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December 2

Proof for Internal/External Division of Line Segment

Hello. What is the proof for internal/external division of a line segment? ${\displaystyle P(x,y)=\left({\frac {sx_{1}+rx_{2}}{r+s}},{\frac {sy_{1}+ry_{2}}{r+s}}\right)}$ such that ${\displaystyle A(x_{1},y_{1})}$ and ${\displaystyle B(x_{2},y_{2})}$ are the first and second points respectively, ${\displaystyle r}$ is the length of the first division, ${\displaystyle s}$ is the length of the second division, and ${\displaystyle P}$ divides line ${\displaystyle AB}$. Thanks in advance. --Mayfare 01:53, 2 December 2007 (UTC)

I'm not quite familiar with your notation and terminology, and you do not state what it is that needs to be proved, but here is my attempt at a sketch.

In general, a line in the Cartesian plane can be described by a linear equation of the form ax + by = c, where the coefficients a, b and c are real numbers, and a and b are not both zero. To find an equation for the line through two distinct points A with coordinates (x₁,y₁) and B with coordinates (x₂,y₂), we have to find expressions for a, b and c such that both points satisfy the equation. A solution is given by

a = −(y₁ − y₂), b = x₁ − x₂, c = x₁y₂ − x₂y₁.

A straightforward computation now establishes that the point P with coordinates given by the expression involving r and s lies on the line through A and B. By looking at the distances between P and these two points, we see that they are in the ratio of r to s, so if r + s is the distance between A and B, then P divides the line segment connecting A and B into subsegments of the required lengths.  --Lambiam 06:09, 2 December 2007 (UTC)

Segment dissection

Is is possibe to dissect a segment into five equal parts under Euclidean rules? If so, how is it done? 72.197.202.36 02:35, 2 December 2007 (UTC)

Same way it is done with trisection. Start by taking your first line. In some arbitrary direction from one endpoint of that line, draw a small segment and multiply the length by five geometrically (easily done by just repeating the length with a compass). Now draw the base of the triangle formed by these two lines, and from each point of the 5x line, draw a line parallel to the base. Those lines will pentasect the original line by the fact that lines parallel to one side of a triangle divide the other sides in a fixed ratio. SamuelRiv 04:33, 2 December 2007 (UTC)

The usual term for division into two equal parts is bisection, while dissection is applied more to hapless frogs and such. I'd prefer the term quinquisect(ion) as the continuation of bisection, trisection and quadrisection, which have Latin number prefixes attached to a Latin-derived form. For Greek-derived terms, which do not suggest division in equal parts but more a distinction into categories, we have dichotomy, trichotomy, tetrachotomy, pentachotomy, ...  --Lambiam 06:25, 2 December 2007 (UTC)

Dissection is a term sometimes used in geometry, but not quite for this. Algebraist 18:24, 2 December 2007 (UTC)

Consider a spherical frog. . .  --Lambiam 20:33, 2 December 2007 (UTC)

Somebody prove Riemann already

COME ON PEOPLE I'M GETTING BORED HERE

its been too long

Also is it 100% certain that Poincare's Conjecture is proven thanks to Perelman's proof of Thurston's geometrization conjecture? Someone mentioned to me that the proof was still under review although that might be wrong. now talk —Preceding unsigned comment added by 81.153.211.188 (talk) 19:07, 2 December 2007 (UTC)

Among the Millennium Prize Problems the Riemann hypothesis was still completely open as of this morning. As to the Poincaré conjecture, all groups who have examined the proof and are competent to judge it, have independently found that it holds. As far as I've heard, the paper has not yet been formally published, a prerequisite for the Millenium Prize. I'm not even sure it has been submitted for publication. For a proof that difficult, there is always a remote possibility that someone notices a subtle but irrepairable gap even after publication; only additional complete and completely formal proof verification by computer – not likely to be attempted any time soon – could provide a major further boost to our confidence in the result.  --Lambiam 20:50, 2 December 2007 (UTC)

Probablility Question - Help Me Win the Elks' Lodge Raffle!

OK. At 2:00 PM, California time (3 hours from this post), I will be heading out with my family for an event at the Elks' Lodge called "Christmas Tree Lane." They take several Christmas trees (about a dozen) and decorate them with donated prizes from sponsors - gift cards for stores, gift certificates for travel, etc. The usual value of prizes on each tree is about $250, often higher. People attending the event buy raffle tickets and then place tickets in a box at the base of each tree. At the end of the event, each box is taken and a winner pulled out. The winner of each box wins the prizes on that tree, as well as tree and decorations.

I intend to buy $20 in tickets. The question is this: What is the best strategy for how to distribute the tickets amongst all the boxes (i.e. which individual raffles to enter)? I know there are lots of variables (the values of each tree, the number of entries for each raffle) but I'm trying to draw up a generic strategy. My aunt Sharrie always puts one ticket in each box, then the remainder in her favorite. I'm leaning more towards finding the least favorite tree (i.e. less attractive prizes/decorations, perhaps the one in the corner with less visibility, etc) and dumping the whole twenty in there in the hopes of a higher probability of winning.

I know it's short notice, but I'd really like to harness the power of the Internet here. Can anyone suggest a strategy? I will post my results tonight after the event.

Thanks,

--KNHaw 19:08, 2 December 2007 (UTC)

That's really short notice, man! I'll give some quick ideas.

The problem can be stated as Maximizing

${\displaystyle \sum _{k=1}^{N}{\frac {y_{k}}{T_{k}}}V_{k}}$

subject to

${\displaystyle \sum _{k=1}^{N}y_{k}=Y}$...

...where

• N is the number of trees
• k=1...N indexes those trees
• ${\displaystyle y_{k}}$ are the tickets you put in box k
• ${\displaystyle T_{k}}$ are the total number of tickets you expect to be in box k

(you may think of it as ${\displaystyle O_{k}+y_{k}}$, where ${\displaystyle O_{k}}$ are the tickets you expect all the people but you to put in that box).

• ${\displaystyle V_{k}}$ is the value you asign to the kth tree. (You may notice I have taken expected utility for granted to state this problem).
• Finally, Y is the total number of tickets you bought.

You may then proceed to find the first-order conditions for your problem. You might want to manage such assignment so that for every box j where you put tickets (i.e. ${\displaystyle y_{j}\neq 0}$), then it must be the case that

${\displaystyle {\frac {O_{j}}{O_{j}^{2}+2Y_{j}O_{j}+Y_{j}^{2}}}V_{i}={\underset {k}{Max}}{\frac {O_{k}}{O_{k}^{2}+2Y_{k}O_{k}+Y_{k}^{2}}}V_{k}}$

You won't put a ticket in one box unless the benefit of doing so is at least equal to the gain of putting it in any other box.

If there are many tickets (the fraction of tickets you have is negligible) then you will put tickets in boxes for which the ratio

${\displaystyle {\frac {V_{i}}{T_{i}}}}$

is the greatest.

Pallida  Mors 19:57, 2 December 2007 (UTC)

As you may have noticed, I have assumed you can "estimate" the number of tickets the rest of the people will put in the boxes. That may have origin in some thumb rule ${\displaystyle U(V_{i})}$. But of course the analysis above can be extended to find Nash Equilibria and more complex equilibrium concepts, if you want to suppose that the other participants will behave according to some strategical motif. Pallida  Mors 20:16, 2 December 2007 (UTC)

Results

Well, I found a tree that seemed to optimize ${\displaystyle {V_{i}}}$ while keeping ${\displaystyle {T_{i}}}$ under control. Specifically, a tree done up with donated restaurant gift cards that turned out to be valued at about $1000 retail (vs. $300 for the others). Also, it drew very little attention, the most popular trees being the ones stocked with liquor (we are talking about the Elks, after all). When the tickets were poured into the rotating drum for the drawing, I noted perhaps only 20% more than the average tree and substantially less than the most popular trees.

Alas, it was not to be. I placed 30 tickets in the box but there were easily 1000+ in the drum when the winning ticket was pulled. So, while I might have managed to optimize my chances and payoff, I still had only a 3% chance of winning. The woman who did win was quite delighted, though, and all the money went to a good cause (with this particular tree, the Muscular Dystrophy Association).

In any case, thanks for the help! --KNHaw 02:35, 3 December 2007 (UTC)

Your words are welcome, KNHaw! And your last remarks on the raffle are celebrated. Pallida  Mors 10:08, 3 December 2007 (UTC)

Hopefully Easy Geometry Term Question...

What is the name or term for this situation: Two lines that are parallel, briefly come together to kiss or touch before veering off again, away from each other, never to meet again. It is some kind of non-Euclidean geometry thing maybe, or something I heard when someone was describing how to draw a Venn diagram? Or the orbits of planets and stars? I don't know, but it is driving me crazy. Thanks Saudade7 17:48, 2 December 2007 (UTC)

The usual definition of parallel lines in general two-dimensional geometries is that two lines are parallel if they do not intersect, so this phenomenon by definition cannot occur. Of course, it's possible for two such lines to meet 'at infinity': in hyperbolic geometry one calls lines 'parallel' if they meet only at infinity and 'ultraparallel' if they meet nowhere, not even an infinity. Other than that, you're going to have to come up with a new definition of 'parallel' to have parallel lines touching each other. Algebraist 18:20, 2 December 2007 (UTC)

Yeah, if two geodesics pass through a point and have the same tangent vector there, then they are the same geodesic. That holds in Riemannian geometry in general, which includes all the common non-Euclidean geometries. —Keenan Pepper 20:07, 2 December 2007 (UTC)

P.S. Why is this question green? —Keenan Pepper 20:07, 2 December 2007 (UTC)

You asked for the name or term. 'Two lines briefly come together to kiss or touch before veering off again, away from each other, never to meet again'. Well, it is called intersecting lines. You requested the lines to be parallel, but they are not parallel if they kiss or touch. Bo Jacoby 21:36, 2 December 2007 (UTC).

If you forget about straight lines and switch to general curves, then you may say they are osculating, which literally means "kissing". Confusing Manifestation(Say hi!) 21:43, 2 December 2007 (UTC)

The question is green because I thought it was pretty. I know that in *our* space parallel lines don't touch, but I know that there are types of geometry where lines that we think of as parallel can touch, or even cross. Since I am not a mathematician I thought this happened in "non-Euclidean geometry" maybe. But I yield to the analytic point that "parallel" by definition means that the lines never touch. I think that maybe osculating *is* the term I was thinking of! I don't need it for anything mathy, I am just describing something in a painting. Keenan Pepper's answer, "if two geodesics pass through a point and have the same tangent vector there, then they are the same geodesic" was mind-boggling, reminding me both of a field trip in 3rd grade to an exhibit of Buckminster Fuller's works, and a B-grade sci-fi movie I once saw about space-lizards. I think Confusing Manifestation's answer about osculation is what I was thinking of, so he is a Mystic, for sure, given that I apparently didn't explain myself well. But MUCH THANKS TO EVERYONE!!! Saudade7 21:52, 2 December 2007 (UTC)

UPDATE: Maybe Unsolved???

To clarify: Is there a mathematical term for the way these two blue lines touch (inside the red box)? When I looked at osculating things they all seemed to be concentric like yolks in eggs)...Thanks for your help... Saudade7 22:39, 2 December 2007 (UTC)

For your picture it just means that the curves are tangent, a particular form of Contact (mathematics), as the curvatures are opposites in sign they are not actually osculating. --Salix alba (talk) 23:13, 2 December 2007 (UTC)

Thank you Salix alba. Saudade7 01:24, 3 December 2007 (UTC)

Tangent is certainly the most correct term, but if you just called those curves 'touching' I think you'd be understood. Note that to a mathematician, a line is not what you might think it is: those blue things are not lines, but curves. Algebraist 01:59, 3 December 2007 (UTC)

Merci beaucoup Algebraist, Alas, I just wanted a cool term. "Touching" is so quotidian! I really remember hearing an actual word once. I am thinking maybe it was in a paper presented by an analytic philosopher called "Can things touch?" Apparently the answer was "No." Oh well. Thanks for your help! Saudade7 02:33, 3 December 2007 (UTC)

If you're after a cool term (and don't mind being wildly imprecise), just call the point where the curves touch a singularity. That word means all kinds of things, and I'm sure it's because it sounds good. Algebraist 03:36, 3 December 2007 (UTC)

You can get precise with the singularities, if you define the curve as a the zero set of a map from R to R, then the you have an example of an A₃ singularity at the origin, which has a normal form ${\displaystyle x^{2}\pm y^{4}}$. The picture also has additional symmetry and there are ways this can be captured in singularity theory by considering Bruce's fold maps. --Salix alba (talk) 12:45, 3 December 2007 (UTC)

Just to make life a little more complicated, there are situations where you could refer to the blue curves as lines (or line segments). For instance, I've seen it done when describing circles in the Poincare half-plane model, which these could very well be. Although I suppose the term geodesic would probably be more appropriate... 134.173.93.150 04:07, 3 December 2007 (UTC)

No, the two tangent circles could not be lines, even in the Poincaré half plane. In the Poincaré half plane, hyperbolic lines correspond to semicircles centered on the horizon line. Other circles not only appear curved; they actually are curved in hyperbolic geometry, i.e., they're not geodesics. Two geodesics cannot be tangent in any Riemannian manifold, as I said before. —Keenan Pepper 05:32, 3 December 2007 (UTC)

A point of osculation is known as a tacnode 86.146.175.166 20:40, 3 December 2007 (UTC)

December 3

Does .999... hold a distinct location on the number line?

Does anyone know if .999... has its own distinct location on the number line, seperate from 1. If it does not I believe that would mean that either no repeating decimal has a distinct location or that there are odd gaps in the continuum. If yes, I believe that would mean that .999... is exactly equal to 1 in all mathematical calculations, but has an inherently different value, which I believe is the case. Can anyone confirm or clear this up for me? —Preceding unsigned comment added by Southcrossland (talk • contribs) 15:22, 3 December 2007 (UTC)

Not sure if this helps, but we have an article on .999... - this "denotes a real number equal to 1". I would conclude that there would be no reason to make a distinction between them on a number line, but deep math isn't my strong suit. Friday (talk) 15:36, 3 December 2007 (UTC)

No, 0.999... and 1 occupy the same location on the number line (unless you move the goalposts and work in a non-standard number system that allows non-zero infinitesimals). I don't follow how you reach your conclusion that "either no repeating decimal has a distinct location or that there are odd gaps in the continuum". Gandalf61 15:54, 3 December 2007 (UTC)

If multiple decimal expressions can hold the same value, than the nature of our number system as consisting of digits of set incremental value seem destroyed, yet mathematically I understand how this occurs.

But that ("digits of set incremental value") is not the nature of our number system, rather just a convenient way we use to represent them - and yes, it does have some features which may seem counterintuitive. -- Meni Rosenfeld (talk) 17:37, 3 December 2007 (UTC)

4, 2+2, 2×2, and 0!+1!+2! all occupy the same spot in the sequence of natural numbers, because, although they are different representations of a number, they represent the same number. Likewise, 1/3 and 0.333... occupy the same spot on the real number line, because they are different representations of the same number. The intuition that different infinite sequences of digits, when interpreted as representations of a real number, necessarily represent different numbers, is wrong, and unjustified. On the contary, if 0.999... and 1 did not occupy the same spot, that would mean that there are truly odd gaps in the continuum, since there is a real number between every two distinct numbers, but there obviously can't be a number between these two. (There is just no space between them, since 1−0.999... = 0.000...; also, since 2×0.999... = 1.999... = 0.999... + 1, we see that an attempt to construct a number "halfway" between 1 and 0.999... is /₂(1+0.999...) = 0.999... again.)  --Lambiam 19:03, 3 December 2007 (UTC)

Yes, I understand different forms of representation. However, I was refering specifically to decimal expressions. Nobody has trouble with accepting formula based equalities or fractional representations. But within a specific system where each digit is meant to have a particular value and be of a particular distance from the previous and subsequent digit, it is hard to reconcile different decimal expressions being equal. It seems to undermind the soundness of our system that any number with repeating 0's or 9's share a single value with another expression, while all other numbers do not. Although I do suspect that ultimately this is logically reconcilable.—Preceding unsigned comment added by 192.154.91.225 (talk) 19:53, 3 December 2007 (UTC)

Basically, the 0.999... issue is, like many things in mathematics, unintuitive at first glance, but works, and breaks a lot of other things if you try to "fix" it. For example, if you try to make them different numbers somehow, then you'll find you'll break subtraction, or division, or the ability to tell when one number is bigger than another. If you really want, consider it an equivalence relation, where you simply identify the "symbol" 0.999... with the "symbol" 1, and treat them as equal from there (in the same way, fractions form equivalence relations, such that 1/2 = 2/4 = 3/6 = ..., another way that you have different symbolic representations of the same value, even in a single representational system). Confusing Manifestation(Say hi!) 21:54, 3 December 2007 (UTC)

<begin OR rant, delete at will>I think the 'unintuitive' character of this result has a lot to do with how real numbers and decimal expansions are taught at school, i.e. (in the UK) more or less not at all. As a result, people end up absorbing the idea that a real number is a decimal expansion, rather than is represented by one. As a result, they quite rightly have great difficulty grasping the simple idea that two different decimal expansions might happen to represent the same real number, while having no problem at all with such claims as "3/5=6/10", or even "3/3=1" which is not far from 0.999...=1. Unfortunately, I'm not yet mad enough to prescribe a proper course in real analysis at high school, so I'm not sure how the current situation can be improved. <end rant> Algebraist 23:02, 3 December 2007 (UTC)

Math Help

I'm having trouble solving the following problems in my math homework:

Problem 1. FACTOR: (x+3)^2-9. I got to ((x+3)+3)((x+3)-3), and I know that (x+6)(x) doesn't work, so I was wondering if I was getting anywhere on this. I know the answer is x^2, but (x)(x) doesn't give me the original problem.

Problem 2. FACTOR: x^12-1. I got (x+1)(x-1)(x+1)(x-1)(x-1)(x^2+x+1) and I was wondering whether this made any sense.

Problem 3. FACTOR: (x+3)^3+27. I got ((x+3)+3)((x+3)^2+(3x+9)+9) and I was also wondering if this made any sense.

I appreciate the help and any assistance you can give me. A touch in the right direction if these answers are wrong would be helpful. I'm not asking for answers, but for help.

S♦s♦e♦b♦a♦l♦l♦o♦s 21:10, 3 December 2007 (UTC)

On problem 1. remember that ${\displaystyle (x+3)^{2}}$ does not equal ${\displaystyle x^{2}+3^{2}}$. You can't distribute the exponent over the addition (only over multiplication). Your factoring is right, it's your simplification of the original problem to check your work where you went wrong.

For problems like number 2, I usually do the even factors first, so you get ${\displaystyle (x^{6}+1)(x^{6}-1)}$. Continue from there. A quick way to check for gross errors in factorings like this... count up the highest powers of x in the factors and see if you get the right total power of x to match the original problem. If you multiply your work back together you'll find you've only got ${\displaystyle x^{7}}$.

For problem 3, you're close but not quite. Remember that ${\displaystyle a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})}$. Do you see your mistake? Also, I'm guessing that you're supposed to simplify. It doesn't make sense to multiply out 3(x+3) but leave the rest unsimplified. Donald Hosek 21:19, 3 December 2007 (UTC)

I have corrected a few misprints in Donald's reply. -- Meni Rosenfeld (talk) 21:39, 3 December 2007 (UTC)

OK, but I'm still confused. Is the answer for problem one x(x+6)after ${\displaystyle (x^{2}+6x+9)-9}$ is simplified?

Is the second one then ${\displaystyle (x+1)(x^{2}-x+1)(x+1)(x^{2}-x+1)(x+1)(x^{2}-x+1)(x-1)(x^{2}+x+1)}$;or can I just leave it as ${\displaystyle (x^{6}+1)(x^{3}+1)(x^{3}-1)}$?

Is the third one then ${\displaystyle (x+6)(x^{2}+3x+9)}$ It doesn't seem to work out when I do ${\displaystyle (x+6)(x^{2}-3x+9)}$, though.

Thanks again for your support! S♦s♦e♦b♦a♦l♦l♦o♦s 21:48, 3 December 2007 (UTC)

1 and 3 are correct. In 2, ${\displaystyle (x^{6}+1)(x^{3}+1)(x^{3}-1)}$ is correct but you should probably factor it further; you have a mistake in your factorization of ${\displaystyle x^{6}+1}$. -- Meni Rosenfeld (talk) 21:55, 3 December 2007 (UTC)

Wait, which one for 3?

I don't see the mistake in 2. S♦s♦e♦b♦a♦l♦l♦o♦s 21:56, 3 December 2007 (UTC)

Oh, didn't notice you gave two expressions for 3. The first one is correct, but you don't need me for that - you can just expand everything and see that it is the same.

As for 2, what did you get is the factorization of ${\displaystyle x^{6}+1}$? Expand it. Does it end up as it should? -- Meni Rosenfeld (talk) 22:01, 3 December 2007 (UTC)

I see: is it ${\displaystyle (x^{2}+1)(x^{4}-x^{2}+1)}$? S♦s♦e♦b♦a♦l♦l♦o♦s 22:11, 3 December 2007 (UTC)

Yes it is. ${\displaystyle x^{4}-x^{2}+1}$ can be further factored into ${\displaystyle (x^{2}-{\sqrt {3}}x+1)(x^{2}+{\sqrt {3}}x+1)}$, but since this involves nonintegers you may not have to do it. -- Meni Rosenfeld (talk) 22:47, 3 December 2007 (UTC)

Corrected your expression. Of course, we can factor further over the complex numbers, but that can't be what's expected (if only because it's too easy!). Algebraist 22:51, 3 December 2007 (UTC)

Of course these are pretty standard factorizations for Algebra I/Algebra II/College Algebra (more the latter two than the first, but given the time in the school year, I'm guessing that this is either a strong High School Algebra I course or possibly high school Algebra II). In that context, we're factoring over ${\displaystyle \mathbb {Z} }$ so all those odd cases are just confusion for the original poster, as entertaining as they might be for us. Donald Hosek 23:44, 3 December 2007 (UTC)

But doing 2 properly is so beautiful! I suppose the OP will have to wait till university like everyone else... Algebraist 00:43, 4 December 2007 (UTC)

Checking nth power residuosity

Is there a way to check if some residue is really an nth power residue?

For example 4^x mod 31 produces

To check for cubic residuocity you raise the residue to (31-1)/3 and mod it with 31 and see if this equals 1.

However in this case all of the values generated by 4 are already cubic residues.

So is there a way to check if the residues really are 3rd powers of 4 mod 31?

By that I mean could you check if they are 9th powers of something? or do some other test.

24.250.132.195 21:29, 3 December 2007 (UTC)ForgotMyLogin

Your question is not clear to me. Mod 31, the only 3rd power of 4 is ${\displaystyle 4^{3}=2}$. -- Meni Rosenfeld (talk) 21:48, 3 December 2007 (UTC)

I used mod 31 because I thought this would simplify the problem. I was asking how I can check any residue generated by 4 to see if it is of the form 4^(3x) mod 31

24.250.132.195 21:57, 3 December 2007 (UTC)ForgotMyLogin

(ec)I don't understand your question very well. Yes, every power of 4 mod 31 is a cube mod 31, since 4 = 16^3. What do you mean by 'really are 3rd powers of 4'? Surely there is only one 3rd power of 4 mod 31, to wit 4^3 = 2? For the last part, I haven't thought about this stuff for ages, but I think the easiest way to more-or-less solve all such problems at once is to find a primitive root mod 31 (3 is one). Given this, the ninth powers mod 31 are exactly 3^9, 3^18, 3^27, 3^6, 3^15, 3^24, 3^3, 3^12, 3^21 and 3^0 (where the exponents are the multiples of 9 mod 30). Hopefully someone for whom number theory is a less distant memory will be able to give a better algorithm. Algebraist 21:59, 3 December 2007 (UTC)

(After reading clarification) Oh, all of {4,16,2,8,1} are of the form 4^(3x) mod 31. Algebraist 22:01, 3 December 2007 (UTC)

Yes that is true, but only 2 is 4^3x where the exponent is less then the order of 5. If you don't use a bound on the exponent then all the residues of 4 would be of the form 4^3x mod 31. The larger exponents don't really count because when the exponent is modded with the order of 4 it becomes equivalent to 4^3x mod 31. —Preceding unsigned comment added by 24.250.132.195 (talk) 22:08, 3 December 2007 (UTC)

Are you worried because (31-1)/3 = 10 is not divisible by 3? This just means that every third power is also a ninth power mod 31. Similarly, every number is a seventh power mod 31. This sort of reasoning works mod p for any prime p. Modulo prime powers p^k, as long as n is not divisible by p, a number is an n'th root mod p if and only if it is an n'th root mod p^k. Modulo composite numbers, you just check mod each prime power. When the power n and the modulus are not coprime, then things are a little more complicated for prime powers, but not too bad. If you want to actually find the roots, there is even a relatively straightforward algorithm, but it does require the modulus to be factored. JackSchmidt 22:32, 3 December 2007 (UTC)

I am trying to find a congruence for the exponent; for example 4^x mod 31 = 2 where x would be 3z+0; 4^x mod 31 = 4 where x would be 3z+1. ect. Normally this would only be possible if the order of 4 was divisible by 3, but I am trying to figure out a way to do this if the order of 4 is not divisible by 3 but p-1 is. —Preceding unsigned comment added by 24.250.132.195 (talk) 00:26, 4 December 2007 (UTC)

I'm still not sure what you want. 4^x =2 mod 31 iff x=3 mod 5, and 4^x=4 mod 31 iff x=1 mod 5. Are those the kind of answers you want? Algebraist 00:40, 4 December 2007 (UTC)

ok 4^1 = 4 mod 31 should return nope;4^2 =16 mod 31 should return nope;4^3 =2 mod 31 should return yep;4^4 =8 mod 31 should return nope; 4^5 mod 31 =1 should return nope; 4^6 mod 31 would just mess everything up. I am just interested in residues generated by 4 during the first cycle (from exponent = 0 to exponent=4). It is assumed the the residue is always generated by the exponent being in this limited range.24.250.132.195 00:58, 4 December 2007 (UTC)ForgotMyLogin

After reading this 31 times I still don't understand what you are asking. It is a true statement that 4 ≡ 4 (mod 31), so why should this "return nope"? What does it mean that a statement returns nope? How, and isasmuch as what, would 4 mod 31 "mess everything up"? Is there a relationship between the example here with nope/nope/yep/nope/nope/mess and the initial powers mod 31 ? Generalizing the definition of quadratic residue, I could imagine the question: Given p, r, and n, does there exist a number x such that x ≡ r (mod n). However, I don't recognize this in your question. Can you formulate the problem in the form: "Given a, b, c, ..., are there x, y, z, ... such that <some relation between these variables> holds?".  --Lambiam 19:50, 4 December 2007 (UTC)

I don't know why I have to keep repeating myself. For an equation A^X mod P; I want to know the divisibility of the X, in this case by 3. Is X divisible by 3 when order of A is ((p-1)/3) for some residue generated by A? what is so hard to understand? To make to problem unambiguous I limit the exponent to be at most the order of A.128.227.194.167 (talk) 22:21, 4 December 2007 (UTC)ForgotMyLogin

I'm sorry to say this, but we are repeating ourselves in our requests for clarification because the way you express the problem is not abundantly clear. You want to know the divisibility of X by 3. I got that. But what is given? A^X mod P is not an equation. What do you mean in this context by the order of A? The order in the multiplicative group mod P? Is it given that P is a prime number? And why "for some residue"? What do you mean by "some residue generated by A"? Do you mean the remainder of A on division by P? Again, if you could take the time to reformulate the question, making explicit (1) which variables are given (and which are not); (2) for which variables a solution is asked; (3) what the constraints are relating the given (input) variables and the output variables; that will in the end save you more time than it costs, and it will save all of us frustration.  --Lambiam 02:02, 5 December 2007 (UTC)

Have you taken a look at Quadratic reciprocity and Cubic reciprocity? I think they're along the right lines. Unfortunately, they're focused on solvability, not the solution itself, and would say "yes, x has a square root mod p" but not "the square root of x mod p is y". Black Carrot (talk) 04:02, 5 December 2007 (UTC)

There's also for the fourth-power case. The general case seems to be Artin reciprocity, but our article is both dense and a stub, so not much use. looks a bit more detailed, but I haven't read it yet. Black Carrot (talk) 04:47, 5 December 2007 (UTC)

December 4

Numbers in matrices

Hello everyone,

Some friends of mine made me notice some amusing fact about matrices and determinants :

Choose any number and multiples of it.

Put these numbers in a matrix, treating each digit as an entry (enough numbers to make a square matrix).

Then the determinant of that matrix is a multiple of the original number. (Doesn't have to be base 10)

Example :

Taking original number 174, multiples 348,522 and 870

${\displaystyle \left|{\begin{array}{ccc}3&4&8\\5&2&2\\8&7&0\end{array}}\right|=3\left|{\begin{array}{cc}2&2\\7&0\end{array}}\right|-4\left|{\begin{array}{cc}5&2\\8&0\end{array}}\right|+8\left|{\begin{array}{cc}5&2\\8&7\end{array}}\right|=3(-14)-4(-16)+8(35-16)=-42+64+152=174}$

and 174 is a multiple of 174.

So, my question is, how come this works ? What are the properties of determinants involved here that make this work ? And does this apparently innocent result hide something of greater importance ?

--Xedi 18:19, 4 December 2007 (UTC)

This is easy to understand using the properties of determinants. If you add a scalar multiple of one column of a matrix to another column, the determinant is unchanged. So we can add 10 times the middle column to the right column:

${\displaystyle \left}$

and then add 100 times the left column to the right column:

${\displaystyle \left}$

and this matrix has the same determinant as the original matrix. There's only one more property of determinants we need to use: if you multiply any column by a scalar, the determinant is also multiplied by that scalar. So recognizing that the right column now consists of multiples of your original number, we can write the determinant as

${\displaystyle \left|{\begin{array}{ccc}3&4&8\\5&2&2\\8&7&0\end{array}}\right|=174\left|{\begin{array}{ccc}3&4&2\\5&2&3\\8&7&5\end{array}}\right|}$

which is obviously a multiple of 174 because the determinant on the right is an integer. —Keenan Pepper 19:10, 4 December 2007 (UTC)

In other words, you can restore the original numbers in one column by multiplying by the matrix

${\displaystyle \left}$

which has unit determinant. It's simple to generalize this to higher numbers of digits and different bases. —Keenan Pepper 19:16, 4 December 2007 (UTC)

One way to see this is to ignore the requirement that the entries must be digits. Just start out with any three (or n) numbers, x,y,...,z as the first row. Since the property you mention (being a multiple of some particular number) is not affected by being multiplied by integers, you can replace x,y,...,z with any multiple without changing the result. Let's just keep it as x,y,...,z. Now when you did the multiplications for your matrix, you did not just multiply x,y,...,z by some number, you also grouped the new answer into "digits". In other words, for (X,Y,...,Z) = a*(x,y,..,z), you replaced X,Y,...,Z by X+1,Y-10,...,Z when you noticed a carry in the "Y" position. Determinants (up to integer multiples) are not affected by adding an integer multiple of one row to another. So we can simply subtract back off X,Y,...,Z from that row without affecting the truth of the statement. Similarly determinants are not affected (up to integer multiples) by rearranging the rows. Subtracting and rearranging, one finally gets a matrix like this:

${\displaystyle {\begin{bmatrix}x&y&\ldots &\ldots &z\\1&-10&0&\ldots &0\\0&1&-10&\ldots &0\\\vdots &&&&\vdots \\0&0&\ldots &1&-10\end{bmatrix}}}$

whose determinant (up to integer multiples) is ${\displaystyle x\cdot 10^{n}+y\cdot 10^{n-1}+\ldots +z}$, or one gets a matrix with a row that is all zeroes and whose determinant is 0, also a multiple of ${\displaystyle x\cdot 10^{n}+y\cdot 10^{n-1}+\ldots +z}$. Of course you can replace 10 with any constant.

At any rate, while writing this, I notice from the edit conflict that User:Keenan Pepper has also given a similar answer working in the other direction, so I suspect this is more than enough. JackSchmidt 19:23, 4 December 2007 (UTC)

Thanks a lot, it's perfectly clear. Couldn't have wished for more. --Xedi 19:57, 4 December 2007 (UTC)

Why dont i get this?

I will start off by saying that this is not a homework question. this is one that the teacher did with us in class and i dont understand it.

Now: I was given the function ƒ(x)=|x+2|-15, and ƒ(x-6)=__. The way i see it, if one was to insert the (x-6) in for the original x, than x=x-6 must be true, which means x=-6 is true. but its not because -6-6≠-6. so how does this work?the juggreserection 20:59, 4 December 2007 (UTC)

Am I right in thinking you are given ƒ(x)=|x+2|-15 and need to find ƒ(x-6) ?

If so, recall what f(x) means : it is the value the function f takes at x.

Let's take a second function g(x) to explain :

Let g(x) = x, then g(3) = 3 = 9

In the same way, g(y) = y or g(rabbit) = (rabbit).

So you also have g(x-6)= (x-6) = x-12x+36

Hope that helps --Xedi 21:12, 4 December 2007 (UTC)

The question was about something slightly deeper, which we are so used to doing without thought that we don't even realize there is something problematic. We know that ${\displaystyle g(x)=x^{2}}$ for any x. In particular, we "know" it is true for x equal to ${\displaystyle x-6}$, so ${\displaystyle g(x-6)=(x-6)^{2}}$. But how can x be ${\displaystyle x-6}$? -- Meni Rosenfeld (talk) 21:16, 4 December 2007 (UTC)

Yes, indeed, when we use f(x) and then f(x+1), we don't think about x and x+1 being the same thing. ${\displaystyle f(\square )=|\square +2|-15}$ and we put whatever we want in the square. --Xedi (talk) 21:38, 4 December 2007 (UTC)

That's indeed a bit tricky, because we use the same letter x to mean different things at different times. This can be clarified by using another letter instead. For example, ${\displaystyle f(x)=|x+2|-15}$ is of course the same as ${\displaystyle f(y)=|y+2|-15}$. Now we can use ${\displaystyle y=x-6}$ to have ${\displaystyle f(x-6)=|x-6+2|-15=|x-4|-15}$. Is this clearer? -- Meni Rosenfeld (talk) 21:13, 4 December 2007 (UTC)

My belated two cents: The confusion arises because function notation looks like a normal equation, but is in fact a different thing entirely. If we have x+y=4 and x=y, we can replace the first x with a y (or vice versa), but only because we know they're equal. In the example f(x)=x+1, we are using x merely as an example. As pointed out above, f(rabbit)=(rabbit)+1 is still the same function. So, when you replace x with x-6, you are not saying that they are equal, you're simply replacing the example with what you actually want to work out. Daniel (‽) 21:55, 5 December 2007 (UTC)

Another way of looking at it is that the original problem, as given by the teacher, has a missing universal quantifier. When we say "ƒ(x) = |x+2|-15", what we really mean, by convention, is that "ƒ(x) = |x+2|-15 for all x" — but to be really rigorous, that "for all" should be explicitly written out. If we did that, we'd see immediately that x in that statement is a bound variable, and is not the same x as the one in "ƒ(x+6)", which is outside the scope of the quantifier, and therefore unbound; in particular, since the x in "ƒ(x) = |x+2|-15 for all x" is bound, we could (and should, for clarity) replace it with any other symbol, such as "ƒ(z) = |z+2|-15 for all z", "ƒ(ξ) = |ξ+2|-15 for all ξ" or even "ƒ(♠) = |♠+2|-15 for all ♠", and it would still mean the same thing. (From a programming viewpoint, you could think of the x in the definition of ƒ(x) as being local to the function, but it's not quite the same thing.)

Of course, they probably don't usually teach predicate logic or variable binding in school, at least not at the level where they're teaching about functions like the one in the example given. This is probably a good thing, since it would probably just confuse most students at that stage — but it's also true, as we've seen here, that not teaching about it can also cause confusion. At the very least, it would've been less confusing if the teacher had used different letters for the two variables, giving the problem as something like "if ƒ(y)=|y+2|-15 (for all y), then what is ƒ(x-6)?". —Ilmari Karonen (talk) 00:33, 8 December 2007 (UTC)

December 5

Extensions, Polynomials, etc

How do you pronounce things like Z and Q(x,y)? Black Carrot (talk) 21:27, 5 December 2007 (UTC)

Generally, I use pauses that make it pretty clear where the brackets are. For example (x+y)-z would have (x+y) pronounced as a phrase, with a pause before -z. If there's much chance of it being ambiguous, I'd say the brackets ('open brackets' etc.). Daniel (‽) 21:49, 5 December 2007 (UTC)

I sometimes do that, and sometimes say things like 'Z adjoin x'. You (might) have to be more careful with the second example, of course, due to the difference between Q(x,y) and Q. Algebraist 22:16, 5 December 2007 (UTC)

In situations where it could be unclear, (in my limited experience) we always said 'Z bracket x' for the first, while reserving adjoin (as in 'Q adjoin x and y') for fields. 134.173.93.150 (talk) 05:34, 6 December 2007 (UTC)

If by ${\displaystyle Q(x,y)}$, you mean that Q is a function of x and y, I think it would be said, "Q of x and y". Strad (talk) 00:14, 6 December 2007 (UTC)

In the context, I think we're talking about the field of rational functions in two indeterminates over ${\displaystyle \mathbb {Q} }$. That's certainly what I was talking about. Algebraist 00:25, 6 December 2007 (UTC)

What's wrong with this proof?

Quite some time ago, I found the following proof on the internet:
Proof that 2=1


1) X=Y  ; Given
2) X^2=XY  ; Multiply both sides by X
3) X^2-Y^2=XY-Y^2  ; Subtract Y^2 from both sides
4) (X+Y)(X-Y)=Y(X-Y) ;Factor
5) X+Y=Y ;Cancel out (X-Y) term
6) 2Y=Y ;Substitute X for Y, by equation 1
7) 2=1  ; Divide both sides by Y


Since I'm fairly sure that 2 != 1, this is probably wrong, but I can't figure out where the mistake is made. Any ideas?
Thanks in advance. Horselover Frost (talk) 23:05, 5 December 2007 (UTC)

Between 4 and 5, you divided both sides by X-Y, which is 0 according to the initial assumption. Dividing by zero makes funny things happen. 69.246.218.176 (talk) 23:10, 5 December 2007 (UTC)

There's also another error: From 6 to 7, you divide by Y without knowing if it is 0 or not. The solution to ${\displaystyle 2y=y}$ is not ${\displaystyle 2=1}$ but ${\displaystyle y=0}$. -- Meni Rosenfeld (talk) 23:13, 5 December 2007 (UTC)

(ec) Once again we see that Misplaced Pages has an article on everything. Check out invalid proof for this plus a number of more cunning ones. Btw, this proof works in the trivial ring, in which you can divide by zero. Fortunately, in this case, 1 does equal 2. Algebraist 23:18, 5 December 2007 (UTC)

Limiting behaviour of Markov Chain

Hi I have a stochastic matrix which represents a Markov Chain. The Markov chain basicly describes the probabilities of a simple game. The game involves 2 people A,B, with 5 counters, at each round theres a probability p that A gains a counter and a probability (1-p) B wins one of A's. I want to find limit of the probability matrix. Here is the matrix:

${\displaystyle {\begin{array}{cccccr}1&0&0&0&0&0\\(1-p)&0&p&0&0&0\\0&(1-p)&0&p&0&0\\0&0&(1-p)&0&p&0\\0&0&0&(1-p)&0&p\\0&0&0&0&0&1\end{array}}}$

(where the column index, from 0, is the number of counters A has) I've found, using eigenvectors, that a solution is 6 rows of (alpha 0 0 0 0 beta) where alpha and beta are chosen arbitrarily but is there any way of finding their exact values? I've not look too much at this topic but i am very interested so any help is appreciated. Thanks 212.140.139.225 (talk) 23:58, 5 December 2007 (UTC)

This exact question (in slightly greater generality) is answered at Gambler's ruin#Coin flipping. Note that you don't need to think about eigenvectors: it's obvious (I suppose formally you'd appeal to a Borel-Cantelli lemma or something) that eventually all the counters are in either A's hands or B's, so the only question is, given that A starts with n counters, what is the probability of A ending up with everything. If one denotes this p_n, one gets some recurrence relations in the p_ns, which are fairly easy to solve, giving the answers in the article I linked. Algebraist 00:17, 6 December 2007 (UTC)

Thanks for you reply. I'm only in my first year of a degree so I haven't come across Borel-Cantelli lemma and i was told the limting matrix can be made up of the eigenvector where the eigenvalue is 1. I see it is obvious that eventually one of the players will win but i thought using eigenvectors may tell me the probability of each player winning. I've had a look at the article you suggested but don't fully understand it, bearing in mind i am only a first year, could you explain the basic idea of recurrence relations? Thanks again 212.140.139.225 (talk) 15:50, 6 December 2007 (UTC)

The lemma I referred to is just the first thing that came to my head for proving rigorously that the game eventually ends: if you can see that this is obvious, then that's certainly good enough for a first year. The problem with using eigenvectors is that this only tells you the possible limiting distributions, which you knew already. You have to do more work to find out the probability of one result rather than the other. Let then p_n be the probability that A wins starting with n counters. We have boundary conditions P₀=0 and P₅=1, since in these cases the game has ended already. For n strictly between 0 and 5, there is a probability p that A will gain a counter (giving him n+1 in total) and a probability 1-p that he will lose one (giving him n-1). We thus have the recurrence relation ${\displaystyle P_{n}=pP_{n+1}+(1-p)P_{n-1}}$. Standard techniques (given in Recurrence relation#solving generally) allow us to solve this relation with these boundary conditions to obtain P_n for all n. Algebraist 18:12, 6 December 2007 (UTC)

December 6

desktop tower defence

I'm trying to figure out the number of possible arrangements of pellet towers. You can build 16 towers (I'm just going with what you start out with). There are 25 x 21 squares on the board, however, the spaces overlap. So if you put a tower in the upper left hand corner (1,1), you could not put another tower in (1,2). How many possible arrangements of the 16 towers are there? 70.171.229.76 (talk) 01:32, 6 December 2007 (UTC)

In case you haven't noticed, there is another, more subtle, restriction: you can't place towers so as to completely block off one of the entrances. As a first supercrude estimate, the answer is between 10 and 2x10. Algebraist 01:42, 6 December 2007 (UTC)

Oh yeah I forgot about that part... Let's just assume for this that that doesn't apply and that the only restrictions are the ones I mentioned above. 70.171.229.76 (talk) 02:17, 6 December 2007 (UTC)

OK, let's first consider the 25*21 grid, and say we were allowed to take any cell, except ones we've chosen. This gives ${\displaystyle {\dfrac {525!}{509!}}=O(10^{43})}$ as an upper bound. To calculate a lower bound, consider the 13*11 non-overlapping grid, which gives ${\displaystyle {\dfrac {143!}{127!}}=O(10^{34})}$. To calculate a better lower bound, consider that every cell we choose removes itself and 8 surrounding cells (adjacent and diagonals), which gives us something ${\displaystyle \prod \limits _{j=0}^{15}\left(525-9j\right)=O(10^{42})}$. Of course, ones on the edge remove 6, and corners remove 4. I'll think about this later. mattbuck (talk) 16:13, 7 December 2007 (UTC)

The bounds I gave are essentially the same as your first two, but I have been taking the towers to be indistinnguishable, so we disagree by a factor of 16!. Algebraist 16:54, 7 December 2007 (UTC)

Oops, my bad, forgot the 1/16! I needed to multiply by. mattbuck (talk) 16:57, 7 December 2007 (UTC)

Find the prime factorization 330

What is the prime factorization 330. —Preceding unsigned comment added by 69.119.211.11 (talk) 04:27, 6 December 2007 (UTC)

The first prime numbers are 2, 3, 5, 7, 11, 13, 17, 19. Try if some of them divide 330, and can be multiplied together to get the prime factorization of 330. PrimeHunter (talk) 04:37, 6 December 2007 (UTC)

See also Prime factorization, in case you don't know what this means.  --Lambiam 14:14, 6 December 2007 (UTC)

cryptography

Can some one please suggest some good books on cryptography. Thanks--Shahab (talk) 08:59, 6 December 2007 (UTC)

For general/historical interest, Simon Singh's "The Code Book" is a great read. However, it's not very technical and there is very little on modern cryptography. 134.173.93.150 (talk) 09:32, 6 December 2007 (UTC)

Books on cryptography might interest you. I'm biased, but if you want a comprehensive formal treatment, Goldreich's Foundations of cryptography might be a good choice. -- Meni Rosenfeld (talk) 16:23, 6 December 2007 (UTC)

More math help

My teacher, as unhelpful as ever, now assigned us new problems. I got several other problems like this using guess and check, but it was going so slow, that I wondered whether there was a better way to do this.

Number 1: FACTOR: ${\displaystyle 21x^{3}+7x^{2}y-35xy^{2}-42x}$ On this one, I factored out the Greatest Common Factor and got ${\displaystyle 7x(3x^{2}+xy-5y^{2}-6)}$. I'm not sure what to do now.

Number 2: FACTOR: ${\displaystyle 0.09x^{3}-0.66x^{2}y+1.21xy^{2}}$ I don't know how to start on this either. Is it prime (irreducable)?

Number 3: FACTOR: ${\displaystyle x^{2}-3xy-6x+18y}$ I also thought this was prime.

I need to understand how to do this. S♦s♦e♦b♦a♦l♦l♦o♦s 22:02, 6 December 2007 (UTC)

On 2, there is at least one factor you can pull out quite quickly, and at the same time you may want to consider pulling out a factor of, say, 1/100, to make things look a bit nicer. For 3, see if you can divide it into two expressions with two terms each, that when you factorise separately they have a common factor. For 1, I *think* it's irreducible from that point, but don't quote me on that. Confusing Manifestation(Say hi!) 22:42, 6 December 2007 (UTC)

OK, so I make 1: ${\displaystyle 9x^{3}-66x^{2}y+121xy^{2}}$? S♦s♦e♦b♦a♦l♦l♦o♦s 23:11, 6 December 2007 (UTC)

I may be stupid, but what is the goal of this and when is a term irreducible? I can factor any arbitrary function of x and y out of any of the terms. I suppose you want the result to be "simple" in some sense, but in what sense exactly? —Preceding unsigned comment added by 84.187.91.68 (talk) 23:56, 6 December 2007 (UTC)

Now, for 2, take out a common factor of x, and note that what remains has a particular, and hopefully familiar, form. It may help to notice that both 9 and 121 are squares, and that 66 is a multiple of their square roots... -- Leland McInnes (talk) 01:02, 7 December 2007 (UTC)

What you're doing in problems 1 and 3 here is factoring by grouping. The tip-off that this is a likely strategy is the fact that you're factoring a polynomial with four terms. The example in the[REDACTED] article, is, unfortunately, needlessly complicated. This should apply for all three problems. After you've pulled out all the GCFs, the next thing to do is look at the first pair of terms and see if you can factor out a common factor (a challenging problem set might require you to try terms one and three if terms one and two don't have a common factor, but that doesn't apply here). We then factor out a common factor from the last two terms. We should get some factor multiplied by a binomial + some other factor multiplied by a binomial with the two binomials matching. Then we can treat those binomials as a GCFs and get binomial times binomial.

Let's look at an example problem, factoring ${\displaystyle x^{2}-3x-xy+3y}$. In this case are first terms are x and -3x. We can factor out an x and get ${\displaystyle x(x-3)}$. Repeating the process on the last two terms we factor out y and get ${\displaystyle y(-x+3)}$ but remember that we can also pull out a negative and get instead ${\displaystyle -y(x-3)}$. That gives us an overall expression of ${\displaystyle x(x-3)-y(x-3)}$.

We finish up by factoring out the (x-3) from each term and end up with ${\displaystyle (x-y)(x-3)}$ (do you see where that x-y came from?) That should be enough to get you through this problem set. Donald Hosek (talk) 01:29, 7 December 2007 (UTC)

Quick update, on problem 1, you can (and should) try the grouping algorithm, just to see what happens when it doesn't apply, which in this case it doesn't. Donald Hosek (talk) 01:35, 7 December 2007 (UTC)

For problem 2, after you've factoring out GCFs, the fact that you've got perfect squares in the first and last terms suggests that you should use the perfect square pattern to try factoring. Donald Hosek (talk) 01:32, 7 December 2007 (UTC)

(cancel indent) Yes, 3x^2+xy-5y^2-6 is irreducible, unless I'm more drunk than I thought. Since it's quite unlikely you'd be set an example which fails to factor nicely, you might want to check you've got the question right. Algebraist 02:10, 7 December 2007 (UTC)

Sorry, I was eating dinner and doing some English homework. I compied the problems number for number, and in the directions, it says that some of the problems may actually be prime. Let me get my sheet out and evaluate everything... I will comment further. S♦s♦e♦b♦a♦l♦l♦o♦s 02:44, 7 December 2007 (UTC)

Yes, Leland, I forgot to factor out x and I noticed it while eating...I finished solving 2. I should have gotten it earlier, but I wasn't paying attention. ... Ah, I see what you are saying, Donald (Allow me to call you that). The example in the[REDACTED] article was overly-complicated. I do see where the x-y came from (excellent example). OK, I see how it doesn't work on 1. S♦s♦e♦b♦a♦l♦l♦o♦s 02:59, 7 December 2007 (UTC)

Thank you everyone on your helpful assistance! S♦s♦e♦b♦a♦l♦l♦o♦s

Years

Why do you guys believe in years? Years are not actually real. Neither are months or weeks. —Preceding unsigned comment added by Bane of Durin (talk • contribs) 23:53, 6 December 2007 (UTC)

A year is the length of time it takes the Earth to complete one full orbit of the sun and is quite "real", just like days, in the sense that it has an astronomical basis. Months are based on the lunar year, i.e. the time it takes for the moon to complete one full orbit of the Earth, but for historical reasons the calendar months are no longer synchronised with the lunar cycle. Weeks are religious in origin, with many religions striking upon a 7 day week. The reason months and weeks are still used even though there is no astronomical basis for them is the same reason hours, minutes and seconds are used. It's convenient to have units of time of varying length, and the week is deeply ingrained into much of human culture as part of the labour cycle. So it's not that people "believe" in these things, they're just lengths of time which are useful when discussing time intervals and for establishing human schedules. Maelin (Talk | Contribs) 00:26, 7 December 2007 (UTC)

Why do you believe in "Why"? And "are" and "or". These aren't real. How silly of you. risk (talk) 01:29, 7 December 2007 (UTC)

December 7

Probability question

This site's business model has intrigued me, and prompted this question. Say every person has to place a bid on an item, between $0.01 and $4.00, with any amount (except fractions of cents) within that range being valid. Assume each bid amount is equally likely to be chosen. How many people must bid for the probability that every possible bid amount has been bid on at least twice (i.e. there are no unique bids) is greater than 50%? I have very little background in statistics, so I have no clue as to how to approach this. Thanks in advance to anyone who can help. -Elmer Clark (talk) 01:14, 7 December 2007 (UTC)

I'm not going to answer this (because I can't), but note that 'every possible bid amount has been bid on at least twice' is not the same as 'there are no unique bids'. If two people bid, both choosing $0.01, then there are no unique bids. Algebraist 02:04, 7 December 2007 (UTC)

I realize that, but I figure this is the factor that will in most cases determine whether there are any unique bids, so I'm considering just it for simplicity's sake. -Elmer Clark (talk) 02:16, 7 December 2007 (UTC)

Alright: Yours is a question obout multinomial distribution: I'm not sure this will be easy to compute. However, you could make some simulations to have a rough idea. With more time, I may be able to make some simulations. Not tonight. Too tired :( Pallida  Mors 03:47, 7 December 2007 (UTC)

Hem... I guess I'm not that tired to let the question sleep for some hours. I started to do some simulations. My first guess is that the number lies between 3000 and 4000 bids (bidders, I mean). Pallida  Mors 05:43, 7 December 2007 (UTC)

A question 8 years in the making

I'm a college student, and I was looking through my old school assignments, and I found an old math assignment from 6th grade. The only problem I had gotten wrong was as follows:

"Jessica had 12 paper fraction pieces, each of twelfths, sixes, fourths, and thirds. He did not have enough pieces to show 1/3 and 2/4 as equivalent fractions. How can he form enough by making one cut with a scissors?"

My answer had been: "Put them in a pile and cut them all in half."

The teacher had circled it in red and wrote: "Nice try. Read the question again! :)"

What was the correct answer, what does the question even mean, and since when is 1/3 and 2/4 eqivalent fractions? --Ye Olde Luke (talk) 02:15, 7 December 2007 (UTC)

I... suspect the teacher intended to check whether you knew how to write 1/3 and 2/4 in common units (that is, as 4/12 and 6/12). I have to admit I find the question a little bizarre. Does the question say that Jessica has 12 paper fraction pieces each of the four types (for a total of 48), or 12 paper fraction pieces distributed in some unspecified way over the four types? Michael Slone (talk) 02:32, 7 December 2007 (UTC)

I worded the question exactly the way it was on the paper. Unfortunately, I don't remember anything about this incident back from 6th grade, so I don't have any more information than what I've already posted. I'm guessing my 6th grade self didn't have a clue either, and merely put down something that might concievably work, even if it's not the best or most efficient answer. Apparently the teacher knew the right answer, so it can't be unsolvable. --Ye Olde Luke (talk) 02:43, 7 December 2007 (UTC)

The answer is clearly 42. Errm... what was the question again?  --Lambiam 04:21, 7 December 2007 (UTC)

This is 6th grade, so the answer shouldn't be too hard to find. If we try solving the problem using both options (12 of each fraction, for 48 all-in-all; or 12 pieces distributed somehow over the four types), which one gives the most reasonable and succinct answer? --Ye Olde Luke (talk) 14:51, 7 December 2007 (UTC)

It's clearly the 12 in total. Otherwise, he would have enough pieces (10) to represent the given fractions as 12ths. Frankly, the question is too murky to be easily solved. Go find your teacher and ask :p mattbuck (talk) 15:31, 7 December 2007 (UTC)

I believe I am able to find a kind of solution, using a bit of imagination to complete the unclear indications of the problem. I suppose it means that we have at our disposal some fractions in twelfths, sixths etc., such that summing up one or more of them we cannot get 1/3 nor 1/2. So a priori we might have 2/3 (not 1/3), 1/4 and 3/4 (but not 2/4), 1/6, 4/6, 5/6 (but not 2/6 nor 3/6), and all fractions of the form k/12 for k=1, 2, ..., 11, excluding 4/12 and 6/12. Too many: we have fifteen of them.

Furthermore, 1/4 + 1/6 + 1/12 equals 1/2, and so does 1/12 + 2/12 + 3/12, so try removing 1/12; for similar reasons, remove 3/12 (1/4 + 3/12 = 1/2); in order to making it impossible to get 1/3, let us also remove 2/12 (for 1/6 + 2/12 = 1/3). So we are left with exactly twelve fractions. (I have not checked other possibilities for being left with exactly 12 fractions none of which sum up to 1/2 or 1/3.)

Cut in half "2/3", so you get 1/3, 1/3, 1/6 and the remaining ones, and the first three ones allow us to get both 1/2 and 1/3.

Sound too fanciful? Goochelaar (talk) 15:56, 7 December 2007 (UTC)

Sounds good. But how does that show 1/3 and 2/4 as equivalent fractions (i.e. 4/12 and 6/12)? --Ye Olde Luke (talk) 01:19, 8 December 2007 (UTC)

Riemannian manifolds

questuion1 : One R^n can had two Riemannian manifolds (M,g) (N,h)?

and

http://en.wikipedia.org/Conformal_map had said:

wiki--------------------------------------

A diffeomorphism between two Riemannian manifolds is called a conformal map

if the pulled back metric is conformally equivalent to the original one.

questuion2 : so conformal map can jest be a map not about angle?

questuion3 : conformal map can denote bijective map f:(M,g)->(N,h) then inverse f^-1(N,h)->(M,g)

between two Riemannian manifolds?

Thanks —Preceding unsigned comment added by 59.104.150.66 (talk) 05:01, 7 December 2007 (UTC)

On a Riemannian manifold the metric tensor can be used to define a Riemannian equivalent of angle - a concept that plays the same role as the normally understood angle does in Euclidean geometry. In the same way as conformal maps on the Euclidean plane preserve the "Euclidean" angle, so we can also define a class of functions between Riemannian manifolds that preserve the "Riemannian" angle - these are also called conformal maps. In fact, you could say that the "Euclidean" conformal maps are a special case of the more general "Riemannian" conformal maps. Conformal maps are invertible, and the inverse of a conformal map is also a conformal map. Gandalf61 (talk) 10:35, 7 December 2007 (UTC)

Solution to a nonlinear ODE

I want to solve:

y''' = C1*y + C2*cos(y)
y(0) = y0
y'(1) = yp1
y''(1) = ypp1

If possible, I want to do it in closed form, but if not the evaluation should be as cheap as possible in terms of computational effort. Any ideas how best I can do it? Regards, deeptrivia (talk) 21:35, 7 December 2007 (UTC)

OK, I'm assuming you mean ${\displaystyle {\dfrac {\partial ^{3}y}{\partial x^{3}}}=c_{1}y+c_{2}\cos y}$, with y(0) = y₀ but I'm confused as to what yp and ypp might mean. mattbuck (talk) 00:06, 8 December 2007 (UTC)

They're just some known constants. deeptrivia (talk) 02:40, 8 December 2007 (UTC)

Since the function on the right is independent of x, why not just integrate with respect to y three times and the constant of integration can be solved for using the initial conditions. A Real Kaiser (talk) 05:16, 8 December 2007 (UTC)

It doesn't work that way, I'm afraid. Try ${\displaystyle y''=y^{2}}$. By double integrating, you get ${\displaystyle -\log y=x^{2}+C_{1}x+C_{2}}$ or, with constants set to 0, ${\displaystyle y=e^{x^{2}}}$ and now ${\displaystyle y''=(d/dx)(2xe^{x^{2}})=(d/dx)(2xy)=2y+4x^{2}y\neq y^{2}}$. The technique fails because you are integrating ${\displaystyle \int (1/y^{2})d(dy/dx)}$, which doesn't work since there is no ${\displaystyle dy/dx}$ term in the integrand. SamuelRiv (talk) 06:34, 8 December 2007 (UTC)

Double Integral

Hello. Can I use a double integral when I add, subtract, multiply, divide, etc. the same number on both sides? For example, does ${\displaystyle (6x^{2}+2x)\times \iint _{}^{}\,{\frac {1-x}{2x}}={\frac {5x}{3x+1}}}$ constitute ${\displaystyle (6x^{2}+2x)\times {\frac {1-x}{2x}}={\frac {5x}{3x+1}}\times (6x^{2}+2x)}$? Thanks in advance. --Mayfare (talk) 23:59, 7 December 2007 (UTC)

I feel a bit confused about quite what you mean, but if you take any ${\displaystyle \int f(x)dx=F(x)}$ then ${\displaystyle g(x)\int f(x)dx=g(x)F(x)}$. Is that what you mean? If so, it works perfectly well for double integrals. mattbuck (talk) 00:10, 8 December 2007 (UTC)

December 8

Shortest distance between two points on the globe

Hi - was wondering if anyone could point me to how I might be able to work out the distance between two points on the globe (going through the Earth's crust) given longitude and latitude? I was previously under the impression that this was termed "straight-line distance", but have also seen "straight-line distance" used to refer to distance "as the crow flies", which is not what I'm after. Would really appreciate it if someone could clear things up... thanks in advance! --220.246.244.132 (talk) 09:06, 8 December 2007 (UTC)

Well, if you can translate the spherical polar coordinates (ρ = Earth's radius; θ = longitude east, or (360 − longitude west); φ = (90 − latitude North), or (90 + latitude south)) into rectangular coordinates, then you can apply the usual distance formula. Does that help? -GTBacchus 10:05, 8 December 2007 (UTC)

The closest distance between two points on a sphere is the great-circle distance. That article has formulas for computing it. --Spoon! (talk) 10:52, 8 December 2007 (UTC)

No, the great circle distance is across the surface, not through the crust. :) kmccoy (talk) 11:25, 8 December 2007 (UTC)

Change from spherical polars into cartesian by ${\displaystyle x=r\cos \phi \sin \theta ,\;y=r\cos \phi \cos \theta ,\;z=r\sin \phi }$, where θ is the longitude and Φ the latitude, then use pythagoras: <math>d = \sqrt{(\delta x)^2 + (\delta y)^2 + (\delta z)^2}. ~~~~ {{subst:Unsigned|1=Mattbuck|2=12:16, 8 December 2007 (UTC)}}