Wikipedia:Reference desk/Archives/Mathematics/2017 September 10

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September 10

Differential equations

I'm having trouble with problems 1c, 1d, 1e, and 2 listed below. I listed 1a and 1b for reference and indented my answers (which are verified as correct). 1e is a multiple-choice question where the first and second answers are verified as incorrect. My previous answers for problem 2 were 6x + 4, 9/4e^{3x + 3} - 9/4e^{-x - 1} - 2x, and 9e³/4e^3x - 9/4ee^-x - 2x, all of which are verified as incorrect (the last two are equivalent).

1a. Verify that the one-parameter family y² - 2y = x² - x + c is an implicit solution of the differential equation (2y - 2) * y' = 2x - 1.

Differentiating y² − 2y = x² − x + c with respect to x, we have the following:

2y(y') - 2y' = 2x - 1

(2y - 2) * y' = 2x - 1

1b. Find a member of the one-parameter family in part (a) that satisfies the initial condition y(0) = 1.

y² - 2y = x² - x - 1

1c. Use your result in part (b) to find an explicit function(s) y = ϕ(x) that satisfies y(0) = 1.

y(x) =

1d. Give the domain of the function ϕ in interval notation.

 

1e. Is y = ϕ(x) a solution of the initial-value problem? If so, give its interval of definition; if not, explain.

• Yes, the interval is (−∞, ∞).
• Yes, the interval is (−∞, 0] ∪ [1, ∞).
• Yes, the interval is (0, ∞).
• Yes, the interval is (−∞, 1].
• No, both functions are solutions of the differential equation, but neither is a solution of the initial-value problem.

2. y = c₁e^3x + c₂e^−x − 2x is a two-parameter family of the second-order differential equation y'' − 2y' − 3y = 6x + 4, where c₁ and c₂ are constants. Find a solution of the second-order initial-value problem consisting of this differential equation and the given initial conditions y(−1) = 0 and y'(−1) = 1.

y(x) =

Thanks in advance for any help. 147.126.10.148 (talk) 05:00, 10 September 2017 (UTC)

For 1c, try to rearrange the equation so that you can apply the quadratic formula to solve for y. For 2, please show what you have tried. Can you find a similar example in your textbook and follow its steps? C0617470r (talk) 06:32, 10 September 2017 (UTC)

Can you explain what you mean for 1c? 147.126.10.148 (talk) 17:41, 10 September 2017 (UTC)

For 1c: $y=\phi _{\pm }(x)=1\pm {\sqrt {x^{2}-x}}$ - there are two explicit solutions. For 1d: if you consider $y,x\in R$ then $x\in [-\infty ,0]\cup [1,+\infty ]$.

For 2: you have a linear system of two equations:

$c_{1}e^{-3}+c_{2}e=-2$

$3c_{1}e^{-3}-c_{2}e=3$

with obvious solution $c_{1}=e^{3}/4,\quad c_{2}=-9/(4e)$. So, the solution is $y=e^{3(x+1)}/4-9e^{-x-1}/4-2x$. Ruslik_Zero 20:09, 10 September 2017 (UTC)

Hex tiling average island size

Given an infinite hex grid, what is the average island size for various numbers of random colorings?

Obviously for 1 colour, there is one "island" of infinite size. For 1000 colours, I suspect that the average island size will be just above 1 (maybe 1.001 - wild guess) For 2 colours, it seems likely that locally one colour will form a single island with patches of the other colour islanding inside it (and maybe first colour inside them). I'm not sure what terms to google to find this out. -- SGBailey (talk) 20:08, 10 September 2017 (UTC)

A related problem is: If a cell being "land" has probability p and being "sea" has probability 1-p, then what is the average island size? It's intuitively clear that for p close the 1 the average size is infinite and for p close to 0 the average size is close to 1. You can ask this for infinite lattices other than the hexagonal one, even higher dimensional ones. The cut-off between infinite and finite average island size is called the Percolation threshold and there has been research on finding its value for various lattices, but afaik little has been done on finding the average size as a function of p (in your case p=1/n) for values less than critical. You might try the references listed in the article, but given the relative youth of the subject I suspect that it will be difficult reading. I'm pretty sure the problem is usually stated in terms of the dual lattice rather than the way you've stated it and you might have to replace p with 1-p, but the general idea should be same. --RDBury (talk) 04:44, 11 September 2017 (UTC)

For n colours, the probability that a random tile has a given colour is n^–1 , and the probability that it has some other colour is 1 – n^–1. Each tile has six neighbours. The probability that any tile is insulated is p = (1 – n^–1)⁶ and the probability that it is not insulated is q = 1 – p. The probability that k – 1 tiles are not insulated is q^k–1. The mean value of k is Σkq^k–1 = (1 – q)^–2 = p^–2. So the mean size of an island is approximately (1 – n^–1)^–12. For n = 1000 it is 1.012. Bo Jacoby (talk) 05:21, 11 September 2017 (UTC).

That reasoning is incorrect for two reasons, though I suspect neither of which will matter a lot for the large-n case.

1. You treat as independent some probabilities that are not. Take a tile A of a given color. The probability that it is not insulated, i.e. that at least one adjacent tile B is of the same color, is indeed $1-(1-p)^{6}$. But then B has probability 0 of being insulated, since A exists. We can easily find there is another tile C with the same color with proba $1-(1-p)^{5}$. If we go down the line we have to take into account which tiles touch which one and that is where the hard stuff kicks in.
2. What this gives is starting from a random tile, what it the average size of the island that includes that tile, not the average size of a random island. Those are different per the bus waiting paradox we don't have an article about, or the closely-related Friendship paradox. Personal story: I was once a member of an association whose members had a characteristic X. They polled members about (among other things) the number of brothers and sisters they had, they found out that the average was higher than national statistics, and speculated in the association's magazine that growing up in large families increased the probability of X. They may have averaged with corrective weights, but more likely averaged all answers without weighting. In which case, they fell prey to the sampling bias that (if the null hypothesis that X was uncorrelated to the number of children) any random family would have an average number of children in the association proportional to the number of children they had, so large families were over-sampled. Put it simply, no member was from a childless family. Tigraan^{Click here to contact me} 11:23, 11 September 2017 (UTC)

• For large n / small p, one can probably find an approximation to whatever reasonably low order one wishes to find. The zero-th order is that it is unlikely there is a neighbor with the same color, so the answer is 1. At the first order, we consider only adjacent tiles, of which an average of 6p are the same color, so the answer is 1+6p. At the second order, we consider tiles at distance 2; there are 12 such tiles, each with good color with proba p, but even if they are the good color they need to be connected; 6 of them can be connected by a single tile (proba p; not exactly that, because "distance 2 is connected" and "distance 1 is the right color" are not independent events, but that will do at that order) and 6 of them give a choice of two tiles (proba (1-(1-p)^2) = 2p, since keeping a lower order would not be justified considering the other approximations). So the answer at order 2 seems to be 1+6p+18p^2. Going further is beyond my capacities.

Again (cf. above) that is the average size of the island a random tile is connected to, not the average size of a randomly-picked island. Tigraan^{Click here to contact me} 11:39, 11 September 2017 (UTC)

The word "approximately" indicate that the formula is not exact. What is the requested average island size according to your analysis? Bo Jacoby (talk) 13:20, 11 September 2017 (UTC).

I don't know. But for any finite k one could (theoretically) make such developments to get a formula valid to a $O(p^{k})$ approximation (see big O notation if you are not familiar with it). Of course it becomes more and more complex as k increases, but it can be done; the trick is to keep the approximations you make in the high-order terms.

Of course, finding an analytical solution would be better, but sometimes there is none known to humans yet. In some cases physicists get along fine with such developments for a long time (example), so it may be enough for the OP's purposes. Tigraan^{Click here to contact me} 16:24, 11 September 2017 (UTC)

The OP has no use of knowing that his question could theoretically be answered when he gets no such answer. You are challenged to improve my approximate formula. Bo Jacoby (talk) 21:14, 11 September 2017 (UTC).

Which approximate formula, Bo Jacoby? If you mean the (1 – p)^–12 thing you wrote earlier, I think to have proven it wrong (see first answer to your post). And since by how much it is wrong is not controlled, the burden is on you to demonstrate that it works with a better accuracy that may asymptotic development of order 2. Tigraan^{Click here to contact me} 10:45, 12 September 2017 (UTC)

What values does your "asymptotic development of order 2" produce? Bo Jacoby (talk) 11:18, 12 September 2017 (UTC).

With p=1/1000, 1+6p+18p^2 = 1.006018 (with an uncertainty of order p^3). I am not sure what the correct term is in English at the place where I used "asymptotic development" - maybe Taylor series is better. Tigraan^{Click here to contact me} 14:28, 13 September 2017 (UTC)

Thanks. I think the word is power series. A taylor series is the case where the coefficients are computed by iterated differentiation. (Taylor's theorem). Bo Jacoby (talk) 03:38, 14 September 2017 (UTC).

For the OP: the field of mathematics you're looking for is percolation theory. But I don't know a more precise pointer for this particular question. (Edit 14:28, 13 September 2017 (UTC): and RDBury got there 12 hours before I did, which somehow I missed -- oops!) --JBL (talk) 16:04, 11 September 2017 (UTC)

Actually I linked to a related article, so your link was only a little redundant. I've made similar oopses so no sweat :) --RDBury (talk) 20:00, 14 September 2017 (UTC)

Thank you all - I'll have a look at the link later. FWIW, the question arose as I was playing "Bouncing Balls" where you fire a coloured ball upwards to a slowly descending mass of hexgrid balls. Any island made of 3 or more evaporates and gets you points. If the mass of balls ever touches the ground you lose. Each level seems to have 2+(int((level+1)/2)) different colours and the cluster sizes are much smaller by level 6 than at level 1. I've never made it to level 7. -- SGBailey (talk) 16:32, 11 September 2017 (UTC)

https://link.springer.com/article/10.1023%2FA%3A1021069209656 Count Iblis (talk) 08:28, 12 September 2017 (UTC)