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January 5

Balls in a bag

If i have a bag of balls, containing 3 red balls, 2 blue balls, and 2x105 orange, what is the chance of picking a red ball?

For a single draw, it's the number of red balls divided by the total number of balls. StuRat 02:51, 5 January 2007 (UTC)[reply]

it is 7 draws —The preceding unsigned comment was added by 81.132.66.73 (talk) 17:17, 5 January 2007 (UTC).[reply]

So do you mean the chances of drawing exactly one red ball out of 7 draws or at least one red ball ? Also, are the balls placed back in the bag after each is drawn ? StuRat 02:27, 6 January 2007 (UTC)[reply]

The probability is vanishingly small; 3 in 200002, or 0.000015. And because you have so many orange balls in the bag, even to get the odds up to 0.1, still a rather small probability (certainly not one to make a safe bet on), would require 199972 draws! Laïka 18:12, 6 January 2007 (UTC)[reply]

Once the ball is taken out it isnt put back in again

Thanks for the help, i just need a very big ball bag —The preceding unsigned comment was added by 81.129.12.4 (talk) 19:42, 6 January 2007 (UTC).[reply]

Elementary questions in a plane

I am interested in the two dimensional plane only. Because my questions are elementary at the second year undergraduate level, perhaps somebody could offer pointers of solutions from the existing literatures.

A curve with endpoints is the continuous image of a closed interval. It is easy to prove that a simple curve C with distinct endpoints is homeomorphic to the closed unit interval [0,1]. Question#1: Is there any homeomorphic automorphism on the plane such that C is carried onto [0,1]?

Let A,B be nonempty closed bounded sets in a plane. The distance from a point x to A is denoted by d(x,A). It is easy to prove that the set C of points x with d(x,A)=d(x,B) is a closed set without any interior point. Question#2: Is it a smooth curve? Thank you in advance. Twma 00:53, 5 January 2007 (UTC)[reply]

Regarding Q2: If I understand you correctly, it is easy to come up with examples where C is a curve that is not smooth, and other examples where C is not a single curve. For example, consider small (closed) circles centered at the corners of a square. If A is the union of two circles at opposite corners, and B is one circle at one of the other corners, C will become two semi-infinite lines that meet at a right angle at the center of the square. This is a non-smooth curve. If, for the same A, B instead is the union of both the remaining circles, C becomes the union of two perpendicular lines that meet at the center of the square. If A is a small circle centered at (0,0) and B is the union of two small circles centered at (0,1) and (0,-1), C becomes two parallel lines. Am I missing something? Did you intend for A and B to be connected as well as closed? Even for that case, it is easy to make up examples that yield non-smooth curves (in the first example above, just connect the two parts of A with a loop that avoids getting closer to B than either of the circles, and A becomes connected without changing C). -- mglg_(talk) 02:21, 5 January 2007 (UTC)[reply]

Actually, without further assumption, it is not true that the set C of points x with d(x,A)=d(x,B) is a closed set without any interior point. Consider two overlapping disks, for instance; the set of points with d(x,A)=d(x,B) includes the entire region of overlap (where both distances are zero). Did you intend to require A and B to be disjoint from each other? —David Eppstein 02:23, 5 January 2007 (UTC)[reply]

Consider A = B = { P }, where P is a single point on the plane......... --CiaPan 14:39, 5 January 2007 (UTC)[reply]

Suppose that A,B must be disjoint. Is it possible to have any interior points? Thank you in advance.

I think so. Take two dense, disjoint subsets of the same segment of a line: in Cartesian coordinates

A=\{(x,0):x\in [0,1]\land x\in \mathbb {Q} \}

and

B=\{(x,0):x\in [0,1]\land (x+{\sqrt {2}})\in \mathbb {Q} \}

. Any point in the perpendicular stripe is then equally distant from A and B — for each point P(x,y) in such stripe

(x\in [0,1],\ \ y\in \mathbb {R} )

you have d(P,A) = d(P,B) = |y|. So

C=[0,1]\times \mathbb {R} .

--CiaPan 07:14, 8 January 2007 (UTC)[reply]

Oops, the conclusion is wrong. Change the symbol:

D=[0,1]\times \mathbb {R} .

Of course for all P in D, P belongs to C, but that means only D is a subset of C, not necessarily D is equal C. Actually for A and B described above, C is the whole plane.
If we want to have C in the stripe shape, we need to make d(P,A) different from d(P,B) for P outside the stripe. To achieve this, extend A and B by single points outside ends of the line segment:

A^{\prime }=A\cup \{(2,0)\},

B^{\prime }=B\cup \{(-1,0)\}

. Then

C^{\prime }=[-1/2,3/2]\times \mathbb {R} .

--CiaPan 07:46, 8 January 2007 (UTC)[reply]

The sets A,B must be nonempty closed bounded and disjoint to start with. Thank you for your help. Twma 06:04, 9 January 2007 (UTC)[reply]

The second question is completely answered. Thanks. For Q1, by homeomorphic automorphism, I mean a homeomorphism from the plane onto itself. Twma 19:33, 5 January 2007 (UTC)[reply]

Since no actual mathematician is stepping up to the plate, I (a humble biophysicist) will offer my two cents, based only on visualizing a rubber sheet: "Well, yes, how could there possibly fail to be such a homeomorphism?" I smell an existence theorem. For example, I think the Riemann mapping theorem could be used to show that there would be an analytic (!) mapping that maps the entire plane except your curve (but plus the point at infinity), to the entire plane except the line (0,1) (but plus the point at infinity). That mapping wouldn't necessarily be defined on the curve itself, though, so it isn't really relevant to your question. The real answer will have to wait for some real mathematician... -- mglg_(talk) 01:04, 7 January 2007 (UTC)[reply]

The answer to your question is certainly yes. (In fact there are hugely infinitely many, since at the very least given one you can smush it around -- yay rubber sheets -- to get another.) That doesn't mean I can write down a formula for it, but it certainly exists. In fact, there should be an isotopy of the plane that maps any such curve onto [0,1]. An isotopy is a homotopy which is a homeomorphism at each moment. This idea, of an isotopy of the ambient space which moves a particular subspace (say a curve) onto another, is called an ambient isotopy, and is the idea of "sameness" that is used to study knots in 3-space. (A better article is here.) Tesseran 16:24, 10 January 2007 (UTC)[reply]

I am sure that the answer is YES based on intuition but I need pointers of acceptable mathematical proofs. Thank you for the above pointers. I collected a lot of definitions. An ambient isotopy is the continuous deformation of a knot or link. This means the aspect of topology, where a knot or link may be bent, twisted, stretched, or pulled, as long as it is not allowed to intersect with itself or be cut. A knot can be untied in the topological sense if and only if it can be smoothly moved through the ambient space until it assumes the shape of a circle. If this can be done, the knot is called the unknot. In mathematical language, knots are embeddings of the circle in three-dimensional space. I may overlook something that I cannot find any existence theorem with proof that a simple curve with different endpoints can be deformed in the plane onto a closed interval. Thank you again but I am looking forward to have more specific pointers. Twma 00:19, 11 January 2007 (UTC)[reply]

symmetry operation in 3 dimensional space

I'm trying to find out what symmetry operation (not an inversion) operates about a point in 3 dimensional space. I've added the table below to clarify what I'm asking for. (Note in the 4d case I've 'assume(d) that all dimensions have to be spatial dimensions.' and orthogonal).

Dimensions (d)           1(number line)   2(plane)       3 (3d space)        4 (?)
.
Inversion in a..          point            point          point            point
.
Reflection in a..         point (d=0)      line  (d=1)    plane (d=2)      'volume' (d=3)
Rotation around a..       impossible       point (d=0)    line  (d=1)      (plane   (d=2)???)
Operation X               impossible       impossible     point (d=0)       ...

Am I missing something obvious (or is it impossible) (Note I've bracketed rotation about a plane in 4 dimensions since I haven't worked out the maths of it..) 87.102.19.164 01:25, 5 January 2007 (UTC)[reply]

This is not a complete answer, since I am not qualified to give one, but here goes anyway: There are reflections in points (i.e. inversions), in lines, in planes, etc, up to dimension n-1, where n is the dimension of your space. There are also rotations around objects of dimension n-2, as you correctly guessed, though one might want to think of them as rotations within a 2D plane rather than around anything. The math of rotations in higher dimensions is simply to apply a 2x2 rotation matrix containing sines and cosines to the 2 coordinates the rotation acts on, after choosing an appropriate coordinate system. By combining such 2D rotations in different planes you can define rotations within k-dimensional subspaces. I don't see an obvious candidate for your operation X, and after adding the different kinds of reflections, and incorporating inversions as one of the reflections, the table structure may no longer suggest (as strongly?) the existence of an operation X. -- mglg_(talk) 02:21, 5 January 2007 (UTC)[reply]

Could "Operation X" be two rotations about orthoginal lines intersecting at the given point ? StuRat 02:43, 5 January 2007 (UTC)[reply]

See point group. All such symmetries can be constructed by composing reflections, where a reflection essentially negates one coordinate in a suitably chosen coordinate system with origin at the fixed point. For example, the composition of two reflections is a rotation. The composition of three can be an inversion (if they are mutually perpendicular) or a rotary inversion (otherwise). In 4D, the composition of four reflections again gives a rotation, but one that simultaneously rotates two planar subspaces. If you like matrices, try these:

{\begin{aligned}{\begin{bmatrix}-1\end{bmatrix}}&\qquad {\text{reflection of line}}\\{\begin{bmatrix}-1&0\\0&-1\end{bmatrix}}&\qquad 180^{\circ }{\text{ rotation of plane}}\\{\begin{bmatrix}0&-1&0\\1&0&0\\0&0&-1\end{bmatrix}}&\qquad 90^{\circ }{\text{ rotary inversion of space}}\\{\begin{bmatrix}0&-1&0&0\\1&0&0&0\\0&0&-1&0\\0&0&0&-1\end{bmatrix}}&\qquad {\text{(double) rotation of 4-space}}\end{aligned}}

Each of these is, and any point symmetry corresponds to, an orthogonal matrix. --KSmrq^T 05:55, 5 January 2007 (UTC)[reply]

Qualifying examination preparation

Hi everyone, I'm in the midst of studying for one of my PhD qualifying examinations, and I've hit a snag on a problem on one of the previous qualifiers; I'm kind of lost as to where to begin. Anyway, here's the statement:

Let G denote the group of all $n\times n$ nonsingular complex matrices (under multiplication). Let H be a finite group and $f:H\rightarrow G$ a group homomorphism. Prove that $\forall h\in H,\ f(h)$ is a diagonalizable matrix.

I don't want you to just give me the answer, I'd like a hint or something. Thanks in advance! –King Bee ^{(talk • contribs)} 04:08, 5 January 2007 (UTC)[reply]

I haven't worked it out, but it's got to be enough to know that some power of h is the identity. It's really all the information you're given. From there it should be pure linear algebra. Melchoir 04:34, 5 January 2007 (UTC)[reply]

Of course! Then

f(h)

is a ~~involutary~~ matrix whose Jordan Canonical Form must be diagonal for all h, since every Jordan block

J_{\ell }

must have the property that

J_{\ell }^{k}=I_{\ell }

. ~~and is thus diagonalizable (Jordan Canonical Form argument).~~ Thank you so much! Maybe you can help me with this one:

Let G be a group with trivial center. Prove that Aut(G) has trivial center.

I know I'm missing something really obvious here. Again, hint over a solution is preferred. –King Bee ^{(talk • contribs)} 04:44, 5 January 2007 (UTC)[reply]

Uh... possibly the representation map helps? This time I'm afraid I also get the feeling that I'm missing the easy solution! Melchoir 04:51, 5 January 2007 (UTC)[reply]

Well, I really feel like it should have something to do with the fact that G is isomorphic to Inn(G), but I'm just not clever enough to make it work. –King Bee ^{(talk • contribs)} 05:06, 5 January 2007 (UTC)[reply]

Okay, this problem was bothering me, so I found a brute-force solution: given a non-identity automorphism, you can construct a group element -- in hindsight, the obvious one -- whose conjugation doesn't commute with the automorphism at a certain point. I don't feel any cleverer for slogging through this method, though, so I hope someone finds a solution using some theoretical machinery. Melchoir 07:03, 5 January 2007 (UTC)[reply]

I see what you're saying. It's kind of annoying when the solution is like that. I'm going to query a few more people today, and if I come up with something else, I'll let you know. Thanks again for your help, I really appreciate it. –King Bee ^{(talk • contribs)} 13:25, 5 January 2007 (UTC)[reply]

Great to hear, good luck! Melchoir 18:45, 5 January 2007 (UTC)[reply]

Thanks. I asked one of the professors around today, and her solution was indeed the brute force one. Neither of us enjoyed it, but hey; sometimes that's how you do things. The exam is tomorrow (!!), so we'll see what happens. Thanks again for your aid. –King Bee ^{(talk • contribs)} 23:25, 5 January 2007 (UTC)[reply]

Proving the Viète formula

In the article for the Viète formula a proof is stated which assumes the following identity:

{{\sin(2^{n}x)} \over {2^{n}\sin(x)}}=\prod _{i=0}^{n-1}\cos(2^{i}x)

The article also states that the identity can be derived from the double angle identity. How can this be done?--Fiber B 06:15, 5 January 2007 (UTC)[reply]

Eh? The article says use

\sin(2x)=2\sin(x)\cos(x),\,\!

which is fairly explicit, and suggests induction over n. This is the n = 1 base case. Try n = 2.

{\begin{aligned}\sin(2^{2}x)&{}=2\sin(2x)\cos(2x)\\&{}=2\left(2\sin(x)\cos(x)\right)\cos(2x)\\&{}=2^{2}\sin(x)\cos(x)\cos(2x)\end{aligned}}

I will leave the general induction step in your able hands. --KSmrq^T 06:53, 5 January 2007 (UTC)[reply]

exponents

when and how were exponents first used? —The preceding unsigned comment was added by 192.133.105.19 (talk) 17:18, 5 January 2007 (UTC).[reply]

Casting of nines

What properties, behavior and/or history is known about casting of nines. Is the term I know it as proper? I originally learned the trick in gradeschool in Madrid Spain. Very few know of this in the United States. It is an error prone method for checking multiplication.

Example:

2537865 × 4589002 = 11646267560730

Add up the individual term digits until you get a single digit answer, whenever you get an answer of 9 just continue any number +9 "simplifies" to 9, so for example 9 + 5 = 14 = 1 + 4 = 5, you don't have to do this neccessarily, I just think its where the name "casting of 9s" comes from

So you add up the digits of the multiplicand: 2+5+3+7+8+6+5 = 36 = 3+6 = 9 Add up the multiplier: 4+5+8+9+0+0+2 = 28 = 2+8 = 10 = 1+0 = 1

Then you multiply 9 × 1 = 9

If the answers adds up to 9 it is most likely correct.

1+1+6+4+6+2+6+7+5+6+0+7+3+0 = 54 = 5+4 = 9

It is most definetly not foolproof, but for a quick check for doing the multiplication by hand... Haloway13 17:56, 5 January 2007 (UTC)[reply]

See Casting out nines. --mglg_(talk) 17:55, 5 January 2007 (UTC)[reply]

In looking at that article I noticed it kind of glosses over proving that the method works. I put a note about that in the article's talk page, but if anybody happens to have a good published reference that they can cite and include in the article that includes a more formal proof the method works, feel free to include it. One useful proof to include would be that for an integer x, x mod 9 is equal to the sum of its digits mod 9. For example, 587 mod 9 = (5+8+7) mod 9. It can be proven using pretty simple math and would be a nice addition to the article. Dugwiki 19:02, 5 January 2007 (UTC)[reply]

I have augmented the talk page, but made no effort to amend the article. --KSmrq^T 14:31, 7 January 2007 (UTC)[reply]

There's a somewhat more technical article on essentially the same thing at Digital root. While I don't think merging them would really be a good idea, certainly there should be prominent links from one to the other. —Ilmari Karonen (talk) 00:36, 9 January 2007 (UTC)[reply]

Approximation of a simple curve by polygonal lines

A polygonal line is obtained by joining consecutive line segments. Let B(e) denote the open disk with center at the origin and radius e>0. Given a SIMPLE curve with dinstinct end points A,B, is it possible to find a SIMPLE polygonal line P with end points A,B such that P is contained in the tube C+B(e)={c+b:c in C, b in B(e)}? The approximation follows easily from uniform continuity but being SIMPLE (injective) is difficult for me to handle. I am interested in plane curves only. Because it is so basic and foundamental, I expect that it is a well known result. Be grateful if somebody could offer me a pointer. Thank you in advance. Twma 19:44, 5 January 2007 (UTC)[reply]

Considering the evolute of a curve may help. If C is your curve and N is a unit normal then the parallels of the curve C+rN, have cusps when the cross the evolute, that is when r is equal to the radius of curvature of the curve. If you choose e so that it is less than the minimum radius of curvature than both edges of the tube will be non singular. This essentially solves the local problem. There is also a global problem to consider as two different parts of the curve can be arbitrarily close. Here it may help to examine the medial axis which can be thought of as a continuous versions of the Voronoi diagrams. If you can choose your tube so that it never crosses the medial axis, then your almost done. --Salix alba (talk) 20:03, 6 January 2007 (UTC)[reply]

What I need is to approximate a bad curve with a nice one. Smooth curves are very nice already. Polygonal lines are the best in my view. Thank you for your help but I am still looking for a better solution. Twma 06:24, 9 January 2007 (UTC)[reply]

Value of money in 1800's`

I am trying to find out what the value of 150,000 dollars would have been in the mid 1800's in contrast to what it is worth today. Can you help me? Barb —The preceding unsigned comment was added by 70.176.52.116 (talk) 19:59, 5 January 2007 (UTC).[reply]

Use this: http://www.minneapolisfed.org/Research/data/us/calc/hist1800.cfm for a rough idea. It's unrealistic enough to assume that everyone experiences the same change in spending power over a year or two, to assume it over 150+ years is highly dubious.86.132.237.148 21:19, 5 January 2007 (UTC)[reply]

Yea, it seems to say that prices were lower in 1942 than in 1800, that can't possibly be right. The basic problem is that people don't buy the same things now as they did in the mid-1800s, so how do you compare the cost of a team of horses and a wagon to a station wagon with 200 horsepower ? Even things that could be bought then which we still buy, like white sugar, aren't valid for comparison, as that was a luxury item then and is a rather basic commodity now. Comparing white sugar then with caviar now might be more meaningful. Also, we need to consider all the modern expenses which didn't exist then, like income tax, cars & insurance, money for movies, music, etc. 01:57, 6 January 2007 (UTC)

Equation of a line

The line has coordinates (-5,6) and (6,-7)

What is the equation and how can it be worked out? Thanks in advance81.132.66.73 20:37, 5 January 2007 (UTC)[reply]

The gradient of the line (m) is y difference divided by x difference, so you have the equation in the form y = mx + c, with m known. Substitute the coordinates of one of the points, which will give an equation to be solved for c. As a check, substitute the coordinates of the other point. 86.132.237.148 21:10, 5 January 2007 (UTC)[reply]

Equivalently, you can substitute the coordinates of both points in the equation y = mx + c, where m and c are still unknown. This gives you two equations in two unknowns. Solve for m and c. This way of approaching the problem generalizes to other cases, like finding a parabola whose axis is vertical passing through three given points, or a plane through three points in 3-space. All you need is a sufficiently general equation for the object some of whose points are given; substitute the coordinates of the points and solve for the unknowns. --Lambiam ^Talk 21:32, 5 January 2007 (UTC)[reply]

OK so can you show me with the numbers in the equation —The preceding unsigned comment was added by 81.132.66.73 (talk) 21:48, 5 January 2007 (UTC).[reply]

No, that would be doing your homework for you, but we can show you with different numbers. Let's find the equation for the line from the point (1,2) to the point (9,8). The slope, m, is the change in y over the change in x, or (8-2)/(9-1) or 6/8 or 3/4. That gives us y = (3/4)x + b. To find b, plug in one point's coords, and solve for b:

y   = (3/4)x + b
2   = (3/4)1 + b
2   = (3/4)  + b
8/4 = (3/4)  + b
5/4 =  b

So, we end up with y = (3/4)x + (5/4). Now, as always, we should check our work by plugging each point's coords into the equation to ensure that both sides are equal:

y = (3/4)x + (5/4)
2 = (3/4)1 + (5/4)
2 = (3/4)  + (5/4)
2 =  8/4
2 =  2

y = (3/4)x + (5/4)
8 = (3/4)9 + (5/4)
8 = 27/4   +  5/4
8 = 32/4
8 = 8

Both sides match, so it checks out ! Now you try it and let us know what you get. StuRat 01:43, 6 January 2007 (UTC)[reply]

Lambiam makes a point that was explored by one of our editors, Stanford emeritus professor Vaughan Pratt, in a paper on fitting algebraic curves and surfaces:

Pratt, Vaughan (July 1987). "Direct Least Squares Fitting of Algebraic Surfaces". Computer Graphics. 21 (4). ACM: 145–152. doi:10.1145/37401.37420. (Proceedings of SIGGRAPH ’87)

He gives as one example the exact fit of a circle through three points. He uses a basis of four polynomials,

(1,x,y,x^{2}+y^{2}),\,\!

for the ideal of all circles, by using the points

P_{1}=(x_{1},y_{1}),\quad P_{2}=(x_{2},y_{2}),\quad P_{3}=(x_{3},y_{3})\,\!

to construct a vanishing (partly symbolic) matrix determinant,

0={\begin{vmatrix}1&x_{1}&y_{1}&x_{1}^{2}+y_{1}^{2}\\1&x_{2}&y_{2}&x_{2}^{2}+y_{2}^{2}\\1&x_{3}&y_{3}&x_{3}^{2}+y_{3}^{2}\\1&x&y&x^{2}+y^{2}\end{vmatrix}}

Expanding by cofactors of the last row gives the coordinates for the specific circle in this basis. For example, if the points are

P_{1}=(2,1),\quad P_{2}=(0,1),\quad P_{3}=(1,2),\,\!

then the determinant gives the polynomial equation

0=-2x^{2}+4x-2y^{2}+4y-2.\,\!

Dividing out an irrelevant factor of −2 and arranging into standard form, this is

(x-1)^{2}+(y-1)^{2}=1,\,\!

a unit circle centered at (1,1).

For those with an interest in this sort of thing, the paper remains an informative read two decades on. The general approach deserves to be known more widely, for its simplicity, generality, and elegance. --KSmrq^T 10:49, 7 January 2007 (UTC)[reply]

Algebra/what is the equation?

If one person can paint a house in 2 days, another person can paint it in 4 days, and a 3rd person can paint it in 6 days, how long would it take the 3 of them to paint it together?

I don't know how to figure this. Please help.

--24.252.201.59 23:12, 5 January 2007 (UTC)--24.252.201.59 23:12, 5 January 2007 (UTC)Deb[reply]

—The preceding unsigned comment was added by 24.252.201.59 (talk) 23:06, 5 January 2007 (UTC).[reply]

Is this your homework? Recall, we don't do your homework for you. You want to think about what the rate of painting is; i.e., how many houses per day, or how many days per house, for each person involved. –King Bee ^{(talk • contribs)} 23:29, 5 January 2007 (UTC)[reply]

Not homework. I have to take a ParaPro test and was just brushing up on a practice test. Can you give me the equation? 24.252.201.59 00:19, 6 January 2007 (UTC)Deb[reply]

Consider that one person paints 1/2 houses/day, second 1/4 houses per day, third 1/6 houses per day, so three of them paint 1/2+1/4+1/6 houses per day. (Igny 00:58, 6 January 2007 (UTC))[reply]

Thanks!!! 24.252.201.59 01:14, 6 January 2007 (UTC)D[reply]