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July 14

A differentiable but not smooth function.

Surely if you have a real valued function which is differentiable but not smooth, for example

$f(x)={\begin{cases}-x^{2},&{\mbox{if }}x<0\\x^{2},&{\mbox{if }}x\geq 0\end{cases}}$

and you define a complex valued function simply by

$g(z)=f(\operatorname {Re} (z))$

g is a complex differentiable function which is not complex smooth. 130.56.84.56 (talk) 08:07, 14 July 2013 (UTC)[reply]

Actually, g wouldn't even be differentiable; unless f is constant. If you look at our article, Cauchy–Riemann equations, you'll see that for f(x + yi) = u(x, y) + iv(x, y) and u, v differentiable; then f is diff. iff u, v satisfy the equations. In your case, both partial derivs. of v are 0 since v is 0. Moreover, u only depends on x and has deriv. wrt x of 0, so it must be constant. If you look at our articles, Smooth function, Holomorphic function, and the stub Looman–Menchoff theorem; complex differentiable, basically, implies complex analytic and smooth.Phoenixia1177 (talk) 08:34, 14 July 2013 (UTC)[reply]

Of course. Differentiable in $\mathbb {R} ^{2}$ does not mean differentiable in $\mathbb {C}$ . 130.56.84.56 (talk) 08:50, 14 July 2013 (UTC)[reply]

I don't understand your response, where you agreeing or objecting? (I'm autistic and that trips me up sometimes)Phoenixia1177 (talk) 09:31, 14 July 2013 (UTC)[reply]

I was agreeing. The limit

\lim _{z\rightarrow 0}{\frac {g(z)}{|z|}}=0

exists, but the limit

\lim _{z\rightarrow 0}{\frac {g(z)}{z}}

does not. 130.56.84.56 (talk) 11:13, 14 July 2013 (UTC)[reply]

Formulas for map projections created by Flex Projector

The page at http://www.flexprojector.com/gallery/gallery.html shows some map projections created by Flex Projector. They only give the coefficient tables that define the projection, and no formula is given to compute the x-y coordinates from long-lat spherical coordinates given the tables supplied. I already know how to compute for pseudocylindrical projections, but what do the parallel curvature value do? Czech is Cyrillized (talk) 15:27, 14 July 2013 (UTC)[reply]

The article mentioned in the text can be found at [1]. That might help. It looks like they divide the latitude into 5° increments and there is one value of Length of Parallels (l), Distance of Parallels from Equator (d), Bending (b) and Meridians Distribution (m). Calculate these for your given latitude, let θ be the longitude. Ignoring bending and Meridians Distribution for a moment the coords will be x=l*θ and y=d. It looks like the bending follows a sine or cosine curve, so you may have y=d+b*cos(θ). Not quite sure how the Meridian distribution works.--Salix (talk): 16:55, 14 July 2013 (UTC)[reply]

What about cubic and quadratic bending? Czech is Cyrillized (talk) 00:08, 15 July 2013 (UTC)[reply]

Then I guess you have a cubic or quadratic equation y=d + α θ + β θ² + γ θ³.--Salix (talk): 04:04, 15 July 2013 (UTC)[reply]

must be a famous problem

Hello. Suppose I have ten counters to distribute as I wish among five squares, numbered 1,2,3,4,5. Then a random number 1-5 is called repeatedly, with known probabilities p1 to p5 (p1+...+p5=1) and if there is a counter on the corresponding square I remove it.

I want to place my counters so that they are all removed as quickly as possible (at first I thought that the optimal strategy was to put all your counters on the square with the maximal probability, but my logic was flawed). Actually I want to minimize the expected time to remove the last counter. Is this question a special case of a well-known problem? I thought it might be something like the secretary problem but it isn't. Can anyone advise? Robinh (talk) 20:35, 14 July 2013 (UTC)[reply]

I don't know whether this problem has been named or studied, but it is an interesting question. I think the optimal placing of counters will depend on the probabilities. If the probabilities are more or less equal then placing all your counters on one square is not a good strategy. But if the probabilities are very unequal - say p1 is close to 1 and p2 to p5 are very small - then placing all you counters on one square may indeed be the best strategy. A quick analysis of the 2 squares and 2 counters case (if stuck, consider a simpler case) shows that the expected time to remove the last counter is 2/p if you put both counters on square 1, and (p^2 - p + 1)/p(1-p) if you put one counter on each square (p is the probability of taking a counter from square 1). If p is less than 1/phi then the "one on each square" strategy is better; if p is greater than 1/phi then "both on square 1" is better. Gandalf61 (talk) 10:20, 15 July 2013 (UTC)[reply]

(OP) thanks Gandalf. I hadn't got round to the 2-square problem yet. But I realized last night that the problem was related to the coupon collector's problem but that doesn't have the element of choice at the beginning as we have here. Best wishes, Robinh (talk) 19:45, 15 July 2013 (UTC)[reply]

Why is it flawed logic to put all your counters on the highest probability square? (phrased that badly as it seems like I'm doubting the maths, I just don't get it intuitively) --Iae (talk) 13:16, 16 July 2013 (UTC)[reply]

Consider the special case where all the probabilities are equal (.2). If we put all the counters on a single square, then we're not done until 10 events occur, each with probability .2. If we instead distribute the 10 counters evenly, then we're still waiting for 10 events to occur, but the probabilities are better than .2: the probability of the first and second counters being removes are 1 each; the probability of the third and fourth counters being removed are each .8 or 1; etc. So if the probabilities are very close to equal, our best strategy is to distribute the counters evenly.--96.245.213.29 (talk) 13:57, 16 July 2013 (UTC)[reply]

(OP). Good to see that it's not just me that got it wrong ;-) re the above, if it helps, make the probabilities 0.20001,0.2,0.2,0.2,0.19999. Robinh (talk) 20:04, 16 July 2013 (UTC)[reply]

Consider a unit length ...

... and I take n cuts at it. What is the probability that I can form a shape from the pieces? (i.e. the cuts are completely random along the length and I wish to be able to construct a polygon by rotating and translating the line segments resulting from the cut, using all of them). I thought it was $1-1/2^{n-1}$ as after the first cut it would be that to prevent the forming of the polygon would have to fall in the same half as the original cut, but it was pointed out to me that if the first cut fell near the end of the length there would be more than half of the line a cut could fall in and still leave a length greater than half the original length.--Gilderien Chat|List of good deeds 21:08, 14 July 2013 (UTC)[reply]

This thread at MathOverflow is mostly about the triangle case (n=2), but has a reference to an American Mathematical Monthly article, and says the general answer is