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February 21

Quick question about log convexity

I've read the article on power series, and it mentions that for more than one variable, the domain of convergence is a log-convex set instead of an interval.

But what is a log-convex set here ? I mean, how do we take the log of a set ? Surely the set can contain negative values for example...

Thanks. -- Xedi (talk) 02:13, 21 February 2008 (UTC)

Sorry, I have no idea what they could mean by log-convexity of a set. The multi-variable stuff was added in June 2004, and the log-convex comment was made in Oct 2004. It does not appear to have been touched since then. I'll ask the author about it.

The domain of convergence is a union of polydisks, but I think perhaps it need not be a polydisk. Some reasonably elementary notes are at Adam Coffman's website, specifically his research notes on Notes on series in several variables. It includes the polydisk result, as well as the continuity, and differentiability of functions defined by a series, and includes the nice "recentering" result familiar for one complex variable allowing analytic continuation. JackSchmidt (talk) 04:25, 21 February 2008 (UTC)

I've added an example and a little more explanation to the article. Terry (talk) 23:07, 21 February 2008 (UTC)

Thanks. I suppose when the "center" is (a₁,a₂,...) then we consider (log|x₁-a₁|,log|x₂-a₂|,...) ? —Preceding unsigned comment added by Sam Derbyshire (talk • contribs) 13:08, 22 February 2008 (UTC)

Natural Logs

I'm being asked to evaluate natural logs and I'm getting very confused. One problem I'm having particular trouble with is lne(-x/3)

Could anyone please help explain to me how to solve problems like these and how to determine how to get started?

Thanks, anon. —Preceding unsigned comment added by 66.76.125.76 (talk) 03:53, 21 February 2008 (UTC)

What is lne(-x/3) ?

Is it Log[ Exp[-x/3] ] ?

or is it Log[-x/3] where Log is Log based e.

202.168.50.40 (talk) 04:34, 21 February 2008 (UTC)

ln is usually a shorthand for log_e --PalaceGuard008 (Talk) 05:07, 21 February 2008 (UTC)

• y = log(-x/3)
• e^y = -x/3
• x = -3e^y

--wj32 ^t/c 05:16, 21 February 2008 (UTC)

If the question is about ln e^−x/3, what can you say in general about ln e^A?  --Lambiam 12:05, 21 February 2008 (UTC)

is math the same as logic?

Is math the same as logic, or why not.

In other words, is there math that isn't logical, but just evaluates truths, for example "experimenting" with numbers to find out "facts" but not using logic to show that these must be necessarily true. After all, the other sciences don't rely on their findings to be NECESSARILLY true, just that they happen to be true... so which is math? —Preceding unsigned comment added by 79.122.42.134 (talk) 10:12, 21 February 2008 (UTC)

You might be interested in logicism and experimental mathematics. Algebraist 10:43, 21 February 2008 (UTC)

You might also be interested in David Hilbert's project to formalize mathematics.--droptone (talk) 12:45, 21 February 2008 (UTC)

Experimental mathematics uses numerical methods to suggest conjectures which are then formally proved (or perhaps disproved if they turn out to be just a numerical coincidence). All mathematical results are proved using the techniques of logic, so, yes, they are necessarily true. But this does not mean that mathematics is the same as logic - logic does not determine what a mathemtician sets out to prove, or why a particular result is considered to be beautiful, deep or interesting. Logic is to mathematics as calligraphy is to the Rubaiyat of Omar Khayyam. Gandalf61 (talk) 14:04, 21 February 2008 (UTC)

Moment

Suppose you have a ladder leaning against a wall, which makes an angle θ with the ground, and a man, who can be treated as a particle, stands a distance l up the ladder. Now his weight will obviously be acting downwards and so working to work out the moment produced, you would have to resolve his weight vector to find the component of it that acts perpendicular to the ladder.

My question is, if you used Pythagoras or trig to determine his perpendicular distance from the ladder's point of contact with the ground to the line of action of the weight vector and then multiplied that by his weight, would that give you the correct value of the moment? Cheers in advance 92.3.49.42 (talk) 12:45, 21 February 2008 (UTC)

Unless I've misunderstood, yes. What I tend to do is extend the "arrow" of the force such that the perpendicular goes through the point you are trying to resolve against, then it's a simple case of force times distance from point. x42bn6 Talk Mess 13:02, 21 February 2008 (UTC)

Yes 92.3.49.42, both those methods are valid ways of calculating the moment around the ladder's point of contact with the ground, and both will yield the same answer. --mglg_(talk) 17:18, 21 February 2008 (UTC)

σ-algebra redux

A revised version of There are no countably infinite σ-algebras, hopefully closer to being correct?

Lemma. If ${\mathcal {A}}$ is an infinite algebra over a set X, partially ordered by inclusion, then ${\mathcal {A}}$ contains an infinite ascending chain $a_{1}<a_{2}<\ldots$ or an infinite descending chain $b_{1}>b_{2}>\ldots$.

Proof: Let ${\mathcal {A}}_{1}={\mathcal {A}}-\{X\}$. If ${\mathcal {A}}_{1}$ has a chain with no upper bound, then that gives us what we wanted. Otherwise ${\mathcal {A}}_{1}$ has a maximal element b₁. If $x\in {\mathcal {A}}_{1}$ and $x\not \leq b_{1}$ then $b_{1}<x\cup b_{1}$ so $x\cup b_{1}=X$ by maximality, and therefore $X-x\leq b_{1}$. That means there are infinitely many subsets of b₁ in ${\mathcal {A}}_{1}$.

Let ${\mathcal {A}}_{2}$ be that collection of subsets, minus b₁ itself. Then ${\mathcal {A}}_{2}$ has a maximal element b₂, and so again contains infinitely many subsets of b₂. Inductively we obtain an infinite descending chain $b_{1}>b_{2}>\ldots$.

Corollary. Every infinite algebra contains an infinite collection of pairwise disjoint sets.

We just take the sequence of differences of elements of the chain.

Corollary. There are no countably infinite σ-algebras.

An infinite σ-algebra is an infinite algebra, and by the above has an infinite subcollection $\{D_{1},D_{2},\ldots \}$ of pairwise disjoint sets. The map $\{n_{1},n_{2},\ldots \}\mapsto \cup _{k\geq 1}D_{n_{k}}$ is then an injection from the power set of N into the σ-algebra.

 — merge 13:58, 21 February 2008 (UTC)

That seems to work, but as Trovatore commented above, you don't need the Axiom of Choice (you may not care about AC, of course, but I do). Since you have no infinite ascending chains, you don't need Zorn to find a maximal element (the top element of any chain is maximal). If you start with a bijection between your (putative) countably infinite sigma-algebra and N and, whenever you have to choose an element, always choose the one of least index (in terms of this bijection), the proof becomes wholly choice-free. Algebraist 16:30, 21 February 2008 (UTC)

Many thanks for the review. You're right of course that there's no need to invoke the ghost of Max Zorn to get maximal elements there! I'm not too concerned about using AC, but I do like to improve my awareness of when and how it's being used and when it is or isn't necessary, so your comments are much appreciated. Although there are other ways to solve the original problem, the results for infinite algebras in general seem more useful than the fact that there are no countably infinite σ-algebras, and it's nice that once you have them the statement about σ-algebras becomes trivial.  — merge 17:19, 21 February 2008 (UTC)

(slightly modified the above) It's worth noting also that for the results on infinite algebras we aren't assuming countability, so for these I believe AC is required (?).  — merge 13:00, 22 February 2008 (UTC)

I think the argument goes through if you just know that every infinite set has a countably infinite subset, which does need a little choice to prove. But you don't need full AC to prove it -- for example, it follows from the axiom of countable choice, which is consistent (for example) with every set of reals being Lebesgue measurable. --Trovatore (talk) 22:23, 22 February 2008 (UTC)

Differential equation gives weird result!

Hi there, I've been given differential equation $dy/dx=e^{-3x}/y^{3}$

I separated variables and integrated, and I eventually came up with the general result $y=(-{4e^{-3x}/3}+D)^{1/4}$ where D is 4C, my initial constant of int.

But... the function is always going to be negative, and I have to take the fourth root of it! Have I messed it up, or am I ok so long as D makes the stuff under the fourth root positive??

Thank you! Psymun (talk) 15:20, 21 February 2008 (UTC)

Your working is correct. Could you not have a result that includes complex numbers? -mattbuck (Talk) 15:42, 21 February 2008 (UTC)

'Spose! I didn't see it coming in this course... although seeing as I did them in the last maths course I possibly should have! Thanks for reassurance! Psymun (talk) 15:47, 21 February 2008 (UTC)

No need to consider complex numbers. The function just isn't defined for $x<{\frac {1}{3}}\log \left({\frac {4}{3D}}\right)$. This isn't any different from solving $2y'y=1\;\!$ and getting ${\sqrt {x+C}}$ for a result. -- Meni Rosenfeld (talk) 16:09, 21 February 2008 (UTC)

Your general solution should be $y^{4}=-{\frac {4}{3}}e^{-3x}+C$. If you're restricting yourself to the real numbers, this is equivalent to $y^{2}={\sqrt {-{\frac {4}{3}}e^{-3x}+C}}$. But if you solve for y by taking the fourth root like you did, you're excluding negative values for y, so your solution isn't the general solution any more. —Bkell (talk) 05:20, 22 February 2008 (UTC)

Is there anything that HAS TO be true but isn't?

Has mathematics ever proved anything, so that it has to be true, but in fact it isn't? For example, I'm thinking of something like proving that something can be done within some constraints, but in fact all the variations have been tried by computer and none of them are successful. So there HAS TO be a way, but there isn't. —Preceding unsigned comment added by 79.122.42.134 (talk) 15:32, 21 February 2008 (UTC)

No, because if it could be proved true but all attempts to give examples fail, then you did it wrong. -mattbuck (Talk) 15:38, 21 February 2008 (UTC)

(edit conflict) No. This state of affairs is impossible based on the any commonly accepted concept of a mathematical proof. If you modify your statement to "Has anything ever been accepted as proven by the mathematical community...", then I do not know. Taemyr (talk) 15:47, 21 February 2008 (UTC)

There are of course things that seem to have been logically proven but aren't true. See paradox, and especially Zeno's paradox. DJ Clayworth (talk) 15:51, 21 February 2008 (UTC)

Do you guys mean to tell me that the universe is 100% consistent 100% of the time, and anything that follows is in fact thus, with no exceptions, ever? I find that a little hard to swallow... —Preceding unsigned comment added by 79.122.42.134 (talk) 16:15, 21 February 2008 (UTC)

We're not talking about the universe, we're talking about mathematics. Mathematics has not been proven to be consistent (and in a certain sense can't be), but is generally believed to be so. But yes, in the universe as in mathematics, the result of a valid deduction from true premises is true (in the jargon, valid deductions are truth-preserving). Algebraist 16:23, 21 February 2008 (UTC)

Indeed. Don't confuse reality with mathematics. And even in mathematics it is possible to have a valid proof that a statement is true and an equally valid proof that the same statement is false (simplest case is that you start with both P is true and P is false as axioms). Unfortunately, that result indicates that you are working in a system that is inconsistent, and so not very interesting or useful. Gandalf61 (talk) 16:28, 21 February 2008 (UTC)

A number of logicians, such as Graham Priest, argue that inconsistent systems can be both interesting and useful. See paraconsistent logic for example. -- Dominus (talk) 00:29, 22 February 2008 (UTC)

Two possibly relevant Einstein quotes [1]:

• One reason why mathematics enjoys special esteem, above all other sciences, is that its laws are absolutely certain and indisputable, while those of other sciences are to some extent debatable and in constant danger of being overthrown by newly discovered facts.
• As far as the laws of mathematics refer to reality, they are not certain; and as far as they are certain, they do not refer to reality.

--mglg_(talk) 17:10, 21 February 2008 (UTC)

Two blog entries that might be relevant are [2] and [3]. – b_jonas 17:46, 21 February 2008 (UTC)

I think it's possible for a mathematical theory to prove something exists, but for it to be provably impossible to find an example (using a different mathematical theory) without either being inconsistent. For example, you might be able to prove that there is an even number above two that isn't the sum of two primes using one theory, and prove that you could never find an example by using a hypercomputer and the original theory with allowance of infinite steps (in other words, your example, but with an infinite number of variations and an infinitely fast computer). — Daniel 01:38, 22 February 2008 (UTC)

Indeed. For example, ZFC proves the existence of a well-ordering of the real numbers R, but (assuming ZFC is consistent) there is no formula that provably well-orders R. Algebraist 21:00, 22 February 2008 (UTC)

There can't be any rules necessary for basic mathematical reasoning that don't work... i.e. disproven by counterexample. Because then they simply wouldn't be rules! Interestingly enough I've been reading about aspects of mathematics which can be logically seen to be true but not provable with the commonly accepted "axioms" or rules of mathematics (Roger Penrose - The Emperor's New Mind, Oxford University Press). See Gödel's incompleteness theorems! Psymun (talk) 22:56, 22 February 2008 (UTC)

Vandermonde determinant and polynomials

Hi, im having trouble doing this problem from a book on galois theory. Its from the start of the book (Galois theory - Ian Stewart) so wont have much to do with real galois theory, mostly just algebra realated to polynomials. The problem is:

Two polynomials $f,g$ over $\mathbb {C}$ define the same function if and only if they have the same coefficients.

If $a_{1},a_{2}\cdots ,a_{n}$ are distinct complex numbers, then the Vandermonde determinant is defined as $D=\left|{\begin{array}{rrrrr}1&1&1&\cdots &1\\a_{1}&a_{2}&a_{2}&\cdots &a_{n}\\a_{1}^{2}&a_{3}^{2}&a_{3}^{2}&\cdots &a_{n}^{2}\\\vdots &\vdots &\vdots &\ddots \\a_{1}^{n-1}&a_{2}^{n-1}&a_{3}^{n-1}&\cdots &a_{n}^{n-1}\end{array}}\right|$ Then there is a hint which doesnt really help me:

Consider the $a_{j}$ as independent determinants over $\mathbb {C}$. Then D is a polynomial in the $a_{j}$, of total degree $0+1+2+\cdots +(n-1)={\frac {1}{2}}n(n-1)$. Moreover D vanishes whenever $a_{j}=a_{k}$ for some $k\neq j$ as it has 2 identical rows. Therefore D is divisible by $a_{j}-a_{k}$ hence it is divisible by $\prod _{j<k}(a_{j}-a_{k})$ now compare degrees.

Thats as far as i got really, any help would be great, thanks. —Preceding unsigned comment added by 137.205.93.126 (talk) 15:50, 21 February 2008 (UTC)

Huh? I'm confused. The statement 'Two polynomials $f,g$ define the same function if and only if they have the same coefficients (up to multiplication by a constant)' is simply false (multiplication by a constant changes the function, different polynomials can define the same function over finite fields), and I can't see any obvious true statement it could be a misprint of. Moreover it seems to have little to do with the Vandermonde matrix. The one standard problem about the Vandermonde determinant is to show that it is non-zero if the a_i are distinct, and that seems to be what the hint is leading you towards proving. Algebraist 16:18, 21 February 2008 (UTC)

Sorry i didn't explain it well. I seemed to have added the up to multiplication bit for no reason as now i think about it it is wrong, ive taken it out of the question and rewote it exactly as it appears in the book. In the book it gives a different proof using calculus, then goes on to say "For a purely algebraic proof see [this problem]". I had the following idea: consider ${\underline {x}}=(x^{0},x^{1},x^{2},\cdots ,x^{n-1})$, the equation $V{\underline {x}}={\underline {0}}$ (where V is the vandermonde matrix) has a solution if and only if $det(v)=D=0$ so showing the vandermonde determinant is non zero if the coefficients in it are distinct, that is the polynomials in the vector solution of $V{\underline {x}}={\underline {0}}$ have different coefficients. Is this in any way correct? Also if you could give me a little help showing the Vandermonde determinant is 0, that would be helpful too. Sorry if my wording is off, im having alot of trouble understanding this. Thanks. —Preceding unsigned comment added by 137.205.93.126 (talk • contribs)

Unfortunately, I was taught Vandermonde determinants by someone that went far too fast, so I don't fully understand them. I have one general tip, though: A matrix has zero determinant if the columns (or, equally rows) aren't linearly independent - in particular, if one row/column is a scalar multiple of another (that's not necessary, but it is sufficient). --Tango (talk) 18:39, 21 February 2008 (UTC)

Looked in my copy of Stewart's Galois Theory (2nd edition). The only problem I can find that mentions the Vandermonde determinant is Exercise 18.4 - but that is in the penultimate chapter of the book, not near the beginning, and it isn't worded like your question. Is your question from a different edition ? Gandalf61 (talk) 20:17, 21 February 2008 (UTC)

Mine is third edition. it appears on page 28, exercise 2.5*. Nothing about it is mentioned in the preface to the third edition though. —Preceding unsigned comment added by 137.205.93.126 (talk) 22:29, 21 February 2008 (UTC)

Okay. I'm still not clear just what the problem is asking. Is it just asking you to show that

$D=\pm \prod _{j<k}(a_{j}-a_{k})$ ? Gandalf61 (talk) 12:09, 22 February 2008 (UTC)

The start of the question wants you to show that D is nonzero. Then [somehow] it wants you to use that to prove that "Two polynomials $f,g$ over $\mathbb {C}$ define the same function if and only if they have the same coefficients."—Preceding unsigned comment added by 137.205.93.126 (talk) 22:29, 21 February 2008 (UTC)

Hmmm. The sentence "Two polynomials $f,g$ over $\mathbb {C}$ define the same function if and only if they have the same coefficients" seems to me to be a definition rather than the objective of the problem. And you can't show that D is unconditionally nonzero, because D is zero if the a_i are not distinct. If you are not clear about what the question is asking, then you start with a big handicap when trying to solve it. Gandalf61 (talk) 15:37, 22 February 2008 (UTC)

I don't think that's a definition. Conceivably it might be possible to have two different polynomials (i.e., polynomials with different coefficients) that happen to give the same value when evaluated at any $x\in \mathbb {C}$. Stating that two polynomials define the same function if and only if they have the same coefficients is equivalent to stating that this never happens, which would require proof. —Bkell (talk) 17:18, 22 February 2008 (UTC)

Well, if you have to prove this, you look at f-g, which is zero everywhere, and so must be the zero polynomial, so f and g must have identical coefficients. But this statement about identical polynomials has no obvious connection with the Vandermonde determinant (as Agebraist has already said above). I suspect it is part of the previous problem that has been copied in error. Stewart often gives exercises that are lists of statements which the student has to mark as "true" or "false"; this statement feels like part of one of these "true or false" exercises. Of course, if someone has the 3rd edition of Galois Theory to hand, they could check this, and maybe also clarify the whole problem and resolve the original posters confusion. Gandalf61 (talk) 17:51, 22 February 2008 (UTC)

The statement is clearly equivalent to the statement "if a polynomial is zero everywhere, it is the zero polynomial", the intention may be to prove that (it's not always true, it's a property special to polynomials over fields of characteristic 0, so it's far from a trivial statement - it shouldn't be all that hard to prove, however). I do seem to remember a connection between the Vandemonde determinant and checking if functions are equal, but as I said before, those lectures didn't make a lot of sense. --Tango (talk) 19:50, 22 February 2008 (UTC)

The characteristic is irrelevant, actually. The usual proof is that a polynomial of degree n has at most n roots; this works over any infinite integral domain. Algebraist 20:10, 22 February 2008 (UTC)

Over a finite field, the number of solutions can be less than n simply because there are less than n elements in the field. Consider $x^{2}+x\in \mathbb {F} _{2}[x]$. --Tango (talk) 20:58, 22 February 2008 (UTC)

But we are not working in a finite field. We are told that f and g define the same function over C. Therefore f-g is zero everywhere in C. Therefore f-g is a polynomial with an infinite number of roots. Therefore by FTA f-g is the zero polynomial. Maybe the point of introducing the Vandermonde determinant is to prove this without using FTA. Gandalf61 (talk) 22:50, 22 February 2008 (UTC)

I now have the book in my hand. The problem is to show that the Vandermonde determinant is nonzero if the a_i are distinct complex numbers. The hint is as given by the OP. The next question is to use the Vandermonde determinant to show that if a polynomial f(t) over C vanishes at all points of C then it is the zero polynomial. The hint 'Substitute t = 1, 2, 3 ... and solve the remaining system of linear equations of the coefficients' is given. Algebraist 20:18, 22 February 2008 (UTC)

(Outdent) Gandalf61: You don't need the FToA to prove that a degree n polynomial has at most n roots. You can do that (over any integral domain) with the factor theorem and induction. The FT tells us that (counting multiplicity) the poly has exactly n roots. Algebraist 14:41, 23 February 2008 (UTC)

Fastest algorithm to find very smooth numbers close to some number

What is the fastest algorithm that could find smooth integers very near some specified integer. In other words which algorithm has the lowest ration of computation/(smoothness of a found number * smoothness of distance of that smooth number from a specified integer). 128.227.1.24 (talk) 20:30, 21 February 2008 (UTC)Mathguy

How would you compare the following two approximations of 123456789:

• 2·3⁹·5⁵  = 123018750
• 2⁷·3⁹·7² = 123451776

Which is better, and why? Do these two approximations have smoothnesses of, respectively, 5 and 7? If not, how do you measure the smoothness of a number. And what do you mean by "smoothness of distance"? —Preceding unsigned comment added by Lambiam (talk • contribs) 22:52, 22 February 2008 (UTC)

I would prefer the first one because its distance to N has a smaller largest factor. I was looking for a better/faster way to find Integers such that (Number-Integer)*(Integer) are smooth. The fastest method I tried so far is sieving. Dynamic programming subset sum solution is much slower than sieving. —Preceding unsigned comment added by 24.250.129.216 (talk) 00:33, 23 February 2008 (UTC)

How smooth is "smooth"? What's the highest prime factor you'll allow? --Tango (talk) 00:37, 23 February 2008 (UTC)

February 22

Transfinite number

What's the name of a transfinite number that can have any finite number subtracted from or added to it, with the result being different than the original number? For example, ω wouldn't be it because there's no ω-1. ℵ₀ wouldn't be it because ℵ₀+1=ℵ₀. — Daniel 02:08, 22 February 2008 (UTC)

The term "transfinite number" refers specifically to cardinals and ordinals, neither of which have that property. You may be thinking of the nonstandard integers. Black Carrot (talk) 02:21, 22 February 2008 (UTC)

Indeed, cardinals and ordinals are extensions of the natural numbers, not the integers, so subtraction isn't particularly meaningful (except for undoing addition). --Tango (talk) 11:49, 22 February 2008 (UTC)

Any nonstandard model of arithmetic has numbers with this property. The OP might also be interested in the surreal numbers. Algebraist 12:41, 22 February 2008 (UTC)

A non-diagram solution?

Is there a way to solve the following question without drawing a diagram, ie using combinatorials, permutations, etc. only? "How many ways are there to seat two students in a row of four desks so that there's at least one empty desk between them?" Imagine Reason (talk) 20:02, 22 February 2008 (UTC)

Well, I just solved it without drawing a diagram. There are two ways with exactly one space between them (the other empty desk can be on the left or the right) plus one way with exactly two. Algebraist 20:08, 22 February 2008 (UTC)

I drew a diagram in my head. Does that count? (I got either three or six, depending on whether you count switching the students.) Maybe a bigger example would help. Let's say we have 10 people and 100 desks. How many ways are there to arrange the people in the desks so that any two people have at least one empty desk between them? This is indeed a standard combinatorics problem, and can be solved without drawing diagrams. It does require imagining the situation in your head, but no more. Black Carrot (talk) 20:23, 22 February 2008 (UTC)

Let's say we're only worried about which desks have a person in them and which don't. There are 10 people, so there must be at least 9 desks empty - one between each pair of people. In particular, each person but the one on the right end must have an empty desk to their right. So, the question is really, how do we arrange 10 people (each attached at their right hip to an empty desk) in 91 desks? Black Carrot (talk) 20:40, 22 February 2008 (UTC)

Obviously order matters in this case, so the answer to the original question is six. I'm not sure it's a simple combinatorial question, however, because the one person who doesn't have an empty desk attached to the hip cannot be placed directly to the left of another student, but I don't see a simple way of preventing that from happening. Imagine Reason (talk) 01:55, 23 February 2008 (UTC)

All have a hip attachment, without exception. Add an extra empty desk to compensate, so we have for example 10 students with attachments + 91 loose desks = 10 people + 101 desks total. Since the rightmost desk is always empty, there is a bijection with the lawful arrangements in the original 100-desk formulation.  --Lambiam 23:57, 23 February 2008 (UTC)

You can decide which seats will contain students first (for example, by ignoring order), and then afterward decide which order to assign the students to the chosen seats. —Bkell (talk) 06:27, 23 February 2008 (UTC)

Yes, but there's no way of doing that without drawing diagrams in your head, right? Imagine Reason (talk) 18:17, 23 February 2008 (UTC)

Depends on how you think, I guess. What's wrong with diagrams? The diagram is a way to help you translate the problem into mathematics which you can then solve. The fact that you used a diagram to get there doesn't make the maths any less mathematical. --Tango (talk) 21:54, 23 February 2008 (UTC)

Here's the thing about rearrangement. I start by putting the students in some order, maybe alphabetical. Then I let them sit wherever they want, keeping to that order. 9 of the students have empty desks attached to them during that. Once I've counted the options without rearrangement, if there are 10 students, I can multiply by 10 factorial to get the number with rearrangement. I start by worrying only about whether a desk is empty or full, then later worry about who it is that's filling it. Black Carrot (talk) 19:28, 24 February 2008 (UTC)

Changing coins

How many ways can you change a five cent coin (nickel) into one cent coins and/or two cent coins?.

The answer is of course 3 ways. (1+1+1+1+1,1+1+1+2,1+2+2)

But I cannot figure out how many ways there are of changing $100 dollar note (10000 cents) into one cent coins and/or two cent coins. Exhaustive counting is way too exhaustive for me. The only thing I could think of is writing a computer program but I dont think that is how mathematicians in the 19th century would solve this problem. 122.107.226.136 (talk) 22:43, 22 February 2008 (UTC)

Hint: Once you choose how many 2-cent coins to use, the number of 1-cent coins is determined. -- Meni Rosenfeld (talk) 22:53, 22 February 2008 (UTC)

I figure out how to do that. If you have n of n+1 cents ($n\in$ even numbers), then you have ${\frac {n}{2}}+1$ ways to do it. In the question, you have 10000 cents, so you have 5001 ways. Visit me at Ftbhrygvn (Talk|Contribs|Log|Userboxes) 02:49, 23 February 2008 (UTC)

Thanks. Now how many ways are there to change $100 dollar note (10000 cents) into one cent and/or two cent and/or five cent coins. This is even more difficult than just one cent and two cent coins. 122.107.226.136 (talk) 03:43, 23 February 2008 (UTC)

Hint: Once you choose how many 5-cent coins to use, the number of ways of changing the remainder into 1-cent and 2-cent coins is determined. Michael Slone (talk) 03:53, 23 February 2008 (UTC)

More ODE madness

Last question of my assignment and I can't do it at all!

The differential equation is ${\frac {\mathrm {d} y}{\mathrm {d} t}}=y(y-1)$ with initial condition $y(0)=2$.

I've been taught solving by separating variables and also the integrating factor method, but I can't see how to apply either here! One attempt got me 1 = -2 which was most disconcerting! I'm really looking for hints just to get me started.

Thanks all Psymun (talk) 23:40, 22 February 2008 (UTC)

What goes wrong with separating the variables and integrating? At worst, it's integration by parts, and there is probably a better method I'm not seeing (in my defence, it is nearly midnight here). --Tango (talk) 23:54, 22 February 2008 (UTC)

It looks like partial fractions to me. --Tardis (talk) 23:57, 22 February 2008 (UTC)

Oh, yeah... Never did like partial fractions! --Tango (talk) 00:14, 23 February 2008 (UTC)

Incidentally, I just checked the method I used (not int. by parts, that was a terrible idea!) by using partial fractions, and it turns out I actually it right - I used a change of variables y->1/y->1-1/y. --Tango (talk) 00:17, 23 February 2008 (UTC)

Right! That's odd, I haven't been taught it, but it won't do any harm for me to look it up. Why did you change the varibles, Tango? Psymun (talk) 10:47, 23 February 2008 (UTC)

Changing variables is a standard method for integrating and is useful in a wide range of cases. It's not really the best way for this problem, partial fractions is better, but I use change of variables more often so that's what I'm more comfortable with. As for how I chose what to change them to - lucky (educated) guess! --Tango (talk) 14:04, 23 February 2008 (UTC)

OP again. I'm sorry chaps, but I'm still a bit confused as to how I get $y(t)$ here - I can't see how integrating as a partial fraction would work here; I know how to integrate rational functions but I haven't been taught it as a method of solving ODEs! I wanted to try separating but I have no independent variable to separate, nor can I get it in the form $dy/dt+P(t)y=Q(t)$. This question is horrible - but thank you for everyone's suggestions so far! What a wonderful resource - I need to look for some questions I can help with! :S Psymun (talk) 10:47, 24 February 2008 (UTC)

To separate the variables, just divide each side by y(y-1) and multiply both sides by dt. --Tango (talk) 12:19, 24 February 2008 (UTC)

Good lord, you may have just saved my bacon! I'm working through it... Psymun (talk) 14:07, 24 February 2008 (UTC)

Ok!! How does $y=\left({1-{\frac {1}{2}}e^{t}}\right)^{-1}$ sound, noting the initial condition? Psymun (talk) 14:38, 24 February 2008 (UTC)

That's what I got. --Tango (talk) 14:55, 24 February 2008 (UTC)

Hurrah!!!! xDDDDD Psymun (talk) 15:18, 24 February 2008 (UTC)

February 23

Finding basic limits

How does one find this limit:

$\lim _{x\to 9}{\frac {x-9}{{\sqrt {x}}-3}}$

When I enter it into my graphing calculator, I can trace the rational curve and the answer should be somewhere between 5.8 and 6.1. But how do I find the exact value? Thanks. Acceptable (talk) 03:12, 23 February 2008 (UTC)

Hint: $(a-b)(a+b)=a^{2}-b^{2}$. PrimeHunter (talk) 03:19, 23 February 2008 (UTC)

$\lim _{x\to 9}{\frac {x-9}{{\sqrt {x}}-3}}$

$=\lim _{x\to 9}{\frac {({\sqrt {x}}-3)({\sqrt {x}}+3)}{{\sqrt {x}}-3}}$
$=\lim _{x\to 9}{\sqrt {x}}+3$
$={\sqrt {9}}+3$
$=6$
Visit me at Ftbhrygvn (Talk|Contribs|Log|Userboxes) 15:15, 23 February 2008 (UTC)

Ftbhrygvn, your desire to help is appreciated, but we prefer not to give full solutions to what may be homework problems. -- Meni Rosenfeld (talk) 15:24, 23 February 2008 (UTC)

An alternative and lazier way to solve this problem is to apply l'Hôpital's rule. Differentiate the numerator and the denominator, so that you get $\lim _{x\to 9}2{\sqrt {x}}$, substitute x for 9 and you get 6. --Taraborn (talk) 18:17, 23 February 2008 (UTC)

question about functions

can we say that any known function,f(x)=y,satisfies the condition,limit[f(x)\x]→1,when x,either approches to infinity or to afixed value,x0?Husseinshimaljasimdini (talk) 11:44, 23 February 2008 (UTC)

I'm not sure I understand the question. Are you asking if there exists such a function? If so, then yes, f(x)=x has that property. If you're asking if all functions have that property, then no, take f(x)=0, for example. --Tango (talk) 13:58, 23 February 2008 (UTC).

no sir.I mean are the all known functions,f(x)=y.I mean all of the functions,satisfy the above properties?in another word.is there any function doesnot satisfies the above properties?Husseinshimaljasimdini (talk) 10:08, 24 February 2008 (UTC)

Yes, as Tango pointed out, f(x) = 0 does not satisfy your condition. There are infinitely many other examples just using f(x) = c where c is a constant. Using y as a variable has no real meaning in this, the properties are fully determined by f(x). -mattbuck (Talk) 10:22, 24 February 2008 (UTC).

OK OK IGOT IT.ok sir,i got it.now can we say that all the functions except f(x)=0,satisfy the above properties?and one more thing about f(x)=0,dont you think that lim[f(x)\g(x)],when both f(x)&g(x)=0,then the limit has aknown value when x→some point.namely,LOHOPITAL RULE.Husseinshimaljasimdini (talk) 10:28, 24 February 2008 (UTC).thank you indeed.thank all of you guys very much.Husseinshimaljasimdini (talk) 11:00, 24 February 2008 (UTC)

$\lim _{x\rightarrow 0}{\frac {0}{x}}=0$, you can check that by l'Hopital or just by observing that ${\frac {0}{x}}=0,\,\forall x\neq 0$. If you want another example, try f(x)=-x, that also doesn't satisfy your property for any values of x (including infinity). You might like to check out fixed point theorem, it's closely related to what you ask, although not exactly the same. If a function has a fixed point other than 0, your property will be satisfied. --Tango (talk) 12:17, 24 February 2008 (UTC)

alittle bit of help

i read here about that figure which has an infinite surface but afinite volume though.can i get some help to locate the article?Husseinshimaljasimdini (talk) 11:49, 23 February 2008 (UTC)

An example of such a surface is Gabriel's Horn. -- Xedi (talk) 13:50, 23 February 2008 (UTC)

A standard example of the lower dimensional case of a finite area and infinite perimeter is any landmass - assuming the coastline is effectively random at every scale, if you measure it with a smaller and smaller ruler you'll get a larger and larger answer (because you take into account smaller and smaller deviations from a straight line) and in the limit, the answer is infinite. --Tango (talk) 13:56, 23 February 2008 (UTC)

That's a good point, there's more than one way to get it. An abstract example of what Tango said is the Koch snowflake. Similar fractals can be defined in three dimensions, giving something with infinite surface and not only finite volume but bounded extent. Black Carrot (talk) 19:18, 24 February 2008 (UTC)

Help needed in understanding 1925 Biometrika paper

I'm reading the paper Tippet, LHC. On the Extreme Individuals and the Range of Samples Taken From a Normal Population. Biometrika vol 17, 364-387, 1925., temporarily available here.

The paper defines the function

${\overline {w}}(n)=\int _{-\infty }^{+\infty }(1-(1-\alpha )^{n}-\alpha ^{n})\,dx$ ,

where $\,\alpha _{p}$ initially was defined as $\int _{-\infty }^{x_{p}}\phi (x)dx$, where $\int _{-\infty }^{+\infty }\phi (x)dx=1$, i.e. it is a cumulative distribution function, and n is an integer greater than one. The function returns the expected value of the range of a sample of size n, taken from the probability distribution defined by $\,\phi (x)$. If $\,\phi (x)=\Phi (x)$ is the standard normal distribution, w is known as the control chart constant d2 in the field of statistical process control.

I posted a related question recently about this function. I have found that the function

$d_{2}(n)=\int _{-\infty }^{+\infty }(1-(1-\Phi (x))^{n}-\Phi (x)^{n})\,dx$

is easily evaluated numerically, using a naive brute force approach such as the following C++ implementation:

double d2(int n)
{
    double x, dx;
    dx = 0.01;
    double sum = 0.0;
    for (x = -12; x <= 12; x = x + dx)
    {
        double alpha = cumulative_normal_distribution(x);
        sum = sum + (1 - pow(1-alpha,n) - pow(alpha,n))*dx;
    }
    return sum;
}

I know that there are far more efficient and elegant ways of evaluating integrals numerically. Nevertheless, this implementation works, and mirrors my simple-minded mental image of what is going on, so I'll stick with it for now. Tippett then defines as equation (10) the following:

$\,\mu _{2}=\sigma ^{2}=2\int _{-\infty }^{+\infty }\int _{-\infty }^{x_{1}}(1-\alpha _{1}^{n}-(1-\alpha _{n})^{n}-(\alpha _{1}-\alpha _{n})^{n})dx_{1}dx_{n}-{\overline {w}}^{2}$

and states (page 372) that

"The second moment, and hence the standard deviation of the distribution, was determined in several cases by equation (10). The work is very laborious, as it involves cubature, and even so, the result can only be given to a few figures. It is believed that the values given in Table IV are correct to the last figure."

The return values square root of $\,\mu _{2}$ should correspond to the control chart constant d3, and Table IV confirms that this is indeed the case. My problem is that I haven't been able to evaluate $\,\mu _{2}$, and I suspect I've misunderstood something, possibly something pretty elementary about double integrals (I have no mathematical training beyond this, and it's a long time ago). Here's my implementation, again using a similar brute force approach. I've highlighted in red the part that I was most unsure of:

//NOTE: Comments in green added by OP after the question was posted, based on Meni Rosenfeld's answers.

double d3(int n)
{ 
    double x_1, x_n, dx_1, dx_n;
    dx_1 = dx_n = 0.01;
    double sum = 0.0;
    for (x_n = -12; x_n <= 12; x_n = x_n + dx_n)
    {
        double alpha_n = cumulative_normal_distribution(x_n); // WRONG - should be doing alpha_1 in the outer loop!
        // Making x_1 (-inf ... +inf) the outer loop, and x_n (-inf ... x_1) the inner loop fixes the problem.
        for (x_1 = -12; x_1 <= x_n; x_1 = x_1 + dx_1) // WRONG - see the code in red: x_1 should be the limit of integration. 
        {
            double alpha_1 = cumulative_normal_distribution(x_1); // WRONG - should do alpha_n in the inner loop!
            sum = sum + (1 - pow(alpha_1,n) - pow(1 - alpha_n,n) - pow(alpha_1 - alpha_n, n))*dx_1*dx_n;
                    // As Meni pointed out there's an error here ^ . The minus sign should be plus.
        }
    }
    double variance = 2*sum - pow(d2(n),2);
    return sqrt(variance);
}

The return values of this function are way off the correct values. My question is: can someone spot the misunderstanding(s) in my interpretation of the paper, the maths, and/or the error(s) in my implementation? Your help would be much appreciated. Thanks in advance, --NorwegianBlue ^talk 16:13, 23 February 2008 (UTC)

I don't see a problem with the way you are calculating the double integral. However, there is something terribly wrong about the function you are trying to integrate. To explain, I'll introduce some notation:

$a(n,x_{1},x_{n})=1-\Phi (x_{1})^{n}-(1-\Phi (x_{n}))^{n}-(\Phi (x_{1})-\Phi (x_{n}))^{n}\;\!$

$b(n,x_{1})=\int _{-\infty }^{x_{1}}a(n,x_{1},x_{n})\ dx_{n}$

$c(n)=\int _{-\infty }^{\infty }b(n,x_{1})\ dx_{1}$

$d_{3}(n)={\sqrt {2c(n)-d_{2}(n)^{2}}}$

In order for any of this to be meaningful, b must exist for any $x_{1}$. For this to happen, at the very least we must have $\lim _{x_{n}\to -\infty }a(n,x_{1},x_{n})=0$ for every $x_{1}$. This is clearly not the case, because $\Phi (-\infty )=0$ and so $a(n,x_{1},-\infty )=1-2\Phi (x_{1})^{n}$. Thus the function is not integrable. Check that you have copied the equations exactly as they appear in the paper, and that the notation means exactly what you think it does; If so, there is possibly a mistake in the paper. -- Meni Rosenfeld (talk) 12:10, 24 February 2008 (UTC)

Hmm. Well, I've checked the equations, and believe my reproduction is correct. I also very much doubt that there is a mistake in the paper, as it was widely cited, and equation (10) was used by subsequent authors to tabulate d3. Therefore, the most likely explanation is that I've misunderstood the notation. I prepared a small .PNG file from the paper with the relevant equations and a figure which explains the notation. x₁ refers to the highest value in a sample of n values, and x_n refers to the lowest, as is shown in the figure. The .PNG is here, the paper is here. Thanks again! --NorwegianBlue ^talk 13:20, 24 February 2008 (UTC)

Note the step from equation (9) to (10), where $+(\alpha _{1}-\alpha _{n})^{n}\;\!$ turns into $-(\alpha _{1}-\alpha _{n})^{n}\;\!$. The plus is correct, and everything falls into place nicely if you make this correction. So it turns out even widely cited papers can have errors. Also, the table you linked to has some errors, for example $d_{2}(1)=0\neq 1$. -- Meni Rosenfeld (talk) 14:36, 24 February 2008 (UTC)

Thanks a lot! The error in the step between equations (9) and (10) was glaringly obvious, once I became aware of it. I've looked through the steps from equation (4) to (9) for more errors, but unfortunately, most of it exceeds my mathematical capabilities. I'm also confused about the notation. What does the S-notation mean? Is appears in places which might suggest summation, is it the same as sigma notation?

Unfortunately, everything did not fall nicely into place. I corrected the minus sign to plus, but the function still does not work correctly. The entire program I've used is here. The section of the paper immediately preceding equation (10) might suggest that equation (10) only works for even values of x. Here's what I found, with the correction applied:

            n   program   correct
            ---------------------
            2    23.94    0.8525
            3     8.65    0.8884
            4    23.90    0.8794
            5    10.01    0.8641
            6    23.86    0.8480
            7    10.70    0.8332
            8    23.82    0.8198
            9    11.15    0.8078

Even values slowly decreasing from 23.84, odd values increasing from 8.65. Strange. A standard deviation of >23 for the range of samples of n=2 drawn from a standard normal distribution is obviously not right! Any suggestions of what else might be wrong? --NorwegianBlue ^talk 16:58, 24 February 2008 (UTC)

Well, there aren't any problems with the mathematics. I have calculated the sum in Mathematica and it gives the correct result. A step of 0.01 is enough to get a few correct digits, and an integration bound of 12 is overkill - 5 will be more than enough. So you need to debug your code - it looks generally okay but I don't speak C++ fluently. Try to define intermediate functions as I have done above, calculate some values of them, and either check them for plausibility or put them here so I can have a look.

The S notation is unfamiliar to me - it is probably, indeed, an archaic version of the Sigma notation. -- Meni Rosenfeld (talk) 17:40, 24 February 2008 (UTC)

Or start throwing cout to test intermediate values of variables in between every single line. x42bn6 Talk Mess 19:49, 24 February 2008 (UTC)

Thanks a million, Meni!!! I implemented your a, b, c and d₃ functions exactly as you wrote them, and the result was reasonably close to the correct one (overestimated it slightly). It turned out that the step size was important. The smaller I made it, the closer I got to the values listed as correct above. I was able to achieve 3 digits using a step size of 1/2048.

So I went back to the original function, and compared the code, to see why it behaved differently. The problem was related to what I had highlighted in red. When I made x_1 the outer loop, and x_n the inner loop, and changed the test in the inner loop to x_n <= x_1, the modified original function gave the same results as the a, b, c and d₃ functions. Thanks again, this was beginning to drive me nuts! By the way, did Mathematica give results with high accuracy? I'm particularly interested in a good approximation of d₃(2). --NorwegianBlue ^talk 20:23, 24 February 2008 (UTC)

Glad to be of assistance, and sorry for not picking up earlier that it should be $x_{n}\leq x_{1}$. Mathematica can give arbitrarily good accuracy, but since this is a double integral it can take some time. $d_{3}(2)=0.8525024664$ should be correct up to the digits specified. I'm running it now for more digits in case you'll need it. A few additional values (n=3 and up) are 0.888368, 0.879808, 0.864082, 0.84804, 0.833205, 0.819831, 0.807834, 0.797051. -- Meni Rosenfeld (talk) 22:04, 24 February 2008 (UTC)

Thank you again for taking the time. You have helped me tremendously, both in increasing my understanding of the maths underlying SPC, and in solving a tough practical problem. --NorwegianBlue ^talk 15:52, 25 February 2008 (UTC)

Gah! I looked at this yesterday, but posted nothing because I didn't see that ordering problem. Shame on the original paper (not the OP!) for writing a double integral as $\int _{a_{1}}^{a_{2}}\int _{b_{1}}^{b_{2}}f(a,b)\,da\,db$! Treating a differential (infinitesimal) as a first-class entity is fine, but writing the normal definite integral operation without nesting it properly leads to madness, since the $\int$ doesn't get a variable name attached to it. --Tardis (talk) 16:39, 25 February 2008 (UTC)

As stated above, my mathematical training is limited, so I need to be spoonfed :-). You are saying that my approach, i.e. making the inner integral the inner loop, would normally be the way to go, right?

Would

$\,\int _{-\infty }^{+\infty }\int _{-\infty }^{x_{1}}(1-\Phi (x_{1})^{n}-(1-\Phi (x_{n}))^{n}{\color {red}+}(\Phi (x_{1})-\Phi (x_{n}))^{n})dx_{n}dx_{1}$

be a more standard way of writing the integral? I was a bit confused about this, and that's why I flagged the x_1<=x_n test (which should have been x_n<=x_1) in the code. --NorwegianBlue ^talk 18:54, 25 February 2008 (UTC)

Yes. We are essentially calculating here the integral of a function which is itself an integral. The inner function is $\int _{-\infty }^{x_{1}}(\cdots )\ dx_{n}$, and thus it should appear as inner in the formula and be implemented as inner. We then calculate the integral of this, $\int _{-\infty }^{+\infty }(\cdots )\ dx_{1}$, thus the formula should be $\int _{-\infty }^{+\infty }\int _{-\infty }^{x_{1}}(\cdots )\ dx_{n}\ dx_{1}$. This would be even more critical if we wanted to calculate, say, $\int _{-\infty }^{+\infty }\int _{-\infty }^{0}(\cdots )\ dx_{n}\ dx_{1}$ - we need to know which is the variable that only goes up to 0, and flipping the order changes this. In our integral, we can understand from the context that $\int _{-\infty }^{x_{1}}$ relates to $x_{n}$ (it wouldn't make much sense for $x_{1}$ to have itself for an upper integration bound). In the implementation, it doesn't really matter in which order you make the loops, since you can change the order of integration as long as you keep the bounds right. The $\int _{-\infty }^{x_{1}}$ bit clearly implies that the bound is $x_{n}\leq x_{1}$, and it's unfortunate it took us so long to notice that. -- Meni Rosenfeld (talk) 23:14, 25 February 2008 (UTC)

Thanks again, and thanks to Tardis for making me aware of exactly why I tripped. I've learned a lot from this thread! --NorwegianBlue ^talk 10:05, 26 February 2008 (UTC)

Mathematics(Algebra)

Railways authorities increased the fare by 6.25%,hence the sale of tickets was decreased by 5%. Even then the income by selling tickets was increased by Rs 15000.Find the original income by selling of tickets?18:07, 23 February 2008 (UTC)117.195.16.73 (talk)

Welcome to the Wikipedia Reference Desk. Your question appears to be a homework question. I apologize if this is a misevaluation, but it is our policy here to not 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 how to solve such problems. Please attempt to solve the problem 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. Thank you. -- Meni Rosenfeld (talk) 18:22, 23 February 2008 (UTC)

Your first step is to formulate the problem in terms of equations. Give each unknown value a symbol (eg the original fare could be F_o and the new fare F_n, etc.), then rewrite all the information in the question as equations using the symbols. Once you've done that, it's just a matter of solving the equations. --Tango (talk) 19:16, 23 February 2008 (UTC)

February 24

derivatives and second derivatives of rotation functions

hi, how do i find the derivative and second derivative of a function that rotates counter clockwise an angle theta? do i just take the derivative and second derivative of (-sin+cos, cos+sin)?

also, how might one prove that a harmonic function composed with a function that preserves dot product is still harmonic? thanks —Preceding unsigned comment added by 199.74.71.147 (talk) 03:33, 24 February 2008 (UTC)

For the first question I assume you mean rotation around the origin in the Euclidean plane, using the mapping

$(x,y)\mapsto (x\cos \theta -y\sin \theta ,x\sin \theta +y\cos \theta )\,.$

To get the n-th derivative with respect to the rotation angle θ, just work out

$(x,y)\mapsto {\frac {d^{n}}{{d\theta }^{n}}}(x\cos \theta -y\sin \theta ,x\sin \theta +y\cos \theta )\,,$

where a pair is differentiated coordinate-wise:

${\frac {d}{d\theta }}(X(\theta ),Y(\theta ))=({\frac {d}{d\theta }}X(\theta ),{\frac {d}{d\theta }}Y(\theta ))\,.$

Using matrix–vector notation, the mapping can be written as

${\begin{bmatrix}x\\y\end{bmatrix}}\mapsto M(\theta ){\begin{bmatrix}x\\y\end{bmatrix}}\,,$

where the 2×2 rotation matrix M(θ) is given by:

$M(\theta )={\begin{bmatrix}\cos \theta &-\sin \theta \\\sin \theta &\,\,\,\,\cos \theta \end{bmatrix}}\,.$

Here you have to find ^d⁄_dθM(θ), which can be done entry-wise:

${\frac {d}{d\theta }}M(\theta )={\begin{bmatrix}{\tfrac {d}{d\theta }}\cos \theta &-{\tfrac {d}{d\theta }}\sin \theta \\{\tfrac {d}{d\theta }}\sin \theta &\,\,\,\,{\tfrac {d}{d\theta }}\cos \theta \end{bmatrix}}\,.$

The second question is not clear. What does it mean for a function to preserve dot products? Normally you expect that to mean f(u·v) = f(u)·f(v), but the dot product turns two vectors into a scalar, so if u and v are vectors, on the left-hand side f is operating on a scalar and on the right-hand side on vectors. How can this be? Also, is it given in which order the harmonic function is composed with the dot-product preserving function? An example might help to clarify this.  --Lambiam 06:26, 24 February 2008 (UTC)

Probably means orthgonal, u·v = f(u)·f(v). With some very minor condition (possibly none), an orthogonal function is an orthogonal matrix. Then it is just asking to show that the laplacian is invariant under orthogonal coordinate changes, which is definitely a very common exercise to give (so likely to be his question). JackSchmidt (talk) 06:44, 24 February 2008 (UTC)

that probably is my question, but i am unfamiliar with most of the terminology you used as my professor is rather scatterbrained in lecture. could you please give me some pointers on where to start if i were to prove this? thanks! —Preceding unsigned comment added by 199.74.71.147 (talk) 08:04, 24 February 2008 (UTC)

Basically you use chain rule. If g:R^2->R is harmonic and f:R^2->R^2 preserves dot products, then f is actually given by an orthogonal matrix (which looks almost exactly like a rotation matrix). Define h:R^2->R by h(x) = g(f(x)). Write out what that means fairly explicitly in terms of the matrix entries of f (just label them f11, f12, f21, and f22; don't use cosines in my humble opinion). Then use chain rule to take the first and second derivatives. The derivatives will involve dot products of rows of f. f is orthogonal, so the dot products will be 0 for the mixed derivatives of g, and 1 for the repeated derivatives of g. In other words, lap(h) evaluated at x is equal to lap(g) evaluated at f(x). JackSchmidt (talk) 08:17, 24 February 2008 (UTC)

how would one avoid using cosines? and also, where do dot products come into the derivatives? I don't see them199.74.71.147 (talk) 20:22, 24 February 2008 (UTC)

(Sorry, I don't have an easy way to explain this; my comments have been meant for other more analytic types to help you out; same continues here). See directional derivative for the dot products. Applying f just changes the direction of a directional derivative. The laplacian is defined in terms of partials, but you get the same definition if you use directional derivatives in directions which are just a rotation of the previous ones. Simply writing it out and taking calc 1 derivatives suffices, it just takes a page. I recommend that method. JackSchmidt (talk) 20:47, 24 February 2008 (UTC)

can you explain a little more about how i get from taking the derivative to having dot products?

There is a dot product in the definition of a directional derivative, is that the one you're looking for? --Tango (talk) 11:44, 25 February 2008 (UTC)

Integral Powers of Trinomials

For expanding binomials, we have the binomial expansion which gives us all the info we need to know about expanding a binomial to (any) power. For example, if I want to find out what is the full expansion of $(x+y)^{25}$, I can do so. My question is, is there a similar expansion formula for $(x+y+z)^{25}$. Is there a shorthand way to write this expansion in terms of the coefficients of the expansion?A Real Kaiser (talk) 05:53, 24 February 2008 (UTC)

Memory of high school maths a bit hazy, but you can break $(x+y+z)^{25}$ into $((x+y)+z)^{25}$, which gives $\sum _{k=0}^{n}{C_{k}^{n}}(x+y)^{n-k}z^{k}$, and then expand $(x+y)^{n-k}$ binomially.

This gives something like $\sum _{k=0}^{n}\sum _{l=0}^{n-k}{C_{k}^{n}}{C_{l}^{n-k}}x^{l}y^{n-k-l}z^{k}$ --PalaceGuard008 (Talk) 06:19, 24 February 2008 (UTC)

You are looking for the Multinomial theorem. You are basically counting how many ways to rearrange "mississippi", or other such combinatorial questions. JackSchmidt (talk) 06:26, 24 February 2008 (UTC)

The coefficient of $x^{i}y^{j}z^{k}$ in the expansion of $(x+y+z)^{n}$ is

${n \choose i,j,k}={\frac {n!}{i!\,j!\,k!}}$

Pascal's pyramid contains the coefficients for the trinomial expansion - it bears the same relation to the trinomial expansion as Pascal's triangle does to the binomial expansion, but it has an extra dimension. So the coefficients of $(x+y+z)^{25}$ are the numbers in the 25th layer of Pascal's pyramid. Gandalf61 (talk) 11:26, 24 February 2008 (UTC)

Thanks guys, I had no idea that a generalization of Pascal's triangle existed.A Real Kaiser (talk) 01:29, 25 February 2008 (UTC)

Curiously enough, I independently discovered this relationship over a year ago. I found it a lot less obvious than Pascal's Triangle. A math-wiki (talk) 07:06, 27 February 2008 (UTC)

Vibrational modes of a drum

I'm interested in creating a few images for Wikipedia of the vibrations of a drum membrane.

It appears these vibrations are related to the Bessel functions - searching on google, I didn't find any precise descriptions of the phenomenon. Somes pages appeared helpful though (for example : Standing waves in a drum membrane...).

So, could anyone explain how it works, and what the equations describing the vibrations are ?

Thanks. -- Xedi (19:25, 24 February 2008 (UTC))

Well, I don't know your mathematical background (have you studied PDEs?), but the simplest thing to do is assume that the membrane is isotropic and homogeneous, and the waves are small enough that it responds linearly. Then you have to solve the wave equation (whose spatial part after separation of variables is the Helmholtz equation) by finding eigenfunctions of the Laplacian for whatever shape your drum is. The solutions for a circular drum do involve Bessel functions. See also Hearing the shape of a drum. —Keenan Pepper 23:13, 24 February 2008 (UTC)

If you want a textbook that discusses this stuff, try Elementary Applied Partial Differential Equations with Fourier Series and Boundary Value Problems by Richard Haberman. —Keenan Pepper 23:17, 24 February 2008 (UTC)

Thanks a lot, the article on the Helmholtz equation is pretty much what I needed. (I'll also have a look at the book) -- Xedi (10:08, 25 February 2008 (UTC))

February 25

1-1

let a,b positive integers.

Is

f(a,b)= 3^a + 10^b

1-1? —Preceding unsigned comment added by 71.97.4.194 (talk) 00:46, 25 February 2008 (UTC)

If you mean one-to-one, then yes, I think it is. Black Carrot (talk) 01:11, 25 February 2008 (UTC)

g(a,b) = 2^a + 3^b fails to be 1-1. g(3,1)=11=g(1,2). It seems like we could do something similar for f, but the numbers get too big. —Preceding unsigned comment added by 71.97.4.194 (talk) 01:25, 25 February 2008 (UTC)

If 3^a+10^b=3^c+10^d, then 3^a-3^c=10^d-10^b. Investigate the properties of differences between powers of 3 and differences between powers of 10. Does that help? —Bkell (talk) 04:48, 25 February 2008 (UTC)

Yes, I tried that, to no avail (yet). It did make me think of using the pigeonhole principle. *How*, I'm not quite sure. 192.91.253.52 (talk) 13:20, 25 February 2008 (UTC)

I was unable to solve the problem, though I approached it in the way Bkell suggested. If a>c then d>b, so I factored it as 3^c*(3^(a-c)-1) = 10^b*(10^(d-b)-1). You get congruences like a=c mod OrderMod( 3, 10^b ), and b=d mod OrderMod( 10, 3^c ) so if b or c is nonzero, this reduces the possible solutions, and in fact you can apply the same argument again with much larger moduli. However, induction does not quite work, because I could not rule out b and c being zero (especially not inductively). When the problem is changed to the g(a,b), in fact very early there is a solution with b or c being 0, g(2,0)=5=g(1,1). While one might decide the arguments are not allowed to be 0 for some reason, during the induction you have to allow that case. At any rate, there are no small solutions for f (0<= a,b,d <= 1000). JackSchmidt (talk) 15:43, 25 February 2008 (UTC)

The original problem states that the exponents are positive integers. —Bkell (talk) 16:33, 25 February 2008 (UTC)

My idea, while not fully fleshed out, is that a difference between powers of 3 is always going to be a string of 0's and 2's when written in base 3, and a difference between powers of 10 is always going to be a string of 0's and 9's when written in base 10. I think there's a contradiction lurking somewhere: it feels like it's going to involve a claim that in order to write a difference between powers of 3 as a sum of things that look like 9000…0 in base 10, you're going to have to use a particular 9000…0 twice. Sorry for the roughness of this idea; it's really just a hunch and needs a lot of work still. —Bkell (talk) 16:40, 25 February 2008 (UTC)

Consider this, if 3^a+10^b=3^c+10^d, then you can rewrite it in most cases as 3^f=((10^h-1)*10^g)/(3^j-1), which must be an integer. Is it possible for this to be an integer, though? (hint: look at 10^h-1 and 3^j-1 under the proper modulus). GromXXVII (talk) 19:33, 25 February 2008 (UTC)

Sure it can be an integer. For example, g=3, j=2 makes ((10^h-1)*10^g)/(3^j-1) an integer for every nonnegative h. JackSchmidt (talk) 19:42, 25 February 2008 (UTC)

I haven't come up with a good way to prove it yet, but I have another direction you might go in. If you write down a given power of 3, the question is whether you can subtract 1 from one digit, and add 1 to another, and get back a different power of 3. If you consider it base 3, the question is whether given a power of 101, you can subtract 1 from one digit and add 1 to another and get back another power of 101. Without carry, powers of 101 are pascal's triangle, which gives a clear enough pattern that it may be possible to prove it false. Black Carrot (talk) 20:50, 25 February 2008 (UTC)

I think I've got it. Take it mod 5, 4, and 20, in that order. The first shows that a=c(mod4), the second shows that that's only possible if either b or d is 1, and the third shows that that can't happen. It's one-to-one. Black Carrot (talk) 07:22, 26 February 2008 (UTC)

Unfortunately, simply looking mod finitely many numbers cannot prove the result. For instance, it is possible for f(a,b) = f(c,d) mod 4, mod 5, and mod 20 (simultaneously), without a=c, b=d: more explicitly, f(1,2) = f(9,10) mod 4, mod 5, and mod 20. JackSchmidt (talk) 16:45, 26 February 2008 (UTC)

Approximation by Smooth Functions

Basically, our teacher is trying to convince us that $C_{0}^{\infty }(\Omega )$ which is the set of all infinitely differentiable functions with compact support on a given set $\Omega$, is dense in $L^{2}(\Omega )$. Now, I know that one of the definitions of a set A being dense in another set B is that the closure of A must be B. Which is the same as saying that B contains A and all of its limit points. So he gave us a theorem in class saying that, for any $f\in L^{2}(\Omega )$, there exists a sequence $\phi _{n}\in C_{0}^{\infty }(\Omega )$ such that $\phi _{n}\rightarrow f$ in $L^{2}(\Omega )$. My question is how does this theorem prove that $C_{0}^{\infty }(\Omega )$ is dense in $L^{2}(\Omega )$? All we have done is shown that every square integrable function f is a limit of a sequence of smooth functions with compact support. We have shown that each f is a limit point but how do we know that those are all the limit points? What if there is a limit point of smooth functions with a compact support outside $L^{2}(\Omega )$?A Real Kaiser (talk) 04:52, 25 February 2008 (UTC)

$L^{2}(\Omega )$ is a closed set - it contains its own limits. Bo Jacoby (talk) 05:36, 25 February 2008 (UTC).

Duh, I seem to have forgotten that little fact. Thanks!A Real Kaiser (talk) 05:52, 25 February 2008 (UTC)

The above comment obscures the point at hand. It is not enough to talk about a set being open or closed; you have to specify the topology you are talking about. Every set is closed in its own subspace topology. Here, the question is whether or not the closure of $C_{0}^{\infty }(\Omega )$ in the metric topology of $L^{2}(\Omega )$ is $L^{2}(\Omega )$. In this topology, it is impossible for the closure of $C_{0}^{\infty }(\Omega )$ to be bigger than $L^{2}(\Omega )$. Perhaps the above post was referring to the fact that $L^{2}(\Omega )$ is complete (is this true actually? ahh...), which would mean that however you embed $L^{2}(\Omega )$, the closure of $C_{0}^{\infty }(\Omega )$ will be exactly $L^{2}(\Omega )$. However, that would not be a requirement for saying that the one is dense in the other. If $A\subset B\subset C$, to say that A is dense in B is to say that the closure of A, in B's subspace topology, is all of B. It would not matter if B were not closed in C, because in the B subspace topology, the closure of A cannot exceed B. (Can someone with more experience please check my post for correctness? Thanks...) J Elliot (talk) 06:08, 25 February 2008 (UTC)

$L^{2}(\Omega )$ is a normed space, the norm being the integral of the absolute square. This norm defines the topology. In this topology the set $L^{2}(\Omega )$ is closed. A function outside $L^{2}(\Omega )$ has no well defined distance to a a function within $L^{2}(\Omega )$ , so it is not a limiting function of a sequence of functions in $L^{2}(\Omega ).$ Bo Jacoby (talk) 14:15, 25 February 2008 (UTC).

The sentence "In this topology [on a space X] the set X is closed" is one of the axioms of a topology, and thus is never a meaningful observation. As was noted above, the only meaningful thing that we can say in this direction (and indeed it is very meaningful) is that L²(Ω) is a normed space (and thus more generally a uniform space), and with respect to this structure L²(Ω) is complete. In particular, this implies that whenever any metric space V is isometrically embedded into L²(Ω) (we might take V=C^∞₀(Ω) with the induced square-integral norm), then any Cauchy sequence in V has a limit in L. Thus it is the completeness of L²(Ω), rather than the meaningless "closedness", which implies that the completion of C^∞₀(Ω) [with the L²-norm] is L²(Ω). This is not necessary to show that C^∞₀(Ω) is dense in L²(Ω), however. Tesseran (talk) 05:26, 27 February 2008 (UTC)

Kaiser, you should take some algebra classes next semester, because my analysis books had been getting nice and dusty before you started posting this semester. :)

BTW, you may seriously want to prove that if f_n are C^oo with compact support, and they form a cauchy sequence under the L^2 norm (so they "converge to something"), then there is some number C such that the L^2 norm of all f_n is bounded by C. In other words, if something is an L^2 limit of compact functions, then it is L^2. This should be easy to prove, other than the fact that it is posed randomly on the net. Once you do this, some of the previous posters talking about completeness should make more sense.

This idea is actually pretty important, even though some people might think my question is silly. The point of using the L^2 norm is to control the L^2 norms of limits. As long as everything in the sequence was L^2, and the sequence converges (is cauchy) according to the L^2 norm, then the limit is L^2. This is why you need to use the Sobolev norms in PDE. You want to take a sequence of approximate solutions to a PDE (or solutions to some approximately equal PDE), and know that the limit is still a solution. If you use L^2 for that, you will often fail, but if you use H^1 (L^2 of the function and its derivative), you will have much better luck, since the limit function at least has an L^2 derivative. JackSchmidt (talk) 15:57, 25 February 2008 (UTC)

I think, Kaiser, that your original post misses the point a bit. B containing A and its limits is nothing like sufficient for A being dense in B (consider $A=[0,1],B=\mathbb {R}$). Instead, what you want is that $\forall x\in B\;\exists \{y_{n}\}\subseteq A\;\lim _{n\rightarrow \infty }\|y_{n}-x\|=0$ (with some metric appropriate to B, which is where topology comes in). The theorem you were given establishes precisely this fact. --Tardis (talk) 16:08, 25 February 2008 (UTC)

Jack, you know what, you are absolutely correct. I am going for a degree in Applied Math and I incorrectly assumed (in my early childhood as a lower division Math student) that Applied Math and Pure Math were two disjoint sets. So I decided to stay away from Pure courses which is exactly why now I have huge holes in my understanding of real, complex, functional analysis, and algebra. I was already thinking about an algebra course next semester but it seems vital that I must take it. Well, lol, don't put those books away. This doesn't mean that I am going to stop asking questions. There are plenty more where these come from. I know that all of you don't have to help me but I really appreciate it. Thanks everyone! :)
Ok, so let me recap. L^2 is the space of all square integrable functions. It has a norm defined on it (the usual L^2 norm). It has an inner product defined on it and it is complete with respect to this norm. Therefore L^2 is a Hilbert space. Being complete means that any Cauchy sequence of L^2 functions converges to an L^2 function. So there are no limit points of L^2 that are outside L^2. Furthermore, if I want to show that the set of all smooth functions (members of $c^{\infty }$) is dense in L^2, then it will suffice to show that if I take any square integrable function, I can always find a sequence of smooth functions that converges to that square integrable function. This is the same as showing that any square integrable function can be approximated arbitrarily well by smooth functions. It is also not possible for a sequence of smooth functions to converge outside the L^2 space because smooth functions are a subset of square integrable functions and the set of all square integrable functions is closed. Is this correct, guys? I want to make sure that my line of reasoning is correct.A Real Kaiser (talk) 20:04, 25 February 2008 (UTC)

GMAT data sufficiency problem

A homeowner must pick between paint A, which costs $6 per liter, and paint B at $4.50 per liter. Paint B takes one-third longer to apply than paint A. If the homeowner must pay the cost of labor at $36 per hour, which of the two paints will be cheaper to apply?

(1) The ratio of the area covered by one liter of paint A to that covered by one liter of paint B is 4:3.

(2) Paint A will require 40 liters of paint and 100 hours of labor.

I first encountered the problem in Arco Math Workbook (2000). Kaplan's 2005 version is similar, but it's got the same problem and explanation, which I cannot agree with:

"Statements (1) and (2) taken together are not sufficient to answer the question. To make an intelligent decision, we need to know which requires more paint and how much more, how long each will take, and we need some info on their labor costs...Using both statements together, we still cannot find the labor costs."

It seems rather clear to me how to find the labor costs.

First, the two statement taken together mean that it'd take 40*4/3 or 160/3 L of paint B.

Second, statement (2) says that paint A is applied at a rate of 40L per 100 hours, or 2/5 L/hr. - but the original problem says it would take a third longer for paint B, which means that paint B is applied at a rate of 2/5 * 3/4 L/hr. or 3/10 L/hr.

This last conversion was the most difficult for me, but you can see it's just multiplying A's rate by 4/3 hr/L.

Thus, it will take (160/3) / (3/10) or 1600/9 hours to use the 160/3 L of paint B to cover the house.

Therefore we know how long it takes. Given the unit labor cost in the original problem, we can figure out how much labor costs for either paint. We also know how much paint we need and how much it costs. Why then would the problem be unsolvable so long as we have statements (1) and (2)?

Imagine Reason (talk) 07:26, 25 February 2008 (UTC)

(1) and (2) on their own aren't enough, you need to information in your first paragraph as well - is that what it means? --Tango (talk) 11:42, 25 February 2008 (UTC)

The first paragraph is a given, while the two statements are optional. I'm saying that if we have the two statements (the first paragraph is always there), we can solve the problem, but the review books say we can't even then. —Preceding unsigned comment added by Imagine Reason (talk • contribs) 17:14, 25 February 2008 (UTC)

Well, it's hard to comment on what the review book says without seeing it, but the problem you've posted is certainly solvable in the way you came up with. Black Carrot (talk) 06:32, 26 February 2008 (UTC)

Clarity

Working in dollars throughout...

Focusing on the given paragraph ONLY... We assume that equal amounts (L) of paint in litres are required no matter whether paint A or paint B is used and that if H hours are needed for paint A, then 4H/3 hours are needed for paint B.

Total cost of using paint A is 6L + 36H and the total cost of using paint B is 4.5L + 48H

Setting these equal to each other gives 6L + 36H = 4.5L + 48H and a solution of L = 8H

Therefore numerically, if L > 8H then paint B is cheaper and if L < 8H then paint A is cheaper.

Conclusion is that additional information is needed.

Incorporating ONLY statement 1 with the given paragraph... We need to make an assumption that the reduced volume of paint A needed (3/4 of the volume required by using paint B) results only in a cost saving on the paint, and NOT that less paint equals less hours of labour: after all, the same area needs covering and we know it takes a third longer with paint B. If this assumption is false and that less paint ALSO equals less labour hours then we have a slightly different problem.

Relative to the L litres required for paint A, paint B requires 4L/3 litres which now costs 6L (dollars).

Case 1.1 Less paint only required.

The expression for the cost of paint A remains 6L + 36H which in all cases is less than the cost for paint B of 6L + 48H.

Paint A is cheaper.

Case 1.2 Less paint and 3/4 of the previous amount of labour hours required (4/3 times the amount are now needed for B compared to A), i.e. relative to the H hours required for paint A, paint B requires 4/3 × 4/3 = 16/9 times the hours (cost = 36H × 16/9 = 64H).

The expression for the cost of paint A remains 6L + 36H which in all cases is less than the cost for paint B of 6L + 64H.

Paint A is even cheaper.

Statement 2 is irrelevant in determining which paint is cheaper to apply; it just gives the information to calculate the ACTUAL costs. All the problem requires though is a general 'which is cheaper', not by how much. So, as far as I can read it from the information provided, choosing paint A is cheaper in all cases if statement 1 is included with the given paragraph; statement 2 being irrelevant, as is the request for any additional information.

However, all this is blindingly obvious from the simple point that statement 1 means that the effective price of paint B is $6.00 per litre, identical to that of paint A. B takes longer to apply at an additional cost of $12 per hour for case 1.1 and $28 per hour for case 1.2 above based on the number of hours needed to apply paint A.

The only ambiguity is in calculating ACTUAL costs for paint B depending on case 1.1 or 1.2 above. Paint A costs $3840: Paint B either costs $5040 or $6640.

Please don't hesitate to contact me if you feel I have made any error in reading this situation! AirdishStraus (talk) 11:23, 26 February 2008 (UTC)

I agree that the wording is ambiguous, but I'm pretty sure it is saying that a gallon of paint B takes a third longer to apply than a gallon of paint A. Regardless, you're right--we only need the first statement to answer the question. I think that's actually what I got the first time I saw the problem, actually, but somehow I let it go. Now I've seen the same nonsensical explanation in a second book, I'm not happy. Imagine Reason (talk) 23:36, 26 February 2008 (UTC)

aquestion about closed surface

i am wondering about something,can we find or create aclosed surface where there all of the points inside that surface has adifferent distances to any of the all points of the surface?i mean here we cannot finde apoint inside the space of that surface has the same distance to at least 2points on the surface?this surface should not be for example like asphere because the center of the sphere is apoint that has asame distances to all of the points on the spherer`s surface.i hope my words is clear.thank you.Husseinshimaljasimdini (talk) 12:04, 25 February 2008 (UTC)

If I understand the question correctly, then I would think not, since you can just choose any two points on the surface and pick the midpoint between them. Well, if the surface isn't convex, you couldn't choose *any* two points, but there there will still be some points - just choose any line that passes through the interior of the region bounded by the surface and the intersection points of that line with the surface should do. I think as long as the surface does bound a region with a non-empty interior, there will be points in the interior equidistant to points on the surface. --Tango (talk) 13:41, 25 February 2008 (UTC)

Interesting question - I think the answer is no. I'll explain in two dimensions, you can easily expand the answer to three dimensions for surfaces..

Suppose I start my asymmetric ring at point A with distance r from the 'centre'.. then increasing r as I turn around the surface eg using polar coodinates (radius,angle) - the first thing I notice is that if I increase r with angle I cannot every decrease it since that would return r to a previously used value.. but to get back to r at angle=0 I must decrease.. Both can't be true so it's impossible..

Such a shape would be a spiral - spirals cannot be closed surfaces.. Did that make sense?83.100.158.211 (talk) 15:00, 25 February 2008 (UTC)

Even a spiral loses: take the midpoint of a small enough chord that it doesn't intersect the next layer (so as to avoid questions of containment). --Tardis (talk) 15:54, 25 February 2008 (UTC)

Didn't understand that - the spiral r=e^angle is single valued in r for all values of angle ?83.100.158.211 (talk) 15:59, 25 February 2008 (UTC)

The OP's request was for an open region $\Omega \in \mathbb {R} ^{n}\ni \forall x\in \Omega \,\not \!\exists (y,z)\subseteq \partial \Omega \;\;y\neq z\land \|y-x\|=\|z-x\|$. It's not enough for the "unique distances" clause to hold for one contained point. --Tardis (talk) 16:50, 25 February 2008 (UTC)

((For the spiral described above every point is at a unique distance, - am I wrong?))

You've confused me now - I understood the OP asked for a closed surface (hence not a spiral) - but said that excluding that 'closed surface' clause a spiral would be such a shape (if only in 2d) I see that a spiral fails for a single unique point in 3D ie a spiral surface has more than one point with a given radius. Was that what your original point was saying?83.100.158.211 (talk) 17:12, 25 February 2008 (UTC)

cf open region with Closed manifold - they asked for closed manifold eg 'closed surface'83.100.158.211 (talk) 17:42, 25 February 2008 (UTC)

I interpreted the question as taking place in $\mathbb {R} ^{3}$ where the closed manifold (ie. closed surface) bounds an open region. --Tango (talk) 17:59, 25 February 2008 (UTC)

still a bit confused about the 'chord' explanation - the answer is a definate 'no' anyway -

another way to look at it is to note that the closed surface must be (at least) double valued (of r in angle) - to be closed (in 2D) {and gets worse in 3d ie infinite values with same r for (angle1,angle2) could go on about being able to draw a loop/ring on the surface of a 3D closed surface at any point }

- which of course means it cannot be single valued as requested. are we answering the same question??? at least we are getting same answer83.100.158.211 (talk) 18:19, 25 February 2008 (UTC)

What function are you saying is multi-valued, $r(\theta )$ or $\theta (r)$? I assume the latter, since that's what we need, but your terminology is a little confusing (to me, at least). The OP (in the 2D case - I'm pretty sure he's actually asking about the 3D case, since you can't have a closed surface in $\mathbb {R} ^{2}$) requires that $\theta (r)$ be multi-valued for some choice of origin within the interior of the region bounded by the surface. That's just a restatement of the question, though, it doesn't prove anything. --Tango (talk) 18:39, 25 February 2008 (UTC)

Yes I wrote "(of r in angle)" is should of course have read "doubled valued in angle for a given r" - my mistake that was very unclear.

$\theta (r)$ is multivalued for not true.. or single valued if the original construct was possible - I think I was showing that when $\theta (r)$=singlevalued the surface cannot be enclosed .

Clearly for a given radius r there must be only one value of theta (2D) or (combiantion) theta,mu (3D) that gives that radius ie if r=fn(theta) then the function theta=fn^-1(r) must have a single value of theta for that value of r eg for a ellipse there are 4 values of theta that give a specific r, so I would have said that (in the case of the ellipse) theta is multivalued.

and multivalued = not possible83.100.158.211 (talk) 19:51, 25 February 2008 (UTC)

My argument for it's impossibility was of this kind:

given that the rate of change of radius with angle is x

x cannot be zero (since this gives a 'flat' region - ie two points have same r)

so x is either positive or negative

x cannot change sign since this would require x=0 at some point.

So x is either always positive or negative.

BUT r=fn(x) = fn(x+2pi) (for a continuous surface) (a full rotation) (equation A)

equation A cannot be true since d fn(x)/dx is not zero and never changes sign.

Therefor assuming that the example function is single valued ie x=fn^-1(r) is single valued for r (PROBABLY DID STATE THIS THE OTHER WAY ROUND at some point sorry)

Then fn(x) <> fn(x+2pi) - it's not a closed ring.

This expands easily to 3D since for a closed 3D surface any plane section through it (through the 'centre') must be a closed ring.

So single valued in x=fn^-1(r) and 'closed ring' are mutually exclusive.

I've expanded a bit on the original explanation (including the differentials) - but it's the same un-proof. I guess I wrote single valued for fn(x) somewhere before when it should have been fn^-1(r) - that would account for any confusion you are getting.

(whether r=fn(angle) is multivalued (or not) (ie non-convex surfaces) doesn't affect this proof)83.100.158.211 (talk) 20:06, 25 February 2008 (UTC)

I think x=0 is allowed, but only instantaneously, and you still couldn't have a change of sign (it would have to be a point of inflexion), so your basic conclusion is correct. Dealing with the 3D case by taking the intersection with a plane is a good idea (although, pedantically, the intersection must be a union of closed rings, but that doesn't affect anything). --Tango (talk) 20:13, 25 February 2008 (UTC)

Yes, forgot about an inflexion -good point.83.100.158.211 (talk) 20:18, 25 February 2008 (UTC)

Geometry Help: Mysterious Equation

So, I'm here studying for a test and one of the formula's in my notes makes no sense to me. I have completly forgotten how to use it and I don't have any examples. The formula is V=BiT or V=BLT. Either or.. does anyone know what the formula stands for.. —Preceding unsigned comment added by 80.148.24.98 (talk) 21:26, 25 February 2008 (UTC)

I expect we're going to need some context. What's the section it's in about? Is there any other mention of V, B, i/L and T near it? --Tango (talk) 21:39, 25 February 2008 (UTC)

No, nothing at all.. We're on a unit about volume and surface area of rectangular prisms and pyramids..... —Preceding unsigned comment added by 80.148.24.98 (talk) 21:42, 25 February 2008 (UTC)

I think you'll have to ask your teacher. I don't see any way we can identify one formula in isolation. The volume of a prism is Volume=Base area * Length, that gives you V=BL, but I don't know what the T could be. --Tango (talk) 22:38, 25 February 2008 (UTC)

The volume of a pyramid is base*height/3, so maybe that extra factor has something to do with the third letter. On the other hand, the volume of a prism is also length*width*height. So yeah, your teacher's probably the only person who knows what she meant. Or one of your classmates. Black Carrot (talk) 22:42, 25 February 2008 (UTC)

T = tallness? —Bkell (talk) 23:48, 25 February 2008 (UTC)

What I meant here, I guess, was "Volume = Breadth × Length × Tallness", though "Tallness" is a strange word to use for that. Maybe BLT? ;-) —Bkell (talk) 23:53, 25 February 2008 (UTC)

Perhaps it's just bad handwriting? V=bh (volume = base area x height). Imagine Reason (talk) 23:49, 25 February 2008 (UTC)

February 26

Measure Theory

Let C be a semi-algebra on a given set $\Omega$ and $\mu :C\rightarrow [0,\infty ]$ such that $\mu (\emptyset )=0$ and $\mu$ is sigma additive.

We then define $\mu ^{*}(A)=\inf \sum _{k=1}^{\infty }\mu (D_{k})$ where $A\subset (\cup D_{k})$ and each $D_{k}\in C$.

I am trying to prove that $\mu ^{*}(E)=\mu ^{*}(B\cap E)+\mu ^{*}(B^{c}\cap E)\,\forall B\in C$ and $\forall E\in P(\Omega )$.

One inequality is easy to show because E is a subset of $(B\cap E)\cup (B^{C}\cap E)$ and $\mu ^{*}$ is monotonic.

But any hints as to how to show $\mu ^{*}(E)\geq \mu ^{*}(B\cap E)+\mu ^{*}(B^{c}\cap E)$ would be appreciated.

$\sum \mu (D_{k})=\sum (\mu (D_{k}\cap B)+\mu (D_{k}\cap B^{c}))=\sum \mu (D_{k}\cap B)+\sum \mu (D_{k}\cap B^{c})$.

Furthermore, $(\cup D_{k})\cap X=\cup (D_{k}\cap X)$. Does this help? -- Meni Rosenfeld (talk) 00:41, 26 February 2008 (UTC)

Grammars

Given {a, ba, bba, bbba} bⁿa|nEN
I calculated that S→aS → bSa → bbSa → bbbSa would be the grammar.
I wanted to make sure that I am on the correct path. NanohaA'sYuri^{Talk, My master} 00:04, 26 February 2008 (UTC)

I don't think that's right, because if "a" is in the set and "S→aS" is a rule, then you can get "aa", which isn't in the set. Am I misunderstanding the question? —Keenan Pepper 05:06, 26 February 2008 (UTC)

One way of reading the rule "S→aS" is the following:

If σ ∈ L(S), then also aσ ∈ L(S) – where L(S) here denotes the language produced using S as the start symbol.

So if bba ∈ L(S), as desired, then you would also get abba ∈ L(S), which would be wrong.

There is a context-free grammar for the language {a, ba, bba, bbba, ...} = {bⁿa|n∈N} with one nonterminal symbol and two production rules. I think the assignment calls for a context-free grammar. In a context-free grammar all rules have a single nonterminal symbol on the left-hand side. What you wrote, "S→aS → bSa → bbSa → bbbSa", does not look like (production rules of) a grammar. Production rules contain only one arrow. It looks more like the begin of a derivation. However, no context-free grammar could have a derivation step of the form . If at some stage of a derivation the string starts with a terminal symbol, like "a" here, then it will start with "a" in all following stages.  --Lambiam 06:40, 26 February 2008 (UTC)

Completing the Square

I was just looking at the article on completing the square and saw in the derivation of the quadratic formula from the equation $ax^{2}+bx+c=0\,\!$ that $\displaystyle {x^{2}+{\frac {b}{a}}x+\left({\frac {b}{2a}}\right)^{2}=\left({\frac {b}{2a}}\right)^{2}-{\frac {c}{a}}\to \left(x+{\frac {b}{2a}}\right)^{2}={\frac {b^{2}-4ac}{4a^{2}}}}$. What I am wondering (although it is probably very simple) is where does the ${\frac {b}{a}}x$ go from the first equation to the next. Much appreciated, Zrs 12 (talk) 02:00, 26 February 2008 (UTC)

Try expanding out $\left(x+{\frac {b}{2a}}\right)^{2}$. Remember, when you expand something like $(L+R)^{2}$, the result is not $L^{2}+R^{2}$, but $L^{2}+2LR+R^{2}$ (the FOIL rule). —Bkell (talk) 02:34, 26 February 2008 (UTC)

Thanks Bkell, I understand it a little better now. Expanding it gives $x^{2}+{\frac {b}{a}}x+{\frac {b}{a}}$, right? Where does ${\frac {b}{a}}x$ go on the right side? Many thanks, Zrs 12 (talk) 03:04, 26 February 2008 (UTC) I got it. When expanded $\left(x+{\frac {b}{2a}}\right)^{2}=x^{2}+{\frac {b}{a}}x+{\frac {b^{2}}{4a^{2}}}$. That was easy enough. I just had to think and write it out. Thanks! Zrs 12 (talk) 03:25, 26 February 2008 (UTC)

Glad to help. —Bkell (talk) 04:23, 26 February 2008 (UTC)

Geometry Help

In a triangular prism with the width 3, and each diagonal side 3 and a length of 7, how would you find the surface area and volume?

--Devol4 (talk) 07:41, 26 February 2008 (UTC)

Triangular prism helps with the volume. Zain Ebrahim (talk) 11:09, 26 February 2008 (UTC)

Well, if I understand the question correctly, your prism can be interpreted geometrically as two equilateral triangles joined by rectangles. To find the volume, just multiply cross-sectional area (the triangle) by the length, and to find surface area, just find the area of all components and sum them. -mattbuck (Talk) 11:10, 26 February 2008 (UTC)

egytian method

divide 16 by 9 —Preceding unsigned comment added by 41.205.189.9 (talk) 15:27, 26 February 2008 (UTC)

16/9. =) –King Bee ^{(τ • γ)} 15:35, 26 February 2008 (UTC)

I assume the OP means the "Egyptian method". Here's one explanation of the process. — Lomn 15:49, 26 February 2008 (UTC)

We do have an article on this, check out Egyptian fraction. After having done the calculations, you can check your work using these calculators. --NorwegianBlue ^talk 18:37, 26 February 2008 (UTC)

maths trick i just found out

1. think of a number and put it in your calculator [make sure you remember it]
2. add 5
3. multiply your answer by 6
4. divide your answer by 2
5. subtract your answer by 15
6. divide your answer by the number you first thought of

• answer is always = 3
• check source at [doxadeocollege.co.za] —Preceding unsigned comment added by 196.207.47.60 (talk) 16:15, 26 February 2008 (UTC)

That is true. It's not really a trick so much as obfuscation - simply writing it out on a piece of paper and you can see the following steps:

• $S_{2}:x\rightarrow x+5$
• $S_{3}:x+5\rightarrow 6(x+5)$
• $S_{4}:6(x+5)\rightarrow {\frac {6(x+5)}{2}}=3(x+5)$
• $S_{5}:3(x+5)\rightarrow 3(x+5)-15=3x+15-15=3x$
• $S_{6}:3x\rightarrow {\frac {3x}{x}}=3$

-mattbuck (Talk) 16:30, 26 February 2008 (UTC)

And what if the number I choose is 0? –King Bee ^{(τ • γ)} 16:36, 26 February 2008 (UTC)

Define 0/0:=3. That's afterall one reason 0/0 is indeterminant, because processes that result in 0/0 could in fact by any real number, or infinity (by appropiate choice of the process). 130.127.186.122 (talk) 17:15, 26 February 2008 (UTC)

How is "obfuscation" not a more specific word for "trick"? In fact, most actual magic tricks use nothing but clever obfuscation. It's so hard to make the hand quicker than the eye, but so easy to find gullible spectators. Take the classic "cup and ball" routine, for instance. You keep the mark's eyes on the cup and ball, and get caught clumsily trying to remove the ball, so that his attention will be on you when your friend picks his pocket. Black Carrot (talk) 17:32, 26 February 2008 (UTC)

First off, it sounds better, and to me, a mathematical trick is some way of simplifying a result or theory. For instance, an easy way to calculate the multiplicity of a pole is to find the multiplicity of the zero at that point of the function's reciprocal - that I would consider a mathematical trick. -mattbuck (Talk) 18:38, 26 February 2008 (UTC)

Trig equation

(Note: this isn't homework, but I'd appreciate a hint rather than being told the answer) I'm having trouble with solving $\!\sin 2x+\sin ^{2}x=1$. We've not covered this in class, so the only things I can see ($\!\sin 2x=2\sin x\cos x=2\sin x\sin(x+{\frac {\pi }{2}})$ or $\!\sin 2x-\cos ^{2}+1$ or some combination thereof) don't seem to be awfully helpful. Any pointers – is there an identity I need that I don't have? Angus Lepper^{(T, C, D)} 20:58, 26 February 2008 (UTC)

Combine the double-angle identity $\sin 2x=2\sin x\cos x\,\!$ with the Pythagorean identity $\sin ^{2}x+\cos ^{2}x=1\,.$  --Lambiam 21:09, 26 February 2008 (UTC)

To 'idiot proof' the above... (almost full solution commented out by Meni Rosenfeld) AirdishStraus (talk) 22:28, 26 February 2008 (UTC)

Why would you give a full solution when the OP explicitly asked you not to? --Tango (talk) 22:41, 26 February 2008 (UTC)

Or call them an idiot? Black Carrot (talk) 23:04, 26 February 2008 (UTC)

Tango is right. I have commented out the solution. -- Meni Rosenfeld (talk) 10:15, 27 February 2008 (UTC)

I agree Tango is right. I got carried away doing the near full solution and had forgotten the 'hint only' request. My apologies. On a seperate note... 'idiot-proofing' something is a mere colloquialism for unambiguous clarification; I never used the word idiot to directly describe anyone and wouldn't ever use it in ANY case. Maybe Black Carrott hasn't come across this term before? AirdishStraus (talk) 11:15, 27 February 2008 (UTC)

I have no doubt that you didn't intend to insult the OP, but using the phrase "idiot-proof" in that way could very easily be misinterpreted, especially if someone isn't a native English speaker, so it's probably best to avoid it. --Tango (talk) 13:06, 27 February 2008 (UTC)

Triangle

The angle bisector of alpha w_α, the median of s_b and the altitude of c h_c of a acute-angled triangle ABC cross in one point, if w_α, the side BC and a circle around the point H_c, which goes through the corner A, have also a common point together.

How can one proof this statement? --85.178.22.17 (talk) 23:18, 26 February 2008 (UTC)

You need to do it one step at a time

1. First find the intersection of the "angle bisector of alpha w_α and the median of s_b" as a function of the parameters of a generalised triangle eg (0,0) (a,b) (c,d) as the points of the triangle.

2. Then find the coodrinates of the crossing point of "the median of s_b and the altitude of c h_c of a acute-angled triangle ABC"

Next you need to show that the two equations abtained above give the same point if the third premise is true.. ie that "w_α, the side BC and a circle around the point H_c, which goes through the corner A, have also a common point together."

3. I'm not sure what H_c is here - but if you do you should be able to generate some sort of parametric equation relating the vertexs of the triangle. {{ Sound's like Hc is the centre of the circle that goes through A and also passes through the intersection of (w_α and the side BC ) - this makes somewhere Hc along a line then}}

Then you need to show that if for instance the equation obtained in 1 is true at the same time as the equation obtained in 3. then equation 2 is also true.87.102.93.245 (talk) 11:43, 27 February 2008 (UTC)

HINT for the "angle bisector of alpha w_α, the median of s_b and the altitude of c h_c" to cross at one point the triangle must be at least iscocoles since we have three lines intersecting at one point and two of them "angle bisector of alpha w_α" "and the altitude of c h_c" already cross at A.. Because the median cannot pass through A, the crossing of the "angle bisector of alpha w_α" "and the altitude of c h_c" must give a line (ie they are the same line) and not just a single point. therefor ac=ab I hope that is enough to get you finished.87.102.93.245 (talk) 12:26, 27 February 2008 (UTC)

February 27

The washer method (disk integration)

I need to find the volume of a solid of revolution. The 2-dimensional area is bounded by y = √x, y = 0, and x = 4, and it is spun around the axis x = 6. I know that my outer radius is 6 - x, but since I'm integrating in terms of y (since the axis is parallel to the y-axis), I should write that as R(y) = 6 - y². But I can't figure out how to express the inner radius r(y) as a function, so that I can plug it into the formula V = π∫_a^b[R²(y) - r²(y)]dy! (I believe a and b = 0 and √6 respectively.) Can anyone offer any help? Thanks, anon. —Preceding unsigned comment added by 70.19.34.80 (talk) 01:10, 27 February 2008 (UTC)

Since the inner wall is a vertical line, the inner radius is constant with respect to y, and in this case is 2. Black Carrot (talk) 03:51, 27 February 2008 (UTC)

Also, if you're integrating with respect to y, it should run from 0 to 2. You should draw some diagrams if you haven't already. Black Carrot (talk) 03:54, 27 February 2008 (UTC)

Chain Rule and Higher Derivative

Let

$z=f(x,y)$

$x=g(s,t)$

$y=h(s,t)$

Then

${\frac {\partial z}{\partial t}}={\frac {\partial z}{\partial x}}{\frac {\partial x}{\partial t}}+{\frac {\partial z}{\partial y}}{\frac {\partial y}{\partial t}}=\left({\frac {\partial }{\partial x}}{\frac {\partial x}{\partial t}}+{\frac {\partial }{\partial y}}{\frac {\partial y}{\partial t}}\right)z$ (1)

${\frac {\partial ^{2}z}{\partial t^{2}}}={\frac {\partial }{\partial t}}{\frac {\partial z}{\partial t}}={\frac {\partial }{\partial t}}\left[\left({\frac {\partial }{\partial x}}{\frac {\partial x}{\partial t}}+{\frac {\partial }{\partial y}}{\frac {\partial y}{\partial t}}\right)z\right]=\left({\frac {\partial }{\partial x}}{\frac {\partial x}{\partial t}}+{\frac {\partial }{\partial y}}{\frac {\partial y}{\partial t}}\right){\frac {\partial z}{\partial t}}$ (2)

Replace (1) into (2)

${\frac {\partial ^{2}z}{\partial t^{2}}}=\left({\frac {\partial }{\partial x}}{\frac {\partial x}{\partial t}}+{\frac {\partial }{\partial y}}{\frac {\partial y}{\partial t}}\right)\left({\frac {\partial }{\partial x}}{\frac {\partial x}{\partial t}}+{\frac {\partial }{\partial y}}{\frac {\partial y}{\partial t}}\right)z$ (3)

${\frac {\partial ^{2}z}{\partial t^{2}}}=\left[{\frac {\partial ^{2}}{\partial x^{2}}}\left({\frac {\partial x}{\partial t}}\right)^{2}+{\frac {\partial ^{2}}{\partial x\partial y}}{\frac {\partial x\partial y}{\partial t^{2}}}+{\frac {\partial ^{2}}{\partial y\partial x}}{\frac {\partial y\partial x}{\partial t^{2}}}+{\frac {\partial ^{2}}{\partial y^{2}}}\left({\frac {\partial y}{\partial t}}\right)^{2}\right]z$ (4)

${\frac {\partial ^{2}z}{\partial t^{2}}}={\frac {\partial ^{2}z}{\partial x^{2}}}\left({\frac {\partial x}{\partial t}}\right)^{2}+{\frac {\partial ^{2}z}{\partial x\partial y}}{\frac {\partial x\partial y}{\partial t^{2}}}+{\frac {\partial ^{2}z}{\partial y\partial x}}{\frac {\partial y\partial x}{\partial t^{2}}}+{\frac {\partial ^{2}z}{\partial y^{2}}}\left({\frac {\partial y}{\partial t}}\right)^{2}$ (5)

Is every step above correct? - Justin545 (talk) 07:07, 27 February 2008 (UTC)

No.

${\frac {\partial z}{\partial x}}{\frac {\partial x}{\partial t}}$

is not the same as

$\left({\frac {\partial }{\partial x}}{\frac {\partial x}{\partial t}}\right)z.$

For example, if z = x = t, the former evaluates to , and the latter to Think of ${\tfrac {\partial }{\partial x}}$ as an operator, and abbreviate it as D. Also abbreviate ${\tfrac {\partial x}{\partial t}}$ as U. Then in your first line of equations you replaced by .  --Lambiam 09:46, 27 February 2008 (UTC)

Recursively defined sets

I want to generate the first several iterations of this set, but I feel I'm not doing it correctly; my notes are sparse on certain details:

Defintion of $C\,$.

$C\subseteq \Sigma$ where $\Sigma \,$ is the set of all strings over $\lbrace 1,2,3\rbrace \,$

Basis step:

$\lbrace 1,2\rbrace \subseteq C$

Recursive step:

$w\in C$ $\rightarrow$ $3\cdot w\in C$

$w_{1},w_{2}\in C$ $\rightarrow$ $w_{1}:w_{2}\in C$

$w\cdot 1\in C$ $\rightarrow$ $2\cdot w\in C$

The dot operator and the colon (:) operator represent character concatentation and string concatentation respectively.

It would follow, in my mind that the first few iterations of this set would go:

{1, 2}

{1, 2, 31, 11, 21}

{1, 2, 31, 11, 21, 32, 12, 22}

e.t.c

My main worry is this:

I assume w(1) and w(2) in the second part of the recursive step start at 1, 1 and go up 1, 2 then 2, 1 then 2, 2 and so on? Or do they have to be seperate values each time? Damien Karras (talk) 08:03, 27 February 2008 (UTC)

If I understand your last question correctly, the answer is that $w_{1}$ and $w_{2}$ don't have to be different. I'll guide you through the first iteration. You know that 1 ∈ C and 2 ∈ C. By rule 1, from 1 ∈ C it follows that 31 ∈ C, and from 2 ∈ C it follows that 32 ∈ C. We now apply rule 2: From 1 ∈ C and 1 ∈ C it follows that 11 ∈ C; from 1 ∈ C and 2 ∈ C it follows that 12 ∈ C; likewise 21 ∈ C and 22 ∈ C. Rule 3 doesn't give us anything new now. So the first iteration will be $\{1,2,31,32,11,12,21,22\}\subseteq C$. -- Meni Rosenfeld (talk) 09:51, 27 February 2008 (UTC)

(after edit conflict) I would interpret your rule #2 of the "recursive step" as being universally quantified over all w₁ and w₂ that are in C. So if 2 is a member of C_n in iteration n, then 22 is a member of the next generation C_n+1. Is that an answer to your question (which I do not fully understand)?

Because of rule #3, you can start the whole process by taking C₀ = {1}. Then the first few iterations give this:

C₀ = {1}

C₁ = {1, 31, 11, 2}

C₂ = {1, 31, 11, 2, 331, 311, 32, 131, 111, 12, 3131, 3111, 312, 1131, 1111, 112, 21, 231, 211, 22, 23}

 --Lambiam 10:10, 27 February 2008 (UTC)

I think what I meant my overall question to be is this: For each iteration, do I pick the value of w, w1 and w2 just once? So the first iteration would be: w = 1, w1 = 1 and w2 = 1 and then I would move to the next iteration.

OR do I consider all possible values of w, w1, w2 for each iteration? In which case I would do, for instance, step one for w = 1 and w = 2, step two would be done for w1 = 1, w1 = 2, w2 = 1, w2 = 2 e.t.c

Furthermore, are the new elements of the set generated on each step? Because if so that would mean that w1 and w2 take on even more values before the end of the total recursive step (unless of course the former point is true). Damien Karras (talk) 10:41, 27 February 2008 (UTC)

I am still confused by your questions, but I have a better idea what it is you are asking. To find $C_{n+1}$, you apply every possible rule to the items in $C_{n}$ and only them. So if you start with $C_{0}=\{1,2\}$, applying every possible rule to $C_{0}$ gives you $C_{1}=\{1,2,31,32,11,12,21,22\}$. You then apply every possible rule to $C_{1}$ to get $C_{2}$, and so on. In each step, you only apply rules to elements you have had in the previous step, not to those you have generated recently. -- Meni Rosenfeld (talk) 11:44, 27 February 2008 (UTC)

Thanks both of you! That's exactly the answer I was looking for, sorry the question was phrased poorly. Just to check will I be right in saying that the start of $C_{2}=\lbrace 1,2,31,32,11,12,21,22,331,332,311,312,321,322,3131,3132,3111,3112,3121...\rbrace \,$? (I won't write the whole thing out because it appears to be massive!) Damien Karras (talk) 12:27, 27 February 2008 (UTC)

This looks correct. Don't forget that $C_{2}$ also has elements such as 131, the concatenation of 1 and 31 which are both in $C_{1}$. -- Meni Rosenfeld (talk) 14:58, 27 February 2008 (UTC)

This is what I get starting from {1, 2}:

C₀ = {1, 2}

C₁ = {1, 2, 31, 32, 11, 12, 21, 22}

C₂ = {1, 2, 31, 32, 11, 12, 21, 22, 331, 332, 311, 312, 321, 322, 131, 132, 111, 112, 121, 122, 231, 232, 211, 212, 221, 222, 3131, 3132, 3111, 3112, 3121, 3122, 3231, 3232, 3211, 3212, 3221, 3222, 1131, 1132, 1111, 1112, 1121, 1122, 1231, 1232, 1211, 1212, 1221, 1222, 2131, 2132, 2111, 2112, 2121, 2122, 2231, 2232, 2211, 2212, 2221, 2222, 23}

 --Lambiam 23:22, 27 February 2008 (UTC)

modular algorithm

how can i solve this problem?

[ x1= a (mod 100) , a= 20 (mod 37) ]

[ x2= b (mod 100) , b= 15 (mod 37) ]

[ x3= c (mod 100) , c= 18 (mod 37) ]

must be ; x2= a.k + y (mod100)

and

x3= b.k + y (mod100)

i need find b and c.. thank you best regards.. Altan B. —Preceding unsigned comment added by 85.98.230.220 (talk) 10:20, 27 February 2008 (UTC)

x1 does not play a role. Eliminating x2 by combining with gives us Likewise eliminating x3 results in We can combine these to eliminate y and obtain or, equivalently,

b(k + 1) ≡ c + ak (mod 100).

Using the (mod 37) congruences given for a and b, and c, we rewrite these as and Substituting that in the equation brings us to:

(37β + 15)(k + 1) ≡ 37γ + 18 + (37α + 20)k (mod 100).

We now multiply both sides by the multiplicative inverse of 37 (mod 100), which is 73 (since This results in

(β + 95)(k + 1) ≡ γ + 14 +(α + 60)k (mod 100).

Bringing γ to one side,

γ ≡ (β + 95)(k + 1) + (99α + 40)k + 86 (mod 100).

Now pick any values for α, β, k, and a new variable n, and compute

γ = (β + 95)(k + 1) + (99α + 40)k + 86 + 100n,

b = 37β + 15,

c = 37γ + 18.

For example, choosing α = β = 0, k = 12, n = −18 gives the solution b = 15, c = 55.  --Lambiam 12:32, 27 February 2008 (UTC)

thank you for your cares, but this solution is not that i want.. 'cause i also know this solving system but, this system isnt exact solution. i will use this formula in the big machine ( this machine has 17 billions as modular ) as you know i cant expect(or according to your solution choosing) all value from my brain. thank you best regards Altan B. —Preceding unsigned comment added by 85.98.230.220 (talk) 14:51, 27 February 2008 (UTC)

Affedersiniz ama yukarıda yazdıklarınızı pek anlayamıyorum.  --Lambiam 19:22, 27 February 2008 (UTC)

shapes and projections

I was wondering about something and I need abig help to correct my understanding to this issue.I am going to put my questions in two parts.part1:assume that we have ashape in two dimensions,like asphere,obviously we can determine the projections of that sphere on (x-y,y-z,z-x)planes, which they are going to be circles no matter how we rotate the sphere, the projections will be always circles.If we apply now the same thing on acone then we would have adifferent projections depend on how we rotate or fix the cone in two dimensions,we can get for example,acircle in(x-y)plane,atriangle in(y-z)plane and another triangle in(x-z)plane.we can determine the projections depend on the shape and its position or you can say the shape and the position vector between the origin and achosen point belong to that shape.Now my question is,can we determine the shape by looking at its projections?if yes,then what is the shape in two dimensions that has the projections,(circle,square and apoint)?p.s:you can improvise another projections.some times we can describe different shapes in 2 dimentions by looking at the projections,for example,assume we have 3identical circles as aprojections,one may say that the shape is atwo dimnensional sphere,or the shape could be 3 identical circles perpendicular to each other and sharing the same center. Part2:if we are able to say that the sphere in three dimensions has aprojection in two dimensions which is asphere too then for the cone in three dimensions what its projection in two dimensions going to be?IF THE ANSWER IS ACONE TOO NO MATTER HOW WE ROTATE OR CHANGE ITS POSITION IN 3 DIMENTIONS THEN IT SHOULD BE ASPHERE, because the sphere is the only shape that maintains its projections in all dimensions>2

Further discussion: Assume we have ashape in 3dimensions where its projections in 2dimensions as follows, (x,y,z)=s1,(x,y,w)=s2,(x,z,w)=s3&(y,z,w)=s4.The projections in one dimension would be,

[s1:(x,y)=a1,(x,z)=a2,(y,z)=a3],[s2:(x,y)=a1,(x,w)=a4,(y,w)=a5],[s3:(x,z)=a2,(x,w)=a4,(z,w)=a6],[s4:(y,z)=a3,(y,w)=a5,(z,w)=a6].We have 6 one dimensional projections .Now assume we have acone in 3 dimentions,we can finde some conditions to make that cone describable,for example ,if the con`s peak is located at the origin and the position vector between the peak and the center of the cone`s makes an angle=pi\4 with each of(x,y,z,w),then a1=a2=a3=a4=a5=a6 and each one of the six projections seems to be having ashape of sector of acircle.on the other hand if,a1=a2=a3=a4=a5=a6 ,shouldnot we get asphere in 3dimensions?my point here is,can we say that all different shapes in 3 dimensions with no preconditions have ashape of sphere???Husseinshimaljasimdini (talk) 12:05, 27 February 2008 (UTC)

Part 1 - you seem to have made a sort of mistake eg if I project a set of 3 circles orthogonal (at right angles) to each other onto the three planes I get a circle with a cross inside - not a filled circle as you might expect from the sphere.(Maybe this is worth thinking about?)

As for your question 'can we determine a shape from it's projections' - it might depend - the answer seems to be yes if we assume the shapes are all quadratics, ... but much more complicated shapes might project as for example 3 circles along one set of coordinate axis (suggesting a sphere) - but produce different shapes along a different set of coordinate axis.87.102.93.245 (talk) 12:39, 27 February 2008 (UTC)

Your question "(if yes),then what is the shape in two dimensions that has the projections,(circle,square and apoint)" - seems impossible - because if the shape projects onto a point it can only be a point or a line - therefor a circular or square projection is impossible...

(I removed a space from your question 2 since the text was going way off the screen..)87.102.93.245 (talk) 12:39, 27 February 2008 (UTC)

(edit conflict) I would guess in "3 identical circles perpendicular to each other and sharing the same center", he means *discs* not circles. Then the projections are the same as for a sphere. An obvious example of shapes that can't be distinguished by their projections are a sphere and a sphere containing another shape - you can't "see through" the sphere, so you'll never detect the contained shape. With suitable restrictions on the shapes (for example, 245's idea of restricting to quadratics) might work, but with general shapes there are all kinds of things that can go wrong. --Tango (talk) 13:21, 27 February 2008 (UTC)

In general, you cannot reconstruct the 3D-shape from three orthogonal 2D-projections. One shape that has three unit disks as projections is the union of these three disks:

{(x,y,z) | }.

Among the convex shapes, one solution is the unit ball {(x,y,z) | }. The largest solid object giving the same three projections is given by the intersection of three cylinders:

{(x,y,z) | }.

This is strictly larger than the unit ball: it contains the point which is outside the unit ball.  --Lambiam 13:18, 27 February 2008 (UTC)

Linear equations

Hello =], I'm just having some problems with a few linear equations and this text-book is terrible example wise. Ive had no trouble with the equations preceding it:
$5(y-2)=12$
$5y-10=12$ (Expand Brackets)
$5y=22$(Add 10 to both sides)
${\frac {5y}{5}}={\frac {22}{5}}$(Divide both sides by 5)
$y={\frac {22}{5}}$


Can the same method be applied to the equation below? As far my answers have been incorrect:
${\frac {x}{2}}+3=5$
${\frac {x}{2}}=2$ (Subtract 3 from both sides)
This looks finished but it is evidential that I have done something wrong, a point in the right direction will be appreciated.


Also something like this, is my answer correct?
$5x+4\leq 0$
$5x\leq -4$ (subtract 4 from both sides)

Then this is where i get lost do i divide or perform some other operation? —Preceding unsigned comment added by 58.165.62.73 (talk) 13:06, 27 February 2008 (UTC)

You are right as far as you've gone in part 2, but you haven't finished. I assume you want to know what x on its own is, and you have x/2. Double each side. As for part 3, yes, you can just divide through, but be careful - if you divide or multiply through by a negative number, you have to invert the inequality. -mattbuck (Talk) 13:13, 27 February 2008 (UTC)

Mattbuck, note than he only subtracted 4 from each side of the inequality. Visit me at Ftbhrygvn (Talk|Contribs|Log|Userboxes) 14:50, 27 February 2008 (UTC)

Yes, but the end result presumably asks for $x\leq {\frac {-4}{5}}$, which requires dividing through, and it's a note about multiplying by negative numbers generally. -mattbuck (Talk) 15:04, 27 February 2008 (UTC)

Sorting a list of points for "in order of closest points"

Hi, suppose I have a list of 3 dimensional points eg (x_n,y_n,z_n) in random order and I want to try to sort them so that closer points are nearer to each other in the list.

The only idea I had was to (after probably writing the numbers in binary form) would be to covert each set of numbers into a single 'string' in which the most significant digits are group together

eg in decimal (1234,6789,-5500) would convert to ( (1,6,-5) , (2,7,-5) , (3,8,0) , (4,9,0) ) and then sorting that in order from left to right... This would put points with 1000<x<=1999 close together in the list..

This works a bit, (but still gives more priority to x than y etc - working in binary reduces this effect a bit)

Can anyone suggest any improvements to this method, or a better method? Note that I only require the list be 'improved' so that nearer points be nearer in the list - it doesn't have to give a perfect sort in perfect order or nearness (that seems impossible anyway) (You can assume I don't need any help with the sort in order of linear magnitude algorhytm)

Any ideas welcome. Just the concept would do.. Thanks87.102.93.245 (talk) 16:58, 27 February 2008 (UTC)

Imagine that your points are uniformly distributed within a cube. How are you going to "walk" around the cube and visit each point? One way is to walk along y=0, z=0 from x=0 to x=N and then back along y=1 z=0 from x=N to x=0 etc. After a while you have visited all of the z=0 plane in a passably efficient manner, but the rest of z hasn't been started. so x=0, y=0, z=1 which is close to x=0, y=0, z=0 is N^2 points away from the that coord. In other words, unless the data has some known clustering that you can take advantage of, there isn't going to be a simple solution to your question. -- SGBailey (talk) 17:56, 27 February 2008 (UTC)

Is your goal to be able to quickly find the nearest points to a given point? If so, see kd-tree. --Trovatore (talk) 18:02, 27 February 2008 (UTC)

right thanks - I already had a sort of octree (if I used binary number) - this 'kd-tree' looks like a possible improvement (though a little more analysis of the data is require - it should give a better list)Thanks.87.102.93.245 (talk) 18:37, 27 February 2008 (UTC)

If you really need a single ordering rather than a data structure, you could try sorting them in their ordering along a Space-filling curve. For 3d variants of the Hilbert curve each comparison of the sorting algorithm can be done by finding the most significant bit at which any of the coordinates of the two compared points differ. —David Eppstein (talk) 18:24, 27 February 2008 (UTC)

Thanks, looks like a good possibilty. I've often wondered if I'd ever find a use for those 'peanut curve' things . Today was that day. I'm not convinced that the hilbert curve is better (than a bsp tree) in terms of points that are just on either side of 0.5 (ie halfway accross) - probably averages out to similarily as good (please correct me if you are more expert than me - which is likely..) Still very interesting though.87.102.93.245 (talk) 18:44, 27 February 2008 (UTC)

Assuming you use in-order depth-first traversal, a bsp tree does not tell you by itself in what order the children of a node should be put. If you partition by cutting up the cube at each level in eight equally sized cubes, it is possible to order the children at each node in such a way that the in-order traversal is the same as traversal along a 3D Hilbert curve, which is the best you can get with a uniform recursive method. But other orderings are not dramatically worse, at most a factor 2 or so in the path length, which is O(n^2/3).  --Lambiam 20:25, 27 February 2008 (UTC)

Rounding conundrum?

I disagree with former teaching colleagues over the correct answer to a rounding problem I posed a few years ago to some 13-year-old pupils, and for my own personal satisfaction I would be interested in people's views to the following simple question:

What is 0.00999 rounded to 2 Significant Figures (2 S.F.)?

The two answers to consider, from myself as head of department and a few of my maths teachers, are 0.010 and 0.0100 (I'm not revealing who claimed which! Yet!). Answer PLUS REASON please if you think you know for sure which is the correct answer! Thanks! AirdishStraus (talk) 18:18, 27 February 2008 (UTC)

It's clearly 0.010, at least according to the convention in our article. The two significant figures are 1 and the subsequent 0. -- Meni Rosenfeld (talk) 18:38, 27 February 2008 (UTC)

As far as I remember from Physics and Chemistry, it should be 0.010. The first two zeros (the one before the decimal and the one right after the decimal) are just place holders. They appear in all the numbers being operated on and I was told that they are not considered a significant value in a problem like this. So we have the one and then the second zero to have two significant digits.A Real Kaiser (talk) 18:48, 27 February 2008 (UTC)

Mmm... obviously the number 0.010 has two significant figures and the number 0.0100 has three, but the issue, I believe, is in the actual process of taking 0.00999 and applying the 'rules'...

'Chop' the number off after the second significant digit and round if the third digit is 5 or greater; write down as your answer the digits you now have up the the 'chop off line'.

So... 0.00999 $\rightarrow$ 0.0099|9 $\rightarrow$ 0.0100| $\rightarrow$ 0.0100

Where, in any article does it say that the resultant answer is further truncated to 2 significant figures? This would mean that a small percentage of all decimals would have a different rule: anything from 0.00995 to 0.00999999... etc. for example for those numbers starting in the third places of decimals. AirdishStraus (talk) 19:47, 27 February 2008 (UTC)

Now I see the problem - which is that significant figures are stupid. For a number that starts with 1, a given number of significant figures represent less accuracy than for a number that starts with 9. For the number in your example, we are faced with the dilemma of treating it as starting with 9, which it does, or starting with 1, like its rounded value. I say the rounded value wins, since we measure the significant figures for it, not for the real value. The quirk is that 0.00999 rounded to 2 sf is 0.010, and rounded to 3 sf it's 0.00999 which is two orders of magnitude more accurate. Again, sfs are stupid. It makes much more sense to discuss relative error, and sfs were never meant to deal with such subtleties. -- Meni Rosenfeld (talk) 20:18, 27 February 2008 (UTC)

I'd think of it this way: the "numbers with two significant figures" are ..., 0.0097, 0.0098, 0.0099, 0.010, 0.011, 0.012, ... and rounding to two significant figures consists in choosing the one of those that (considered as an exact real number) is closest to your number, in this case 0.010. A rounding algorithm, like the one you gave, which produces two significant figures in most cases and three in a few boundary cases I would describe as buggy. Think of it also from the perspective of someone seeing the answer 0.0100; they'll most likely assume that it was selected from the list ..., 0.00998, 0.00999, 0.0100, 0.0101, ... and not from the list ..., 0.0098, 0.0099, 0.0100, 0.010, 0.011, ... (which doesn't even make mathematical sense, since two of the entries are the same). Not that there's anything particularly sensible about the standard convention, it's just standard. As Meni said, the whole concept is rather badly broken, and if you really care about accuracy you have to do a proper error analysis and give explicit error bars. -- BenRG (talk) 20:24, 27 February 2008 (UTC)

(outdent) Just a thought. While strictly speaking, rounding to 2 sig figs does give 0.010, a partial way to sidestep the noted discontinuity of accuracy is to look at the unrounded value and note at which place (tenths, hundredths, etc.) the number loses reliability, then round so as to remove all digits at and past that place. So if the third "9" is not reliable, quote the result as 0.0100, and so on.

The above posting has a good point which I reflected in my advice: rounding should not comply to rote symbolic manipulation rules of digit strings but should reflect the precision and reliability in obtaining the value to be rounded.

This discussion also relates to Benford's law. Baccyak4H (Yak!) 20:38, 27 February 2008 (UTC)

Guys... thanks for the contributions (others still welcome). I'm the one who believes 0.0100 is the correct answer as opposed to the rest of my (now ex-) department who favoured 0.010. The reasons are that it follows the given rule with no deviation (i.e. no 'special cases') and that it gives an indication of the accuracy of the original number. 0.0100 declared as having been rounded to 2sf means something that HAD TO BE originally between the bounds of 0.00995 and 0.009999999.... etc. whereas 0.010 could have come from between bounds of 0.00995 and 0.0105 Once again, many thanks for the various inputs. AirdishStraus (talk) 22:06, 27 February 2008 (UTC)

The correct answer is definitely 0.010. If you round to 2sf, you have to have 2sf at the end, otherwise it's nonsense. 0.0100 clearly has 3sf. Your reason for rounding to 0.0100 sounds to me like simply "3sf is more precise than 2sf", which is obviously true, but irrelevant, since you asked for 2sf. --Tango (talk) 22:30, 27 February 2008 (UTC)

You're probably right, rounding in such a way as to accurately reflect the precision of the measurement is best, but that only works if people know how to interpret the data. If I see "0.010" I know that means the real value is between 0.0095 and 0.0105. Conventions have to be followed whether they are good conventions or not, because otherwise people have no idea how to interpret what you say (you could devote a paragraph at the beginning of your paper to explaining the obscure rounding system you've used, but your readers would probably prefer you just to use the standard method). —Preceding unsigned comment added by Tango (talk • contribs) 22:35, 27 February 2008 (UTC)