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May 17

Life density

In Conway's Game of Life, what's the maximum (or maximum known) density for stable (or oscillating) periodic patterns? For oscillators, I'll accept either the maximum or the minimum density in the cycle as "the density".

Mere curiosity here, no higher purpose. —Tamfang (talk) 03:06, 17 May 2011 (UTC)[reply]

This is mentioned in the Life Lexicon and its been proved the max density is a half. --Salix (talk): 08:50, 17 May 2011 (UTC)[reply]

Some nice examples can be found at LifeWiki: Agar.--Salix (talk): 08:55, 17 May 2011 (UTC)[reply]

It says this was only proven for stable patterns, but conjectured for oscillating patterns. -- Meni Rosenfeld (talk) 08:34, 23 May 2011 (UTC)[reply]

Thanks! —Tamfang (talk) 06:46, 15 June 2011 (UTC)[reply]

EXPAND MULTINOMIAL EXPRESSION

HOW DO I EXPAND THESE BINOMIAL EXPRESSION - (1+2X-X^2)^5 —Preceding unsigned comment added by 202.63.228.211 (talk) 03:42, 17 May 2011 (UTC)[reply]

1+10x+35x^2+40x^3-30x^4-68x^5+30x^6+40x^7-35x^8+10x^9-x^10 Wikinv (talk) 04:56, 17 May 2011 (UTC)[reply]

Giving the worked-out expression doesn't answer the question "how do I..", and more generally doesn't help the questioner solve similar (homework?) problems in the future. AndrewWTaylor (talk) 11:47, 17 May 2011 (UTC)[reply]

LIKE SO.

(1+2X-X^2)^5

( 1+2X-X^2 )^5

( 1+2X-X^2 )^5 —Preceding unsigned comment added by 72.179.51.84 (talk) 13:07, 17 May 2011 (UTC)[reply]

http://www.wolframalpha.com/input/?i=(1%2B2X-X^2)^5 Bo Jacoby (talk) 14:53, 17 May 2011 (UTC).[reply]

Polynomial expansion should be a good place to start. You could work up the powers finding

(1+2X-X^{2})^{2}

first then multiplying that by

1+2X-X^{2}

etc. Or let

y=2x-x^{2}

and use the binomial theorem on

(1+y)^{5}

or indeed jump straight to the Multinomial theorem.--Salix (talk): 21:50, 17 May 2011 (UTC)[reply]

Making fake fields with known covariance

I have a problem where I would like to simulate instances of a random data field given known statistical properties.

Specifically, $y({\vec {x}},t)$ is a random variable with a time-averaged expectation of zero at all locations, and

\mathrm {Cov} (y({\vec {x}}_{1},t),y({\vec {x}}_{2},t))=C({\vec {x}}_{1},{\vec {x}}_{2})

\mathrm {Cov} (y({\vec {x}},t),y({\vec {x}},t-t_{0}))=T({\vec {x}},t_{0})

Where $C({\vec {x}}_{1},{\vec {x}}_{2})$ and $T({\vec {x}},t_{0})$ can be assumed to be known for all locations and times.

How would one go about creating simulated fields for $y({\vec {x}},t)$ (over a set of grid locations) that obeyed the specified properties? I suspect (but am not entirely sure) that specifying C and T is not sufficient to completely constrain the statistical properties of the underlying probability distribution. If so, what other assumptions / approximations might be necessary. Dragons flight (talk) 10:17, 17 May 2011 (UTC)[reply]

You might find Correlation and dependence#Correlation and linearity useful or perhaps Normally distributed and uncorrelated does not imply independent. Dmcq (talk) 12:42, 17 May 2011 (UTC)[reply]

One way to do it is to use a Gaussian process. HTH, Robinh (talk) 20:44, 17 May 2011 (UTC)[reply]

Mobius transformations and concentric circles

Hello everyone,

In a course I'm taking, the lecturer has stated that if we have 2 circles which do not intersect at any point in the complex plane (lines also considered as circles), then we can take a mobius map mapping these 2 circles to concentric circles (e.g. |z|=1, |z|=R). Now it is obvious that if we can map to 2 concentric circles then our original 2 circles cannot touch, but it is not obvious to me why there necessarily exists such a transformation. I am aware that there is always a mobius map between any 2 circles on the Riemann sphere, but why is there necessarily one which will also make our 2 given circles concentric? (I mean 'concentric' in the sense that when they are seen in the complex plane, they are concentric around the origin.) Could anyone give me a very brief proof or explanation?

Thanks very much 131.111.185.74 (talk) 19:46, 17 May 2011 (UTC)[reply]

I think any pair of circles must be concentric on the Riemann sphere. You could rotate the sphere so that infinity lies in the centre of one circle, projecting to the plane would ensure that they images were concentric.--Salix (talk): 06:11, 18 May 2011 (UTC)[reply]

Sphere rotations only have 3 degrees of freedom. But you'd need 3 to map the first circle to the unit circle and 2 more to map move the center of the second to the origin. I'm pretty sure it can be done but group of Möbius transformations has 6 degrees of freedom and you need all but one of them; one degree of freedom use used up by a nontrivial subgroup that leaves both circles invariant. You know any circle can be mapped to the unit circle, and inversion swaps the inside with the outside, so you can assume the first circle is already the unit circle and the second circle is inside it. The Möbius transformations that preserve the unit circle are basically (though not quite) the isometries of the hyperbolic plane so it's really a matter of showing that there is an isometry that takes any circle to a circle with center at the origin. In other words what you would actually be showing is that circles on the plane are actually hyperbolic circles when considered as curves in the hyperbolic plane. See Poincaré disk model but the article does not mention this.--RDBury (talk) 08:25, 18 May 2011 (UTC)[reply]

The standard way to prove this is to put the circles into a normal form. I would map the first circle to the x axis such that the common diameter of the two circles goes to the y axis, and then use the remaining real scaling degree of freedom. Sławomir Biały (talk) 13:03, 18 May 2011 (UTC)[reply]

Probability Question

Is it possible to calculate the chance of getting a certain empirical probability relative to a given theoretical probability? For example, if I have a theoretical probability of about 25.5% but I get the result only 8.3% of the time. Do you understand what I am asking at all? Ryan Vesey (talk) —Preceding undated comment added 22:06, 17 May 2011 (UTC).[reply]

You're not being very clear, but let me guess: You took a random sample—in effect performed some Bernoulli trials—from a population where success occurs 25.5% of the time, and in your sample you got a success 8.3% of the time, i.e. about 1 out of 12. Certainly the probability of getting exactly one success, or at most one success, in 12 independent trials, with probability 25.5% of success on each trial, can be computed easily. It's a routine exercise. See binomial distribution. On the other hand, you could have had 2 successes, or at most 2 successes, in 24 trials, in which case the probability would be different.

But you shouldn't be making people guess what you mean. Michael Hardy (talk) 02:13, 18 May 2011 (UTC)[reply]

Well, I think the meaning of the question is reasonably clear, even though the terminology is incorrect -- Ryan is asking about the chance of getting a certain observed frequency given a certain probability distribution. It's a little harsh to demand that people who are confused about something be able to state their confusion in correct terminology. (The answer that Michael gave is the correct one, though: the probability is given by a binomial distribution.) Looie496 (talk) 02:34, 18 May 2011 (UTC)[reply]