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May 17

does anyone know how to solve these?

1. [(2+1)(2²+1)(2⁴ + 1)(2⁸ + 1)(2¹⁶ + 1) + 1]/ 2³³

2. a rectangle has its sides of length sinx and cosx for some x. what is the largest possible area it can have?

3. ROW 1 - 1

  ROW 2 - 3 5

  ROW 3 - 7 9 11

  ROW 4 - 13 15 17 19  etc...

what's the number at the end of ROW 80?? —Preceding unsigned comment added by 122.50.137.185 (talk) 05:58, 17 May 2009 (UTC)

When asking questions on the reference desk, please tell us what progress you have made so far towards the solution and where you are stuck. This allows us to give answers that are most helpful, best directed towards your level, and takes less of our time guessing what it is you need help with.

For the first question, you need to expand $(x+1)(x^{2}+1)(x^{4}+1)\cdots (x^{2^{k}}+1)$. It may help to write this out by hand for a few small k to get an idea of what is going on. To expand this for general k, I see two appraoches; the easier one is to multiply by x - 1 (and then remember to divide back by x - 1 later), and the other uses the fact that any integer from 0 to $2^{k+1}-1$ inclusive can be uniquely written as a sum of powers of 2 up to $2^{k}$, a property of the binary writing system.

For the second question, what is the area of the rectangle with sides $\sin x$ and $\cos x$? From there you can solve the problem several different ways: the most direct is to use calculus to maximize the function; or observe that the area function is symmetric about a value of x, which must be its maximum; or use a trigonometric identity to simplify the area function into an easier-to-use form with a very obvious maximum. Or you can just guess the answer, as there's really only one reasonable guess.

For the third question, I'll need to know where you are stuck before I can help. Eric. 131.215.159.91 (talk) 08:45, 17 May 2009 (UTC)

For the third question, observe that all the numbers are odd, so add one to every number and divide by 2. Does that make the pattern easier to see? Can you then extend that pattern to row 80? Then you just double it and subtract 1. -- SGBailey (talk) 22:03, 17 May 2009 (UTC)

For the first question it will help to know that 2⁰ = 1 and 2¹ = 2. Then you have some multiplying to do which can all be done by additions since 2^a x 2^b = 2^a+b. Similarly if you need to divide then 2^c / 2^d = 2^c-d.

For the second question draw a rectangle with a diagonal whose length is 1. Apply what you know about the definitions of sine and cosine for a right triangle.

For the third question write down and look at the rows in order on a single line [ROW1][ROW2][ROW3][ROW4]. Can you see 1) what happens at each step along the chain, and 2) how many numbers of the chain go to make up each row? BTW Row 4 is complete as given, the "etc." means there are successive Rows 5, 6 and so on. Cuddlyable3 (talk) 11:34, 19 May 2009 (UTC)

The literal answer to the OP is Yes! Cuddlyable3 (talk) 11:36, 19 May 2009 (UTC)

Fairly simple $\nabla$/inequality problem

Hi there -

I've got a very straightforward question here, it's short and simple but I still can't work out how to solve it and it's becoming very annoying, I was wondering if anyone could give me a hand (although it seems so short I wonder whether there's even anything to help or if I'm just missing the blindingly obvious!) -

Consider the cone C in $\mathbb {R} ^{3}$ defined by $x_{3}^{2}=x_{1}^{2}+x_{2}^{2}$, $x_{3}>0$ Find a unit normal $n=(n_{1},n_{2},n_{3})$ to C at $x=(x_{1},x_{2},x_{3})$ such that $n_{3}\geq 0$ - fine so far, I used the grad function and just took the upwards-pointing normal.

Show that if $p=(p_{1},p_{2},p_{3})$ satisfies $p_{3}^{2}\geq p_{1}^{2}+p_{2}^{2}$ and $p_{3}\geq 0$ then $p\cdot n\geq 0$.

This is where I get stuck - intuitively I can see why it's true, since p is 'trapped' inside the cone, so must be at an angle < ${\frac {\pi }{2}}$ with the cone, but I'm not sure how to prove it formally. I'm trying to avoid using the $a\cdot b=|a||b|cos\alpha$ definition, rather the $a\cdot b=\sum a_{i}b_{i}$ one if possible - although naturally if the former is significantly simpler then I'd take either! It doesn't seem like it should take much work, I just can't see how to transfer the 'x's and 'p's and keep the inequality correct. Thanks a lot! Delaypoems101 (talk) 06:21, 17 May 2009 (UTC)

For the norm n we know $n_{1}^{2}+n_{2}^{2}\leq n_{3}^{2}$ (actually, we have equality, but we don't need that). So we have

$n\cdot p=n_{1}p_{1}+n_{2}p_{2}+n_{3}p_{3}\geq n_{3}p_{3}-|n_{1}p_{1}+n_{2}p_{2}|=n_{3}p_{3}-|(n_{1},n_{2})\cdot (p_{1},p_{2})|$.

We can do two different things here. One is to use the fact that the cone is rotationally symmetric to simply; we can take p_2 = 0, for example, so we get $n\cdot p\geq n_{3}p_{3}-|n_{1}p_{1}|$ and we are nearly done. A more elegant approach is to use the Cauchy-Schwarz inequality to say

$|(n_{1},n_{2})\cdot (p_{1},p_{2})|\leq {\sqrt {n_{1}^{2}+n_{2}^{2}}}{\sqrt {p_{1}^{2}+p_{2}^{2}}}\leq n_{3}p_{3}$.

Eric. 09:05, 17 May 2009 (UTC) —Preceding unsigned comment added by 131.215.159.91 (talk)

LaTeX fraction spacing

If, I have, say:

${\begin{array}{rclr}{\dfrac {1}{2}}&=&{\dfrac {1}{4}}+{\dfrac {1}{4}}\\&=&{\dfrac {1}{8}}+{\dfrac {1}{8}}+{\dfrac {1}{8}}+{\dfrac {1}{8}}\end{array}}$

You can see the fractions start to bunch themselves together over multiple lines.

Is there a way to fix this? I don't want to use \\\\ because I have a huge document and want to avoid finding out which lines are problematic (some lines, obviously, don't bunch together - only when \dfrac is involved). Is there a length command I can adjust via \setlength? x42bn6 Talk Mess 23:08, 17 May 2009 (UTC)

You should be using "align", not "array", for this sort of thing. Here it is:

${\begin{aligned}{\dfrac {1}{2}}&={\dfrac {1}{4}}+{\dfrac {1}{4}}\\&={\dfrac {1}{8}}+{\dfrac {1}{8}}+{\dfrac {1}{8}}+{\dfrac {1}{8}}\end{aligned}}$

Also, one can add more space between lines as follows:

${\begin{aligned}{\dfrac {1}{2}}&={\dfrac {1}{4}}+{\dfrac {1}{4}}\\[6pt]&={\dfrac {1}{8}}+{\dfrac {1}{8}}+{\dfrac {1}{8}}+{\dfrac {1}{8}}\end{aligned}}$

Michael Hardy (talk) 23:59, 17 May 2009 (UTC)

That was the last thing I wanted to hear... I suppose now I have to find a way of going through all 168 occurrences of the array environment in my document and analyse things... Thanks, anyway. I did find an alternative which is \addtolength{\extrarowheight}{...} which you can set and unset after each problematic environment (or do globally). x42bn6 Talk Mess 02:03, 18 May 2009 (UTC)

Using the "array" environment, one can still add space between the lines:

${\begin{array}{rclr}{\dfrac {1}{2}}&=&{\dfrac {1}{4}}+{\dfrac {1}{4}}\\[12pt]&=&{\dfrac {1}{8}}+{\dfrac {1}{8}}+{\dfrac {1}{8}}+{\dfrac {1}{8}}\end{array}}$

Michael Hardy (talk) 10:36, 18 May 2009 (UTC)

May 18

Decomposition of a symmetric matrix

Hi there - I'm trying to show that for any non-zero vector vi , any 3×3 symmetric matrix T_ij can be expressed as $T_{ij}=A\delta _{ij}+Bv_{i}v_{j}+(C_{i}v_{j}+C_{j}v_{i})+D_{ij}$ for some numbers A and B, some vector C_i and a symmetric matrix D_ij , where $C_{i}v_{i}=0,D_{ii}=0,D_{ij}v_{j}=0$. (Summation convention implicit here).

I've been told to show that the above statement is true by finding A, B, C_i and D_ij explicitly in terms of T and v - and I've been fiddling around for a while now, not achieved much more than $C_{i}={\frac {T_{ij}v_{j}}{T_{ii}}}$ and a few expressions in A/B/etc by dotting T with c and v. Can anyone suggest where I might go next? 131.111.8.102 (talk) 08:36, 18 May 2009 (UTC)

I guess that the best way should be to understand geometrically the problem; however, while waiting for a deeper answer from other people, here are at least some remarks. I'm using the same summation convention.

1. In order to simplify notations you can also assume that $v:=(v_{1},v_{2},v_{3})$ is a unit vector, that is $v_{i}v_{i}=1$. For a general non-zero $v$, just normalize it.

2. If there is a solution, it is unique; in particular, we find from the equations:

(2.1) $\scriptstyle T_{ij}v_{i}v_{j}=A+B$;

(2.2) $\scriptstyle T_{ij}v_{j}=Av_{i}+Bv_{i}+C_{i}=(A+B)v_{i}+C_{i}$;

(2.3) $\scriptstyle T_{ii}=3A+B$;

so, equivalently

$\scriptstyle A={\big (}T_{ii}-T_{ij}v_{i}v_{j}{\big )}/2;$

$\scriptstyle B={\big (}3T_{ij}v_{i}v_{j}-T{ii}{\big )}/2;$

$\scriptstyle C_{i}=T_{ij}v_{j}-T_{kj}v_{k}v_{j}v_{i}$

(the last seems to differ from your expression of C_i: how did you get it?). Hence you also have $D_{ij}$ by the initial equation. Thus, if there exists that decomposition, it is unique, and the value of $A,B,C_{i},D_{ij}$ are the one we found. Last step, you should plug these value into the equations and verify that they actually solve the problem (this seems OK to me, but you better check it). --pma (talk) 14:46, 18 May 2009 (UTC)

Addendum: in fact, you have the analogue unique decomposition of a symmetric matrix $T$ in any dimension $n>1$; in eq (2.3) you only have an $n$ instead of 3. For the computation you might find it easier (as I do) to use a matrix notation. You want numbers $A$ and $B$, a vector $C$ and a symmetric matrix $D$, such that for all vectors $x$ there hold:

$\scriptstyle Tx=Ax+B(v\cdot x)v+(C\cdot x)v+(v\cdot x)C+Dx;$

$\scriptstyle (C\cdot v)=0;$

$\scriptstyle Dv=0;$

$\scriptstyle \mathrm {tr} (D)=0.$

Assuming as before $\scriptstyle \|v\|=1$, the unique solution is given by

$\scriptstyle A:={\frac {1}{n-1}}{\big (}\mathrm {tr} (T)-(Tv\cdot v){\big )}$

$\scriptstyle B:={\frac {1}{n-1}}{\big (}n(Tv\cdot v)-\mathrm {tr} (T){\big )}$

$\scriptstyle C:=Tv-(Tv\cdot v)v$

$\scriptstyle Dx:=Tx-Ax-B(v\cdot x)v-(C\cdot x)v-(v\cdot x)C.$

Ask for further details if you need it. --pma (talk) 17:35, 18 May 2009 (UTC)

I re-approached the question and noticed an error in my C - thankfully I concur with you! The question also claims the A/B/C/D space is exactly the correct dimension to parametrize an arbitrary 3x3 symmetric matrix - why is that? I can't see a way to justify this - why would the space created by A-B-C-D be of the correct size? I can see how the fact that any such matrix has a single exact decomposition would imply that the space uniquely defines a map so I can see -how- it's the correct size, but I can't actually see -why-: what's the reason behind this? Does anyone have any insight, algebraic or geometric or any? Thanks!

Oh, and thankyou very much for the help pma, I was just making a big mess of things because I didn't spot an easy way to obtain 1 last equation so I was trying everything complicated, not very carefully apparently! I guess I just need to be more cautious when solving these sorts of things - thanks very much for the help again, 131.111.8.96 (talk) 18:30, 18 May 2009 (UTC)

At least, a dimensional check is easy. First, A and B are constants and C is an n-vector orthogonal to v: hence the (A,B,C)-space has dimension 1+1+(n-1)=n+1. What is the dimension of the space of all n×n symmetric matrices D with null trace and with the non-zero vector v in the kernel? If you observe that for an orthogonal matrix U the matrix UDU^T is still symmetric, with null trace, and with Uv in the kernel, you should realize that one can assume v=e₁=(1,0,..,0) with no loss of generality. In this case it is evident that D is fixed by n(n+1)/2 coefficients with n+1 independent linear conditions (all entries in the first column together with the trace of D are zero). So the D-space has dimension n(n+1)/2-(n+1)=(n+1)(n-2)/2, and the (A,B,C,D)-space has the right dimension n(n+1)/2 of the space of n×n symmetric matrices. In other words, with the assumption v=e₁ (that you can have after an orthogonal change of basis as said above), the four terms decomposition of the symmetric matrix T is just the sum of: a multiple of the identity; a (symmetric) matrix with support in the corner (1,1) that is, whose coefficients are zero except possibly the one of index (1,1) ; a symmetric matrix with support in the first frame (that is, whose coefficients are zero except possibly those in the first column or in the first row); a symmetric matrix with zero trace and support in the lower (n-1)×(n-1) square (that is, the first row and the first column are zero). --pma (talk) 21:16, 18 May 2009 (UTC)

how many spheres diam D1 can pack on the surface of a larger sphere diam D2

If you have spheres of diam D1, and a sphere of diam D2, D1< D2, how many small spheres can "pack" on the surface of the larger sphere ? If you have an answer, you might consider adding it to the article "sphere" thanks Cinnamon colbert (talk) 12:52, 18 May 2009 (UTC)

Please expand this question. A sphere is 3D, the surface of a sphere is 2D. I can imagine asking the question "How many circles can one fit on the surface of a sphere?", but what does packing spheres on the surface mean? In the event that you mean circles on a sphere's surface, what are you defining as the radius of the circles: The great circle distance, the circle centre to the circumference linear distance, the distance across a "2d chord" to the centre of a 2D circle whose centre is not on the sphere's surface, some other option...? -- SGBailey (talk) 14:30, 18 May 2009 (UTC)

sorry; if you have a sphere (say a basketball) you can put a pingpong ball on the surface, so the two are touching. The question is, how many pingpong balls can touch the surface of the basketball - this is an extension of the kissing number ( http://local.wasp.uwa.edu.au/~pbourke/geometry/kissing/ ) which addresses spheres of the same size —Preceding unsigned comment added by Cinnamon colbert (talk • contribs) 16:46, 18 May 2009 (UTC)

The problem is usually stated in terms of disks: the circle that is "under" the pingpong ball with respect to the basketball's gravity.

No general answer is known (though an upper bound is easy). For some small n, proofs exist that a specific arrangement is the densest packing of n disks on a sphere, and thus for bigger disks the packing-number is less than n; for all other n, all we have is best known packings. Some solutions are linked at [1]. —Tamfang (talk) 16:59, 18 May 2009 (UTC)

thanks, tamfang, that looks great, cinn col —Preceding unsigned comment added by 65.220.64.105 (talk) 17:45, 18 May 2009 (UTC)

On a similar note, I've been trying to show topologically (looking at you if you're bored PST! ;)) that in $\mathbb {R} ^{n}$ there exist $(1+c)^{n}$ disjoint closed unit balls inside any closed ball of radius 3.001 (or an analogous ratio - does the 3.001 specifically matter?) for some constant c > 0 (using the Euclidean metric) - can anyone suggest a starting point? Cheers muchly 131.111.8.96 (talk) 18:44, 18 May 2009 (UTC)

Try seeing what happens when you double the size of each of the sphheres, that should cover every single point inside the large sphere possibly a number of times for some points. What is the ratio of the volume of all the spheres to the large sphere? This seems to indicate that just over 2 for the large sphere should be good enough but it seems rather counter intuitive to me so I'd check the logic carefully. Dmcq (talk) 09:54, 19 May 2009 (UTC)

This was wrong, see WP:RD/MA#Spheres in an n-dimensional sphere below. If the big sphere has diameter greater than 1+sqrt(2) then the number goes up as a power of n. For that number it just goes up linearly. Dmcq (talk) 11:57, 22 May 2009 (UTC)

Still no simple answer to what i thought would be a simple question: if you put a small sphere onto the surface of a larger sphere (in the commonsense usage, onto just touching) is, given the diameters of the 2 spheres, there a simple formula that says how many small spheres can coat the surface of the larger sphere . I don't need a guaranteed math best result; this is more of an engineering type question (and maybe I should go there) - I'm happy with a formula that is within ~90% of bestCinnamon colbert (talk) 13:50, 19 May 2009 (UTC)

If D1<<D2, a reasonable estimation should be: the number N of 2 dimensional disks of diameter D1 that you can pack in a disk of diameter 2(D1+D2), which has the same surface area of a sphere of diameter D1+D2: thus, N≈4c((D2/D1)+1)², where c=0.906.. is the highest density for 2 dimensional unit disks. The idea is that we are looking at the traces of the small balls on the sphere of diameter D2+D1, like it were a planar disk packing.--131.114.72.215 (talk) 14:06, 19 May 2009 (UTC)

Consider the spherical triangle formed by the centers of three discs, all tangent to each other. That triangle contains 1/6 of each of three discs. Using spherical trigonometry, find the angle α of the triangle, which will exceed π/3. If my algebra is right, the number of such discs that will fit on a sphere is no more than ${\frac {2\pi }{3\alpha -\pi }}$. —Tamfang (talk) 06:30, 20 May 2009 (UTC)

Oops, wrong wrong wrong. —Tamfang (talk) 14:47, 20 May 2009 (UTC)

I think I have it right this time. Let r be the angular radius of a disc; β = asin((√3/2) tan r); the number of discs ≤ 12β/(6β-π). —Tamfang (talk) 15:13, 20 May 2009 (UTC)

What is the ratio of the area of a circle to a square

What is the ratio of the area of a circle to a square?

Why is answer arctan(1/2) + arctan(1/3) ? 122.107.207.98 (talk) 13:23, 18 May 2009 (UTC)

Actually, the answer is 5, or 17, or any other number. It really depends on the relative size of the two. Is this homework? --Stephan Schulz (talk) 14:00, 18 May 2009 (UTC)

Since the answer is supposed to be arctan(1/2) + arctan(1/3) = π/4, we can deduce that it is a ratio of the area of a circle with radius r to a square with side 2r. — Emil J. 14:24, 18 May 2009 (UTC)

So the ratio of the area of a square to the area of an inscribed circle. --Tango (talk) 18:13, 18 May 2009 (UTC)

Please, people. When someone posts what is clearly a homework answer, you should not produce the question for them. 79.122.45.200 (talk) 18:29, 18 May 2009 (UTC)

Saying that the answer is arctan(1/2) + arctan(1/3) is certainly not a good hint at how to answer the initial question (unless maybe the problem was approached via some particular geometric argument that you haven't told us about). It's really quite a separate question. Recall the formula for the tangent of a sum:

$\tan(\alpha +\beta )={\frac {\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }},$

and as a consequence

$\arctan x+\arctan y=\arctan {\frac {x+y}{1-xy}}$

(except that here you have to qualify the statement because of the "multiple-valued" nature of inverse trigonometric functions).

So plug arctan(1/2) + arctan(1/3) into that formula and by the time you're done simplifying you've got something really simple.

As for the ratio of areas, you should know how to find the area of a circle and how to find the area of a square (if you know the radius of the circle and the length of the side of a square). Michael Hardy (talk) 18:59, 18 May 2009 (UTC)

90 Degree Angle

Is a 90 degree angle acute, obtuse, or fall into its own category, as a right angle? I know that anything bigger that 90 is obtuse, and anything smaller is acute, but does that mean that a 90 degree angle is a right angle? —Preceding unsigned comment added by 99.58.205.224 (talk) 23:01, 18 May 2009 (UTC)

Yes, 90 degrees is called a right angle. That is a separate category inbetween acute and obtuse. --Tango (talk) 23:30, 18 May 2009 (UTC)

Ah, the acute, right and obtuse angles!! Great times, when I heard about this stuff for the first time. What I found exciting about maths at that age was that everything we learned at school, we would find there out, in the same day. --pma (talk) 06:13, 19 May 2009 (UTC)

As a maths teacher I get the young students to count how many right angles they can see from where they are sitting. Only then do I throw Pythagoras and the 3-4-5 triangle at them. Cuddlyable3 (talk) 11:04, 19 May 2009 (UTC)

May 19

Lifting prime ideals to integral extensions

While working on homework I ran into the following obstacle: suppose A is an integrally closed integral domain, and B is a ring, integral over A. Suppose p is a prime ideal of A. I believe that it follows that $pB\neq B$; I am looking for a proof of this fact. Can anyone knowing of a proof provide a few small hints? I have Marcus's Number Fields, which has a proof under the assumption that A is a Dedekind domain, which unfortunately is not good enough. Eric. 131.215.159.91 (talk) 08:37, 19 May 2009 (UTC)

You might check if "lying over" proves this. If q is a prime ideal of B lying over p, then p is contained in q, so pB is contained in q, so pB is properly contained in B. JackSchmidt (talk) 14:18, 19 May 2009 (UTC)

Unfortunately my professor's proof that there exists a prime lying over p starts by assuming that pB is not B. The article Going up and going down is of interest but doesn't have a proof. Eric. 131.215.159.91 (talk) 15:41, 19 May 2009 (UTC)

Some course notes reduce to the case of local rings first. At any rate, it appears to have complete proofs so might be useful. You can also just exhaustively check commutative algebra books for proofs until you find one you like. I think Kaplansky's Commutative Rings is likely to have good proofs. JackSchmidt (talk) 16:04, 19 May 2009 (UTC)

Thanks for the link. It had not occurred to me to localize first; if I still have trouble then I'll read the rest of the proof that you linked to. Eric. 131.215.159.91 (talk) 19:38, 19 May 2009 (UTC)

Assume for contradiction that pB = B. Then there exist $a\in p$ and $b\in B$ such that ab = 1. As B is integral over A, b is a root of a monic polynomial with coefficients from A. However, b = a⁻¹ is in the fraction field of A, hence $b\in A$ as A is integrally closed. Thus p = A, a contradiction. — Emil J. 16:13, 19 May 2009 (UTC)

Do you think it is still true if A is not integrally closed? JackSchmidt (talk) 16:23, 19 May 2009 (UTC)

I don't think so. The argument goes both ways, doesn't it? If $a\in A$ is not invertible in A and a⁻¹ is integral over A, we can find a maximal ideal p containing a, and put B = A[a⁻¹]. Then p is a prime ideal of A, and pB = B. — Emil J. 16:37, 19 May 2009 (UTC)

I'm talking nonsense: whenever a has an inverse in an integral extension of A, it has an inverse in A, regardless of whether A is integrally closed (it does not even have to be a domain). If ab = 1 and bⁿ + c_n−1bⁿ⁻¹ + … + c₀ = 0 with c_i from A, we can multiply by aⁿ to obtain 1 + c_n−1a + … + c₀aⁿ = 0, hence ad = 1 where d = −c_n−1 − … − c₀aⁿ⁻¹ is in A. — Emil J. 10:44, 20 May 2009 (UTC)

Thanks for the clarification. I only have a limited repertoire of domains, and I determined that such an a must be a unit in the integral closure. However, looking at specific units in quadratic integers, it was always annoyingly simple to convert from a to a⁻¹, usually no harder than to convert from i to 1/i = −i. The reason I asked is because the Lying Over proof does not require (as far as I can see) that A be integrally closed, only that B be integral over A (both domains). I can't tell if "lying over" is much too heavy of machinery for the problem though. JackSchmidt (talk) 12:33, 20 May 2009 (UTC)

I'm not sure what you mean by "the Lying Over proof". Also, it turns out that my assumptions of integrally closed and integral domain were unnecessesary. Eric. 131.215.159.91 (talk) 06:11, 21 May 2009 (UTC)

The proof I gave using LO. JackSchmidt (talk) 13:58, 21 May 2009 (UTC)

I also have a counterexample (courtesy of a friend of mine) for the case B not integral over A. Take $A=\mathbb {C} [x,y],p=(x,y),B=\mathbb {C} [x,y,z]/(xz+y-1)$. Then the minimal polynomial for z is $xZ+y-1\in A[Z]$ which is not monic. (Or simpler yet, take A any integrally closed integral domain that has a nonzero maximal ideal p, and B the fraction field of A.) Eric. 131.215.159.91 (talk) 19:48, 19 May 2009 (UTC)

I should clarify that by pB I mean the smallest ideal of B containing p. It is not necessarily true that all elements of pB are of the form ab with a in p and b in B; in general they are sums of elements of such form. Eric. 131.215.159.91 (talk) 19:27, 19 May 2009 (UTC)

That's a good point, I forgot that one cannot express the ideal so easily. This is a serious problem, I don't know how to fix the argument. — Emil J. 10:44, 20 May 2009 (UTC)

For future reference, here is an outline of the solution. Localize A and B at $A\setminus p$ to make $A_{p}$ and $B_{p}$. Then p is the unique maximal ideal of A_p. Let m be any maximal ideal of B_p. Then m lies above a maximal ideal of A_p, which must be p (we have used that B / A is integral). Then the assumption pB = B gives a contradiction. Eric. 131.215.159.91 (talk) 06:11, 21 May 2009 (UTC)

Name that theorem...

"A sequentially weakly lower semicontinuous coercive functional on a reflexive Banach space has a minimizer." A slight generalization appears in Giusti (2003), Direct methods in the calculus of variations, World Scientific. I'd like something to link from an article, so a theorem name or article that contains the theorem would be welcome. Otherwise, I would be happy to add the theorem to an article if a suitable target can be found. Sławomir Biały (talk) 20:49, 19 May 2009 (UTC)

What do you want exactly? I'm not sure if one can link this classic theorem to a name of an author. Usually, it is called the direct method in the calculus of variation: proving existence of a minimizer by showing that a minimizing sequence is compact in some topology for which the functional is sequentially lower-semicontinuous. A reference for it is more easily found in books rather than articles: e.g. Giaquinta & Hildebrandt "Calculus of Variations I", or Struwe's "Variational Methods", &c. --pma (talk) 22:10, 19 May 2009 (UTC)

Yes, I'm aware that one form or another of this result appears in books. And, yes I'm aware that it is absolutely fundamental in the calculus of variations. However, our article calculus of variations is primarily devoted to Euler-Lagrange type calculations, and addresses neither necessary nor sufficient conditions for the existence of a minimizer. Would it be reasonable, then, to start an article direct methods in the calculus of variations? Or has anyone on Wikipedia already done this, and put it in a less obvious place? Sławomir Biały (talk) 22:55, 19 May 2009 (UTC)

Exponential functions multi-valued?

My math teacher says the following is wrong, but can't explain why. I'd appreciate it if someone here could point out the error for me... For g(x) = x^1/2, R -> R, y = g(x) -> x = g^-1(y), so y = x^1/2 -> y^2 = x, which is multi-valued at least for positive real x. Any exponential function a^x, for x = 1/2, is a^1/2. Therefore, exponential functions are multi-valued at at least one point. (More than that, actually, because of x^1/4, etc., at least a countable infinity number of times.) (I'm only in Pre-Calc, by the way, sorry if this should be obvious.) —Preceding unsigned comment added by 68.46.107.173 (talk) 22:50, 19 May 2009 (UTC)

The question is not quite clear (which may be because I'm not aware of the term "multi-valued"). First, you are certainly not talking about exponential functions (which are expressed by x:->a^x, i.e. the parameter is the exponent, not the base). I think you are looking at the inverse "function" of a parabola, and expect it to have two branches, one above the x-axis and one below, so that sqrt(4) should be both 2 and -2. Well, what is wrong is the assumption that the inverse of a function is a function. A function by definition only has one output value for every input value. That implies that only strictly monotonic functions have an inverse that is also a function - for other function, the inverse is only a relation. --Stephan Schulz (talk) 23:12, 19 May 2009 (UTC)

Yeah, I was looking at the inverse of y = x^2 there in the beginning.

Also, oops, I didn't mean to say function. I knew that was a relation.

What I was saying was that if all for all positive x > 0, y = x^1/2 is multi-valued (by that I mean it maps to more than one output) then so should exponentials. For example, when x = e in y = x^1/2, y = ±√e. (right?) I meant that there would be a point in e^x that is the same relation, y = e^1/2, when x = 1/2. Does x^1/2's not being a function make the argument invalid? I'm trying to say that because exponentials share the same things that make x to some fraction with an even base not functions, it seems like they shouldn't be functions either. 68.46.107.173 (talk) 23:52, 19 May 2009 (UTC)

Exponentials are single-valued, the inverse of an exponential (a logarithm) is multi-valued. The notation isn't ideal, but in this context e^1/2 means the positive square root, so it is single-valued. --Tango (talk) 00:01, 20 May 2009 (UTC)

Actually, that isn't quite right. Logarithms are only multivalued when you are working with complex numbers, any positive real number has a unique real logarithm (in a given base). Exponentials where the base in a complex number is also multi-valued. See Exponential function for a rather better explanation that I seem to have managed! --Tango (talk) 00:06, 20 May 2009 (UTC)

Alright, then. So you're saying that the reason e^x only yields positive values is because it's conventionally defined that all roots in the exponent of an exponential function are the principal root -- not based on what the combination of symbols 'e^x' taken by itself would entail, but an external definition? 68.46.107.173 (talk) 03:45, 20 May 2009 (UTC)

Ah, perhaps I misunderstood what you were asking before (when I wrote my reply below). Take $e^{1/2}$ as a concrete example. This expression is meaningless until we define what a fractional exponent means. By definition, $e^{1/2}={\sqrt {e}}$, which is the positive square root. Likewise, by definition, $e^{1/3}={\sqrt[{3}]{e}}$, which is the principal cube root. I don't think this is an "external definition"—it's the definition of fractional exponents, as far as I am aware. You argue that $y=x^{1/2}$ is equivalent to $y^{2}=x$. But this isn't true, because that's not how fractional exponents are defined. [For example, $(-2)^{2}=4$, but $4^{1/2}\not =-2$.] I suppose fractional exponents could be defined in this way, and then fractional exponents would be multi-valued. This would complicate things, though, so the "principal root" definition of fractional exponents is used instead of your definition. This is a subtle point—either way you go, you have to define what a fractional exponent means separately, because the "repeated multiplication" definition of positive integer exponents doesn't make sense for a fraction. —Bkell (talk) 06:44, 20 May 2009 (UTC)

Another way to answer this question is to note that the "squaring" function, $f(x)=x^{2}$, is a function, meaning that each input x has exactly one output, but it is not one-to-one, meaning that several different inputs can map to the same output. (It is also not onto, meaning that there are some real numbers that are never produced as output—namely, the negative numbers—assuming that we are restricting the domain of the function to real numbers only.) So, when we look at it from the other direction, mapping outputs of the squaring function back to the inputs that produced them, we find that some outputs (the positive real numbers) correspond to more than one input, and other outputs (the negative real numbers) correspond to no inputs at all. The number zero is the only output that corresponds to exactly one input. So the "inverse" of the squaring function is not a function. This is because the squaring function is not both one-to-one and onto. The properties of injectivity ("one-to-one-ness") and surjectivity ("onto-ness") are not necessarily preserved under the operation of finding an inverse to a relation. You might find it interesting to work out the conditions needed for a one-to-one relation to have a one-to-one inverse, and other similar questions. It might help to draw pictures like the ones in the injective function article. —Bkell (talk) 03:11, 20 May 2009 (UTC)

See Exponentiation#Roots of arbitrary complex numbers and multivalued function. An expression such as z^1/2 where z is a complex number is defined either as the multiset of solutions to the equation x²=z, or as a so-called principal value. Both options are found in the litterature, but none of them are satisfactory, because the multiset of solutions is usually not considered to be a number, and the principal value of z^1/2 is not a continuous function of z. The usual rescue is to avoid using noninteger powers of (complex) numbers which are not positive reals. Bo Jacoby (talk) 09:54, 20 May 2009 (UTC).

You can define the unambiguously single-valued exponential function

$\exp x=\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}=\lim _{n\rightarrow \infty }\left(1+{\frac {x}{n}}\right)^{n}$

where n is an integer and the superscript means repeated multiplication, and then define $y^{x}=\exp(x\log y)$, where log is the real log, the multivalued complex log, or a branch of the multivalued log. So there are at least four different meanings of the superscript notation in common use: repeated multiplication (also used in discrete algebras where the others don't make sense), the single-valued continuous one where y is a positive real (also used in things like e^A where A is a matrix), the multivalued continuous one where y is any nonzero complex number, and (at least in numerical computing) the discontinuous complex one using the standard branch of the log function with imaginary part in (−π,π]. Fortunately it's usually obvious which one is meant. -- BenRG (talk) 12:44, 20 May 2009 (UTC)

May 20

Goldbach conjecture

Can someone explain why this doesn't solve the Goldbach conjecture? I know nothing of maths.

"1 - All of the prime numbers other than 2 are odd, 2 being the only even prime number. Further, the even number 4 = 2 + 2.

2 - The sum of any two of the odd prime numbers is always an even number.

3 - All combinations* of the odd numbers ≥ 3 [whether prime or not] summed in pairs produces all of the even numbers ≥ 6.

3 - While just the prime odd numbers in sequence is a sequence with gaps as compared to that of all of the odd numbers; nevertheless, all combinations of the odd prime numbers ≥ 3 summed in pairs produces all of the even numbers provided that there are enough primes preceding the gaps.

4 - That requirement is that π(N) ≥ R(N) = G·Ln(N) where N is the first number in the gap, π(N) is the number of primes less than or equal to N, R(N) is the number of preceding primes needed to assure clearance of the gap, and G is the number of sequential non-primes in the gap. This requirement is comprehensively satisfied by all of the prime numbers and gaps because of the sufficiently smooth nature of π(N)." 86.8.176.85 (talk) 02:00, 20 May 2009 (UTC)

Do you care to tell us where you found this? —Tamfang (talk) 06:34, 20 May 2009 (UTC)

most likely here: [2] --pma (talk) 06:57, 20 May 2009 (UTC)

A friend of mine who does maths as a hobby was explaining the Goldbach conjecture to me, and told me he was sure this doesn't solve it but unsure why. Just wondering if anyone could tell me what's wrong with it... 86.8.176.85 (talk) 19:03, 20 May 2009 (UTC)

This is basically an argument that Goldbach's conjecture is plausible because there are "enough primes" available to make sums from. The primes look randomly distributed, and if you pick a bunch of random odd numbers distributed similarly to the primes then it's exceedingly unlikely that there will be any large even number you can't make with them (and the small even numbers have been tested). Goldbach's conjecture#Heuristic justification has more details. To prove the conjecture you would have to completely rule out the alternative, not just judge it implausible. The plausibility argument is obvious to modern mathematicians, so the author of this paper hasn't done anything new. I don't know how obvious it was in 1742 when Goldbach made the conjecture. -- BenRG (talk) 19:40, 20 May 2009 (UTC)

The term "plausibility argument" is standard for this sort of thing, but in cases like the Goldbach conjecture it rather understates the case.

The Goldbach conjecture is true. It's a mortal lock. In a corresponding case in the physical sciences we would have no squeamishness about expressing it that way. It's only the availability of proof-from-axioms in mathematics, that makes us worry about the fact that we don't have one (from any widely accepted axiom set) in the case of Goldbach. --Trovatore (talk) 20:33, 20 May 2009 (UTC)

No, this is wrong. The physical sciences don't deal with philosophical absolute truth. Math does. And we do have conjectures that check out for extremely large sets of numbers, but do eventually yield under an even larger counterexample (for a trivial example, use "all integers are smaller than 10^{((((((((((((10!)!)!)!)!)!)!)!)!)!)!)!)!}. --Stephan Schulz (talk) 08:49, 21 May 2009 (UTC)

The notion that mathematics deals in apodeictic certainty is an error that goes back to the ancients, but an error nonetheless. The differences between mathematics and the empirical sciences is one of degree, not kind. You always need axioms, and the axioms could be wrong.

If you think that the argument is simply that Goldbach holds up to very large values, then you haven't understood the argument. Review the link Ben provides above. --Trovatore (talk) 09:56, 21 May 2009 (UTC)

Can you explain in what sense axioms can be "wrong"? One way is obviously that they're inconsistent. Another would be that they do not correctly describe some physical reality (like Euclidean geometry vs. relativistic space), but I think we all agree that a physical interpretation is not a requirement.

I have always been under the impression that mathematics aspires to make certain deductions. While the statement "π is irrational" might not be an absolute truth, the statement "from the ZFC axioms it follows that π is irrational" should be. If ZFC turns out to be inconsistent, the latter will still be an absolute truth, as will "from the ZFC axioms it follows that π is rational".

Another issue that may be relevant here is that even if some statement is true, we humans will never be able to know this with certainty. Usually we are convinced when we see a proof, but it's possible the proposed proof contains a mistake which we have overlooked. We can overlook it again when we re-read the proof, and it's possible (even if extremely implausible) that every human or computer that has ever checked the proof has overlooked it. -- Meni Rosenfeld (talk) 11:10, 21 May 2009 (UTC)

Obviously I can't do justice here to the realist–formalist debate. I'll limit myself to noting the following:

• Most people are realists about at least small natural numbers; they think of the number 2 as being a real, though non-physical, object, and if a collection of axioms proves that 2+1=5, then at least one of those axioms is wrong (assuming that 2,1,5, and + are expected to have their usual interpretations).
• The existence of a formal, syntactic proof of a formal sentence from ZFC, assuming it's short enough to actually write down, is indeed about as absolute as things get (maybe less so than cogito ergo sum). However it's not very interesting. It gets more interesting if you interpret it semantically, as "this statement is true in every model of ZFC". But only if you're willing to make an ontological commitment to models of ZFC, and in that case you're back to realism. (By the way, this sort of "realism about models" is enough to guarantee that CH has a determinate truth value, at least given that the powerset axiom is actually true, which is itself a claim that a realist-about-models must consider well-specified.)
• I disagree that the essence of mathematics is to make deductions. The essential purpose of mathematics is to discover what the truths are about mathematical objects. In the case of the Goldbach conjecture, we know the truth, at least if knowledge is interpreted as justified true belief. --Trovatore (talk) 23:34, 21 May 2009 (UTC)

While I find very instructive your point of view, your statement about the Goldbach conjecture seems a bit optimistic to me. For sure we think that it is true, but it seems to me that this belief is based on the fact that all that it is known about the problem suggests a very reasonable statistic model where this conjecture is true, with extremely high probability. So in a certain sense this belief is based on our lack of knowledge of the problem. Don't you think that a new result may change our state of knowledge and force to abandon that model (in particular, the independence assumptions, that are based on a principle of indifference)? --pma (talk) 01:26, 22 May 2009 (UTC)

Oh, it's conceivable in principle, certainly. But that's my point — this is not different, in principle, from our other sorts of mathematical knowledge. It's also conceivable in principle that a new result could change our state of knowledge and cause us to reevaluate some of our axioms. --Trovatore (talk) 01:52, 22 May 2009 (UTC)

No doubt that this is a point of view that one can not ignore. The similarity with physics is very attractive indeed. But my concern is mainly on the status of certainty of Goldbach conjecture in particular. So, if one estimates the number of ways an even number n writes as sum of 2 primes, and one assumes "enough independence" (in some technical sense) about the distribution of primes, one finds that the number of ways is even divergent. This independence is a very reasonable assumption, because there is not a single result, nor any similar conjecture, that may suggest the contrary. As a consequence, we strongly believe that the CG is true. But does it means that we should believe with the same certainty that nobody will disprove it in the next years, or just find a good reason to change the above statistical model? I feel a bit dumb, but I am inclined to say no. The CG has very high probability to be true, given the current state of art, but it seems to me vaguely improper to argue from this, that any new change of the state of art is most likely one going in the direction of the proof. After all, what mathematicians usually do, is to find unexpected results, able to completely change a perspective. If today we look at primes like a "swarm of bees", it is maybe just because they have not yet done the next step into the knowledge.--pma (talk) 14:17, 22 May 2009 (UTC)

As to the probabilistic argument about the validity of Goldbach conjecture, let's recall that although very meaningful, it is far from an absolute statement. As I see it, it says that, given our current knowledge about prime numbers, it is extremely unlikely that a counterexample exists. So, it is rather a description of the current state of knowledge of the problem, than a definitive claim on the trueness of the GC. Indeed, it is based upon a model that may be completely changed after a single observation --of course, that would be ipso facto a very deep one. (To make an example, there was a time when the knowledge of geometry was such, to make very plausible the conjecture that the edge and the diagonal of a square be in rational ratio. Why to assume there is a non-rational ratio, if all you know is rational?). Indeed I do not see how the probability that a given conjecture is true may have a meaning outside a relative, Bayesian point of view.

Moreover, the fact is that we are more curious to see a proof of a mathematical fact, than to know whether it is true. Suppose the archangel Michael appears to a number theorist claiming that GC he's trying to prove is true. Then, he would be a bit embarassed... while certainly glad of the visit, I don't think he would give up working on it. --pma (talk) 22:47, 21 May 2009 (UTC)

Parimutuel betting

As a novice, I have been trying to understand the mathematics of parimutual betting. I can understand the (theorectical) situation where bettors are only allowed to bet on the winner. After deduction of the running costs, the total money bet by all punters is divided up among the winning bettors in proportion to how much they have bet.

But how is the money divided up when other types of bet are also allowed? The win, place, show, and more exotic bets? 89.242.109.25 (talk) 12:21, 20 May 2009 (UTC)

I imagine you would create a separate pool for each type of bet (win, place, show etc.), pro-rata costs across pools (i.e. take same percentage cut from each pool), then divide net amount left in each pool between the folks who placed winning bets of that type. I can see a problem with more exotic bet types though - what happens to the trifecta pool, for example, if no-one correctly predicts first, second and third places ? Gandalf61 (talk) 14:43, 20 May 2009 (UTC)

From reading the article (and the talk page), I get the impression that separate pools for each bet type are typically used. However another possibility (perhaps never employed) is to have a single pool, and then fix the ratio of the payouts. For example, the winners on the "win" bet could receive 3 times that of any "place" bet winners, who receive twice that of any "show" bet winners. From there it's simple algebra to figure out the payouts for each bet. -- 128.104.112.117 (talk) 18:59, 24 May 2009 (UTC)

transcendental equation

May somebody help me solve this equation? (Maple's numerical methods don't work, apparently) --Taraborn (talk) 14:03, 20 May 2009 (UTC)

sin(7.147*sin(x)) / (7.147*sin(x) = 1/sqrt(2)

And this one:

cos(x)*sin(9.3462*sin(x)) / (9.3462*sin(x)) = 1/sqrt(2)

Thanks. --Taraborn (talk) 14:03, 20 May 2009 (UTC)

I get 0.195957 and 0.147234 for the smallest positive solutions; in each case the function is even, of course, and they have periods π and 2π. Is that what you needed? --Tardis (talk) 14:24, 20 May 2009 (UTC)

No, both equations don't have any solutions at all. You can find this by plotting the function and checking that they don't intersect where they should for a solution to exist. - DSachan (talk) 19:36, 20 May 2009 (UTC)

Erm, it looks to me like each of those equations has solutions, where Tardis said. There is no simultaneous solution to both equations, but separately each one is OK. —Bkell (talk) 21:56, 20 May 2009 (UTC)

What are the solutions then? The solutions given by Tardis are not correct. Alright, sorry, everything seems to be alright. I goofed up a bit. Solutions given by Tardis are correct ones. - DSachan (talk) 22:29, 20 May 2009 (UTC)

does anyone know how to solve this????

(18 + 308^1/2)^1/2

this equation can be written in the form of a^1/2 + b^1/2, where a and b are whole numbers and a is greater than b.

i need the values of a and b.

plz hhhhheeeeelllllpppp.... —Preceding unsigned comment added by 117.197.245.126 (talk) 14:49, 20 May 2009 (UTC)

As far as I know how it works here, people will help you sooner, if you show that you have made some effort.

I write the equation for you:

(18 + 308^1/2)^1/2 = a^1/2 + b^1/2

Now, what is your next step, to find a and b? (the funny thing is that a and b are hidden in your user number) --84.221.208.46 (talk) 14:57, 20 May 2009 (UTC)

It may help to see if you can put √308 into the form c√d, where c,d are whole numbers. —Tamfang (talk) 15:20, 20 May 2009 (UTC)

ok, i've done it till (18 + 2(77)^1/2)^1/2 now what???? —Preceding unsigned comment added by 122.50.130.33 (talk) 16:15, 20 May 2009 (UTC)

Well, what you want is

(18 + 308^1/2)^1/2 = a^1/2 + b^1/2

that is, if you square both sides,

18 + 308^1/2 = (a^1/2 + b^1/2)².

So why don't you try expanding the square at the RHS,

(a^1/2 + b^1/2)² = a + b + 2(ab)^1/2,

and then try writing two equations:

18 = integer term on the LHS = integer term on the RHS = a + b;

308^1/2 = square root on the LHS = square root on the RHS = 2(ab)^1/2.

Then get rid of the square root in the second equation. Try it, it's not that difficult; you'll get

a + b = 18

ab = 77.

Now either you find the solution at the first glance (remember that a and b are integer numbers), or you first compute as usual

(a - b)² = (a + b)² - 4ab.

84.221.208.46 (talk) 16:35, 20 May 2009 (UTC)

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)

1. 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.
2. 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)

Spheres in an n-dimensional sphere

Resolved

 – Dmcq (talk) 07:08, 22 May 2009 (UTC)

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 $\scriptstyle R<1+{\sqrt {2}}$, 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 $\scriptstyle R\geq 1+{\sqrt {2}}$ 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(x_N,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-x_i||<2 (because if on the contrary there were a point x in B(0,R-1) such that ||x-x_i|| 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(x_N,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<asup>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)

May 22

HOW HE DID IT?

On a web I come across a problem that runs as follows: (http://www.cimt.plymouth.ac.uk/resources/help/h1alge1.pdf )

"Han Sin, a Chinese general, devised a method to count the number of soldiers that he had. First, he ordered his soldiers to form groups or 3, followed by groups of 5 and then groups of 7. In each case he noted down the remainder. Using the three remainders, he was able to calculate the exact number of soldiers he had without doing the actual counting."

Please help me to know HOW HE DID IT? Kasiraoj (talk) 08:01, 22 May 2009 (UTC)

Possibly a chinese remainder theorem has something to do here...? --CiaPan (talk) 08:07, 22 May 2009 (UTC)

Hardly a general if he only did those numbers, sounds like the equivalent of a lieutenant or possibly a major with a small company. Unless the Chinese had rather inflated ranks then. I suppose it's possible he was in something like intelligence where you don't need so many people below you to get high rank, it would explain also why he counted in such a strange way that gave little information away as it was being done. Perhaps it was an early form of this Zero knowledge average, the wiki article Zero-knowledge protocol doesn't seem to have much on this sort of thing. Dmcq (talk) 10:31, 22 May 2009 (UTC)

Well, all solutions would be congruent modulo 105, so the army must be small enough that a rough estimate is accurate to the hundreds. This should be achievable with sizes within an order of magnitude of 1000. So General could be correct. Taemyr (talk) 23:39, 23 May 2009 (UTC)

I'd have thought a general would command a lot more. It's hard to find from wiki but I know of a colonel who was in charge of a few thousand troops and a general is a few ranks above that. I'm surprised he got so far, he refused staff officer training the first time because his wife wanted him at home. Dmcq (talk) 13:24, 24 May 2009 (UTC)

The lowest rank of general (a Brigadier or Brigadier General) typically commands a brigade (hence the name) of a few thousand men. Sometimes they are commanded by colonels, too, though, so that's probably what the colonel you knew commanded. --Tango (talk) 15:15, 24 May 2009 (UTC)

Hmm, I was going to suggest looking at our article about Garner's algorithm, but we don't have one. Maybe someone can fix that. It should be in Knuth vol. 2 and places like that. Anyway, yes, the story is obviously supposed to be an illustration of the Chinese remainder theorem. 207.241.239.70 (talk) 10:35, 22 May 2009 (UTC)

probability book?

Anyone got a favorite introductory probability book for math nerds? I'd like something mathematically solid (free of hand-waving; e.g. it should introduce sigma-algebras at the appropriate place) but basically application-oriented rather than foundational, discussing common distributions and how to calculate stuff with them, etc. It would be good to also have some coverage of basic statistics, stochastic processes like Brownian motion, and maybe some trendy subjects like machine learning. Thanks. 207.241.239.70 (talk) 10:32, 22 May 2009 (UTC)

My favorite is Standard Probability and Statistics Tables and Formulae by Zwillinger and Kokoska. Warning: The book is less a text than a "handbook" -- it assumes that the reader has a solid foundation in mathematics. Wikiant (talk) 10:42, 22 May 2009 (UTC)

Thanks, I may look for that, but I think I want a textbook with exercises. I'm starting to feel slightly clueful about the subject through surfing wikipedia, but that's usually a bogus feeling, since the understanding hasn't really sunk in. 207.241.239.70 (talk) 11:54, 22 May 2009 (UTC)

What about this one by Grinstead & Snell, available for free download from the website of Dartmouth College? --NorwegianBlue ^talk 13:51, 22 May 2009 (UTC)

Nice!!!! Online books are the greatest. This one is a little more elementary than I had in mind, but it's probably a good way to get started anyway. Thanks! 207.241.239.70 (talk) 14:29, 22 May 2009 (UTC)

Maximum Volume of Cone made from cut sector of a circle.

Let's say we have a circle with a radius R cm. A sector can be cut out to form a cone,by joining the straight sides of the sector together. What is the angle of the sector to be cut out such that the maximum of volume of the cone can be achieved? I have already arrived a few formulas of the cone.

Radius of cone's base= $\pi R\theta /180$
height of cone =${\sqrt {R^{2}-(\pi R\theta /180)^{2}}}$

is differentiation needed? if yes, can you please explain in more detail as i have not learnt it yet. —Preceding unsigned comment added by Invisiblebug590 (talk • contribs) 11:43, 22 May 2009 (UTC)

It's not clear to me that you need to use calculus for this, but yes, differentiation is a standard way to solve maxima and minima problems. You've got formulas for the cone's base and height. So you should be able to find a formula for the volume as a function of theta. Notice that if theta is very small, the cone is thin and needle-like with small volume. And if theta is very large, the cone is flat like a slightly pointed pancake; again the volume is small. If you draw a graph between those, there is a maximum somewhere, and the derivative is zero at the maximum. So you have to differentiate the function and find where the derivative is equal to zero. 207.241.239.70 (talk) 11:51, 22 May 2009 (UTC)

I'd rather declare some symbol for the cone's base radius, say r, and write

$r=R{\tfrac {\theta }{2\pi }}$

where θ is the sector's angle in radians. Then the cone's height

$h={\sqrt {R^{2}-r^{2}}}$

so its volume

$V={\tfrac {1}{3}}\pi r^{2}\cdot h={\tfrac {\pi }{3}}r^{2}{\sqrt {R^{2}-r^{2}}}$

The volume approaches zero as r becomes zero (because the $r^{2}$ term becomes zero) or when r approaches R – these are two cases described by 207.241.239.70 above. Between these limits volume is non-zero, positive continuous, and it reaches a maximum somewhere. We may further simplify expressions if we note, that the volume V is maximum iff its square V² is maximum. So we write:

$V^{2}=\mathrm {some\ constant} \cdot r^{4}\cdot (R^{2}-r^{2})=\mathrm {some\ constant} \cdot (R^{2}r^{4}-r^{6})$

If we consider r growing from 0 to R, the difference in parens grows as long as the growing speed of $R^{2}r^{4}$ is greater then that of $r^{6}$. When the two become equal the expression gets its maximum value and falls back as $r^{6}$ grows faster than $R^{2}r^{4}$.

A 'growing speed' is simply a derivative, which can be easily calculated for the expression above – see rules at derivative#Computing the derivative and List of differentiation identities#Derivatives of simple functions.

The equation $(V^{2})^{\prime }=0$ transforms to $(R^{2}r^{4})^{\prime }=(r^{6})^{\prime }$

When you substitute appropriate expressions for derivatives you'll get a simple equation with r. Solve it, and then calculate θ from the first equation above. --CiaPan (talk) 14:41, 22 May 2009 (UTC)

I calculate that the angle to be cut out for greatest conic volume is about 66 degrees. An interesting (to me, anyway) extension is to use the removed part of the original disc to make a second cone, the objective now being to maximise the combined conic volume. The expression to be maximised is rather more complicated than before, so it's easiest to get an answer numerically. It's obvious that there'll be two maxima, as either portion can be deemed to be cut out or left behind. What's maybe surprising is how flat the volume function is, with the value within about 1% of the maximum for all angles between 90 and 270 degrees.→86.164.72.176 (talk) 18:06, 23 May 2009 (UTC)

working out foot pound ratio for powder in grams

Hi

Im working on a experiment involving a small cannon, and need to know foot/ pound ratio for grams of gunpowder. I learned from a documentary, that 660 pounds can launch 2700 pounds 23 miles. How far could 30 grams of powder move a 165 pound projectile? i know its only a matter of 4-6 feet, but I would like to know exactly. can someone help me with these numbers? Rob —Preceding unsigned comment added by 79.68.219.232 (talk) 17:53, 22 May 2009 (UTC)

You want to figure out the kinetic energy that it takes to heave the 2700 pound shell 23 miles. Assume that's the maximum range, which means the gun is fired at a 45 degree angle, so the shell exits with equal horizontal and vertical velocity components. Is that enough for you to finish the calculation? Keep in mind that the big cannon may convert its gunpowder energy to the shell much more efficiently than your small cannon. 207.241.239.70 (talk) 20:35, 22 May 2009 (UTC)

I imagine that real-world issues like air resistance and cannon "efficiency" (proportion of energy that goes into kinetic energy versus heat, noise, recoil, friction, etc.) would make any such calculation without extensive experimental work hopelessly incorrect. At high speeds, at least, I bet that neglecting air resistance would cause you to be wrong by a factor of two or more. And efficiency probably varies wildly from one cannon to another, as 207.x.x.x said. The only feasible way I see to get a good estimate is to take your cannon and carefully experiment with different payload masses and different powder quantities and try to work out a reasonable estimator. (Also, your type of gunpowder might be different from the gunpowder mentioned in the documentary.) Eric. 131.215.159.91 (talk) 21:36, 22 May 2009 (UTC)

For example, assuming 100% efficiency and no air resistance and some other unreasonable approximations tells me that 660 pounds of explosive can send 2700 pounds of payload a distance of 239.5 km (148.8 miles), which is much larger than the actual 23 miles. Eric. 131.215.159.91 (talk) 21:48, 22 May 2009 (UTC)

May 23

A LaTeXnicality

I used \renewcommand to make the "enumerate" environment label each item with a capital letter followed by a period, thus:

\renewcommand{\labelenumi}{\Alph{enumi}{.}}

\begin{enumerate}

\item so there

\item and so on

\item blah blah

\item \label{thisgap}

\end{enumerate}

We will see that any gap of the kind described in (\ref{thisgap}) corresponds to a 

I expected a sentence that says

We will see that any gap of the kind described in (D) corresponds to a

but instead I got this:

We will see that any gap of the kind described in (4) corresponds to a

How can I get it to show a (D) rather than a (4)? Michael Hardy (talk) 02:17, 23 May 2009 (UTC)

This does the job:

\documentclass{article}
\usepackage{enumerate}
\begin{document}
\begin{enumerate}[A.]
\item so there
\item and so on
\item blah blah
\item \label{thisgap}
\end{enumerate}
We will see that any gap of the kind described in (\ref{thisgap}) corresponds to a
\end{document}

 — Pt _(T) 22:01, 23 May 2009 (UTC)

Probability of dots on a circumferance

I've read a popular book that mentions in passing calculating the randomness of dots on a circumferance of a circle. The details given are where seven dots lie within an arc of 24 degrees, and four other dots lie outside that range. 24/360 is 1/15. The formula given for calculating the probability is p = (1/15)^7 (14/15)^4 11!/(7!4!) = 1.5 x 10^-6.

Can someone explain how this formula was obtained please, so that I could use it with different numbers of dots?

The formula was used to test the periodicy of the recurring peaks in a time series, by having the time run repeatedly around the circle rather than along the more usual straight line. Are there any other formula for testing the randomness of peaks in a time series? 78.149.172.201 (talk) 13:02, 23 May 2009 (UTC)

Look up Binomial distribution here on wikipedia or in that text book. $(1/15)^{7}(14/15)^{4}{\frac {11!}{7!4!}}=(1/15)^{7}(1-1/15)^{4}{11 \choose 7}$ —Preceding unsigned comment added by Taemyr (talk • contribs) 13:52, 23 May 2009 (UTC)

This looks like a case of testing a hypothesis suggested by the data. A better statistical test would take into account the precise locations of the 11 points. Michael Hardy (talk) 11:06, 24 May 2009 (UTC)

May 24

Margin vs Markup

I'm having some problems wrapping my head around the difference between margin and markup. I understand that margin is the percentage of the sale price that's profit and that markup is the percentage of the cost to make up the selling price. I know this information but I'm having difficulty comprehending it.

If the cost of an item is $100 and I want to make 50% profit, do I calculate the margin or markup? If I calculate markup I get a $150 selling price while calculating margin gives me a $200 selling price. In other words, does that 50% refer to margin or markup?

I'm pretty confused. Help! - Pyro19 (talk) 04:12, 24 May 2009 (UTC)

To me, "50% profit" sounds as if the profit is 50% of the selling price, so that you have a 50% margin, or equivalently, a 100% markup. Michael Hardy (talk) 11:03, 24 May 2009 (UTC)

value of pi

What is the value of pi?--Lightfreak (talk) 08:35, 24 May 2009 (UTC)

Check Pi#Numerical value. —JAO • T • C 08:57, 24 May 2009 (UTC)

The value of pi is approximately:

3.14159265358979323846264338327950288419716939937510

See this site for a value of pi to a greater degree of accuracy. Let me note that the value of pi to a finite number of decimal places is really not of much interest, except of course for computational problems. In that case however, a calculator or a computer is used. In other words, the value of pi has no real mathematical meaning, but rather there are some interesting properties of pi. These properties of pi can be determined without knowing the value of pi. According to normal number for example, it is not known whether pi is a normal number. In other words, if you give me a finite string of digits, is it true that this finite string appears consecutively in the decimal expansion of pi? Although this is thought to be true, no proof exists. There are other properties of pi, however, which are known and are relatively simple to derive. For example, pi is not an algebraic number and it is irrational. Proving such facts has a much greater merit than computing the value of pi. However, if you search "value of pi" on google, you should get relevant values, or even click on the link suggested by JAO above. I hope that this answers your question. --PST 09:09, 24 May 2009 (UTC)

If you just type pi as a query to google it gives a value to 8 decimal places at the start of the results. You can also convert units and do other simple maths these, e.g. 1+sqrt(2). Dmcq (talk) 12:34, 24 May 2009 (UTC)

Nonlinear Homogeneous Differential Equations

Can a differential equation be nonlinear but is still homogeneous? --Yanwen (talk) 15:19, 24 May 2009 (UTC)

What's your definition of homogeneous differential equation? If you mean, for instance: "a differntial equation such that for all solution u(t) and any real number λ, the function λu(t) is also solution", consider e.g. ű(t)=f(u,u') where f(x,y) is a nonlinear function on R² and 1-homogeneous, that is f(λx,λy)=λf(x,y), like for instance f(x,y) = xy²/(x ² + y²).--84.221.81.90 (talk) 17:05, 24 May 2009 (UTC)

ü