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May 13

What are the benefits to humans of using a tree structure?

Such as building one on paper.

Or accessing one in a book or on a webpage.

Take the example in the article tree structure for example. What would the benefits be of having an entire encyclopedia rendered into a tree?

The Transhumanist 02:20, 13 May 2009 (UTC)

In the old days before computers, they used to render trees into encyclopedias. Does that help? 67.122.209.126 (talk) 10:06, 13 May 2009 (UTC)

Very funny. :) The Transhumanist 00:05, 14 May 2009 (UTC)

Second difference equation

What functions f: R → R satisfy f(x+2)-2f(x+1)+f(x)=0 for all x in R? Affine times 1-periodic works... Then? --Carminat (talk) 09:54, 13 May 2009 (UTC)

Homework? This is a linear recurrence. The first thing you could do is figure out the generating polynomial. 67.122.209.126 (talk) 10:16, 13 May 2009 (UTC)

For any 1-periodic functions a(x) and b(x), the function f(x) = a(x) + xb(x) solves the equation. This is the general form of a solution. — Emil J. 14:13, 13 May 2009 (UTC)

Carminat's guess is quite correct, except that I'd say "affine plus 1-periodic". Michael Hardy (talk) 21:52, 13 May 2009 (UTC)

Oh.... I see. Carminat was asking whether there are any other solutions besides affine plus 1-periodic. Notice that

$f(x+2)-2f(x+1)+f(x)=0\,$

implies that

$f(x+2)-f(x+1)=f(x+1)-f(x),\,$

so that every time you go one step to the right, you add the same amount. That makes it a straight line as long as you move only in steps of size 1. So affine plus 1-periodic is all there is. Michael Hardy (talk) 22:03, 13 May 2009 (UTC)

No, it isn't, see my post above. For example, the function $f(x)={\begin{cases}0,&x-\lfloor x\rfloor <1/2,\\x,&x-\lfloor x\rfloor \geq 1/2\end{cases}}$ solves the recurrence, and cannot be written as a sum of an affine and 1-periodic function. — Emil J. 12:14, 14 May 2009 (UTC)

Also, affine times 1-periodic functions do solve the equation (though they do not exhaust all solutions either), so it is reasonable to assume that Carminat really meant "times" when he wrote "times", not "plus". — Emil J. 12:21, 14 May 2009 (UTC)

OK, so MH's argument proves that for all x the restriction of f to x+Z is an affine function; but as you say it possibly depends on x, so you got your general form f(x) = a(x) + xb(x), where a and b are 1-periodic because x+Z=x+1+Z. Nice! --pma (talk) 14:01, 15 May 2009 (UTC)

Sigh. Haste makes waste. Michael Hardy (talk) 00:01, 18 May 2009 (UTC)

Does mathematics exist in an empty universe?

I am struggling with this question for a long time and at last decided to ask it here. The subject line says it all. Suppose the universe was absolutely empty. Will mathematical concepts make any sense in such a universe?

thanks for your help,

Yes they would. Mathematics centers around formal systems, which are just a set of axioms and rules of inference. They can certainly 'exist' in some platonic sense.

No they wouldn't. Formal systems with some degree of consistency and usefulness require a certain amount of creativity to develop. They also wouldn't make "sense", in the sense that they wouldn't have any kind of useful real-world interpretation.

That is, of course, just the tip of the iceberg on the debate of whether mathematics 'exists' in a platonic sense outside of our universe. You're welcome. --COVIZAPIBETEFOKY (talk) 14:05, 13 May 2009 (UTC)

Assuming that no empty universe exists, the answer is both yes and no, because a false statement implies any statement. So both: "If the universe is empty then mathematics does exist", and: "If the universe is empty then mathematics does not exist", are logically true. See material conditional. See however also counterfactual conditional. Bo Jacoby (talk) 15:18, 13 May 2009 (UTC).

It is an interesting question in the philosophy of mathematics, and I'm not sure it has a universally accepted answer. You can make the question more extreme, as well - if there was no universe (rather than there being a universe, it just being empty) would there be mathematics? Mathematics certainly exists outside the universe to some extent - you can discuss mathematical concepts that don't model any physical thing and they can even be quite useful in solving problems where the problem and solution are described in terms of other, more physical, mathematics. Whether it can be divorced from the universe entirely it a difficult question. I'm tempted to say it can be, but I'm not sure of all the philosophical implications of that, having never studied it in any depth. --Tango (talk) 15:30, 13 May 2009 (UTC)

No, mathematics is just a set of (sets of) axioms and their logical theory. It exists all on its own, without the need for any physical reality at all. If this is relevant is akin to the old question if a falling tree makes a noise if nobody is present to listen. --Stephan Schulz (talk) 15:38, 13 May 2009 (UTC)

If one identifies mathematics with the logical implications of certain axiomatic choices, then one can certainly argue that those axiomatic choices are chosen based on physical observations. One could then argue that a universe which is empty or sufficiently different from our own would neither construct nor find useful the mathematics of our universe. Dragons flight (talk) 15:52, 13 May 2009 (UTC)

That's certainly the view I'm erring towards, but I don't think it is universally held. --Tango (talk) 16:02, 13 May 2009 (UTC)

"God made the integers ;-)". I agree that the choice of axiom systems we are interested in is often guided by our experiences with physical reality. But the existence of these theories is unaffected. I'd make the point that logic as the language of maths is universal (or should that be "suprauniversal"?), and you can easily (*) use that to argue about all possible finite or enumerable axiomatizations. (*) for suitably weird values of "easy".... --Stephan Schulz (talk) 16:04, 13 May 2009 (UTC)

LOL. "But the existence of these theories is unaffected." Does a theory "exist" even if no intelligent being in your entire universe can concieve of it.  ;-) Personally I'd lean towards no. Dragons flight (talk) 16:13, 13 May 2009 (UTC)

Definitely yes! No question! I can write a program that enumerates all finite axiom sets, and all enumerable ones in the limit. Thus, the existence of these axiom sets (and hence the implied theories) is inherent in this one program. --Stephan Schulz (talk) 16:19, 13 May 2009 (UTC)

That's just begging the original question, your program does not exist in the empty universe either. — Emil J. 16:32, 13 May 2009 (UTC)

In addition to the problem that you aren't in the relevant universe, I'd love to know how you plan to enumerate axioms that you are incapable of conceiving. Dragons flight (talk) 16:40, 13 May 2009 (UTC)

Easy. I enumerate all strings over a suitable alphabet. That includes all valid axiomatizations (and a lot of garbage, ok). The harder part is to enumerate the infinite sets... --Stephan Schulz (talk) 16:47, 13 May 2009 (UTC)

And what if your alphabet is incapable of expressing certain axioms? The judgment that it is "suitable" is still constrained by a set of conceivable things. Dragons flight (talk) 16:55, 13 May 2009 (UTC)

That comes down to the thesis of Turing and Church (and the whole sheebang coming with it), which implies that any alphabet with 3 characters (2 real ones and a blank) is enough to encode any problem "efficiently". It's certainly enough to describe mathematics as we know it. --Stephan Schulz (talk) 17:35, 13 May 2009 (UTC)

As discussed at Church-Turing thesis, though it is established that "computability" can be expressed in many forms, the underlying conception of what computation is remains a physically motivated assumption (or in some views an axiom). The underlying issue has never actually been expressed in a way that is provable. Hence people in a different universe could very well have a different intuition for what computation is and hence arrive at different program for expressing / exploring the axioms of mathematics. If they understand computation differently, then they may ignore (or be unable to express) axioms we take for granted. Dragons flight (talk) 18:18, 13 May 2009 (UTC)

You might as well ask whether, in a hypothetical universe, the English language and the color blue exist. It comes down to some philosophical argument about whether the English language, mathematics, and the color blue can exist without anyone actually speaking English, without any object actually colored blue, and without any person actually studying mathematics. There are philosophical arguments on both sides about the sense in which these abstract objects exist, and there is not any definitive answer. — Carl (CBM · talk) 19:23, 13 May 2009 (UTC)

Indeed. If you are a Platonist or a formalist then you believe that mathematical objects and results are not dependent on the contents of "reality", so hypothesising an empty universe has no effect on the "existence" of mathematics. If you are a intuitionist or a constructivist then you believe that mathematical objects are the result of physical (or at least mental) processes; as these cannot exist in a empty universe then an empty universe contains no mathematics. Gandalf61 (talk) 11:07, 14 May 2009 (UTC)

If a question cannot be asked experimentally, you do not need to answer it theoretically (said Werner Heisenberg). Bo Jacoby (talk) 12:29, 14 May 2009 (UTC).

I suspect that the original question is made ambiguous in first place by the unspecified meaning of the verb to exist in the question itself. Therefore let's say: A) mathematics A-exists in an empty universe and B) mathematics does not B-exist in an empty universe, where the meaning of A-existence and B-existence is chosen in such a way that statements A) and B) are true by definition. --pma (talk) 15:13, 14 May 2009 (UTC) ;-)

The question you are asking is about a philosophical view called mathematical platonism. Some people believe in it, others don't. 207.241.239.70 (talk) 04:22, 15 May 2009 (UTC)

Triangle areas

If a triangle has semiperimeter s and interior angles a, b, and c, what is its area? --Lucas Brown 42 (talk) 16:20, 13 May 2009 (UTC)

Please do your own homework.

Welcome to the Wikipedia Reference Desk. Your question appears to be a homework question. I apologize if this is a misinterpretation, but it is our aim here not to do people's homework for them, but to merely aid them in doing it themselves. Letting someone else do your homework does not help you learn nearly as much as doing it yourself. Please attempt to solve the problem or answer the question yourself first. If you need help with a specific part of your homework, feel free to tell us where you are stuck and ask for help. If you need help grasping the concept of a problem, by all means let us know.  Buffered Input Output 16:42, 13 May 2009 (UTC)

This is not homework - this is purely for personal interest. --Lucas Brown 42 (talk) 16:52, 13 May 2009 (UTC)

It should be easy to derive from Heron's formula and the law of sines. — Emil J. 16:57, 13 May 2009 (UTC)

(ec) You can use the law of sines together with the semiperimeter to get the side lengths and then use $\mathrm {area} =AB\sin {c}$ to get the area. --Tango (talk) 16:59, 13 May 2009 (UTC)

Do you mean "$\mathrm {area} =AB\sin {c}/2$?" --Lucas Brown 42 (talk) 17:03, 13 May 2009 (UTC)

Yes. And the result is ${\frac {2s^{2}\sin a\sin b\sin c}{(\sin a+\sin b+\sin c)^{2}}}$. — Emil J. 17:12, 13 May 2009 (UTC)

Um... no, I'm using the new fangled definition of area that is twice as big as the old one, get with the times! ;) --Tango (talk) 18:00, 13 May 2009 (UTC)

Note to the literal-minded: Tango's last comment was with tongue in cheek. The formula that says area = AB sin c is for the area of a parallelogram given the lengths A and B of its adjacent sides and the angle c between them. The area of the triangle is half of that, since the triangle is half of the parallelogram. Michael Hardy (talk) 21:48, 13 May 2009 (UTC)

The winking emoticon was supposed to indicate that. What actually happened was that I was concentrating so much on getting the mathrm bit right in the latex that I forgot to include the fraction (which was the whole reason for writing it in latex in the first place!). --Tango (talk) 21:52, 13 May 2009 (UTC)

I don't see any winking emoticon or any hint of one. Michael Hardy (talk) 01:15, 14 May 2009 (UTC)

Oh.......... that thing. I hadn't noticed it until I looked for it. Michael Hardy (talk) 01:15, 14 May 2009 (UTC)

Thanks! --Lucas Brown 42 (talk) 17:18, 13 May 2009 (UTC)

Finding an approximation to a linear system

I have this equation-
Ax = y
Where A is a rational 10*10 matrix, and x and y are vectors in R¹⁰. I know A and y, I don't know what x is equal to. I also know that there is no x where Ax equals exactly y.
I want to find the vector x' such that Ax' is as close as possible to y. Meaning that (Ax' - y) is as close as possible to (0,0,0,...0).

(If it helps I can add that I also know that A has no rows/columns full of 0's).

You want the Moore–Penrose pseudoinverse; check in particular the Applications to linear systems. --pma (talk) 17:35, 13 May 2009 (UTC)

Maybe I'm daft, but shouldn't a linear system of 10 equations in 10 variables always have at least one exact solution? --Stephan Schulz (talk) 17:49, 13 May 2009 (UTC)

I would have thought so... If A is singular, there wouldn't be a unique solution, but there should be a solution. --Tango (talk) 18:05, 13 May 2009 (UTC)

No. If A is not invertible, then the map $x\mapsto Ax$ is not surjective, so there exist y's such that no x solves the equation. Aenar (talk) 18:12, 13 May 2009 (UTC)

Indeed, I stand corrected. Thank you! --Tango (talk) 18:34, 13 May 2009 (UTC)

Me too, after thinking of the trivial example of A=(0), y=(1). But what's the correct form of my solid pseudo-knowledge about systems of linear equations and their solutions? --Stephan Schulz (talk) 20:23, 13 May 2009 (UTC)

The number of solutions to a linear system is always 0, 1, or infinite. For A m×n: if m is greater than, equal to, or less than n, you get those numbers of solutions for almost every $\mathrm {A} x=y$, respectively. AllMost combinations are possible, though, for at least some choices of y. If A has full column rank (implying $m\geq n$), the $\infty$ case is ruled out (the transformation is injective); otherwise that case is guaranteed to occur for some y. If A has full row rank (implying $m\leq n$), the 0 case is ruled out (it's surjective); otherwise it is guaranteed for some y. (Note that for $m=n$ the only choices are "unique solution" (invertible A) or "0, 1, or many solutions, depending on y" (singular A).) --Tardis (talk) 21:10, 13 May 2009 (UTC)

That "All" was incorrect — at least one combination of shape and solution count can't happen. In particular, if A hasn't full column rank, it has a null space and so can't have unique solutions for any y; some y may have 0 solutions (unless A has full row rank), and all the rest have many. This implies that "wide" matrices ($m>n$) can't have unique solutions. I don't think that "tall" matrices have any such categorical exclusions, but I'm suspicious of the asymmetry. --Tardis (talk) 21:21, 13 May 2009 (UTC)

The question is asymmetric (Ax=y and xA=y are very different equations), so the solutions being asymmetric isn't that surprising. --Tango (talk) 21:49, 13 May 2009 (UTC)

That is quite true, but I think the asymmetry really arises because the forward transformation from x to y always gives exactly one y, regardless of how well-defined (or well-behaved) its inverse is. Meanwhile, I messed up again: for no y does a singular A yield a unique x (because, again, it must have a null space). --Tardis (talk) 17:26, 14 May 2009 (UTC)

For more advanced functional analysis kinf of thing, look at the Fredholm alternative (Igny (talk) 21:53, 13 May 2009 (UTC))

OK, let's go back to the original request that was very clear. Given an m times n matrix A (a singular 10x10 matrix in the OP), and given y in R^m, we want a vector x in Rⁿ minimizing ||Ax-y||. This is an elementary convex problem; it always has a solution, that you find differentiating wrto x the quadratic polynomial ||Ax-y||²=(Ax,Ax)-2(Ax,y)+(y,y) : by convexity, the minimizers are exactly the critical points, thus solve the square linear system: A^TAx=2A^Ty. Moreover, if the system has a kernel, the solution is of course not unique: to make a canonical choice one chooses the vector x of minimal norm in the affine space of the solutions of A^TAx=2A^Ty, that is, among all minimizers of ||Ax-y||. This choice exhibites x as linear function of y: that is y=A⁺x. The corresponding matrix A⁺ is the so called Moore–Penrose pseudoinverse of A; you can do the same in the context of Hilbert spaces and linear operators. The details are given in the links I provided, just click on it! --pma (talk) 06:44, 14 May 2009 (UTC)

7 tens

How many three digit numbers having 7 tens. —Preceding unsigned comment added by True path finder (talk • contribs) 18:43, 13 May 2009 (UTC)

for the ones starting with a 1, the possibilities are "170, 171, 172, 173, 174, 175, 176, 177, 178, 179" and you have the same number of possibilities for all the starting digits, of which there are nine (1, 2, 3, 4, 5, 6, 7, 8, or 9), so count the list and multiply it by 9.

The reason there aren't 10 starting digits is because "070" and so on aren't three-digit numbers. 79.122.61.98 (talk) 20:00, 13 May 2009 (UTC)

May 14

how many possible soduko variations are there

how many possible soduko variations are there - I mean different solved final solutions... how much of this needs to be revealed is a different question... —Preceding unsigned comment added by 79.122.61.98 (talk) 01:21, 14 May 2009 (UTC)

6,670,903,752,021,072,936,960 total configurations and 5,472,730,538 unique configuration after eliminating rotations, reflections, permutations, and relabelings. See: Sudoku. Dragons flight (talk) 01:39, 14 May 2009 (UTC)

I didn't expect the numbers to have such big prime factors – respectively

• 2^20 3^8 5 7 27704267971
• 2 11^2 23 983243

—Tamfang (talk) 05:28, 18 May 2009 (UTC)

On the other hand there are just 2³ = 8 variant spellings of the name, all of which get Google hits, with "sodoko" being the least common at 13,200. -- BenRG (talk) 02:05, 14 May 2009 (UTC)

geometry and algebra

i need some hints to solve these geomotry questions:

1. two congruent triangles (30°, 60°, 90°) are placed so that they overlap partly and their hypotenuses coincide. if the hypotenuse is 12 cm, find the common area of the triangles.

2. a circular arch of width 24m and height 9m is to be constructed. what is the radius of the circle of which the arch is an arc?

3. ABC is a right triangle with hypotenuse AB and AC=15cm. altitude CH divides AB into segments AH and HB, with HB=6cm. what's the area of ABC?

also this algebra question:

if x² - xy + y² = a and x³ + y³ = b, then xy = ? —Preceding unsigned comment added by 122.50.138.137 (talk) 05:05, 14 May 2009 (UTC)

Some hints: 1) draw the prolongations of the shorter edges and look for symmetry. 2) and 3): draw the complete picture and look for similar right triangles. For the algebra question: factorize b. Ask for details if you need further explanation --pma (talk) 07:01, 14 May 2009 (UTC)

For the first one, you've got a triangle with two 30-degree angles and one 120-degree angle. The side opposite the 120-degree angle has length 12. You can use the law of sines to find the other lengths and then use Heron's formula to find the area.

Or you could find the height by thinking about the tangent of 30°, and then say that it's (1/2)×base×height. Michael Hardy (talk) 23:02, 15 May 2009 (UTC)

Limit of sequence

Hi all, I'm having difficulty with this problem, and I was hoping someone could give me a hint. I'm trying to show that $\lim _{n\to \infty }{\sqrt[{n}]{3^{n}+5^{n}}}=5$

Looking at the real function,

$y={\sqrt[{x}]{3^{x}+5^{x}}}$

this can be rewritten as

$ln(y)=ln(3^{x}+5^{x})/x$

But showing how the limit is determined here has me stymied. What's the right thing to do? - RedWordSmith (talk) 07:56, 14 May 2009 (UTC)

This: $5\leq {\sqrt[{n}]{3^{n}+5^{n}}}=5{\sqrt[{n}]{1+(3/5)^{n}}}\leq 5{\big (}1+(3/5)^{n}{\big )}$, and have a sandwich --pma (talk) 08:11, 14 May 2009 (UTC)

Alternative solution, using ln(y) = ln(3^x+5^x)/x:

Apply l'Hôpital once, then factor 3^x out of both the numerator and the denominator. Apply l'Hôpital a second time. Simplify. --COVIZAPIBETEFOKY (talk) 12:40, 14 May 2009 (UTC)

...but that method leaves the student with no clue as to why the result is what it is. Whereas pma's method is more instructive and can be generalised to $\lim _{n\to \infty }{\sqrt[{n}]{a^{n}+b^{n}+c^{n}+...}}$ where a, b, c ... are any set of positive real numbers. Gandalf61 (talk) 12:53, 14 May 2009 (UTC)

That's odd. I seem to understand it just fine. Both methods lead one to the same conclusion: the largest number under the radical prevails, while all of the rest ultimately become so insignificant that they don't matter in the limit.

Besides, while my method may not appear as insightful to the generalization you describe (it's not that it can't be generalized; it's just a bit more difficult to do), I thought it had the advantage of being less creative thought and more computational, which is what you want ideally if, say, you have to take a test on the material. --COVIZAPIBETEFOKY (talk) 13:15, 14 May 2009 (UTC)

If I gave students the original problem in an analysis class, it would be long before l'Hôpital's rule had been covered. Moreover, relying on l'Hôpital's rule to prove that the limits of sequences exist will prevent one from developing the skill with estimates. it's a bad habit to get into. If I gave a test on the material, I would not permit the use of l'Hôpital's rule on this problem in the first place. — Carl (CBM · talk) 13:26, 14 May 2009 (UTC)

l'Hopital's rule was never covered at all in my first course in analysis (which is where this kind of question would have appeared) and its use was pretty much always forbidden. As a result, I learnt how to do limits from first principles and how to use intuition to get a good guess at what it is going to be and have thus never seen the need to use l'Hopital since. l'Hopital is basically a shortcut to working out the limit using Taylor expansions - I prefer the Taylor expansions themselves, it's generally quicker (you only need the first term). --Tango (talk) 14:53, 14 May 2009 (UTC)

My view is equivalent to the idea that you should use a theorem, if and only if you can prove it yourself without reference to the proof prior to doing so. The sandwich theorem is easy to prove for anyone who has even a little understanding of the epsilon-delta definition of a limit. This is not to say that L'hôpital's rule (don't forget the circumflex on the "o"!) is difficult to prove. However, many students use it freely without knowing its proof. This all links back to the idea that you will most certainly understand something well, if you have discovered it yourself. People who apply L'hôpital's rule without checking the indeterminate, clearly have not understood the rule well, nor do they have a good intuition of it. However, there is a large number of people who do understand the rule well, and have an intuition of it, and therefore should have the right to use it at liberty. However, I would strongly encourage that one at least attempts to discover other methods of calculating limits, because there will be cases when L'hôpital's rule does not apply. Nevertheless, I firmly agree that pma's solution is the easiest and most original solution to this problem. --PST 03:54, 15 May 2009 (UTC)

Rate of change

Water is pumped out at a rate of 3 m³min^-1 from a hemispherical water tank with radius=13 m. When the depth of water is x m, the volume of water inside the tank is ( π/3 )x²(39-x) m³.At what rate is the water level dropping when the depth of water is 8 m? —Preceding unsigned comment added by Dansonncf (talk • contribs) 11:04, 14 May 2009 (UTC)

Clue: The flow is A·dx/dt where A is the surface area and dx/dt is the rate of water level dropping. Bo Jacoby (talk) 12:21, 14 May 2009 (UTC).

Very nice clue, Bo. Some people overlook that and do it by more cumbersome methods. Michael Hardy (talk) 00:54, 17 May 2009 (UTC)

Thanks Michael! Bo Jacoby (talk) 08:39, 19 May 2009 (UTC).

The OP's calculation of the water volume is correct. The surface area of the water is π(x⁴-338x²+28561). Cuddlyable3 (talk) 15:30, 14 May 2009 (UTC)

Shouldn't it be the derivative of the volume? —Tamfang (talk) 05:23, 18 May 2009 (UTC)

Totally real fields

If F is a totally real number field, is it always possible to find an element which is mapped to a negative number by one real embedding and to a positive number for all the others? For real quadratic fields, this is true (for Q(sqrt(d)), take 1+sqrt(d)), but what about higher-dimensional fields? --Roentgenium111 (talk) 14:10, 14 May 2009 (UTC)

Yes. The embeddings e satisfy e(q + rx) = q + re(x) for any rational q, r, hence the question is equivalent to whether there exist an element which can be mapped to two different numbers, which is in turn equivalent to whether there exist at least two different embeddings. They do, unless F is Q. — Emil J. 14:24, 14 May 2009 (UTC)

I need an element x such that the e(x) are pairwise different. But this can be achieved by taking an x in F that is not contained in any proper subfield of F, i.e. a primitive element. Thank you for your help! --Roentgenium111 (talk) 14:48, 14 May 2009 (UTC)

Question from a mechanics past paper

A particle P of mass 0.5 kg moves upwards along a line of greatest slope of a rough plane inclined at an angle of 40◦ to the horizontal. P reaches its highest point and then moves back down the plane. The coefficient of friction between P and the plane is 0.6.

Show that the magnitude of the frictional force acting on P is 2.25N, correct to 3 significant figures.

So I know I have to use F = μR but I'm having problem getting R. R is equal to the component of the particles mass perpendicular to the plane. I know I should get R = 0.5g cos(40) but when I work it out I get R = 0.5g / cos(40). Can anyone see what I'm doing wrong? —Preceding unsigned comment added by 212.120.248.41 (talk) 22:55, 14 May 2009 (UTC)

I get 2.25. Your formula R=mg cos (40 degrees) is correct. Is μ the coefficient of friction? Are you remembering to convert degrees to radians if your computer's cos function expects radians? Are you remembering that $g=9.85m/s^{2}$? 207.241.239.70 (talk) 04:27, 15 May 2009 (UTC)

I know that R=mg cos (40 degrees) cos I've seen the answer. it's just that i don't understand why. when i resolve the forces i get a right-angled triangle like this with the top angle of 40 degrees:

|\
|  \
|    \
|      \
|        \
W       R

So cos(theta) = adjacent/hypotenuse

So cos(40) = W/R

So R = W / cos(40), not R = W cos(40)

Where am I going wrong? --212.120.248.41 (talk) 16:47, 15 May 2009 (UTC)

First, the plane was described as being at 40 degrees to the horizontal, your diagram is to the vertical. Resolving a force into two components at right angles must give each of them as less than the original, so you can't divide by sin or cos - these must be multipliers. Think of it this way - if the plane was horizontal, i.e. at zero degrees, the perpendiculat component would be equal to mg, i.e. mg cos 0.217.43.211.26 (talk) 19:51, 15 May 2009 (UTC)

Yes but what i have drawn isn't the plane itself. Its the weight of the particle and the component of the weight perpendicular to the plane. Since the plane is at 40 degrees to the horizontal, I believe then that the angle between the weight and the perpendicular component is also 40 degrees. Right? —Preceding unsigned comment added by 212.120.248.41 (talk) 15:39, 17 May 2009 (UTC)

Think of the case where the plane is flat (i.e. zero degrees from the horizontal). Then the vertical component of the force is just mg. Now tilt it a tiny bit from flat, like 0.01 degrees. The vertical component is now slightly less than mg, i.e. mg*z where z is close to 1. We can see that z is either cos theta or sin theta, but which is it? Clearly z is cos theta rather than sin theta, since theta is close to 0, and z is close to 1 instead of close to 0. 67.122.209.126 (talk) 01:54, 18 May 2009 (UTC)

The diagram should be like that

|\
|  \
|    \
|      \
|        R
W

W is along the hypotenuse. Dauto (talk) 23:10, 19 May 2009 (UTC)

May 15

shape of solution space

The solution space of x+y=a is a straight line through (0,a) and (a,0). The solution space for $x^{2}+y^{2}=a$ is a circle center at 0,0 with radius ${\sqrt {a}}$. What about $x^{3}+y^{3}=a$? What about higher powers?

The general problem I'm interested in is positive integer solutions for systems of n equations with n unknowns: ${\begin{cases}x_{1}+x_{2}+\cdots +x_{n}=a_{1}\\x_{1}^{2}+x_{2}^{2}+\cdots +x_{n}^{2}=a_{2}\\x_{1}^{3}+x_{2}^{3}+\cdots +x_{n}^{3}=a_{3}\\\cdots \\x_{1}^{n}+x_{2}^{n}+\cdots +x_{n}^{n}=a_{n}\\\end{cases}}$

Is there any method of solving systems like that? I do know various constraints on the variables, if that helps. Thanks 67.122.209.126 (talk) 08:08, 15 May 2009 (UTC)

From the sums of powers of the unknowns, $a_{k}=\sum _{i=1}^{n}x_{i}^{k}$ for k=1,2,...,n, you first compute the coefficients of the n 'th degree polynomial $p(x)=\prod _{i=1}^{n}(x-x_{i})$ in which the unknowns $x_{i}$ are roots. See Power sum symmetric polynomial. From the coefficients the roots are then computed numerically by the Durand-Kerner method. The roots are usually not positive integers but might even be nonreal complex numbers. Bo Jacoby (talk) 08:22, 15 May 2009 (UTC).

Thanks. In this particular problem (given where the data came from), the solutions are definitely integers. Finding them numerically seems reasonable. I think what I'm looking at is a bunch of intersecting shells in the first "2ⁿ-tant" (quadrant, octant, 16-tant, etc) that start out as a hyperplane, then part of a sphere, and get more and more cube-like as the exponent gets higher, and I need to find the points of intersection. 67.122.209.126 (talk) 01:58, 18 May 2009 (UTC)

logarithm

hey, does anyone how to solve this queston???

log_aa - log_aa² + log_aa³ - log_aa⁴ + .......

(please explain it in the simplest way possible, i'm new to this thing, and if possible, tell me any identities of logarithms (relevant or irrelevant to this question, i don't care)

thanx —Preceding unsigned comment added by 122.50.131.217 (talk) 09:42, 15 May 2009 (UTC)

sory, i have to make a change to this question.

the series ends at 99 terms, i.e. it ends at +log_aa⁹⁹ —Preceding unsigned comment added by 122.50.131.217 (talk) 09:59, 15 May 2009 (UTC)

Note that $log_{a}(a^{n})$ equals n so your series becomes:

1 - 2 + 3 - 4 + 5 + ....... + 99

Famously performed by Gauss, we may note the following:

1 + 99 = 100

3 + 97 = 100

5 + 95 = 100

In this manner, we add all odd numbers with another term in the series (which you can verify exists) such that the sum is 100. Adding in this way, we obtain (25 x 100) since there are 25 odd numbers between 1 and 49 (inclusive) - note that adding odd numbers greater than 50 in this manner will repeat sums. Now, we have taken care of all odd numbers. Similarly, for the even numbers:

- 2 - 98 = - 100

- 4 - 96 = - 100

- 6 - 94 = - 100

In this manner, we add all even numbers with another term in the series (which you can verify exists) such that the sum is - 100. Again, by similar reasoning, since there are 24 even numbers between 2 and 48 inclusive, and - 50 has no other number associated to it, the sum is (24 x (-100)) - 50 = - 2400 - 50 = - 2450. The other sum we calculated was 2500. Since, 2500 - 2450 = 50, the sum is 50.

Let me note that the method I have provided is delibrately long. There is a much shorter way to obtain the answer to this problem (by pairing numbers as I have done to get the answer). See if you can find it (if it is obvious (which it should be), do not worry about it). Note however, this method will not work for infinite sums, contrary as it seems. In general, different rearrangements of an infinite series may yield different sums, unless certain conditions are imposed. See Convergent_series#Conditional_and_absolute_convergence for more details.

Regarding logarithms, I recommend looking at the article Logarithm which provides many of the basic identities that you will probably use, should you continue such problems. Hope this helps. --PST 10:25, 15 May 2009 (UTC)

We may also look at our 1 − 2 + 3 − 4 + · · · article, section Divergence for partial sums sequence of this series. --CiaPan (talk) 11:09, 15 May 2009 (UTC)

Fatou's Lemma and liminfs

I am working on a problem that uses Fatou's Lemma. I have all but the very last detail finished. I have two sequences f_n and f_n' and they are equal almost everywhere. I need the equality of $\int \varliminf f_{n}'\,dx=\int \varliminf f_{n}\,dx$ and I think it is true. I know that if two functions are equal almost everywhere, then their integrals are equal. I assume that if I have two sequences of functions where for each n, the functions are equal almost everywhere, then their liminfs will be equal almost everywhere also, but I'm not exactly seeing this. I could see something weird happening where f_1 = f_1' except on some set of measure 0, f_2 = f_2' except on some other set of measure 0, and so on. Can any one help me out a bit? Thanks StatisticsMan (talk) 15:19, 15 May 2009 (UTC)

A countable union of null sets is a null set. Therefore there is a null set Z of the domain X such that for all x in X\Z and for all n in N you have f_n(x)=f '_n(x)(so here Z is just the union of the Z_n:={x: f_n(x)≠f '_n(x)} )--pma (talk) 17:03, 15 May 2009 (UTC)

Good point. Thanks. StatisticsMan (talk) 17:11, 15 May 2009 (UTC)

Irrationality of Pi

I pretty much get the proofs that pi is irrational, in that I can see that they're valid proofs, but I don't really understand why they work, and I can't see how anyone could have come up with them. Why is pi irrational? For comparison, the proof that e is irrational seems pretty reasonable to me. Black Carrot (talk) 21:07, 15 May 2009 (UTC)

You might consider the Hermite-Lindemann transcendence theorem to be a more direct approach. 207.241.239.70 (talk) 22:37, 15 May 2009 (UTC)

It was in the 18th century that π was first proved to be irrational, and I think the way it was done was that someone found a continued fraction expansion of the arctangent function, and consequently of particular values of the arctangent function, such as π/4. Some types of continued fractions (such as those in which every numerator is 1) terminate if the number they represent is rational, and this one did not terminate. Michael Hardy (talk) 22:56, 15 May 2009 (UTC)

May 16

Correcting linear least squares coefficients for small samples

I performed a linear regression on a large population and got y=a0*x+b0. Then I performed a linear regression on a sub-sample of (similar) members and got another equation: y=a1*x+b1. How do I correct the sub-sample's regression to take into account that the size of the sub-sample is very small (compared to total population). Right now I have a1'=(a1-a0)/sqrt(#members in sub-sample)+a0 and b1'=(b1-b0)/sqrt(#members in subsample)+b0. Is there a more accurate adjustment method? Especially one that uses variance as well? 70.171.0.134 (talk) 04:03, 16 May 2009 (UTC)Mathnoob

I find your question very unclear, and I think I'm much better at seeing through opaque writing on this sort of thing than are most of those who post here, and I have a Ph.D. in statistics. So first we have to work on what it is you're trying to ask. Usually when you fit a line based on a small sample from a large population, you don't know the data for the whole population, but only for the small sample, in such cases, one bases estimates of properties of the whole population upon the data from the small sample. But you seem to say that you do know the whole population. What, then, is the purpose of the sample? If you're trying to quantify uncertainty about the population when all you have to go on is the small sample, there are standard ways of proceeding, including, e.g., a confidence interval for the slope of the fitted line. Is that sort of thing what you have in mind? Michael Hardy (talk) 00:50, 17 May 2009 (UTC)

Maybe I can simply the problem. Suppose I have 1 million marbles where each marble weights 5 grams on average. Each blue marble weights 50 grams on average and there are 1000 blue marbles, and each red marble weights 1 gram on average and there are 100 red marbles in the set. And I want to compute the averages of blue/red marbles that are closest to their true averages. However in my case, instead estimating averages I want to estimate linear regression equations.70.171.0.134 (talk) 02:24, 17 May 2009 (UTC)Mathnoob

You lost me with this sentence:

And I want to compute the averages of blue/red marbles that are closest to their true averages.

I have no idea what you mean by that. Michael Hardy (talk) 10:41, 18 May 2009 (UTC)

Let me try... *wipes dust off crystal ball*. I think the OP is trying to say that s/he has a large population K, of whose members some small fraction have a (boolean) property P, and that s/he's trying to estimate the correlation between the (continuous) properties x and y conditioned on P. (I'm not a statistician, so I apologize if my terminology is also a bit off.) Furthermore, the OP apparently would like a more stable estimator than simply applying linear regression to the subpopulation J = {x ∈ K: P(x)}, which would seem to make sense if that subpopulation is very small. In particular, I suspect s/he might like to start by testing the null hypothesis that the linear relationship between x and y is independent of P.

From a Bayesian viewpoint, what the OP seems to want is to use the posterior distribution of regression coefficients for the general population to define a prior distribution for the corresponding coefficients in the subpopulation, but adjusted in some way so as to account for their degree of belief in the similarity of the populations. In general, I suspect the problem isn't entirely well posed in a mathematical sense, yet such problems do seem common enough in practice to merit at least some consideration. I suppose that, for all I know, there might be some standard statistical formula for this, but I kind of doubt it. —Ilmari Karonen (talk) 15:25, 18 May 2009 (UTC)

I guess some (very special) kind of maximum likelihood estimation is called for. Pallida  Mors 04:41, 19 May 2009 (UTC)

Showing an element is irreducible

Hi, I am trying to show ${\sqrt {10}}$ is irreducible in $\mathbb {Z} +\mathbb {Z} {\sqrt {10}}$. I am trying the obvious method: Assume

 (1) $(a+b{\sqrt {10}})(c+d{\sqrt {10}})={\sqrt {10}}$

which gives

 (2) $ac+10bd=0$

and

 (3) $ad+bc=1$.

I can show little bits. If a = 0, then it must be that d = 0 and c is not zero by (1) and you get $b=\pm 1$. Similarly, if b = 0, $a=\pm 1$. The last case is a and b are both nonzero. By the reasoning above, we must have c and d nonzero also for if one were 0 it would imply a = 0 or b = 0.

But, here I am stuck. I thought about comparing parity (right word for even/odd right?). That is, solve (2) above for a to get

 (4) $a={\frac {-10bd}{c}}$

then plug this in to the second to get

 (5) ${\frac {-10bd^{2}}{c}}+bc=1$

Then, think about even/oddness, so for example assume c is odd. Then, a must be even by the equation (2). By (5), ${\frac {-10bd^{2}}{c}}$ is even if c is odd. Since the equation adds up to 1, the bc part must be odd so b is odd. Any way, so this is what I'm trying here and I'm not getting anywhere.

I thought about using the norm also, but I don't see how that helps and it's not introduced until the next chapter of the book so I don't think that is intended. I would be interested in knowing how to do this problem by any method, even if it is not "allowable" for this problem. Thanks StatisticsMan (talk) 14:47, 16 May 2009 (UTC)

Remember that an element that's irreducible is still divisible by units. For example, $19+6{\sqrt {10}}$ is a unit in the ring $\mathbb {Z} +\mathbb {Z} {\sqrt {10}}$, since

$\left(19+6{\sqrt {10}}\right)\left(19-6{\sqrt {10}}\right)=1,$

and therefore:

${\sqrt {10}}=\left(19+6{\sqrt {10}}\right)\left(-60+19{\sqrt {10}}\right).$

The simplest way to proceed is using the norm. Define

$N\left(a+b{\sqrt {10}}\right)=|a^{2}-10b^{2}|.$

Then $N(xy)=N(x)N(y)$ for any $x,y\in \mathbb {Z} +\mathbb {Z} {\sqrt {10}}$. Moreover, $x$ is a unit if and only if $N(x)=1$. The norm of ${\sqrt {10}}$ is 10, so any two non-unit divisors would have to have norms 2 and 5. But norm 2 is impossible (to see this, think about $a^{2}-10b^{2}$ in mod 10), and therefore 10 is irreducible. Jim (talk) 15:56, 16 May 2009 (UTC)

Another way that is slower but doesn't use norms is with statistics man's eqn 3:ad+bc=1. So there's a lot of relative primeness.You get c=kb and a=md with km = -10.Then (a+bsqrt10)(c+dsqrt10) is (md+bsqrt10)(kb+dsqrt10)=sqrt10, or mdd+kbb=1=kbb - (10/k)dd. but then exactly one of the two factors (a+bsqrt10), c+dsqrt10 has an inverse you can construct so it's a unit.I think(hope) this works.Best wishes, Rich (talk) 23:01, 16 May 2009 (UTC)

Proof that n orthogonal vectors form a basis for $\mathbb {R} ^{n}$?

How can one show that n orthogonal vectors form a basis for $\mathbb {R} ^{n}$?

I can see how to show that they're linearly independent, so I suppose what I'm really asking is how can you show that if n vectors are linearly independent they span $\mathbb {R} ^{n}$? If anyone can find a proof online I'll be happy to just read that, you don't need to explain it yourself, I just can't actually find a proof anywhere :) Thanks a lot! —Preceding unsigned comment added by 131.111.8.97 (talk) 15:07, 16 May 2009 (UTC) 131.111.8.97 (talk) 15:08, 16 May 2009 (UTC)

Have you looked in elementary linear algebra books? The result you are looking for is closely tied to the proof that the dimension of $\mathbb {R} ^{n}$ is n. There are several ways to prove it, depending on which path to the final result you want to take. I am certain there is a proof in David Lay's linear algebra textbook. — Carl (CBM · talk) 15:19, 16 May 2009 (UTC)

P.S. A simple proof, but not aesthetically pleasing, is to take your n orthogonal vectors and place them into the columns of a matrix. Because the vectors are independent, the matrix is invertible. Thus the linear transformation from $\mathbb {R} ^{n}$ to $\mathbb {R} ^{n}$ obtained from the matrix is surjective, which means that the columns of the matrix span $\mathbb {R} ^{n}$. — Carl (CBM · talk) 15:21, 16 May 2009 (UTC)

Thanks very much Carl, I just figured out a proof for myself but it's good to have a couple under the belt! 131.111.8.97 (talk) 15:31, 16 May 2009 (UTC)

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)

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 approximation should be: the number of 2 dimensional disks of diameter D1 that you can pack in a disk of diameter D2, thus ≈c(D2/D1)², where c=0.906.. is the highest density for 2 dimensional unit disks. Somehow better is maybe c((D2+D1)/D1)². The idea is that we are looking at the traces of the small balls on the sphere of diam D2+D1, like it were a planar disk packing.--131.114.72.215 (talk) 14:06, 19 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)