Wikipedia:Reference desk/Archives/Mathematics/2009 May 21

= May 21 =

concentric circles
q1. there are two concentric circles and a chord of the bigger circle is a tangent to the smaller circle. the length of the chord/ tangent is 6cm. it is known that the radii of the two circles are integers. what's the radius of the outer circle???

i'm totally stuck! all i've drawn is a perpendicuar from the centre which divides the chord into two parts of 3cm length each. but then what??

and please help me with this one:

q2. a mixture of wine and water is made in the ratio of wine: total = k:m. adding x units of water or removing x units of wine (x is not equal to 0), each produces the same new ratio of wine: total. what is the value of the new ratio??

as the ratios are equal, so k:m + x = k - x:m (adding and removing x units produces the same new ratio). i solved it and i got x = k - m. but if i put this value in any of the equations, i get the ratio as 1:1. but the answer, as given in the book, is 1:2. —Preceding unsigned comment added by 122.50.130.33 (talk) 04:01, 21 May 2009 (UTC)


 * For the second question,Your mistake is in the part where you remove wine - it is k-x/m-x...:)Rkr1991 (talk) 04:45, 21 May 2009 (UTC)


 * Draw a radius of the bigger circle which ends at the chord's end. You get a right triangle with two radii and a half of the chord as its sides. Try to apply Pythagorean theorem to it.
 * It'a wine to total, not wine to water ratio, as Rkr1991 noted. --CiaPan (talk) 06:44, 21 May 2009 (UTC)


 * in case u didn't read the question carefully, CiaPan, i have got only one measurement, that of half the chord. the other measurements, i.e. the hypotenuse of the triangle and its height, are unkown. so, how am i gonna apply pythagoras here??? —Preceding unsigned comment added by 122.50.130.33 (talk) 08:52, 21 May 2009 (UTC)


 * In case you didn't read your excersise text: you have one more information: 'both radii are integers', suppose that means their lengths are integer numbers of centimeters. Additionally, we know they are positive integers. The Pythagorean theorem tells you something about their correlation, which probably can be satisfied by only a few pairs (possibly by only one pair) of positive integer numbers.... HTH. --CiaPan (talk) 10:25, 21 May 2009 (UTC)
 * Here's another hint, in case you're not sure the values you've found are indeed the only solution: The smaller radius, r, is an integer, and the larger radius, R, is an integer greater than r. What, then, is the smallest value R could be? What happens if you plug this into the equation given by the Pythagorean theorem? -- Meni Rosenfeld (talk) 10:40, 21 May 2009 (UTC)
 * Take a look at the introduction to Pythagorean triple (not the whole article!) Note that each triple corresponds to a right triangle with side lengths that are integers. I bet you will find the first (smallest) triple is helpful. Cuddlyable3 (talk) 22:07, 25 May 2009 (UTC)


 * Hm. I'm stuck on q2 too.  Hints anyone? —Tamfang (talk) 00:43, 4 June 2009 (UTC)

Spheres in an n-dimensional sphere
In a previous question there was an aside asking how many n-dimensional spheres could fit inside one with three times the diameter, did it go up as the power of n? I said that the number would go up proportional to an nth power for a big sphere more than twice the smaller ones as can be seen by doubling the size of the smaller spheres. This would have to cover every point in the large sphere or another small sphere could be fitted in. One could fit in at least floor((2+e)/2)n.

However I've now been thinking about the size of a sphere that holds a simplex of touching n-dimensional spheres in it. The big sphere would have diameter 1+sqrt(2n/(n+1)) as far as I can figure out which tends to 1+sqrt(2). However it only holds n+1 smaller spheres which does not go up as the nth power. And I can'rt see how to fit more small spheres in, the won't fit in the centre nor on the outside of a face where they would form another simplex. And 1+sqrt(2) is definitely more than 2. So where on earth am I going wrong thanks? Dmcq (talk) 15:20, 21 May 2009 (UTC)


 * It does go up pretty fast. As for n-dimensional sphere packing, who knows?  There may be a known answer for n/=3, I'm not sure.  See curse of dimensionality for how fast the volume increases. 207.241.239.70 (talk) 18:27, 21 May 2009 (UTC)


 * I agree with your latter argument, while the former is not very clear to me. How is it? For sure, if $$\scriptstyle R<1+\sqrt2$$, the number of disjoint unit open n-dimensional Euclidean balls that you can pack into an n-dimensional ball of radius $$R$$ is bounded uniformly wrto the dimension (by the way, I wrote few lines here after noticing some interest here on the topic). If $$\scriptstyle R\geq 1+\sqrt2$$ the number of balls certainly diverges with the dimension, but I do not see how to prove that there is an exponential growth. Maybe for large R is not that difficult but I do not have ideas. It is possible that there is exponential growth only for $$R$$ large enough ($$\scriptstyle R\geq 3 ??$$), whereas for smaller R there is only a polynomial growth, who knows... As to $$R=3$$, for sure the most dense packing of unit balls in a ball of radius 3 has a number of unit balls at least 1 plus the kissing number (and if we knew that the maximizing configuration may be obtained putting one of the unit balls concentric with the containing ball, then there would be equality). In any case, I don't know if it is known whether the kissing number grows exponentially. In general it seems to me a topic where immediately one encounters open and possibly old difficult questions; so before start thinking on it one should check the state of art... --pma (talk) 19:58, 21 May 2009 (UTC)


 * Thanks very much for that reference to the packing problem for n-dimensional spheres. That shows my second bit of reasoning is okay so I must have a problem with the first argument. I better explain what I was thinking a bit better. Given a number of balls of diameter 1 if you can find a point which is more than a unit distance from the centres of all the balls then another ball can be put there. Around a unit ball all the points which cant be the centre of another unit ball form a ball with double the diameter of the unit ball. Now suppose we have a big ball of diameter D in n-dimensional space then its n-dimensional volume is Dn times greater than that of a unit ball. It is also (D/2)n times greater than a ball of diameter 2. For D>2 this last quantity goes up as a power of n, and even if the whole volume of the balls of diameter 2 were disjoint from each other and inside the big ball there is still space left over if there are less than (D/2)n of them. Therefore it must always be possible to put in the integer part of (D/2)n unit n-dimensional balls + 1 extra in the big ball. Applying this to a big ball of diameter 1+sqrt(2) which in n-dimensional space contains n+1 unit balls we have instead ((1+sqrt(2))/2)16=20.32.. which is more than 16+1=17.


 * Well, it seems to me that this way one can only get a bound from above on the number of balls... Certainly there are at most Dn n-dimensional balls of diameter 1 in a ball of diameter D, but how to use a volume comparison to bound the number of balls from below? For instance, a ball of radius 2 has a volume 2n times a ball of radius 1; nevertheless we can only pack 2 unit balls in it. Maybe for some large number R and some c>1 it is true that there are at least cn unit balls into the ball of radius R; however it doesn't seems obvious to me, even if R=1000 and c=1.001 say. The first idea is trying c=2 with 2n balls of radius 1 in a cubic disposition: but this way it seems to pack them you need a radius $$\scriptstyle 1+\sqrt n$$ in dimension n, that diveregs, so there is no R this way :-/  pma (talk) 00:06, 22 May 2009 (UTC)


 * Problem solved. Isn't it marvellous what a good nights sleep can do? I hadn't excluded centres for the unit balls which were close to the outside of the big bal so some could poke through. So adding the radius of the unit ball to the big ball what I'd shown was that an n-dimensional ball with a diameter of more than 2+1=3 contains a power of n times balls. However a change to the argument can I believe be used to show the bound can be reduced to 1+sqrt(2) so that is a very tight bound. If all the unit balls are arranged around a small ball diameter d then the centres of the unit balls have to be a unit apart as before. All the centres will be on a sphere of radius (d+1)/2 and we can apply the argument about if the unit balls were expanded to radius 1, how much of this sphere would be covered? Each of these would cover a cap on the sphere and less than half of the sphere would be covered if we have (d+1)/2 * sqrt(1) > 1, that is if the triangle subtended at the cnetre of the inner ball by the centre of the sphere of radius 1 and its edge is less than a right angle. If less than half of the sphere is covered by the cap then I believe the area covered by the cap compared to the total area goes down as the nth power as the dimension goes up. Working this out gives d = sqrt(2)-1 so the outside diameter of a sphere containing unit balls around the inner ball would be sqrt(2)-1+2 = 1+sqrt(2). So for anything bigger than this the number of unit balls that could be fitted inside goes up as the power of n. Thanks very much for the help. Dmcq (talk) 07:08, 22 May 2009 (UTC)


 * Good night?? Don't mention...I tried to understand your first argument, and it seemed completely OK to me too!!! Anyway, I completely agree now with your last answer: very elegant!! You say: For R>3 and n given, let N be the greatest number such that there exist N disjoint open n-dimensional Euclidean unit balls: B(x1,1), B(x2,1).. B(xN,1) included into the ball B(0,R). Since N is the greatest possible, there is no room for another ball, therefore for any point x in B(0,R-1) there exists an index i, $$\scriptstyle 1\leq i\leq N$$ such that ||x-xi||<2 (because if on the contrary there were a point x in B(0,R-1) such that  ||x-xi|| is greater than or equal to 2 for all i=1,..N, that point x would be the center of a unit ball disjoint from the other balls, and still included in B(0,R) (that was the neglected point, where I too fell in the trap). Therefore, the balls  B(x1,2), B(x2,2).. B(xN,2) cover the ball B(0,R-1), which implies 2n N>(R-1)n so N>cn with c:=(R-1)/2>1. Compliments, very elegant. The doubling argument reminds me of the Vitali covering lemma. Still I do not see completely the case you said,  1 + sqrt(2) < R < 3... --pma (talk) 15:08, 22 May 2009 (UTC)


 * Thanks, I think I saw something like it in the past so I can't claim a full credit. Anyway in the business about 1+sqrt(2) what I considered was the hypersurface of the ball with diameter greater than sqrt(2), i.e. radius 1/sqrt(2). We want to put unit balls with their centres that surface, i.e. radius 1/2 balls. Now consider how much of that hypersurface would be covered if we doubled the radius of a ball to 1 and still had its centre on the hypersurface. If the radius of the hypersurface is 1/sqrt(2) then half of the hypersurface would be covered because we have a right angled triangle. The argument about filling up all the space can be applied to the hypersurface, and there is no nasty boundary to take care of. So this shows we can always fit two unit balls into a ball of radius sqrt(2)+1 where the 1 is from the extra radiuses of the unit balls. That is not a terribly wonderful result! However if the hypersurface has radius greater than 1/sqrt(2) then less than half the hypersurface is covered. The question then is how does the amount covered change with the dimension? There's a formula in Spherical cap, I haven't worked it out exactly but it certainly looks to me like the ratio of the cap size to the total hypersurface size will tend to be proportional to h n-1 where h is radius of cap/radius of hypersurface because it'll be dominated by the outer rim. With a hypersurface radius r > 1/sqrt(2) h=sin(2*asin(1/2r)). So the cap takes up a space which goes down as an nth power. So we need an nth power of caps on the hypersurface before we can be certain of covering every point of it. Another unit ball can be put wherever there is an uncovered point. I had a look at that Vitali covering and it seemed reasonable enough until it got to the infinite bit and it not working and I couldn't quite see the point unless one could choose the subset easily. Oh well I think I'll have to leave it till another time :) Dmcq (talk) 19:10, 22 May 2009 (UTC)

joint probability distribution terminology
I don't really know anything about probability so I hope I'm not using these terms incorrectly. Let's say event X occurs with probability p(X) and event Y occurs with probability p(Y). If we naively presume them to be uncorrelated, the joint probability $$p(X\cap Y)$$ would be p(X)p(Y). But they are actually correlated, and the observed (by experiment) joint probability is z. My question is, does the ratio
 * $$p(X\cap Y) \over p(X)p(Y)$$

have a name? I'd guessed that this was the correlation coefficient but that turns out to be rather different, as far as I can tell. Thanks. 207.241.239.70 (talk) 18:21, 21 May 2009 (UTC)
 * I don't know a special name for this ratio, but for future reference, the correct term is that the events are independent rather than uncorrelated. -- Meni Rosenfeld (talk) 20:33, 21 May 2009 (UTC)
 * Thanks. Does the ratio actually come up in probability calculations much?  Maybe I'm on the wrong track.  Basically I want to compute a lot of these ratios for a large, static data set, and use them to make crude estimates of the corresponding joint probabilities, as part of a query optimizer for the data.  I see from the article you linked to that the fancy way to do this involves copulas.  I will see if I can use that approach. 207.241.239.70 (talk) 00:34, 22 May 2009 (UTC)


 * If it helps any, your numerator can be rewritten as p(X|Y)p(Y). That makes your expression equivalent to $$p(X|Y) \over p(X)$$. This is the ratio of the conditional probability of X to the unconditional probability of X. Wikiant (talk) 10:53, 22 May 2009 (UTC)