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May
31 |
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A quick and elementary question from Hubbard's Teichmuller Theory : Volume I
The Cantor-Lebesgue function has derivative 0 almost everywhere but is continuous and strictly increasing. The issue is that for a function on $R$ to have distributional derivatives it must be absolutely continuous; simply having a well-defined almost everywhere derivative is not enough. |
May
31 |
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A quick and elementary question from Hubbard's Teichmuller Theory : Volume I
Suppose you are on $R^1$ and $l$ is just the point $\{0\}$. Then the function $f$ which is 0 when $x<0$ and 1 when $x\ge 0$ has distributional derivative 0 away from 0. So $f'$ is defined almost everywhere and $f'$ is the distributional derivative for any function compactly supported in $R^1-\{0\}$. But if $\phi$ is any $C_c^\infty$ supported function which is 1 in a neighborhood of $0$, then $-\int f\phi'=\phi(0)=1$ rather than 0 if $f'$ were actually the distributional derivative on all of $R$. You can come up with clever ones so that $f$ is continuous: see the Cantor-Lebesgue function. |
May
31 |
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A quick and elementary question from Hubbard's Teichmuller Theory : Volume I
Yep, got a surprising little more mileage out of that book. The OP and I should thank you for asking for the book back, otherwise the book wouldn't have been right next to me when I saw this question and this answer would never have happened. |
May
31 |
answered | A quick and elementary question from Hubbard's Teichmuller Theory : Volume I |
May
31 |
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Why is mechanical differentiation so hard to get right?
Well, it doesn't seem to be an issue with the derivative at all. Wolframalpha seems to believe that the given expression is $x^2/2$ when $x>0$, for instance. |
May
30 |
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Why is differentiating mechanics and integration art?
@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 |
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Why is differentiating mechanics and integration art?
@Deane A bit tongue in cheek, but: division of real numbers is harder than multiplication because it's secretly still multiplication, just with the additional step of inverting. Multiplication by $x$ is a continuous function, but multiplication by $x^{-1}$ is no longer continuous near zero. But if you look at, say, the circle group, multiplication and division are clearly equally difficult! |
May
30 |
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Why is differentiating mechanics and integration art?
Ryan --- he's saying that computing the derivative is simpler because it only depends on local information at a point. Integration depends on information about every point on a fixed interval, simultaneously. Thus differentiation is fundamentally simpler than integration, simply from the definitions. Doesn't it seem a lot easier to organize the cutting down of a forest one tree at a time than having to cut them all down simultaneously? |
May
30 |
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Why is differentiating mechanics and integration art?
That should be $n^{318}$. How shameful! |
May
30 |
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Why is differentiating mechanics and integration art?
@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 |
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Why is differentiating mechanics and integration art?
1. u-sub is a statement about derivatives pushed into integration via the fundamental theorem of calculus: it's just a restatement of the chain rule. You can still prove a version of u-sub for absolutely continuous functions, but it takes more work. 2. You still need to check continuity at each point, but you can basically check the points one-by-one. But in any case, every absolutely continuous function is continuous, but plenty of continuous functions are not absolutely continuous. Indeed I would guess "most" continuous functions are not absolutely continuous. Isn't that an asymmetry? |
May
30 |
answered | Why is differentiating mechanics and integration art? |
May
30 |
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Why is differentiating mechanics and integration art?
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
15 |
answered | A Bijection Between the Reals and Infinite Binary Strings |
May
12 |
answered | Best online mathematics videos? |
May
1 |
awarded | Yearling |
Dec
1 |
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Ingenuity in mathematics
That's interesting to hear, Elizabeth. I've almost always had the opposite experience, where they don't believe me and spend a great deal of energy arguing their point. |
Nov
30 |
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What is the difference between hard and soft analysis?
Mariano, the most likely reason people find your comment downvotable is that barbarous is essentially a synonym for barbaric in English, i.e. savage, violent, cruel, brutal, uncivilized, and uncultured. So, the obvious interpretation of your statement is something to the effect of "wouldn't it be nice if someone translated that ugly German text into a language actually spoken by civilized people?" I assume this was not your intention, of course! It is also somewhat common for people to say that German is an ugly language, probably a sentiment that has carried over from the 1940s. |
Nov
27 |
awarded | Nice Answer |
Nov
26 |
answered | Nice Classes of Non-Closable Operators |