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May 11

Curvature question

some time ago, I asked here what the curvature of the torus {cos(u),sin(u),cos(w),sin(w)}, (u & w varying from 0 to 2pi) was. David Eppstein answered that the curvature was zero. What is the curvature of {Acos(u),Asin(u),cos(w),sin(w)} if A does not equal 1? Thanks, Rich (talk) 04:19, 11 May 2009 (UTC)[reply]

Notice that the various notions of curvature are concepts in Riemannian geometry, and as such, they refer to a Riemann structure. So your question is ambiguous if you do not specify what curvature you mean, and what metric you endow your torus. The simplest is the flat, corresponding to the torus as a quotient of the 2 dimensional Euclidean space; in this case the various curvatures are zero at any point. The metric induced from R⁴ by the embedding (Acos(u),Asin(u),cos(w),sin(w)) is also isometric with the torus as quotient of R². ~~then they are not identically zero, and can be easily computed because the two principal curvatures at any point are 1/A and 1~~. If you referred to the total Gaussian curvature, then it is 0 for the torus, no matter what the metric is, by the Gauss-Bonnet theorem.--pma (talk) 09:14, 11 May 2009 (UTC)[reply]

I refer to the metric induced from R4 by the embedding that was given. So what are the computations?Rich (talk) 19:45, 11 May 2009 (UTC)[reply]

~~OK, sorry. If you agree that κ₁ and κ₂ are 1 and 1/A, just use the definition of curvature that you want :~~ mean curvature = (κ₁+κ₂)/2, Gauss curvature =κ₁κ₂, scalar curvature=2κ₁κ₂. Check the links. --pma (talk) 21:12, 11 May 2009 (UTC)[reply]

No problem, and thanks!Rich (talk)

Well in fact I think I was not very concentrated, and I was still thinking of the surface as embedded in R³, sorry. Indeed, the metric on your torus induced from R⁴ is obviously isometric to the flat metric on the torus as a quotient of R², for the same reason that the metric on {(x,y) in R²: x²+y²=r²} induced by the scalar product of R² and the quotient Riemann structure on R/2πrZ are isometric --the isometry being given by the exponential map. --pma (talk) 18:41, 12 May 2009 (UTC)[reply]

So the matrix of curvatures is the zero matrix then??Actually, is there a matrix of curvatures that'sbeen set up for this?-if so is it a 4x4 matrix or what? You seay it's isomorphic to theflat torus; so let me tweak it a bit more:What's the curvature of (Acosu,Bsinu,cosw,sinw}?Thanks, Rich (talk) 20:52, 12 May 2009 (UTC)[reply]

Still zero, and it will remain zero as long as no coordinate in your embedding depends on both u and v. For a 2D surface like yours the intrinsic curvature is given by a single number at each point, the Gaussian curvature. -- BenRG (talk) 01:58, 14 May 2009 (UTC)[reply]

ok, what about the weingarten matrix?(just remembered name)Is it zero? What size matrix is it for this surface, 1x1,2x2,4x4? Thanks, Rich Peterson99.169.203.133 (talk) 21:22, 14 May 2009 (UTC)Rich (talk) 21:47, 15 May 2009 (UTC)[reply]

do number of consecutive misses of coin toss follow normal distribution? What is the standard deviation?

Out of n coin tosses (say, 100) what is the number of times you would expect more than j consecutive misses, for example, j=2, more than 2 consecutive misses, j=3, more than 3 consecutive misses, etc. Do number of consecutive misses follow the normal distribution, and what is the standard deviation? This isn't homework, I'm just interested. Thanks :) —Preceding unsigned comment added by 94.27.149.212 (talk) 13:49, 11 May 2009 (UTC)[reply]

I'm not sure what you mean by a "miss", so I'll just assume that a "miss" is some event that occurs with probability p and is independent for each coin toss. I'm also not sure what your question is asking. Do you mean:

What is the probability that the first j tosses are misses? (See geometric distribution.)
What is the probability that the first j tosses are misses, and the j + 1-th toss is a hit? (See geometric distribution, which gives the probability distribution of j.)
In n tosses, what is the probability that some j consecutive tosses are misses?
In n tosses, let j be the largest number of consecutive misses that occur. What is the distribution of j?
In n tosses, count how many times a streak of j or more consecutive misses is ended. [bold by OP] What is the distribution of that count?
In n tosses, count how many tosses are misses preceeded by j - 1 other misses. What is the distribution of that count? (This question differs from the preceeding in that streaks of j + k misses are counted with multiplicity k + 1.)

It looks to me like you're asking the second to last question above, but I don't think I understand you correctly. Eric. 131.215.159.91 (talk) 21:29, 11 May 2009 (UTC)[reply]

Yes, how embarrassing. By "misses" I should have said "tails", since a hit is a heads. The question IS the second-to-last one, which you can tell because I said what is the NUMBER of times you would expect more than j consecutive misses. I even gave some examples. WWhen I ask about whether it follows the normal distribution, what I'm really asking is whether I can use standard deviation as we usually do for normal distributions...

Maybe I should make up the kind of answer that I'm looking for. THe following is just BS showing the kinds of answers that I want:

“

For 100 guesses:

number of streaks of 1 tails:

average of 50, standard deviation of 5, distribution follows the bell curve. (So with 67% chance the 100 guesses will leave you with between 45-55 streaks of 1 or more tails, with 95% chance it will leave you with between 40-60 streaks of 1 or more tails, and with 99.7% chance you will get between 35-65 streaks.)

number of streaks of 2 or more tails (sample ...HTTTHTTH... includes 2 such streaks):

average of 30 out of 100, distribution still follows a bell curve, standard deviation of 4. (so with 67% chance there will be between 26-34 streaks of 2 or more tails, with 95% chance there will be between 22-38 such sreaks, with 99.7 there will be between 18-42 such streaks...)

number of streaks of 3 or more tails:

an average of 20 out of 100, distribution is a bell curve, standard deviation is 3.

How many streaks of 4 or more tails? 10 on average, bell curve distribution, standard deviation 2.5
How many streaks of 5 or more tails? 6 on average, bell curve distribution, standard deviation 1.8
How many streaks of 6 or more tails? 3 on average, bell curve distribution, standard devetation 1.5

”

But these are just guesses, not a real answer, just BS. What is the real answer please? Thanks. 79.122.55.34 (talk) 08:01, 12 May 2009 (UTC)[reply]

"Do number of consecutive misses follow the normal distribution"? The "number of" something is a nonnegative integer, but a normally distributed random variable also assumes fractional and even negative values. So the answer is "no". This does not mean that the normal distribution is not a useful approximation to distributions of counts. For example, the number of counts on a Geiger counter has a (poisson) distribution which is usefully summarized by its mean value and standard deviation, and approximated by a normal distribution. Bo Jacoby (talk) 08:25, 12 May 2009 (UTC).[reply]

That answer doesn't make any sense. For example the number of heads out of 100 fair coin tosses follows a normal distribution, with a mean of 50. The fact that you can't have a fractional or negative number of heads has nothing to do with it. Your answer isn't helpful in the least. 79.122.55.34 (talk) 09:33, 12 May 2009 (UTC)[reply]

Sorry, but, yes, Bo's answer does make sense, even though you may not have found it helpful or it may not have answered the question that you intended to ask. And sniping at the folks who do attempt to answer your questions will only discourage everyone else from responding - remember that the people who read your questions, try to understand them and perhaps spend time writing a response are doing you a favour here. Gandalf61 (talk) 10:27, 12 May 2009 (UTC)[reply]

And the number of heads out of 100 fair tosses does not follow a normal distribution. It follows the Binomial distribution. One of th things that immediately tells us that it is not a normal distribution is the fact that the distribution is discreet. Not every bell shape curve is a normal distribution. Taemyr (talk) 11:03, 12 May 2009 (UTC)[reply]

I apologize for my tone. You have to understand, I've been trying to get understand this simple stuff for a week, to no avail. If the number of heads out of 100 fair tosses doesn't follow a normal distribution, does that mean I am in error when I try to apply the 68-95-99.7 rule to it using standard deviation? It is getting so frustrating trying to get an answer that I'm just running a simulation with a million times n flips (in my example 100), and seeing how many times each number of runs of j tails in a row comes up. (here is an example output from my simulation doing just 10,000 runs of n=128: "10: 0:9445 1:545 2:10". This means that out of 10,000 runs of 128 flips, in about 95% of them there were no runs of 10 (or more) tails, in a small percentage (about 5%) there was a single run of 10 (or more) tails in a row, and 2 times out of 10,000 there were two separate runs of ten or more tails in a row. As you go from j=10, to lower and lower numbers, obviously the number of runs you find on average increases. For example for n=128 and j=3, my result is: 3: 0:1 1:1 2:23 3:112 4:374 5:736 6:1268 7:1670 8:1888 9:1603 10:1181 11:662 12:304 13:119 14:44 15:10 16:4. It's clear that the mode in this case is 8 runs. As for the actual distribution, I will have to wait until my simulation over 1,000,000 runs, as I think 10,000 runs might not be exact enough. But I think it's totally stupid for me to have to do this, absent some easy closed formula, which I have not been getting help with, hence my frustration. Sorry again about the latter. 79.122.55.34 (talk) 11:16, 12 May 2009 (UTC)[reply]

It should rather approximately follow a Poisson distribution with mean

\lambda ={\frac {n-j}{2^{j+1}}}

.

The larger n / j is, the better this approximation (you've done enough simulations to test roughly how accurate this approximation is). It's possible that a different value of lambda gives a better approximation, but in any case the above lambda is the true mean of your distribution. For a fixed j, your distribution divided by n converges to a Gaussian distribution as n goes to infinity; this is the same Gaussian distribution that the above Poisson distribution, when divided by n, converges to as n goes to infinity. The convergence is about the same speed regardless of the value of j. Thus for large enough n, your distribution can be approximated by a Gaussian, but the approximiation with a Poisson is probably much better, especially if j is small. Eric. 131.215.159.91 (talk) 20:17, 12 May 2009 (UTC)[reply]

The OP asked for the standard deviation, which is the square root of λ. In order to obtain a normal distribution you

subtract the mean value,
divide by the standard deviation,
let n go to infinity.

Bo Jacoby (talk) 14:58, 13 May 2009 (UTC).[reply]

How did you find the standard deviation? I don't think that it is exactly

{\sqrt {\lambda }}

. Eric. 131.215.159.91 (talk) 17:28, 13 May 2009 (UTC)[reply]

The Poisson distribution article says that the mean and the variance are both equal to λ, and the standard deviation is the square root of the variance. Bo Jacoby (talk) 18:37, 13 May 2009 (UTC).[reply]

Sorry, I mean, how do we know the standard deviation of the OP's distribution is

{\sqrt {\lambda }}

? Sure the Poisson distribution has that standard deviation, and the Poisson distribution is a reasonable approximation for the OP's distribution, but we don't know that they have to have the same standard deviation. Eric. 131.215.159.91 (talk) 20:19, 13 May 2009 (UTC)[reply]

Take, for example, n = 2j, in which case the approximation will actually be rather poor (as n / j is so small). Then the standard deviation of the OP's distribution will be

{\sqrt {\lambda -\lambda ^{2}}}

, which is not equal to

{\sqrt {\lambda }}

for positive j. Eric. 131.215.159.91 (talk) 20:27, 13 May 2009 (UTC)[reply]

An approximation to the OP's distribution can be obtained like this. There are n − j + 1 consecutive sequences of j tosses out of a total of n tosses. (If n = 10 and j = 5, the n − j + 1 = 6 consecutive sequences are 1..5, 2..6, 3..7, 4..8, 5..9, 6..10). If these sequences are considered approximately independent, then the probability that such a sequence is all misses is approximately P = 2^{− j}, and the number of such sequences has a binomial distribution with mean value ± standard deviation = (n − j + 1)·P ± √((n − j + 1)·P·(1 − P)). When j is not very small, the factor (1 − P) can be omitted and the distribution turns into the poisson distribution with mean value ± standard deviation = (n − j + 1)·P ± √((n − j +1)·P).

Change of variables in 3 dimensions - please help!

Hi there - I'm revising for my exams in a few weeks and decided to look at some of the older past papers for my uni course, but they tend to deviate off the syllabus a fair bit so I haven't been taught how to approach a problem like this systematically, I'm using mostly guesswork and logic, so it'd be very handy to know a more standard tactic to solve a question like this (although I imagine much of it like regular integration is experience...)

(i) 'The domain S in the (x,y) plane is bounded by $y=x$ , $y=ax(0\leq a\leq 1)$ and $xy^{2}=1(x,y\geq 0)$ . Find a transformation $u=f(x,y)v=g(x,y)$ such that S is transformed into a rectangle in the (u,v) plane.'

I figured I may as well aim to transform into the unit square if I could, since it's only a short step translating from a rectangle in the (u,v) plane to the unit square. The only problem which struck me here was the fact a rectangle needs 4 'bounds' whereas the given area in the (x,y) plane has only 3 bounds. Through, as I said, mostly fairly unconfident guesswork, I came up with $u={\frac {xy^{2}(y-ax)}{{\frac {1}{\sqrt {x}}}-ax}},v={\frac {xy^{2}(y-x)}{{\frac {1}{\sqrt {x}}}-x}}$ - would this work? It gives u=1 and v=1 along the $xy^{2}=1$ boundary and 0 along the y=x and y=ax boundaries, so would that be okay? In general, what's a good method for approaching this sort of question systematically?

(ii) 'Evaluate $\int _{D}{\frac {y^{2}z^{2}}{x}}\,dx\,dy\,dz$ , where D is the region bounded by

$y=x$ , $y=zx$ , $xy^{2}=1$ $(x,y\geq 0)$

and the planes $z=0$ , $z=1$ .'

Obviously a change of variables is the way they want you to go, and I'm happy with the actual process of changing variables, the Jacobian, the volume integral etc, but how do I actually work out -what- variables to change to? It's not like it's a symmetrical area which points towards spherical polars or anything like that, so I'm not sure how to proceed!

Thanks very much for the help! Typeships17 (talk) 19:47, 11 May 2009 (UTC)[reply]

Try

u={\frac {y}{x}}

and

v=xy^{2}\,

. Dauto (talk) 23:16, 11 May 2009 (UTC)[reply]

Just giving an answer isn't going to help the OP much. How did you come up with that answer? --Tango (talk) 23:29, 11 May 2009 (UTC)[reply]

Seems prety obvious to me. Dauto (talk) 23:31, 11 May 2009 (UTC)[reply]

Obviously it wasn't obvious to the OP. --Tango (talk) 23:36, 11 May 2009 (UTC)[reply]

Yes, obiously not. But I found it just by inspection and you asked me how did I get the answer. I don't know what else to say except that it seemed obvious to me. Dauto (talk) 02:23, 12 May 2009 (UTC)[reply]

I am inclined to agree with User:Dauto in this situation. Usually with questions of this nature, some algebraic work has to be done. However, in this case, the question is best answered by inspection (correspond this to the case of linear equations - equations such as 2x + 7 = 18 are hard to solve in the head of a beginner, but solving an equation such as x + 3 = 6 is so simple, that it is nearly impossible to explain how to solve it). As for the integral, it is easy once you use the Jacobian matrix. --PS T 04:36, 12 May 2009 (UTC)[reply]

If the method is "by inspection" then write "by inspection", that way the OP knows there isn't anything complicated going on and can work out what you've done from themselves. --Tango (talk) 15:32, 12 May 2009 (UTC)[reply]

Thanks very much for the help guys, that's great! :) Typeships17 (talk) 11:00, 12 May 2009 (UTC)[reply]

"Dauto" is not expressing himself very well. u = y/x is contant along straigh lines through the origin. When u = a then you're on the lower boundary of the region, the line y = ax. When u = 1, then you're on the upper boundary. As u goes from a up to 1, observe how the line moves. The look at v = xy². It is equal to 1 on the curved boundary, and as it shrinks down to 0, notice how the curve moves. Michael Hardy (talk) 21:04, 12 May 2009 (UTC)[reply]

I thought I expressed myself pretty well, thank you. I said all I wanted to say, no less, no more. Dauto (talk) 22:05, 13 May 2009 (UTC)[reply]