Wikipedia:Reference desk/Mathematics

May 10

Tried proving it - didn't work

This is slightly arbitrary.

I was given an A4 envelope from an accountancy firm. On the front of the envelope was the following expression.

$n(1+x)^{5}=2n$

Is there any way of proving that this expression is true or false? I tried, but didn't get anywhere. Dividing both sides by n and then 5th rooting each side, I was able to work out the value of x.

$n(1+x)^{5}=2n$
$(1+x)^{5}=2$
$1+x=1.148698355$
$x=0.148698355$

However, when using the above value of x to work out a value for n, the answer I end up with it 2n = 2n

Aiyda 20:24, 10 May 2007 (UTC)

I'm not sure what you're trying to prove - the statement isn't always true or always false; it's true or false depending on what the value of x and n are. Thus, you can solve for x and/or n to find what values make it true. What you've done is to find a value of x for which the statement is true for all n. (When x=0.148698355, the statement is always true, regardless of what the value of n is, since 2n=2n is true for all values of n.) There's also a value of n for which the statement is always true, regardless of the value of x - I'll leave you to find that on your own. Chuck 20:32, 10 May 2007 (UTC)

n(1+x)^5 = 2n

(1 + x)^5 = 2 provided n ≠ 0

5 possible solutions

x = 0.1487 provided n ≠ 0

x = -1.9293 - 0.6751i provided n ≠ 0

x = -1.9293 + 0.6751i provided n ≠ 0

x = -0.6450 - 1.0925i provided n ≠ 0

x = -0.6450 + 1.0925i provided n ≠ 0

Plus of course

x = anything and n = 0

202.168.50.40 03:19, 11 May 2007 (UTC)

Exponentiation with compass and straightedge

Can this be done? If so, how? Thanks --87.194.21.177 23:00, 10 May 2007 (UTC)

No. The only operations possible with Compass and straightedge are addition, subtraction, multiplication, division and square roots. -- Meni Rosenfeld (talk) 23:27, 10 May 2007 (UTC)

I thought there were only four operations, though I knew Descartes wasn't sure if they were four or five (because of the square root). His exact quote (well, translated from French): "Just as arithmetic consists of only four or five operations, namely, addition, subtraction, multiplication, division, and the extraction of roots, which may be considered a kind of division..." A.Z. 02:37, 11 May 2007 (UTC)

This edit is a test. A.Z. 22:38, 23 September 2007 (UTC)

So, considering the operations at hand, if an exponent has a finite base two representation, then, sure, it can be done. Root⁴(one) 02:52, 11 May 2007 (UTC)

In case I need to explain... Say you need to exponentiate $m$ to binary 1101.0011 (13.1875 to us in decimal land), you'd find:

$m^{8},m^{4},m,{\sqrt {\sqrt {\sqrt {m}}}},{\sqrt {\sqrt {\sqrt {\sqrt {m}}}}}$, multiply them all together (which adds the exponents), and you have your $\,m^{13.1875}$.

Root⁴(one) 03:15, 11 May 2007 (UTC)

This is clever, but it steps outside the system. When we say we can add two numbers, the two numbers are given inside the system as lengths, and we produce a length as the result. If we are asked to compute β^α, where both numbers are given as lengths, we cannot do so. --KSmrq^T 05:08, 11 May 2007 (UTC)

What if we were given three lengths: α, β and 1. Could we not then use 1 (and various divisions and multiplications of it) to measure the other two? (Don't you have to have a 1-length to construct a square root anyway?) - Rainwarrior 06:32, 11 May 2007 (UTC)

Yes, we are assuming that a 1 length is given. However, if I understand correctly what you meant by "use 1... to measure the other two", that is only possible if they are rational. If you allow any construction starting from 1, you can reach α and β iff they are constructible. -- Meni Rosenfeld (talk) 08:33, 11 May 2007 (UTC)

If you have a and b as lengths, you can construct the length (ab)^1/2, so if a unit length is given, you can do square roots in general. However, in the standard interpretation of compass-and-straightedge construction, no unit length is given, and then it is not possible to construct one. From the simple mantra "addition, subtraction, multiplication, division and square roots" it would appear possible (a/a), but actually multiplication and division are together one ternary operation: from a, b and c, construct a×b/c.  --Lambiam^Talk 11:05, 11 May 2007 (UTC)

(kill indent) So, just be sure, if I have a unit length "1" and a length "m" based off that unit length, I can essentially construct m^p for the restrictions of p I gave above, Correct? However it is not possible for not for me only possessing arbitrary lengths "m" and "n". I suppose I didn't factually answer the original question.

We've had a discussion on compass constructions within the last few months. I swear Compass and straightedge constructions#Constructible points and lengths is just as confusing as it was then. Is there not some way like:

If your number of interest β is constructible by recursively applying any of the following formula on measurements/lengths $\{m,n,p\}$ previously constructed.

1. $m'$ <- $m+n$
2. $m'$ <- $m-n$
3. $m'$ <- $(mn)^{1/2}$
4. $m'$ <- $mn/p$
5. ... other formulas...

Then you can also construct a length measuring β by compass and straight edge constructions. Or would that task be arduous, a long winded and almost useless list, not necessary, or original research? Thoughts? Root⁴(one) 15:57, 11 May 2007 (UTC)

Obviously, such a list would have to use better wording than above. Root⁴(one) 16:00, 11 May 2007 (UTC)

I think you are right about exponentiation. Given u and v, you can construct u^1−pv^p for those values of p that have a terminating binary expansion. Setting u := 1 gives the desired result. In general, if C(x₁, ..., x_n) is an expression in n variables using only the operations of addition, subtraction, multiplication, division and square root, then, given one length u and n lengths v₁, ..., v_n, the length uC(v₁/u, ..., v_n/u) is constructible. (Warning: I have not actually proved this.) The preceding statement for exponentiation is an application of this for C(x) = x^p. A clarification as suggested may be in order. Is there no authoritative source that has a concise and clear formulation? But further discussion of this should be taken to Wikipedia talk:WikiProject Mathematics and/or Talk:Compass and straightedge constructions.  --Lambiam^Talk 17:37, 11 May 2007 (UTC)

May 11

Additional Mathematics

Most secondary school students here take a subject called Additional Mathematics (A Maths for short), where we learn mathematics topics that are harder than those we learn in Elementary Mathematics (E Maths). For example, in E Maths trigonometry, we learn sin, cos and tan, but in A Maths trigonometry, we learn sec, cosec and cot. If you fail E Maths and hence don't take A Maths, you will learn all the A Maths topics, such as binomial and differentiation, when you go to JC.

Since Wikipedia doesn't have an article on Additional Mathematics, my question is: do other countries' education systems have Additional Mathematics, what is it called there and what mathematics topics do they learn?

--Kaypoh 08:11, 11 May 2007 (UTC)

In South Africa, Additional Mathematics is called Additional Mathematics (or Add Maths). I did it in 2001 and 2002 (Grade 11 and 12). The course covers Maths of Finance (interesst rates, annuities etc), Algebra, Co-ordinate geometry, Statistics (although you only needed 2 of the previous 4 to pass) and mainly calculus. The calculus section covered about 90% of the stuff we did in first-year at University.Zain Ebrahim 08:38, 11 May 2007 (UTC)

In the US it may vary from state to state and even from district to district. In my case, we had math in the following order:

Pre-Algebra
Algebra 1
Geometry
Algebra 2
Trigonometry
Pre-Calculus
Calculus

StuRat 08:56, 11 May 2007 (UTC)

The mathematics curriculum seems to have no international standard. Web searches may be your best research tool. For example, here is the core mathematics curriculum for the state of New York in the United States; and here is the mathematics framework suggested by the state of California, on the opposite coast. And the Asia Pacific Network of Education Knowledge Bank has a great deal of international information, covering many countries. You may also find some interesting reading here, especially with regard to mathematics teaching in France. --KSmrq^T 10:39, 11 May 2007 (UTC)

In the UK, additional maths is called Further Maths. I'm not sure about what level you are talking about, though; Further Maths is Sixth Form (US Junior/Senior years), and tends to be very high level stuff (calculus of hyperbolic functions, the Argand diagram) Some school also run Pure Maths or Statistics course, which go beyond normal level maths but are far more focused on a single topic; since you say that additional maths covers cosec/cotan functions, I'm guessing it must be a lower level (at GCSE, the UK currently only has one maths course, although it is divided into Foundation, Intermediate and Higher, and there are plans to split it into core maths and further maths). Laïka 11:14, 11 May 2007 (UTC)

While the IB has four levels of mathematics, of which the highest is again Further Maths. Unlike the UK syllabus, this contains an element of genuine pure mathematics (rather than bucketloads of calculus), including some group theory, graph theory and analysis. Algebraist 11:30, 11 May 2007 (UTC)

Since, according to the Sixth Form article, students are usually 18 when they enter the sixth form and take their A Levels after that, this would make the sixth form equal to the second year of JC, after which we take our A Levels at the age of 18. Additional Mathematics is taken during the last two years of secondary education (at the age of 14-16), and is an O Level subject.

The topics you mentioned, such as calculus of hyperbolic functions, are only taught at H3 Maths in JC, which is reserved for the best students. Most students who take Additional Mathematics in secondary school will take H2 Maths in JC, while those who did not take Additional Mathematics will take H1 Maths in JC (the H1 Maths syllabus is roughly equal to the Additional Mathematics syllabus - only that the students learn it at JC instead of secondary school).

I should note that while Further Maths, and H3 Maths in JC, are reserved for the best students, most students here take Additional Mathematics. Only students who fail Elementary Mathematics do not take Additional Mathematics - although students who consistently fail Additional Mathematics will also be allowed to drop it. In fact, because the education system is so competitive, Additional Mathematics is compulsory in some schools.

Perhaps I should post the list of topics we learn in Additional Mathematics?

--Kaypoh 15:13, 11 May 2007 (UTC)

Students are 16 when they enter Sixth form; they graduate at 18 to go on to uni/vocational traing. Laïka 15:26, 11 May 2007 (UTC)

Then sixth form would be the equivalent of JC here. JC's a two-year course leading to the A Levels.

The Additional Maths Syllabus:

1. Sets
2. Simultaneous Quadratic Equations
3. Indices, Surds and Logarithms
4. Quadratic Equations and Expressions (using b² - 4ac to find nature of roots)
5. Remainder and Factor Theorems
6. Solving Cubic Equations
7. Matrices
8. Cooordinate Geometry
9. Linear Law
10. Functions
11. Trigonometric Functions (sec, cosec, cot and the four quadrants)
12. Simple Trigonometric Identities and Equations (tan = sin/cos, sin² + cos² = 1)
13. Circular Measure (about radians)
14. Permutations and Combinations
15. Binomial Theorem
16. Differentiation and Its Techniques
17. Rates of Change
18. Higher Derivatives (using dy/dx to find max/min points)
19. Differentiation of Trigonometric Functions
20. Differentiation of Exponential and Logarithmic Functions (we've learnt up to this point)
21. Integration
22. Applications of Integration
23. Kinematics
24. Vectors
25. Relative Velocity

--Kaypoh 09:41, 12 May 2007 (UTC)

Limits for cosh and sinh

Is there an elegant limit of a sequence for cosh and sinh that is reminiscent of this limit for the exponential?

$e^{x}=\lim _{n\to \infty }\left(1+{x \over n}\right)^{n}.$

I could make use of say, cosh(x) = 1/2 exp(x) + 1/2 exp(-x) and substitute the expressions above, but I am looking for something else. Any ideas? --HappyCamper 17:57, 11 May 2007 (UTC)

I'm having some trouble accessing it, but I believe [1] has an answer to that. Black Carrot 23:14, 13 May 2007 (UTC)

Non-iterative PRNG

Is there a deterministic PRNG where one can calculate the n^th value (for a very large and entirely arbitrary n) without calculating the preceding values? All the PRNGs I'm aware of are iterative, meaning generating the gagillionth number means generating all gagillion-1 numbers before it too. My intended use is for simulation, and I'm not terribly concerned with periodicity or cryptographicly-acceptable randomness (if a few hundred values in a row starting at some arbitrary point are random enough to look random to a human, that's sufficient). Thanks. -- Finlay McWalter | Talk 21:20, 11 May 2007 (UTC)

Do you consider the digits of pi to be random? There's a formula for calculating arbitrary digits of it. --Carnildo 22:01, 11 May 2007 (UTC)

You can modify any PRNG that accepts seeds, such as LFSR to do this by using successive seeds i.e. the first number is the one generated from the seed "1", the second from the seed "2" ect. You may want to iterate it a certain number of times in order to make it seem random if you use certain PRNGs. In some cases, just starting from a high seed would work. — Daniel 22:31, 11 May 2007 (UTC)

Another approach, which does yield crypto-quality randomness, is to use a block cipher in counter mode. (See also this page for further details.) —Ilmari Karonen (talk) 22:36, 11 May 2007 (UTC)

If you are satisfied with a linear congruential generator, then you can compute the n^th successor of any value on the cycle in O(log n) steps. Although not the best (see the article), LCGs are "good enough" for many applications, including yours. The basic idea is to generalize the relationship (omitting the "mod M" everywhere – this is tacitly understood)

V_j+1 = AV_j + B

to

V_j+2^k = A_kV_j + B_k.

Clearly, you can use

A₀ = A, B₀ = B.

Further,

V_j+2^k+1 = A_kV_j+2^k + B_k

= A_k(A_kV_j + B_k) + B_k = (A_k)²V_j + (A_k+1)B_k,

so, moreover,

A_k+1 = (A_k)², B_k+1 = (A_k+1)B_k.

You can compute and store the vectors A_• and B_• ahead of time for up to the largest possible value of log₂(n) you might want to consider, or recompute them each time on the fly. To compute V_j+n from V_j, use a "binary" method similar to exponentiation by squaring. Let us know if you need more details.  --Lambiam^Talk 22:47, 11 May 2007 (UTC)

Why not just use 'n' as the starting value for a PRNG? I mean, use an iterative one, but pretend the last iteration's result was 'n'. A generator like (An+B)modC wouldn't be very good at this (subsequent values would always be AmodC apart), but you could start there and modify it somehow, like using a bitwise XOR to make these evenly spaced values look random, like if it was more like ((An)modC)xorB) it might be good enough for your purposes. - Rainwarrior 06:52, 12 May 2007 (UTC)

Absolutely. Most (all?) generators will let you provide the starting value, or seed, as it is called, so you need not make your own. You can do like this in C:

#include <stdlib.h>
#include <stdio.h>
#include <math.h>

int main() {
  for (int i=1; i<=10; i++){
    srand(i);
    printf("%u ", rand());
  }
  printf("\n");
 exit(0);
}

Each time we go through the loop, srand will give the PRNG a seed value and rand will give you the random number for that seed. —Bromskloss 08:13, 12 May 2007 (UTC)

This idea of using successive seeds was already presented by Daniel above. It is equivalent to the application of an (ideally strongly mixing) hash function.  --Lambiam^Talk 08:39, 12 May 2007 (UTC)

May 12

Nim Variation

In Nim, given X = {x₁,x₂,x₃...x_n}, a move consists of reducing some x_i by an amount k. I'm pretty sure that if a move consisted of reducing any selection of an odd number of xs, each by an amount k, the game would remain unchanged. That is, any position with a winning strategy would keep that winning strategy, and no position without a winning strategy would gain one. So, just like in Nim, a position would have a winning strategy iff its heaps didn't Nim-sum to 0. I'm having trouble proving it, though. The first part is easy; from each position you'd still be able to reduce any 1 pile by an amount k, so anything I've assumed is a winning position reduces easily to an alleged losing position. The second part I can't get; how do I prove that if the Nim-sum across X is 0, it isn't 0 if you reduce any odd number of xs by equal amounts? Or equivalently, if n is odd, how do I prove that reducing all xs by equal amounts must change the Nim-sum? Counterexamples are also welcome. Black Carrot 05:25, 12 May 2007 (UTC)

An equivalent statement is: for odd n, the nim-sum (x₁+d) ⊕ ... ⊕ (x_n+d) is not equal to x₁ ⊕ ... ⊕ x_n – unless, of course, d = 0. Let p be the lowest bit position in the binary representation of d that has a 1 digit. For each i, x_i+d agrees with x_i in all lower position, but differs in position p. Then the nim-sum also agrees in all positions below p, while we find a difference at position p. An even number of differences in the same position would always cancel in the nim-sum; an odd number never does.  --Lambiam^Talk 08:51, 12 May 2007 (UTC)

Algorithm assistance

Consider the group < σ₁, σ₂, ..., σ_k >. What is the simplest way of determining the group's order? —The preceding unsigned comment was added by 149.135.125.155 (talk • contribs) 06:31, 12 May 2007 (UTC) - Please sign your posts!

That would depend on what else you know about it, wouldn't it? Given no information, it seems like counting it would be simplest. Black Carrot 06:47, 12 May 2007 (UTC)

You are only given the generating set and no other information. The equivalent question to ask would be to determine its elements, and an algorithm to do that isn't so immediately apparent either. —The preceding unsigned comment was added by 149.135.125.155 (talk • contribs) 07:04, 12 May 2007 (UTC) - Please sign your posts!

Do you mean checking it to see if it's in ascending or descending order ? This can be done with a single pass, using lines like this (FORTRAN used in the example):

ASC_FLAG = .TRUE.
DSC_FLAG = .TRUE.
DO I = 2,N
  IF (SIGMA(I-1) .GT. SIGMA(I)) THEN ASC_FLAG = .FALSE.
  IF (SIGMA(I-1) .LT. SIGMA(I)) THEN DSC_FLAG = .FALSE.
ENDDO

StuRat 06:59, 12 May 2007 (UTC)

No, you are given a generating set of a group, and I would like an indication of a simple algorithm to determine the order of the group generated by that generating set. —The preceding unsigned comment was added by 149.135.125.155 (talk • contribs) 07:04, 12 May 2007 (UTC) - Please sign your posts!

There's a survey on algorithms for groups, including permutation groups (what I assume you mean by using sigmas to denote the generators) at [2]. It doesn't really have enough detail to tell you how to solve the problem, I think, but maybe it will give you a better idea where to look. —David Eppstein 07:07, 12 May 2007 (UTC)

I know where I'd look in texts, but the problem is, I don't have ready or simple access to them. I was hoping someone would outline some algorithm in a general way or point to some specific instance on the internet somewhere. I suppose what the effective question is: is there something simpler or easier than implementing the well-known Schrier-Sims algorithm? —The preceding unsigned comment was added by 149.135.125.155 (talk • contribs) 07:09, 12 May 2007 (UTC) - Please sign your posts!

How large is that order going to be? In the thousands, millions, or more? Also, do you want this algorithm to be incorporated in a larger program? Depending on the circumstances and the magnitude of the problem, it may be acceptable to just generate the group and count the elements. If you have a non-humungous one-shot problem, you can use Magma's online calculator.  --Lambiam^Talk 08:04, 12 May 2007 (UTC)

I personally won't be using it for very large orders, but I want a basic algorithm that works and that I can replace with something better later. It's all very well and good to say "generate the group", but I need an algorithm to do that as well -- I can only think of very inefficient ones.

There is very inefficient and then there is very inefficient. Would this work for you? Below Q is a variable containing a stack or a deque – the order in which elements are "dequeued" is immaterial. Variable S contains a representation of a set (whose elements are permutations) that allows (1) a fast test for membership, and (2) fast addition of a new member. A hash table will do if it can handle collisions; pragmatically the operations then take O(n) time, where n is size needed for representing a permutation. Counter C is at all times the size of S.

Initialize (S, C) := ({id}, 1); Q := [id]; (where id is the identity permutation)

While Q is not empty:

Dequeue σ := head(Q);

For i := 1, ..., k:

Compute σ' := σ o σ_k;

If σ' ∉ S:

Set (S, C) := (S ∪ {σ'}, C+1); Enqueue σ' onto Q;

End If

End For

End While

This takes time O(knc), where k is the size of the set of generators, n the size of the underlying set, and c the order to be computed. If you don't have to do this thousands of times a second, and knc is not more than a million or so, then this naive algorithm is fine on a reasonably fast computer. Instead of an explicit stack, you can also use a recursive function with side effects. (Disclaimer: I haven't tested the above algorithm.)  --Lambiam^Talk 11:32, 12 May 2007 (UTC)

Great! I think this will work! I will test it tomorrow and will let you know how it goes.

Yes, it looks like it indeed works. Thanks very much!

statisticians (life story , summary of their contribution , detailed exposition of two applications)

i want to know about three statisticians ? —The preceding unsigned comment was added by 59.180.45.86 (talk) 09:13, 12 May 2007 (UTC).

Thomas Bayes would be a good one to do.

Sure. I was born in Sydney, went to the University of Sydney ... oh, I assume you mean important statisticians? Take your pick from Category:Statisticians. Confusing Manifestation 11:18, 12 May 2007 (UTC)

... or see list of statisticians. Gandalf61 11:21, 12 May 2007 (UTC)

Statistics has a shorter list of the more important contributors. --Salix alba (talk) 11:29, 12 May 2007 (UTC)

I agree Thomas Bayes is a good choice. Florence Nightingale is another one; an interesting story, and most people are unaware of her contributions to statistics. I'd also like to suggest William Sealy Gosset, aka Student. Together they cover quite different aspects of statistics.  --Lambiam^Talk 11:42, 12 May 2007 (UTC)

If A not equal to B

If A is never equal to B, and A - B = N, is N ever equal to 0? I don't think so but I'm not sure, thanks, Jeffrey.Kleykamp 16:53, 12 May 2007 (UTC)

I believe that is correct, N ≠ 0. StuRat 17:42, 12 May 2007 (UTC)

What manner of things are A and B? There are some types of mathematical object and operations called minus for which this is not true, but assuming you're dealing with numbers or somesuch thing, you're fine. Algebraist 17:46, 12 May 2007 (UTC)

Thanks Jeffrey.Kleykamp 17:53, 12 May 2007 (UTC)

Here is a proof of the statement. By checking all steps, unwarranted assumptions on the nature of the objects involved should become apparent. First, let us simplify the statement to be proved a bit, by removing the temporal aspect. ("2+2 is always 4" means the same as "2+2 = 4".) The statement is then: If A ≠ B and A − B = N, then N ≠ 0. Well, if A − B = N, then N = A − B, and since we may replace equals by equals without change of meaning, we can rephrase the statement as: If A ≠ B and A − B = N, then A − B ≠ 0. Notice now that N is no longer referenced, so the assumption A − B = N is no longer required, and we can remove it. We have now simplified the statement to: If A ≠ B, then A − B ≠ 0. Appealing now to the fact that a statement is equivalent to its contrapositive, we obtain the equivalent statement:

If A − B = 0, then A = B.

Thus far, no properties of the objects involved, including the operations, have been used. We now set out to prove the above statement, starting by assuming the antecedent A − B = 0, and working towards the consequent A = B. So assume that

A − B = 0.

Rewrite the left hand side, using the fact that A − B is the same as adding the additive inverse −B of B to A:

A + (−B) = 0.

Add B to both sides:

(A + (−B)) + B = 0 + B.

Rearrange the left hand side, using the associativity of addition:

A + ((−B) + B) = 0 + B.

The sum of the additive inverse of a number and the number itself equals (by definition) the neutral element of addition, which is 0:

A + 0 = 0 + B.

Using, at both sides, the fact that 0 is the neutral element of addition operation, we arrive at the desired conclusion:

A = B.

We have used no other properties than that + is the operation of an Abelian group whose neutral element is denoted as 0, and for which A − B is shorthand notation for A + (−B), where −B is the inverse of B. These assumptions hold for all number systems in use, and many other algebraic structures, such as rings.  --Lambiam^Talk 18:23, 12 May 2007 (UTC)

To hi-jack the question - in $\mathbb {F} _{2}$, the field of two elements, what is (0 - 1) equal to? Is it 1? I'm trying to use the axioms for a field to prove it, but I'm not seeing it. It's ridiculously simple, isn't it? Icthyos 18:21, 12 May 2007 (UTC)

The following is an ugly proof, but if there are exactly two elements 0 and 1, then either 0 − 1 = 0, or 0 − 1 = 1. In the first case, we find (see above) that 0 = 1, which contradicts the fact that there are two elements. Conclusion: only the possibility 0 − 1 = 1 is left. A nicer proof is based on figuring out more constructively what 1 + 1 is, which can be done from first principles, or by appeal to the theorem that for prime p the field F_p, viewed as a ring, is the same as Z/pZ.  --Lambiam^Talk 18:36, 12 May 2007 (UTC)

May 13

Hard-to-Read Formulae

In the image [3](warning: seminaked females), I can't make out most of what's written, and I don't recognize some of the symbols I can see. Could someone translate or transcribe for me? Black Carrot 05:47, 13 May 2007 (UTC)

I see a rotation matrix blocksum a 2x2 identity (IIRC a quaternion rotation matrix?) on the woman to the leftmost (can't make out the rest), Maxwell's equations(?) on the woman second from the left, and SU(3) otimes SU(2) otimes U(1) on the rightmost woman on the bottom line. HTH.

As a quaternion rotation matrix that would be a rotation by 2θ about the i axis. I also see a Feynman diagram on the rightmost woman, Newton's second law on the third woman (first line), and what looks like the definition of the Christoffel symbols on the first woman (third line). [The fourth line there looks like a differential equation involving Christoffel symbols, but I can't place it.] Tesseran 10:32, 13 May 2007 (UTC)

Fourth woman, first line is the Canonical commutation relation. Second line might be a form of Schrodinger's equation. Third woman, fourth line is the wave equation. I think the matrix on the first uses cosh and sinh, making it a Lorentz transformation, which would fit since the rest of the first woman seems to be about relativity: we have something involving m^2c^2, and the last line is the Geodesic equation. Algebraist 12:32, 13 May 2007 (UTC)

You might want to warn people about that pic, as some might be taken aback by it. StuRat 13:47, 13 May 2007 (UTC)

Good call. Label attached. Black Carrot 21:17, 13 May 2007 (UTC)

It's no racier than you'd commmonly see on billboards in liberal countries. —Tamfang 22:22, 15 May 2007 (UTC)

Perhaps not, but if the subjects were turned around, that would definitely be seen as an affront, by some, who would see it as no less than a full frontal assault on their values. StuRat 01:58, 16 May 2007 (UTC)

Not full frontal ... —Tamfang 17:27, 16 May 2007 (UTC)

Modules

I'm trying to do the following: Given R, a commutative ring with 1, and I and J two ideals of R, prove that R/I and R/J are isomorphic as R-modules if and only if I=J. Give an example to show that it is possible to have R/I and R/J isomorphic as rings even if I≠J.

Going from I=J to the isomorphism is trivial, but I have basically no idea how to go the other way. In any discussion of algebraic widgets, as soon as "quotient widgets" arise, my level of understanding plummets through the floor. What does it mean to consider R/I and R/J as R-modules? Maelin (Talk | Contribs) 08:34, 13 May 2007 (UTC)

First, DON'T PANIC. You already know a great deal about (commutative) rings, quotient rings, ideals, and modules. The one (commutative) ring to rule them all (sorry, Tolkien) is the integers. For an ideal, take multiples of 12; this is closed under addition, negation, and multiplication by any integer. Then the quotient ring is "clock arithmetic", the integers modulo 12. A module is just a vector space whose scalars forgot how to divide. (In general, the scalars for a module also need not have a multiplication that commutes; but here that's not the case.)

So what can it mean for the integers modulo 12 to be (almost) a vector space? Since it is called an R-module, where here R = Z, we know that the scalars are the integers. Thus the numbers (or equivalence classes) 0 through 11 must be the "vectors". And, sure enough, we can add them and scale them.

Play with these examples a little, and ask again if you want more specific hints. --KSmrq^T 10:43, 13 May 2007 (UTC)

There is a natural action of R on itself by left multiplication; this action descends to an action of R on R/I by left multiplication. (Explicitly, r acts on x+I by sending it to rx+I, using the additive notation for cosets. You might prefer the notation r acts on [x] sending it to [rx].) An R-module is by definition an object [an abelian group] with an action of R on it. What this means for you is that a proposed isomorphism between two R-modules needs to not only be a bijection between the two underlying objects --- that bijection must preserve the additional structure given by the action of R on the objects. (This is a loose, wordy explanation, since you should have the formal definition written down somewhere. If not, please say so.)

An analogous, perhaps simpler example: consider the group G = Z x Z₂ (where Z₂ is the cyclic group of order 2). Consider two subgroups: H = 2Z x Z₂, and K = Z x 1 (that is, K is simply the first factor of the direct product above). You should convince yourself that G/H and G/K are isomorphic (what familiar group are they isomorphic to?) Yet these quotients both have a natural G-action on them, and as G-sets they are very different.

To start with, you could look at the "kernel" of an action, which is in this case those elements of G that act trivially in a given action [on a given G-set]. If the kernels of two actions are different, then the two actions cannot be the same. As an example, the group of integers acts on the set of real numbers as follows: let the integer n act by multiplication by (-1)ⁿ, sending the real number x to the real number (-1)ⁿx. First check that this is a well-defined action; then see if you can identify which integers act trivially. Tesseran 10:53, 13 May 2007 (UTC)

(after edit conflict) R/I is a ring, and therefore an abelian group (by forgetting multiplication). We can define a "left" scalar multiplication operation • : R × R/I → R/I by r•(a+I) = (ra)+I. That is all we need by way of ingredients for R/I to be a module over R, and it is easy and straightforward to verify that the four listed conditions are met. (Actually we can likewise define a "right" scalar multiplication (a+I)•r = (ar)+I, giving us a bimodule.) For a homomorphism h : M → N to be also a module homomorphism (where M and N are R-modules over the same ring R), it has to respect scalar multiplication: h(r•x) = r•h(x).  --Lambiam^Talk 10:54, 13 May 2007 (UTC)

Okay, I think I can see R/I and R/J as R-modules now, but I have no idea how to prove that the isomorphism implies I=J. I tried using the fact that, since R/I and R/J are isomorphic, there exists a bijective homomorphism φ : _R(R/I) -> _R(R/J) , and then by the first isomorphism theorem, _R(R/I) / ker(φ) ≈ im(φ), and since φ is bijective, ker(φ) = I and im(φ) = _R(R/J), so _R(R/I) / I ≈ R/J, but this looks very suspiciously like not telling me anything useful. Am I going in the complete wrong direction? Maelin (Talk | Contribs) 13:08, 13 May 2007 (UTC)

Tesseran has already given you the idea, I think. Consider the kernel of the R-action on R/I, that is, the set of r in R such that r(a + I) = 0 + I for all a in R. Can you work out what this kernel is? Now use the isomorphism between R/I and R/J as R-modules to show that both actions must have the same kernel. Algebraist 13:14, 13 May 2007 (UTC)

The kernel is {0}, but so what? What does that have to do with the isomorphism theorem? And what does it matter whether the actions have the same kernel when they're different actions? *getting increasingly frustrated at impenetrable, unintuitive concepts using impenetrable, unintuitive notation* Maelin (Talk | Contribs) 13:34, 13 May 2007 (UTC)

The kernel is not {0}! Consider a special case: R=Z, I = 4Z say. Z acts on Z/4Z by multiplication. Which elements of Z send everything to 0 by this action? Algebraist 13:40, 13 May 2007 (UTC)

This is why I hate this damn coset notation and in fact the entire concept. Is the kernel I then? Maelin (Talk | Contribs) 13:45, 13 May 2007 (UTC)

If you think so, prove it. Aside: I myself have always found ring quotients (unlike group quotients) very intuitive, since they generalise the natural idea of modular arithmetic on Z. To quotient by an ideal I is just to declare all elements of I to equal 0, and accept all consequences. Modules, on the other hand, have always baffled me. Algebraist 13:55, 13 May 2007 (UTC)

Suppose r is in I. Then ra is in I, so ra+I = 0+I, so r is in the kernel. Conversely, suppose r is not in I. Consider a=1. Then ra+I = r+I ≠ 0+I since r is not in I, so r is not in the kernel. Therefore the kernel must be I. QED. Right? Maelin (Talk | Contribs) 14:03, 13 May 2007 (UTC)

Correct Algebraist 14:19, 13 May 2007 (UTC)

Okay, but I don't understand how I can use the isomorphism theorems. I haven't seen any results anywhere that allow me to derive an equality from any kind of isomorphism that look even remotely related to this problem. Should this be a proof by contradiction, maybe? I can't even see how finding the kernel of the R-action on R/I is relevant. Maelin (Talk | Contribs) 14:23, 13 May 2007 (UTC)

Because as I claimed above, the isomorphism between R/I and R/J (as modules) implies the actions have the same kernel. The isomorphism theorems are (I think) useless here, which is unsurprising since you are given an isomorphism and asked to prove an equality, which the isom. theorems can't give you. Algebraist 14:59, 13 May 2007 (UTC)

So you're saying that the kernel of the R-action on R/J must also be I, because of the isomorphism? That isn't obvious to me at all. Maelin (Talk | Contribs) 21:21, 13 May 2007 (UTC)

(cancelling indents) It wasn't obvious to me either (though it is now). As I said earlier, I have little intuition for modules. It is however, obviously a result that might be true. So what you do is, you try to prove it. If you succeed, you've learned something, and hopefully improved your understanding of modules. If you fail, then work out why you failed, use this to construct a counterexample, and you've improved your understanding of modules. Algebraist 09:19, 14 May 2007 (UTC)

That's really the issue, you see. I haven't the faintest idea how I might prove that. I can't see any way to connect the kernel of the R-action on R/I and the R-action on R/J even with the isomorphism. What is the basic idea of the proof that the kernels are the same? Maelin (Talk | Contribs) 11:46, 14 May 2007 (UTC)

Okay, I've found a proof that goes via another way. Now I'm trying to find the counterexample to show that it's not true for R/I ≈ R/J as rings. I know the integers can't offer a counterexample, since the only ideals in Z are nZ, and Z / nZ = Z_n, and no two of them are isomorphic without having the same n. I also know the rationals, reals, etc won't, since they're all fields and fields don't contain any nontrivial ideals. I suspect it will be some polynomial ring, but I have only a vague grasp of ideals in them and virtually no idea what quotient rings look like or how they might be isomorphic. Can anybody help? Maelin (Talk | Contribs) 14:21, 14 May 2007 (UTC)

Again, Tesseran has done this already up there (more or less). Aside:proof that the actions of R on A and B have the same kernel, given that A and B are isomorphic R-modules: let r be such that ra=0 for all a in A. Then Φ(ra)=0 for all a. But Φ(ra)=rΦ(a) and Φ is surjective so rb=0 for all b in B. Thus (r in kernel of action on A) implies (r in kernel of action on B) and the reverse holds by the same argument using Φ^-1. Thus the kernels are equal. Algebraist 16:34, 14 May 2007 (UTC)

To answer your earlier question, the idea, as so often when doing proving very low-level facts in algebra, is not to have any ideas. The closest thing to an idea needed is to prove two sets are equal by showing each contains the other, but since that's the definition of set equality, I'm not sure that counts. Algebraist 16:38, 14 May 2007 (UTC)

Transformation

Hi there, I have a question on transformation.

What transformations must you apply to y=x^2 to create the new graph? List the transformations in the order you would apply them.

a) y=-x^2+9

b) y=(x-3)^2

c) y=(x+2)^2-1

d)y=-2(x-4)^2+16

this is not for homework. I just want to know.

thank you —The preceding unsigned comment was added by 76.64.55.160 (talk) 14:27, 13 May 2007 (UTC).

Homework or not, this is straight out of some textbook. I suggest the book's section on curve sketching and transformations as a fruitful source of information. Alternatively, you might find playing around with an online graphing calculator (there are loads of them) aids your understanding. Algebraist 15:11, 13 May 2007 (UTC)

Try GCalc 2: [4]. StuRat 18:08, 13 May 2007 (UTC)

What's the graph in the first place? You'll have to use the concepts of transformations taught in your textbook. Splintercellguy 03:52, 14 May 2007 (UTC)

Minimum value

Hi there, I have a question about minimum value.

How can you find the minimum value of the graph of y=3x^2-9x-30 and use it to express the equation in vertex form and can you show the work so I can understand. thanks. by the way, it is not homework. —Preceding unsigned comment added by 76.64.55.160 (talk • contribs)

What do you mean by vertex form? I've never heard the term before. Algebraist 15:19, 13 May 2007 (UTC)

Don't worry, Google told me. This is what was taught to me under the name Completing the square. Algebraist 15:23, 13 May 2007 (UTC)

Transformation 2

What transformation must be applied to the graph of y=x^2 to produce the graph y=2x^2-12x+7 and can you justify you reason for the question? thanks. this is not homework. —The preceding unsigned comment was added by 76.64.55.160 (talk) 14:38, 13 May 2007 (UTC).

Observe that 2x²-12x+7=2(x-3)²-11. This has the same form as one of your previous questions. Algebraist 15:17, 13 May 2007 (UTC)

As an aside, can someone explain why some usernames are blue and some are red? I used to think it was denoting status, in that some had earned the right to be in some sort of in group, but this questioner's ID is in both colours in the various posts. -- 86.132.167.85 19:03, 13 May 2007 (UTC)

One links to Special:Contributions/76.64.55.160, which is not empty, the other to User:76.64.55.160, which does not exist.  --Lambiam^Talk 20:08, 13 May 2007 (UTC)

To be precise, the "Special:Contributions" link would be blue even if the list was empty. In fact, all links to "Special:" pages (even Special:Therereallyisnosuchpage) are always blue, period. But yes, usually blue links mean the target page exists, and red links that it doesn't. —Ilmari Karonen (talk) 20:20, 13 May 2007 (UTC)

In my case, it's because I don't write on my User page. The page hasn't officially been created, so it's red, a broken link. Black Carrot 20:55, 13 May 2007 (UTC)

Effect of "reroll twice" on probability distribution

Objects are to be randomly generated by rolling a die and consulting a table n times. Most of the table entries will generate one object per roll, but a "nothing" entry (P=x) is also present, as is a "reroll twice" entry (P=y). Since rerolls-twice can in turn generate more rerolls-twice, there is no upper bound on the number of objects, and the expected number of objects is not simply 2y + 1-x-y. What is the probability distribution? NeonMerlin 15:25, 13 May 2007 (UTC)

Too lazy to do full analysis, but if y>1/2, then with positive probability you sit rolling dice for ever (see Branching process). If y=1/2, then unless x=1/2 also the expected number of objects is infinite, though the actual number is almost surely finite. If y<1/2 (hopefully the case!) then the expected number E is finite and satisfies E=2yE +1-x-y, i.e. E = (1-x-y)/(1-2y). No idea about the actual distributions, I'm afraid. Algebraist 16:03, 13 May 2007 (UTC)

In case you want this, the positive probability I alluded to above is ${\frac {2y-1+{\sqrt {1-4y(1-y)}}}{2y}}$. Algebraist 16:11, 13 May 2007 (UTC)

The following is not a closed formula, but a recurrence relation that allows you to calculate the probabilities. If p_n stands for the probability of observing n objects, and xy < 1/4, then

p₀ = 2x / (1 + √(1−4xy)) ;

p₁ = (1−x−y) / (1−2yp₀) ;

p_n = yS / (1−2yp₀) for n > 1, where S = Σ_{k=1,...,n−1} p_kp_n−k .

If xy = 1/4, x = y = 1/2, and then p₀ = 1, so p_n = 0 for n > 0.  --Lambiam^Talk 20:00, 13 May 2007 (UTC)

So that's when the table is initially to be consulted once? NeonMerlin 01:48, 17 May 2007 (UTC)

Yes. For k lookups (say), the probability of n items will be $\sum _{a_{1}+a_{2}+\dots +a_{k}=n}p_{a_{1}}p_{a_{2}}\dots p_{a_{k}}$ (with the sum over all ordered k-tuples of nonnegative integers summing to n). Algebraist 16:41, 17 May 2007 (UTC)

Gun barrel bore

Why is 9mm such a popular barrel size for handguns? Why not 10mm or 8mm? Thanks, WSC

The reason is historical, not mathematical. Other sizes have also been used. See 9 mm Luger Parabellum, List of handgun cartridges, and Cartridge (firearms). nadav 17:20, 13 May 2007 (UTC)

Our cartridge page is not for victorian censored children. I read : "The alteration of the military flint-lock to the percussion musket was easily accomplished by replacing the powder pan by a perforated nipple, and by replacing the cock or hammer which held the flint by a smaller hammer with a hollow to fit on the nipple when released by the trigger.".

Nothing changes. You may find an analogy with the width or railways, originating from Roman waggons. For guns, ask yourself who manufactured iron tubes and why when the gun was invented. -- DLL ^{.. T} 19:10, 15 May 2007 (UTC)

(related to projective varieties) Is a quotient ring of a graded ring a graded ring as well?

Hello,

Let $R$ be a graded ring (see [5] for a definition)

Let $I$ be some ideal in that ring $R$ Can the quotient ring $R/I$ be interpreted as a graded ring as well?

This is actually the reason why I'm asking : I am using a book where S stands for all polynomials in $n+1$ variables over a field $k$, where $Y$ is a projective variety in $\mathbb {P} ^{n}$; and where $I(Y)$ is the ideal generated by all homogeneous polynomials vanishing on all points in $Y$. They let $S(Y)$ denote the quotient ring $S/I(Y)$ Then they speak of $S(Y)_{(}x_{0})$ , which stands for the localisation with the multiplicative set$\{1,x_{0},x_{0}^{2},\dots \}$ . And then they speak of $S(Y)_{(}x_{0})$ , which stands for "the degree zero part" of $S(Y)_{(}x_{0})$ . Now I do not understand how one can speak of "degree zero part" without the ring being graded?

As you can see, I'm really confused. Could it be that my book is a bit sloppy with definitions. I hope someone with some experience in this matter will manage to get me out of this darkness. Thank you and greetings, Evilbu 18:24, 13 May 2007 (UTC)

You'll need for your ideal to be homogenous, i.e. the generators all homogenous elements of the ring. If this is the case, then the quotient ring inherits grading from the starting ring. Michiexile 18:37, 13 May 2007 (UTC)

Okay, that is good, since I(Y) is a homogeneous ideal right? But do go on, please, how would you "grade" that new quotient ring? I mean : how would you decomposition that quotient ring? The answer may seem straightforward at first, but it isn't, since there are several representants of the same quotient class.Evilbu 19:25, 13 May 2007 (UTC)

I don't know about your specific case, but we know $R/I=\sum (R^{n}+I)/I$, assuming that I is homogenous. I can include the proof that this is true if you like. nadav 01:36, 15 May 2007 (UTC)

May 14

Equation

How can you find this equation where the vertex is (4,6) and the x-intercepts are (8,0) and (0,0)? Please answer the question. thank you. —The preceding unsigned comment was added by 76.64.53.59 (talk) 01:22, 14 May 2007 (UTC).

We don't do homework, but your question is very ambiguous. There are an infinite number of equations that can accept the information you have given. x42bn6 Talk Mess 01:59, 14 May 2007 (UTC)

I think it's safe to assume you're talking about a Parabola. The article is very informative. At least, bits of it are. Given the vertex and the x-intercepts, there are several ways to work out the equation in some form or other. One makes use of the base equation y = x^2 and adjusts from there. It starts out with vertex at (0,0), opening up. To move the graph horizontally and vertically, we introduce the vertex (h,k), making (y-k) = (x-h)^2. To stretch the graph out to the right shape, and flip it over if necessary, we introduce a constant I'll call b, making (y-k) = b(x-h)^2. To find b, plug in h, k, x, and y (any x and y will do, like one of the x-intercepts) and solve for b. Black Carrot 02:45, 14 May 2007 (UTC)

Perhaps this is the same person who asked about minimum value, above. If so, then a parabola is expected. But the questioner must respect the rules of this page (stated at the top), which explicitly prohibit homework solutions. What we can do is offer guidance, as Black Carrot has done. We can also confirm reasoning.

Besides, the way I personally would fit a parabola through three points is likely not appropriate as a homework answer. I might fit a straight line through (0,0) and (4,6), fit another through (4,6) and (8,0), then blend them to produce a parabola, in the manner of Lagrange interpolation.

${\begin{aligned}p_{0,4}(x)&={\frac {4-x}{4-0}}0+{\frac {x-0}{4-0}}6,\\p_{4,8}(x)&={\frac {8-x}{8-4}}6+{\frac {x-4}{8-4}}0,\\p_{0,8}(x)&={\frac {8-x}{8-0}}p_{0,4}(x)+{\frac {x-0}{8-0}}p_{4,8}(x).\end{aligned}}$

But if you hand this in for homework, you will have learned nothing, the teacher will immediately suspect cheating, you will be asked to explain the answer, and you will fail. And besides, it takes no advantage of the special nature of the three points given. The advantage for me is that I can extend this method to fit a polynomial of degree n−1 through n data points, so it's a simple and handy method to remember. --KSmrq^T 03:31, 14 May 2007 (UTC)

When he said, "vertex", I thought he might have meant something like an absolute-value graph (y=-|x+1|?). x42bn6 Talk Mess 14:52, 14 May 2007 (UTC)

A good point. My analysis would extend to that case nicely, though I'm afraid KSmrq's might not. Black Carrot 05:25, 15 May 2007 (UTC)

No worries; omit the blending step and my algorithm delivers a piecewise linear fit. A slight variation will yield B-spline curves instead of interpolating curves (as in the de Boor algorithm), and a combination of the two can deliver a family of different types of curves, including piecewise cubic interpolants with first derivative continuity. --KSmrq^T 10:07, 15 May 2007 (UTC)

Additional MathsGCSE question on Vectors

(Throughout thisquestion i and j denote unit vectors parallel to standard set of x-y axis)

A bodyof mass 2kg is in equilibrium under theaction of three coplanar forces P,Q and R where

P=(3xi-2yj)N Q=(2yi+4xj)N R=(-11i+4j)N

(i)Calculate the values of x and y

Cone-Pine1991 14:14, 14 May 2007 (UTC)

Do your own homework. However, a hint: P+Q+R=0, allowing you to find the values of x and y. x42bn6 Talk Mess 14:50, 14 May 2007 (UTC)

Proof of Cone and Sphere Surface Areas

Hello. Can anybody please tell me the proof of the right cone's and the sphere's surface area? I am only in middle school, so please keep the explanation simplified. Thanks. --Mayfare 18:01, 14 May 2007 (UTC)

Are you familiar with calculus? It could be tough to explain otherwise. -- Meni Rosenfeld (talk) 18:58, 14 May 2007 (UTC)

For the cone, I suggest you first take a look at Cone (geometry). If you're wondering why the lateral area is $\pi rs$, the intuitive explanation is: You start with a circle of radius r (and perimeter $2\pi r$), and you take it along a distance of s while shrinking it to a point. The average perimeter is $\pi r$, and you multiply the perimeter with the length to obtain the area. -- Meni Rosenfeld (talk) 19:10, 14 May 2007 (UTC)

For the cone there is an elementary derivation, although it uses a bit of handwaving. Let r be the radius of the disk making up the base, and s = √(r²+h²) the slant height (the distance from any point on the circle that is the boundary of the base to the apex of the cone), where h is the standard height. The formula A = πr² + πrs gives the area as the sum of two terms. The first, πr², is simply the area of the base disk, which is flat. The second, πrs, then must be the area of the surface formed by the lines to the apex, which is curved in 3D space. Its intrinsic curvature, however, is 0: we can cut this surface open along a line from any point on the circle to the apex, and then flatten it. (Here is where the handwaving comes in; I have not actually given a proof that the surface is flattenable.) The result is a circular sector, like a pizza slice. The disk of which this is a segment has radius s. The arc length of the curved boundary is the length of the original circle, 2πr. The relationship between the arc length a and the angle θ of the sector is a = sθ, or θ = a/s. The area of the sector is ¹/₂θs². Using θ = a/s and a = 2πr, we have θ = 2πr/s, so the area is ¹/₂θs² == ¹/₂(2πr/s)s² = πrs. Q.E.D.  --Lambiam^Talk 22:40, 14 May 2007 (UTC)

And for that you already need that the area of a circle is πr², which I don't think can be proven elementarily without handwaving either. Anyway, I think someone should point out that without calculus (or something like it), it's not clear how we can even define surface areas of curved objects, let alone actually calculate them! One can assign area to flat polygons easily enough by cutting them up into triangles, but how do you extend this to circles, let alone non-flat surfaces? Algebraist 00:05, 15 May 2007 (UTC)

(edit conflicted) Archimedes famously found the surface area of a sphere using only the area formulae for the cone and for the conic frustum. He inscribes a solid composed of frustums (and cones on the two ends) in a sphere. Projected into two dimensions, the solid looks like a an n-gon inscribed in a circle. He finds the surface area of the solid and then lets n go to infinity. The details of the proof are involved, but I think the intuition is clear. As a historical note, Archimedes did not have access to limits the way we do, so his proof used a proof by contradiction: he showed the s.a. of a sphere cannot be strictly more or strictly less than "four times its greatest circle." Archimedes was so proud of his result that according to legend, he had asked for a picture of a sphere inscribed in a cylinder to be carved opon his tombstone. See if you can find the connection. (I took this info from Dunham's The Mathematical Universe) nadav 00:18, 15 May 2007 (UTC)

This is what I meant by something like calculus above. The classical greeks (who were far more rigorous in their mathematics anyone else until at least the mid 19th century) would not have been happy with any handwaving, and developed a cunning method for this sort of problem. However, the method (being based as you say on contradiction) could only be used to prove results and not find them. This is especially telling in the first use of the method, when Eudoxus proved that the area of a circle is proportional to the area of the square on its diameter, a result that is blindingly obvious as long as circles have area at all. Thus the purpose of the technique was to put results on a sound footing, not to discover them. For this Archimedes used an entirely different method, with more handwaving than a busload of queens. Algebraist 00:37, 15 May 2007 (UTC)

Strange image; I wonder if this is what you mean? Or perhaps a bus in Queens, NY?

As for the mathematics, see area of a disk. --KSmrq^T 02:13, 15 May 2007 (UTC)

Lol. I assume he meant the first one. I love that metaphor. Did you come up with it yourself, Alg? I reserve the right to steal it since it's published under GFDL (but I'll give you credit). nadav 02:42, 15 May 2007 (UTC)

KSmrq, your first image fits the waving part, but this one is what first came to mind for me... —David Eppstein 03:47, 15 May 2007 (UTC)

Having seen both this (recommended) and this (an acquired taste I never acquired), I had two obstacles: (1) where to find the bus (well done!), and (2) although anyone with the chutzpah to be on that bus would hardly be offended, I didn't want to feed the dark side (intolerance) of humanity. --KSmrq^T 10:55, 15 May 2007 (UTC)

I was indeed thinking of the main purpose of existence of Her Britannic Majesty, Elizabeth the Second, by the Grace of God of the United Kingdom of Great Britain and Northern Ireland and of Her other Realms and Territories Queen, Head of the Commonwealth, Defender of the Faith, Lord of Mann, Duke of Normandy, Sovereign of the Most Noble Order of the Garter, Fellow of the Royal Society, freewoman of the City of Philadelphia etc etc, but am happy for its meaning to be ruthlessly reinterpreted per the GFDL. Algebraist 15:05, 15 May 2007 (UTC)

I have elsewhere described KSmrq's second example as "absolutely the vilest film ever made". Fortunately, it's not representative of anything but itself. JackofOz 02:40, 16 May 2007 (UTC)

May 15

Table of Integrals

Historically, how is the first Table_of_integrals created? It's like a chicken and egg situation. To solve an integral, you need a table. To create a table, you need to solve that integral. So you can neither solve an integral nor create a table of integrals. 202.168.50.40 01:37, 15 May 2007 (UTC)

You don't need a table. When someone sets out to learn calculus, they do the integrals without this help. Integration is merely the inverse process of differentiation, which is not hard to do. So if you know a lot of derivatives, and understand some basic properties of integrals (such as linearity), then you can already integrate a lot of the functions you usually come across. nadav 01:50, 15 May 2007 (UTC)

Of course, then come more powerful methods such as substitution, integration by parts et cetera. -- Meni Rosenfeld (talk) 08:23, 15 May 2007 (UTC)

I really hope this is a joke... I knew engineers, scientists etc were given tables of integrals, but I always assumed it was "here's a bunch of integrals done for you by people who know maths," not "here's a bunch of integrals that have come into being without human agency" Algebraist 14:54, 15 May 2007 (UTC)

The first time Moses went up the mountain, he came back with the Tables of Multiplication. The second time he came back with the Tables of Integrals. Scientists are still trying to figure out where the Tables of Logarithms came from.  --Lambiam^Talk

I take it addition is the product of feeble human intellect, then? Algebraist 17:49, 15 May 2007 (UTC)

Addition was derived from the multiplication tables. For example, $6+6=6*2=12$ and $8+4=4*2+4=4+4+4=4*3=12$. The prominent theory regarding the Tables of Logarithms is that they were dug up from the ruins of an ancient alien civilization with transfinite intelligence. -- Meni Rosenfeld (talk) 19:16, 15 May 2007 (UTC)

Fractal Artwork

This is part Math Desk, part Humanities. I've been looking around a bit, and I've noticed that everybody who does art with fractals seems to favor either the abstract, or the obviously self-similar. There are, of course, the many fractal-based images of ferns and mountains and occasionally snail shells, some of them gorgeous, some of them even realistic, but the closest I've found to an attempt to capture concrete nonfractal objects with fractal techniques was a gallery of rather unconvincing fractal "faces". Can anyone point me towards someone who's doing that kind of work? Or, if nobody is, do you think there would be a market for it? Black Carrot 05:32, 15 May 2007 (UTC)

Could the reason that these "faces" were rather unconvincing be that fractals are not a good approach to capturing non-fractal objects?  --Lambiam^Talk 10:27, 15 May 2007 (UTC)

See "fractal compression"; though I don't know who, if anyone, is actively pursuing research in this area today, as mainstream compression eventually produced better compression than Barnsley's initial optimistic claims. But you may be interested in neighboring topics such as cellular automata and Lindenmayer systems. A group of researchers at Calgary has produced some beautiful results. --KSmrq^T 11:37, 15 May 2007 (UTC)

BERMUDA TRIANGLE

What is BERMUDA TRIANGLE?
Where can I find pictures of it?

Donlesnar 08:55, 15 May 2007 (UTC)

Try Bermuda Triangle. --CiaPan 09:11, 15 May 2007 (UTC)

linear programming

Hi there just doing some pre-uni work on my course coming up to get up to the required level. I am currently teaching myself linear programming and have come across this gem of a question.

tasks 1,2,3,4,5 are to be undertaken by individuals A,B,C,D and E. Each person has to do one task. Each task has to be undertaken by one person. The table below shows who can do which task - y=yes

	A	B	C	D	E
1	y		y
2			y	y
3	y				y
4		y		y	y
5		y		y

Formulate as an LP the problem. How would i go about doing this on LINDO

Thank you very much for you time, sorry its a bit long winded but i really want to get on top of this before uni :) chemaddict

This is a special case of the assignment problem, which is a special case of the transportation problem, which is a special case of the minimum cost flow problem, which is a special case of linear programming. That gives you as many ways to solve this problem as you could wish for, and if you learn all of them, you'll know as much linear programming as any sane pre-uni student could want, IMO. Algebraist 14:50, 15 May 2007 (UTC)

I rejigged your table. Hope you don't mind. Algebraist 15:17, 15 May 2007 (UTC)

LP problems have an objective function to be maximised or minimised, in the case of the assignment problem it would be the total effectiveness or time (respectively) of the allocation of jobs to people. Your problem doesn't have such an OF, but it does have multiple feasible solutions (eg A1 B4 C2 D5 E3 or A1 B5 C2 D4 E3) - there is no criterion to select the best. .. 81.151.82.52 16:46, 15 May 2007 (UTC)

That's not a problem. Just set each job-person allocation to have effectiveness 1 (or whatever) so you're maximising the number of people allocated to jobs. Then both solutions given are optimal. Algebraist 17:48, 15 May 2007 (UTC)

Jordan Normal Form & the Structure Theorem

I'm attempting to do this:

Calculate the Jordal Normal Form for the following matrix A by calculating the invariant factor matrix of $\textstyle xI-A\in M_{4}(\mathbb {C} [x])$ .

$A={\begin{bmatrix}-2&0&0&1\\1&1&0&1\\2&0&1&-2\\-1&0&0&0\end{bmatrix}}$

What are the characteristic and minimal polynomials of A?

I've taken xI - A and reduced it to its invariant factor matrix, which I got as a diagonal 4x4 matrix with 1, 1, x-1, (x-1)(x+1)² down the leading diagonal (and zeroes elsewhere). Earlier today I asked my lecturer and he said the way to do this is to then find the invariant factor decomposition. Going from my notes, I -think- that the invariant factor decomposition is

$A\cong {\mathbb {C} \over {\langle 1\rangle }}\oplus {\mathbb {C} \over {\langle 1\rangle }}\oplus {\mathbb {C} \over {\langle x-1\rangle }}\oplus {\mathbb {C} \over {\langle (x-1)(x+1)^{2}\rangle }}\cong {\mathbb {C} \over {\langle x-1\rangle }}\oplus {\mathbb {C} \over {\langle (x-1)(x+1)^{2}\rangle }}$,

where I'm using the angle bracket notation <f(x)> to denote "the ideal generated by f(x)". I think the primary decomposition is then

$A\cong {\mathbb {C} \over {\langle x-1\rangle }}\oplus {\mathbb {C} \over {\langle x-1\rangle }}\oplus {\mathbb {C} \over {\langle (x+1)^{2}\rangle }}$.

This would make the primary factors (x-1)¹, (x-1)¹, (x+1)², and so the minimum polynomial is μ(x) = (x-1)(x+1)². And I'd hazard a guess that the characteristic polynomial is χ(x) = (x-1)²(x+1)², which would mean the JNF of A is a (mostly) diagonal 4x4 matrix with 1, 1, -1, -1 down the leading diagonal, and a 1 underneath the first -1 (and zeroes elsewhere).

Have I done this correctly? If I've made a mistake at any step of the process, please point it out to me. Thanks. Maelin (Talk | Contribs) 12:53, 15 May 2007 (UTC)

I think your argument is correct (I knew this stuff properly for a brief period around exam time last year). I can confirm that your answer is right, since (0,1,0,0), (0,0,1,0) are linearly independent 1-eigenvectors, and (1,-1,0,1) is the unique evec with val -1, while the fact the trace is zero means the last diagonal entry must be another -1. Algebraist 14:39, 15 May 2007 (UTC)

May 16

infinitey

can infinitey be equal to infinitey —The preceding unsigned comment was added by Sivad4991 (talk • contribs) 00:10, 16 May 2007 (UTC).

Inasmuch as infinity is some uniquely defined mathematical object, it is (like any other such object) equal to itself. As you can read in the article, there are actually many kinds of infinity in mathematics, such as the cardinal number ℵ₀. Indeed, ℵ₀ = ℵ₀. Another infinity is c, the cardinality of the continuum. This is different: c ≠ ℵ₀.  --Lambiam^Talk 01:04, 16 May 2007 (UTC)

(edit conflict) Yes, and they can be unequal, too. More correctly, there is no single number "infinity", as I hint at in the answer to your next question. There are many "numbers" (note that they are not real numbers, but usually part of some extension of the set of real numbers) that posses infinite properties, and what you mean when talking about infinity differs depending on the circumstances. For example, if you are talking about the $infinity$ in an infinite sum or limit such as

$\lim _{x\to \infty }{\frac {1}{x}}=0$

we're talking about a convenient notation to represent a definition that involves lots of upside-down As and back-to-front Es, without technically invoking any properties of "infinity". On the other hand, if you talk about the number of integers being infinite, you're talking about the cardinality of a set, and you use the symbol known as aleph null, which is the starting point for a whole bunch of big numbers. There are many more infinities, and it takes some work to formally define them, and define what it means for one of them to "equal" another (in at least one case, we even know that there are two infinities that may or may not be equal, but that which we can never prove one way or the other - see continuum hypothesis). Confusing Manifestation 01:07, 16 May 2007 (UTC)

Actually, Confusing, that's a somewhat, um, confusing way to put it. What we actually know is that certain specific axiomatizations of set theory, such as ZFC, can never prove it one way or the other. We don't know that it can't be settled by other axioms that can intuitively be seen to be true of the objects of discourse of set theory (the von Neumann universe). --Trovatore 01:16, 16 May 2007 (UTC)

Fair enough. It seemed to me that the poster probably wasn't at a point where they were likely to follow the finer points of set theory like that, so I was trying to keep it simple. Unfortunately, with infinity it's hard to keep things simple. Confusing Manifestation 02:14, 16 May 2007 (UTC)

black hole

Ive heard that black holes are objects that have an infinite mass but is it even possible for an abject to have an infinite mass. And my teacher said that negative nubers arent number that arent shown in nacher is that true for infinate to. thanks --Sivad4991 00:22, 16 May 2007 (UTC)

A black hole doesn't have infinite mass, but it does have a theoretically infinite density at its centre - note that the word infinity is used in several different forms in mathematics and science, and in this case it's referring to the "kind of infinite" you get when you divide a non-zero number by zero, also known as a singularity, although in the case of black holes it's more a case of "inside the event horizon we can't really see what's going on, but mathematically we can describe things outside the event horizon as though there were a singularity at the centre, so that's what we do".

For the same reason, you could argue for there being no numbers at all in nature, just human mathematics using them to describe what nature does. The argument is strongest for "infinity", and to a certain extent for negative numbers - but there are things that seem to need negative numbers, such as electric charge, although the choice of which charge is negative and which is positive was fairly arbitrary (and probably would have been better the other way around for some applications). Confusing Manifestation 00:58, 16 May 2007 (UTC)

Interesting math topic for a presentation

I'm in high school and I have to give a math presentation very soon. Thus, I want to know of some interesting math topics. My presentation can be on any mathematical topic. An example of an interesting topic would be Pascal's triangle. —The preceding unsigned comment was added by Metroman (talk • contribs) 02:28, 16 May 2007 (UTC).

For the love of everything self-referential, please use Interesting number paradox.--Kirby♥time 02:48, 16 May 2007 (UTC)