit only a finite table as sufficient, so this method doesn't go very far unless you want to use infinite series, where everything becomes truly mechanical except for figuring out which closed-form function you have. | |
| May 30, 2011 at 2:27 | comment | added | Michael Lugo | But why is differentiation the "direct" problem and integration the "inverse" problem, and not the other way around? | |
| May 30, 2011 at 2:09 | history | edited | Thierry Zell | CC BY-SA 3.0 |
shortening the answer to avoid redundancy with other answers.
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| May 30, 2011 at 2:02 | history | answered | Thierry Zell | CC BY-SA 3.0 |