Wikipedia:Reference desk/Archives/Mathematics/2014 March 30

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March 30

Continuously differentiable function

Hi,

This is a homework question, but I'm hoping just for a hint or two, or confirmation I'm doing it correctly. We were given a piecewise function, defined by f(x)= x^3 if x>=0, -x^3 if x<0. The first part was to prove that f was differentiable, and then it asks what order is f continuously differentiable. I've been finding the derivatives of f by using the definition at x=0, and I've been checking continuity of the derivative function using the definition of continuity, i.e. left side limit=function value=right side limit. Then I repeat for the next derivative (f', f⁽²⁾x, f⁽³⁾x). This seems very tedious- is there another way of doing it? Once I get to f⁽³⁾x=0, I can conclude that f is continuously differentiable of infinite order, right? Or am I missing something...

Thanks in advance! 27.32.104.185 (talk) 12:11, 30 March 2014 (UTC)

If you get f⁽³⁾(x)=0 in this case then you're doing something wrong. Also, for real functions, differentiable, twice differentiable, thrice differentiable, ..., infinitely differentiable and analytic are all different, of which your problem is partial example. For complex functions things are (ironically) a bit simpler.--RDBury (talk) 16:05, 30 March 2014 (UTC)

You might be taking an excessively tedious approach. The derivatives are (or should be) easy to find whenever x ≠ 0 without resorting to the tedious approach, and you can make more intuitive inferences at x = 0 from this approach. —Quondum 18:25, 30 March 2014 (UTC)

Sorry, I meant f⁽⁴⁾(x)=0. I know apart from x=0 the power rule suffices, but what other way could you find f'(0), etc other than using the definition of the derivative on both sides? Can we assume f'(0)=lim f'(x) as x approaches 0? Could you explain what you mean by intuitive inferences? 129.94.8.150 (talk) 00:53, 31 March 2014 (UTC)

I mean that simply by plotting the derivative of successive orders you'll quickly get a feel for what's what. And you would not be correct to say that f⁽⁴⁾(0)=0. To see that, you need to find f⁽³⁾(x). —Quondum 02:24, 31 March 2014 (UTC)

• The approach is perhaps excessively tedious in the sense that an advanced mathematician looking at this would immediately know the answer without having to do anything -- but the tedious approach is what you are expected to do as a student. But if you are coming to the conclusion that the function is continuously differentiable of all orders, you've made a mistake -- f⁽³⁾ is not continuous. Looie496 (talk) 02:37, 31 March 2014 (UTC)

We don't know what your teacher asked for in terms of work shown. They might prefer you to show an epsilon-delta proof of the derivative for all points in the domain (but I doubt it). Anyway, the "least work" is to show a that a certain derivative of f is not continuous. Demonstrating lack of continuity at a point rigorously can be bit tricky, but in this case your teacher probably doesn't want that either. As for "can we assume.." -- no, not in general, unless you've already shown that f' is cts at x=0. SemanticMantis (talk) 15:32, 31 March 2014 (UTC)

Thanks, I got it. And I realized the simple mistake I've been doing as I differentiated it. Thanks everyone. 27.32.104.185 (talk) — Preceding undated comment added 05:29, 1 April 2014 (UTC)

Another Curios "Coincidence" Between The Superellipse x³ + y³ = 1 And The Gaussian function

As before, my question is whether there's more to this than meets the eye, especially given the general relationship between Gaussian integrals and superellipses:

$\int _{0}^{1}{\sqrt[{2}]{1-x^{2}}}~dx~=~{\frac {\pi }{4}}$

$\int _{0}^{1}{\sqrt[{3}]{1-x^{3}}}~dx\approx {\sqrt {\frac {\pi }{4}}}~=~\int _{0}^{\infty }e^{-x^{2}}dx$

The error in that last approximation is of about 1/344 . For yet another "coincidence" between the two, see here. My only explanation so far is that all three functions (Gaussians, cubic superellipses, and hyperbolic secants) are bell-shaped, and therefore can be brought close to each other in a manner similar to that of Laplace's method. — 79.113.195.131 (talk) 14:37, 30 March 2014 (UTC)

Name for operation on convex polytopes

The cone of a convex polytope P in a vector space V can be defined as a the subset of R×V given by {(t, {1-t)p): p∈P, t∈[0, 1]}. I found it useful to generalize this as follows: Let P₁ and P₂ be convex polytopes in vector spaces V₁ and V₂. Define a convex polytope Q in R×V₁×V₂ by Q={(t, {1-t)p₁, tp₂): p₁∈P₁, p₂∈P₂, t∈[0, 1]}. This reduces to the cone of P₁ when P₂ is a single point. More generally, If P_i has v_i vertices, f_i faces and dimension d_i then Q has v₁+v₂ vertices, f₁+f₂ faces and dimension d₁+d₂+1. Does the operation (P₁, P₂)→Q (up to affine variations in the definition) have a name or standard notation? --RDBury (talk) 16:20, 30 March 2014 (UTC)

This looks like a join (topology). --Mark viking (talk) 16:31, 30 March 2014 (UTC)

Thanks, it looks like topologically it is the join. So I'll go with "polytope join" unless someone has a more standard name. I also found prismatoid which is slightly similar but not the same. The combinatorial equivalence class of the join depends only on P₁ and P₂ but for the prismatoid it depends on how they are oriented. --RDBury (talk) 19:17, 30 March 2014 (UTC)