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July 28

Solve this pair of two equations

Solve for x and y using two equations given below.
x² + y² = 25
x³ + y³ = 91
I have been finding the value of x and y using above two relations for a number of days but I failed. At last I gave up and put this one here. Please suggest me different ways to solve this. Also suggest me some simple tricks (or handy methods or videos) for solving problems like this. Thank you! Publisher54321 (talk) 03:54, 28 July 2013 (UTC)

You could use Newton's method to solve it numerically, but it turns out that it might be easier to just graph the two. They happen to intersect at a point where both x and y are integers. I don't know of a way to reduce these two equations to one except by substitution - first solving for y from the first equation yields $y={\sqrt {25-x^{2}}}$, which we can substitute into the second equation to obtain $x^{3}+(25-x^{2}){\sqrt {25-x^{2}}}=91$. We move the $x^{3}$ term to the right, square both sides, and solve.-Jasper Deng (talk) 04:00, 28 July 2013 (UTC)

By observation, x = 3, y = 4 and x = 4, y = 3 are both valid solutions. You can easily deduce that solutions must satisfy |x| ≤ 5 and |y| ≤ 5, plotting the curves will give you a good idea of the number of solutions. EdChem (talk) 04:13, 28 July 2013 (UTC)

Indeed (although it's more correct to say they must be on the circle with r=5 and center at the origin than that) - the only way to solve it algebraically is to then try to factor that sixth-order polynomial via synthetic division - it might take a long time, but that's the only algebraic way I can think of.--Jasper Deng (talk) 04:18, 28 July 2013 (UTC)

• Addendum: After plotting both equations, I see there are in fact four solutions. These first two points with integer coordinates aren't that hard, but there are two others that have to be found using the quartic formula.--Jasper Deng (talk) 05:45, 28 July 2013 (UTC)

The equations are symmetric: if (x,y) is a solution then so is (y,x). Therefore each equation can be written in terms of the product p=xy and the sum s=x+y, and knowing p and s you find x by solving the equation x² − sx + p = 0. The equations are

0 = s² − 2p − 25 = s³ − 3ps − 91
= 3s(s² − 2p − 25) − 2(s³ − 3ps − 91)
= s³ − 75s + 182 = (s − 7)(s² + 7 s − 26)

Bo Jacoby (talk) 11:13, 28 July 2013 (UTC).
The equations are solved like this.
p = (s² − 25)/2
x² − sx + p = 0
x² − sx + (s² − 25)/2 = 0
x = (s + (50 − s²)^1/2)/2
y = s − x
(s − 7)(s² + 7 s − 26) = 0
There are six solutions: the following three and their symmetricals.
s₁ = 7 ; x₁ = (7 + (50 − 7²)^1/2)/2 = 4 ; y₁ = 7 − 4 = 3
s₂ = (3×17^1/2 − 7)/2 ; x₂ = (s₂ + (50 − s₂ ²)^1/2)/2 ; y₂ = s₂ − x₂
s₃ = (− 3×17^1/2 − 7)/2 ; x₃ = (s₃ + (50 − s₃ ²)^1/2)/2 ; y₃ = s₃ − x₃
Bo Jacoby (talk) 21:53, 28 July 2013 (UTC).

Really, six solutions? The graph of the two only suggests the existence of four, which makes two of those solutions extraneous.--Jasper Deng (talk) 23:20, 28 July 2013 (UTC)

Only four of the solutions are real. Sławomir Biały (talk) 00:07, 29 July 2013 (UTC)

Bézout's theorem guarantees 6 solutions (including multiples and complex).RDBury (talk) 00:46, 29 July 2013 (UTC)

Solutions: (3, 4), (4, 3), (4.61313, −1.92847), (−1.92847, 4.61313), (−4.84233+3.3088i, −4.84233−3.3088i), (−4.84233−3.3088i, −4.84233+3.3088i) Bo Jacoby (talk) 05:08, 29 July 2013 (UTC).

I'm sure it was one of the last two that the original poster was looking for. :) Naraht (talk) 15:37, 29 July 2013 (UTC)

i imagine so. :-) StuRat (talk) 06:11, 1 August 2013 (UTC)

The OP was looking for tricks. It is luck that these equations are solvable by square roots. Algebraic equations 0=f(x,y)=g(x,y) are solved by eliminating y getting 0=h(x) where h is a polynomial. In our case h(x) has degree 6, as pointed out by Jasper Deng, and so there are six solutions. Bo Jacoby (talk) 05:14, 2 August 2013 (UTC).

July 29

Inconsistency in definition of figurate numbers.

There seems to be an inconsistency in the manner by which figurate numbers are defined. Triangle, Square and Hexagonal are generated by determining the figure that matches their respective grids with a minimum side equal to the figurate number and then the other figurate numbers, 5, and 7 up seem to be generated so they sort of match the hexagon (center entry and then wrap perimeters of the shape n times until you get to the size. This means that for the figurate shapes larger than 4, they are basically equal to x*Tsubn +1 where x is the number of sides and Tsubn is the nth triangular number (in the current sense). (or no center entry and they are just x*Tsubn)

However this would be viewed as very odd for Triangular and Square. The triangular numbers would go 1, 4, 10, 19, 31, etc and the square numbers would go 1, 5, 13, 25, 41. Is there any way to get any real consistency here other than to define the triangle, square and hexagon ones in the current manner and just ignoring the rest? This isn't really dealt with in Figurate numbers.Naraht (talk) 15:56, 29 July 2013 (UTC)

There are two separate families of two-dimensional figurate numbers - polygonal numbers and centered polygonal numbers. Is it possible you are confusing the two types ? For example, the sequence of centered triangular numbers does indeed start 1, 4, 10, 19, 31 ... Gandalf61 (talk) 16:18, 29 July 2013 (UTC)

I think that covers the difference, however from an external standpoint, I think the polygonal numbers "feel more right" for the triangle and square, and the centered polygonal number for a hexagon. For example, the hexagonal numbers (which are "cornered") just look odd to me.Naraht (talk) 16:57, 29 July 2013 (UTC)

Differentiation under the integral sign without absolute convergence

From the article Differentiation under the integral sign#Other problems to solve there's a clever little trick:

$\int _{0}^{\infty }\;{\frac {\sin \,x}{x}}\;dx\to \int _{0}^{\infty }\;e^{-\alpha \,x}\;{\frac {\sin \,x}{x}}\;dx$

The idea is that differentiating under the integral sign gives an easier integral, and then setting α = 0 lets you evaluate the original integral. But as for the validity of this method, the article just says "The first integral is absolutely convergent for positive α but only conditionally convergent when α is 0. Therefore differentiation under the integral sign is easy to justify when α > 0, but proving that the resulting formula remains valid when α is 0 requires some careful work." Unfortunately the article doesn't give a lot of clues about when differentiating under the integral sign is valid. In fact this sentence is the only reference to "absolute convergence" in the entire article, so even the requirements for the more basic case are not expounded at any point. In the more complicated case of α = 0, what is the "more careful work" that the article is hinting at? (This is related to an earlier question on which I got a helpful response but I think is sufficiently different to justify a new question.) ManyQuestionsFewAnswers (talk) 21:50, 29 July 2013 (UTC)

There are actually two limits that need to be commuted with the integral. First, for $\alpha >0$, you need to justify that differentiation under the integral sign is justified. This can be shown by the dominated convergence argument, and requires that (for each fixed $\alpha$) the difference quotients

${\frac {1}{h}}{\frac {\sin x}{x}}(e^{-(\alpha +h)x}-e^{-\alpha x})$

be bounded by an absolutely integrable function of x (which in this case is true, so differentiation under the integral is justified). The second limit needed is that

$\int _{0}^{\infty }{\frac {\sin x}{x}}\,dx=\lim _{\alpha \to 0^{+}}\int _{0}^{\infty }e^{-\alpha x}{\frac {\sin x}{x}}\,dx.$

Since the first integral doesn't converge absolutely, dominated convergence doesn't work and you have to use a direct $\epsilon \delta$ argument (or emulate the proof of Abel's theorem). Sławomir Biały (talk) 23:01, 29 July 2013 (UTC)

This is a very helpful reply thanks. I will go away and try to put some flesh on this skeleton argument to see if I have absorbed it correctly - would be too greedy of me to ask others to fill it in! I do think the Differentiation under the integral sign could really be improved by inclusion of some of this material, especially the dominant convergence argument. 23:21, 29 July 2013 (UTC) — Preceding unsigned comment added by ManyQuestionsFewAnswers (talk • contribs)

July 31

Software for making images

Hello. What software was used to make the red and gold model on tetrahedron? Is it free? If not, what free software would you guys recommend to produce similar images? Thanks, Robinh (talk) 21:39, 31 July 2013 (UTC)

I am having trouble identifying which one is red and gold, but I recommend clicking on it and contacting the editor who created it on their talk page. Apteva (talk) 21:41, 31 July 2013 (UTC)

Probably Robinh refers to File:Tetrahedron.png (it has red sides, silver bars on edges and gold balls on vertices) or File:Compound of two tetrahedra.png (similar, plus yellow sides). Both pictures were made with Stella software, as their Licensing sections say. --CiaPan (talk) 05:27, 1 August 2013 (UTC)

In addition to Apteva's good advice, Stella (software) has been used to make a number of polyhedral images used in WP articles, such as the stella octangula image in the tetrahedron article. --Mark viking (talk) 22:27, 31 July 2013 (UTC)

If you're looking to make similar looking images, but not necessarily of polyhedra, the programs POV-Ray and Blender are a good possibility. The downside of those programs is that they leave the mathematical heavy lifting to you. -- 205.175.124.72 (talk) 22:59, 31 July 2013 (UTC)

(OP) I should have clarified that I meant free as in beer. Stella looks the best, but is expensive. Blender and POV-ray both look good. I don't mind doing the heavy lifting, so long as the final result is mathematically exact. Anyone got any further advice? Would blender be better than POV-ray? thanks, Robinh (talk) 01:51, 1 August 2013 (UTC)

if you are looking for free, you might also check out antiprism--it is open source and free, but not quite as do it yourself as Povray or Blender. --108.210.201.111 (talk) 03:22, 1 August 2013 (UTC)

I think the image was made with POV-Ray, File:Poly.pov looks like it might contain the source.--Salix (talk): 03:38, 1 August 2013 (UTC)

August 1

Brouwer's Fixed Point Theorem

In our article on Brouwer fixed-point theorem under the proof by homology section, I had two questions regarding the content there. The first, it says that the function F is a retraction, but I don't see why the boundary points would have to be fixed; I don't see that that is crucial to the proof though. Second, it says that the case for n > 2 is more difficult and requires homology, and while that proof makes sense, I don't understand why we can't use the same idea since π_n(Sⁿ) = Z for n > 0, unless the closed unit ball has a non-trivial group for higher n. Any help would be greatly appreciated:-)Phoenixia1177 (talk) 06:20, 1 August 2013 (UTC)

Every boundary point is fixed for F, that follows directly from the definition of F: if x is a boundary point, then whatever f(x) is (provided by definition, that f(x) ≠ x), the ray from f(x) through x meets the boundary at x, thus F(x) = x. --CiaPan (talk) 07:42, 1 August 2013 (UTC)

Wow, I feel really really stupid now. I flipped the direction in my mind as x going through f(x) to the boundary! I had a feeling I was getting something wrong since that's not the first time I've seen that version of the proof. At any rate, is there a similarly simple reason why we can't use the higher homotopy groups?Phoenixia1177 (talk) 07:54, 1 August 2013 (UTC)

Same 'direction flip' problem was explained at the article's talk page here in 2006. Anyway I do not understand the connection between a retraction and a group homomorphism, either, so I'm unable to help you with the central step of the proof. --CiaPan (talk) 08:52, 1 August 2013 (UTC)

As for why not use "the same idea" in higher dimensions, I'd say that the homology approach is the same idea. In the dimension 1 case, since the fundamental group is abelian, the fundamental group is the same as the dimension 1 homology group. So when you want to generalize to higher dimensions you can choose to generalize to higher homotopy or to higher homology. But homology is simpler and suffices to make the argument work, so using the homotopy groups is unneccesary. Staecker (talk) 13:43, 1 August 2013 (UTC)

That makes sense; the wording of going from fundamental group to homology makes it sound like there is a reason we couldn't use the higher homoptopy groups, this is what was throwing me for a loop (ha!). I'm guessing the reason the n = 2 case uses the fundamental group is because historically, and in order learned (usually?), the fundamental group would come first?Phoenixia1177 (talk) 13:59, 1 August 2013 (UTC)

That's what I'd assume. In my experience as a student and teacher, people generally learn fundamental group, then homology groups, then (if at all) homotopy groups. Most working mathematicians know basic things about the fundamental group and homology groups. Most nontopologists (and some topologists) know very little about the higher homotopy groups. Staecker (talk) 16:59, 1 August 2013 (UTC)

Speaking of what working mathematicians know and homology, brings me to another question. If you grabbed a mathematician, at random, would they be familiar with all the various cohomology theories out there? In general, I've always been curious, how many mathematicians have a working knowledge of various topics (like Galois Cohomology, or K-Theory, or Model Categories, etc.). Algebraic Topology and Algebraic Geometry, to me, seems to become "a nightmare in the key of category theory" after a point, so I've always wondered just how common a detailed understanding of it is.Phoenixia1177 (talk) 04:20, 2 August 2013 (UTC)

August 2

For those that enjoys probability puzzles

An initially non pregnant woman has three boyfriends (A,B and C). In a night of utter drunkenness, she went with one of them back to her room and had unprotected sex. The very next day , all three of her boyfriends where killed when their jet plane crashed into the twin towers in New York. None of their bodies were recovered. Each boyfriend has a style in lovemaking, Boyfriend A will kiss her one third of the time and use a condom half the time. Boyfriend B will kiss her half the time and use a condom one third of the time. Boyfriend C will kiss her definitely and never use a condom. Given that the woman is now pregnant from that night's sexual encounter and she remember being kissed. The kissing and using condom are events that are independent for boyfriend A and B. What is the probability for each of the boyfriend of being the father of her child? 220.239.51.150 (talk) 10:32, 2 August 2013 (UTC)

Insufficient information. Assuming uniform a priori distribution, 1/9, 2/9 and 2/3, respectively.--208.169.64.170 (talk) 12:12, 2 August 2013 (UTC)

Odds for A is 1/3×1/2= 1/6. Odds for B is 1/2×2/3=1/3. Odds for C is 1×1=1. So the credibilities are as 1:2:6 in accordance with the previous user. The result reflects the information given, which is therefore sufficient. Uniform prior distribution follows from the principle of insufficient reason. Bo Jacoby (talk) 14:12, 2 August 2013 (UTC).

My answer using Mathematica
kiss["A"]=1/3;
kiss["B"]=1/2;
kiss["C"]=1;
nocondom["A"]=1/2;
nocondom["B"]=1/3;
nocondom["C"]=1;

Next we do a Sum of person X where X={A,B,C}

probkissnocondom=Plus@@Map[1/3*kiss[#]*nocondom[#]&,{"A","B","C"}]
4/9

prob[x_]:=1/3*kiss[x]*nocondom[x]/probkissnocondom

Map[ {#,prob[#]} & , {"A","B","C"} ]
{ {A,1/8} , {B,1/8} , {C,3/4} }
220.239.51.150 (talk) 11:48, 3 August 2013 (UTC)

The probability of B not using a condom is 2/3, not 1/3.--63.136.113.135 (talk) 11:55, 3 August 2013 (UTC)

You are right. Re runing the mathematica code again, gives the result { {"A", 1/9}, {"B", 2/9}, {"C", 2/3} } 220.239.51.150 (talk) 03:46, 4 August 2013 (UTC)

orbifold questions

1) Given a manifold and a group acting on it, is the process of making an orbifold out of them a functor? Like, does the manifolds fundametal group nicely transform into the orbifolds fundamental group?

2) Is a projective space, gotten from taking the orbits of nonzero scalar matrices as the points(the orbits are lines thru origin, deleting origin) an orbifold, at least under the string theory definition of orbifold?

3) More generally, in the mathematical dfinition and/or int the string theory definition, can an orbifold be nontrivially a manifold? (I mean not just taking the trivial group and acting on the manifold with that)Thanks, Rich76.218.104.120 (talk) 12:44, 2 August 2013 (UTC)

August 4

math

what is the greatest common factor of32 and 28 plase ans — Preceding unsigned comment added by 39.53.158.255 (talk) 16:12, 4 August 2013 (UTC)

Why don't you read Greatest common divisor and see if you can't work it out for yourself? Rojomoke (talk) 16:58, 4 August 2013 (UTC)

$32=2^{5}$

$28=2^{2}\cdot 7$

The greatest common factor of 32 and 28 is $2^{2}=4$. Widener (talk) 17:38, 4 August 2013 (UTC)