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February 6

Max Quad Area

How do I calculate the maximum area of a real quadrilateral given sides a, b, c, and d. The diagonals p and/or q are the variables. While we're at it, how about the minimum area as well. Thanks, hydnjo (talk) 00:05, 6 February 2012 (UTC)

The area is given in quadrilateral. Try the formula with the two opposite angles when they add up to two right angles - that gives a cyclic quadrilateral.. Dmcq (talk) 00:27, 6 February 2012 (UTC)

Umm... How do I calculate the maximum area of a convex quadrilateral given sides a, b, c, and d. The diagonals p and/or q are the variables. This isn't homework. hydnjo (talk) 03:33, 6 February 2012 (UTC)

Use Bretschneider's formula. Looie496 (talk) 04:01, 6 February 2012 (UTC)

@Looie496: Can't use as the interior angles are variable along with the diagonals. hydnjo (talk) 19:03, 6 February 2012 (UTC)

And as for max and min when you have sides A, B, C, and D and can arrange them in any order, there aren't many ways to arrange the sides, so try them all. I get three ways to arrange them, with the rest being mirror reflections and or rotations of one of those three (and thus with the same area):

ABCD
ABDC
ACBD
ACDB mirror and rotation of         ABDC
ADBC mirror and rotation of                ACBD
ADCB mirror and rotation of  ABCD
BACD mirror and rotation of         ABDC
BADC mirror and rotation of  ABCD
BCAD mirror and rotation of                ACBD
BCDA            rotation of  ABCD
BDCA            rotation of         ABDC
BDAC            rotation of                ACBD
CABD            rotation of         ABDC
CADB mirror and rotation of                ACBD
CBAD mirror and rotation of  ABCD
CBDA            rotation of                ACBD
CDAB            rotation of  ABCD
CDBA mirror of                      ABDC
DABC            rotation of  ABCD
DACB            rotation of                ACBD
DBAC mirror and rotation of         ABDC
DBCA mirror of                             ACBD
DCAB            rotation of         ABDC
DCBA mirror of               ABCD

StuRat (talk) 04:51, 6 February 2012 (UTC)

@StuRat: Assume sides a, b, c, and d are successive and of arbitrary length. hydnjo (talk) 19:03, 6 February 2012 (UTC)

There is a formula using p and q in the quadrilateral article but you are far better off looking at the angles in the formula pointed at above and ignoring p and q and just consider that the cos is squared so always positive or zero. Dmcq (talk) 09:12, 6 February 2012 (UTC)

@Dmcq: I understand that the maximum area is achieved when the quadrilateral with successive sides a, b, c, and d of arbitrary length is arranged to be a cyclic quadrilateral. I just don't know how to make angle ab + angle cd = 180º (or angle bc + angle da = 180º). Also, assume a convex quadrilateral. hydnjo (talk) 19:03, 6 February 2012 (UTC)

Well consider stretching out one diagonal as far as one can, then one side or the other of the diagonal forms a straight line, the angles at the end must be less than two right angles as a triangle is formed (or equal in which case an infinite radius circle goes through the points as they form a line). Then try doing the same with the other diagonal, since the other angles are now less than two right angles there must have been a point where each pair was two right angles. Dmcq (talk) 19:39, 6 February 2012 (UTC)

OK, got it. First find the area of the cyclic quadrilateral using Brahmagupta's formula and then find the interior angles from:
area = 1/2 (ab + cd) sin(angle ab). Thanks for your help. hydnjo (talk) 20:17, 6 February 2012 (UTC)

Category theory homework problem - subobject classifier and showing a morphism is self-inverse

Hello all, I'm working on the following Topos/Category Theory problem: you'll have to forgive me for my inability to LaTeX commutative diagrams, but I have in mind a commutative rectangle here composed of 2 commutative squares side by side, with both commutative squares being pullbacks. I'm really struggling with this one so if you think you can help please do.

Let $\epsilon$ be a category with finite limits and a subobject classifer Ω, "True" map $T: 1 \to \Omega$ (where 1 is the terminal object) and let $f: \Omega \to \Omega$ be a monomorphism in $\epsilon$. By considering the pullback square of f and T, which pulls back to an object U and morphism $u: U \to \Omega$ and the unique morphism $U \to 1$ say, and the pullback square of morphisms T and u to some object V (and the composite of these 2 pullback squares, 'joined' by u), show that $f \circ f$ is the identity morphism: 1_Ω.

I hope that made some sense, apologies if not. Now I've spent a while playing around with morphisms to try and get somewhere with this but I haven't managed anything so far and I was hoping you might be able to help: I've got an extremely hard problem sheet to hand in and i'm having trouble with a number of the questions so I'd be very grateful for any thoughts you could give whatsoever. Really, so far the only ideas I have are to use the fact that the 'composite' of 2 pullbacks in the way I've described is also a pullback square, and to use the fact that the 'subobject classifier property' applies to all monomorphisms, therefore we could apply the property to f to find some unique corresponding morphism $\chi_f: \Omega \to \Omega$ to get a third pullback square and then perhaps use that somehow, but otherwise I've really been unable to make progress. Could anyone please help? I'm trying to learn this subject so I would be very grateful for hints rather than answers (but please don't be too vague as I'm finding the problem rather difficult! :)) - many many thanks in advance, Totenines99 (talk) 00:23, 6 February 2012 (UTC)

Does an adjective exist to describe this situation?

Hello,

Is there an adjective to describe a planar entity that can expand infinitely along both dimensions of the plane it's contained within? (For example, a planar square that could hypothetically have infinite width or height would be such an entity.) All I can think of is "unbounded", which isn't sufficiently impressive-sounding for my purposes.

Any suggestions would be appreciated!

Hiram J. Hackenbacker (talk) 14:21, 6 February 2012 (UTC)

Unbounded is probably the answer. See articles about 'Unbounded <something>', especially Unbounded set. --CiaPan (talk) 15:07, 6 February 2012 (UTC)

you might also use infinite, although this means an object which actually is infinite, while unbounded denotes one that may be arbitraily large, but not necessarily infinite --CiaPan (talk) 15:10, 6 February 2012 (UTC)

I wouldn't suggest "unbounded" for this- the unit square seems to have the property you describe, but it is certainly bounded. But really I don't have any idea what exactly your property really means. Does a triangle in the plane have your property? What's an example of a set in the plane without this property? Staecker (talk) 17:08, 7 February 2012 (UTC)

I should have been a bit more specific - I was thinking of a dynamic entity, i.e., one with dimensions that can change (for example, as a function of time) existing on an unbounded plane. A square or a triangle with static dimensions would not possess the property in question, while one that could increase its planar dimensions in an unconstrained way (possibly reaching a point at which it would occupy an infinite area on its resident plane) would. Such an entity is purely theoretical (at least, as far as I know), but I'm curious as to whether there's a way to describe it that's a bit more precise than simply saying that it can expand in an unbounded fashion within the plane on which it lies. I think "unbounded" may end up being the best choice of adjective, but I was hoping there might be a more interesting term from topology or one of the more esoteric fields of mathematics. Hiram J. Hackenbacker (talk) 00:45, 8 February 2012 (UTC)

t-test help please

I'm trying to figure out whether the claim in [1] that white students are more likely to get into Oxford University than black students is justified or not. I created tables where the sample values are either 0 or 1, with sample sizes and success rates as described in the article (for x₁, 1089 1s and 2799 0s; for x₂, 13 1s and 54 0s). I'm trying to do a one-tailed heteroscedastic unpaired t-test. As my raw data I got:

$n_1=3888, \bar{x}_1=0.280093, s_1^2=0.280072, n_2=67, \bar{x}_2=0.19403, s_2^2=0.193468$.

Then I calculated:

s_pooled = 0.054402,t = 1.581967,degrees_freedom = 69 using the formulae in [2].

Am I correct so far? And if so, what do I do now? It Is Me Here ^{t / c} 15:38, 6 February 2012 (UTC)

You want to use Pearson's chi-squared test, not Student's t-test. Qwfp (talk) 16:38, 6 February 2012 (UTC)

Sorry, could you be a little more specific? It Is Me Here ^{t / c} 20:45, 6 February 2012 (UTC)

You have four categories (whites that got in, whites that didn't get in, blacks that got in, blacks that didn't get in) with a certain number of people in each and you want to see if those numbers could plausibly come from a particular distribution (namely, two binomial distributions with the same probability of success). That is what the chi-squared test is for. You need to work out the expected number in each category and compare that to the actual number in each category in the way described in the article Qwfp's linked to. --Tango (talk) 21:07, 6 February 2012 (UTC)

OK, so I got

O

13	1089
54	2799

$\hat{p}=\frac{13+1089}{13+1089+54+2799}=0.278635 \rightarrow X \sim \mbox{Bin}(n,0.278635)$

E

18.66852	1083.331
48.33148	2804.669

Χ² = 2.427138

Is that correct? How do I get degrees of freedom and test whether the value is significant? It Is Me Here ^{t / c} 12:00, 9 February 2012 (UTC)

That looks about right. The degrees of freedom is the number of cells (4) minus the number of parameters in your probability distribution that you got from the data (in this case, 3: the probability of someone getting in, the number of white applicants and the number of black applicants), so that's 4-3=1 (there's always 1 degree of freedom for this kind of 2-by-2 contingency table). You test for significance in the same way as for any hypothesis test - look up the critical value for the confidence level you are interested in (it's always a one-tailed test for chi-squared) and compare it to the test statistic you have calculated. If your value is greater than the critical value, you reject the null hypothesis. --Tango (talk) 22:19, 10 February 2012 (UTC)

Consider the Bayesian approach. See Bayesian inference#Binomial distribution parameter.

The probability for obtaining 1089 out of 1089+2799 is

$p_1\approx\frac{1089+1}{1089+1+2799+1}\pm\sqrt{\frac{\frac{1089+1}{1089+1+2799+1}\frac{2799+1}{1089+1+2799+1}}{1089+1+2799+1+1}}$

or

$p_1\approx 0.280206\pm 0.007200$

The probability for obtaining 13 out of 13+54 is

$p_2\approx\frac{13+1}{13+1+54+1}\pm\sqrt{\frac{\frac{13+1}{13+1+54+1}\frac{54+1}{13+1+54+1}}{13+1+54+1+1}}$

or

$p_2\approx 0.202899\pm 0.048067$

So

$p_1-p_2\approx 0.280206-0.202899\pm\sqrt{ 0.007200^2 + 0.048067^2}$

or

$p_1-p_2\approx 0.077307\pm 0.048603$

Zero is only 0.077307/0.048603=1.59058 standard deviations away from p₁ − p₂. That is not significant. Bo Jacoby (talk) 00:43, 11 February 2012 (UTC).

Froda's Theorem Flawed ?

The proof assumes that all points of discontinuity in the interval can be enumerated: Let x1,x2,...xn be the points of discontinuities of f. This assumes that the set of points of discontinuities is countable (since if they can be enumerated with the natural numbers) yet this is exactly what needs to be proven. One cannot assume the theorem is true and claim it shows the theorem is true. A correct proof would be to assume the set of points of discontinuity is not countable and show that this would contradict that f is continuous. If I can either develop or find a proper proof of the theorem I will submit it for review. Feedback would be appreciated. - Noah Vieira — Preceding unsigned comment added by Vieiranoa (talk • contribs) 16:47, 6 February 2012 (UTC)

It works from that there can only be a finite number of discontinuities greater than a given amount in a monotone function between two points on a closed interval because the monotone function can only change by a finite amount. So for each size the number bigger is countable and you just count the lot in decreasing size order. Dmcq (talk) 19:51, 6 February 2012 (UTC)

This makes sense, but I was a little confused too. I think that the number of jumps larger than some alpha being finite must have to do with the fact that, for a monotone function, (finite) left and right limits must exist for every point in the domain. Reading over the definition again, I conclude that e.g. Tan(x) is not monotonic on [-pi/2, pi/2]. Is that right? SemanticMantis (talk) 01:11, 7 February 2012 (UTC)

It is monotonic, but it isn't defined at the ends of that closed interval. The interval being closed is important because it means the function is bounded by the values at the two ends. Dmcq (talk) 11:44, 7 February 2012 (UTC)

Anyway I thought Froda's theorem didn't depend on the function being monotonic but what's there talks about monotonicity and looks a bit silly. It should talk about essential discontinuities as well where either the left or right limits don't exist - there the ones that can be uncountable. That turns it into a decent sized theorem. So yeah I think something there is flawed. Dmcq (talk) 12:02, 7 February 2012 (UTC)

The proof that the discontinuities of a monotone function are countable is much easier. Just pick a rational number within each jump. Sławomir Biały (talk) 12:06, 7 February 2012 (UTC)

True - but that's not the theorem I believe. I just had a look at the talk page to say something like this and somebody else has said it and more nine months ago with no reply. Dmcq (talk) 12:10, 7 February 2012 (UTC)

I'm not disagreeing. It would be strange to refer to this almost trivial result by a moniker. Sławomir Biały (talk) 12:30, 7 February 2012 (UTC)

Maple advice

Hi. I'm working on a presentation, and I'd like to make animated graphs. I've only used Maple a little bit, and in looking at the documentation, I'm a bit overwhelmed by my options for making plots. I'd like to describe the graphs I'm trying to create, and see if anyone can suggest the easiest Maple packages to use to do it. If there's another program that would be even easier, I'd be open to that, but I only have until Friday to learn it and make it happen.

The idea is this: I've got a curve in space given by three parametric equations, and a point on the curve.

1. As two points independently approach my fixed point, the line joining them approaches the tangent line to the space curve at my fixed point.
2. As three points independently approach my fixed point, the plane containing them approaches the osculating plane to the space curve at my fixed point.
3. As four points independently approach my fixed point, the spherical surface containing them approaches the osculating sphere to the space curve at my fixed point.

I've got the parametric equations for my space curve already, and I know how I want the sets of 2, 3 and 4 points to approach the fixed point. I know how to write down the equations for all the lines, planes and spheres that I need.

What I don't know is how to make it all happen with Maple, or another user-friendly software package. Can anyone help me with this? Thanks very much in advance for any tips. -GTBacchus^(talk) 20:59, 6 February 2012 (UTC)

Use the animate3d, for example:

with(plots):

animate3d([x,y,x^2+k*y^2],x=-1..1,y=-1..1,k=-1..1);

will give you an animate where k is the parameter. You can add commands like frames=100 to tell Maple how many frames to make, i.e. how slowly to animate:

animate3d([x,y,x^2+k*y^2],x=-1..1,y=-1..1,k=-1..1,frames=100,style=wireframe);

— Fly by Night (talk) 17:22, 7 February 2012 (UTC)

Sums of unit vectors of prime factions of pi?

Consider the set V = (Vp where p is prime and V is the unit vector from 0,0 to (cos(pi/p),(sin(pi/p)). Is the following statement true: The only sum of vectors in Vp (each vector occurs a non-negative number of times) (which go from 0,0 to xt, yt) where xt and yt are rational are the multiples of V2?Naraht (talk) 21:24, 6 February 2012 (UTC)

It is true. Instead of vectors in R², consider these objects as complex numbers in the complex plane. Then each V_p represents a primitive 2p^th root of unity ζ_2p. A rational vector in R² corresponds to an element of the field Q[i] = Q(ζ₄). All the elements ζ_2p are linearly independent from one another over Q (see cyclotomic field), so there's no non-trivial rational linear combination of them which is contained in Q[i], unless it's just a multiple of ζ₄. Rckrone (talk) 03:10, 7 February 2012 (UTC)

Thanx!Naraht (talk) 21:06, 9 February 2012 (UTC)

February 7

Trains allegedly haul 1 ton for nearly 500 miles on a gallon. How to calculate?

I saw on a commercial about a rail freight corporation that trains will haul 1 ton of cargo nearly 500 miles on a single gallon of fuel.

Would anyone please demonstrate how this statistic is calculated? Thank you. --70.179.174.101 (talk) 07:41, 7 February 2012 (UTC)

For one ton of cargo, you'd use:

(MILES TRAVELED)/(GALLONS OF FUEL USED)

However, I suspect that they didn't count the fuel needed to move the train itself:

(MILES TRAVELED)/(GALLONS OF FUEL USED BEYOND THAT FOR EMPTY TRAIN)

If so, then this statistic is rather useless. After all, almost all of the fuel used in a car or plane is also used to move the car or plane itself, and not it's cargo. StuRat (talk) 08:55, 7 February 2012 (UTC)

The formula used must be

(MILES TRAVELED)×(TONS OF CARGO)/(GALLONS OF FUEL USED)

Bo Jacoby (talk) 10:27, 7 February 2012 (UTC).

I would expect they mean miles per gallon for tons of cargo and the gallons is total over their whole operation so in fact it is quite a bit better than one might think. Trains have lower air resistance and 'tire' resistance, they go at an even speed for long distances and are quite good at recovering energy when braking, basically a diesel electric locomotive is a huge and extremely efficient Toyoto Prius. Dmcq (talk) 10:34, 7 February 2012 (UTC)

Diesel-electric trains certainly are efficient, but while the locomotive may employ dynamic braking, power will typically be dissipated in brake grid resistors, not stored in batteries as on a Prius. Electric trains connected to overhead lines or a third rail may return energy to the system during braking. (Freight trains usually try to avoid stop-and-go traffic.) -- ToE 01:24, 9 February 2012 (UTC)

We do have an article on Hybrid trains, but that are not currently in common use. -- ToE 01:39, 9 February 2012 (UTC)

Yes it looks like the ones with batteries are only being used in any great number for shunting rather than long haul and one would only get power returned for the others with overhead lines where it would just need an electric train anyway. Dmcq (talk) 02:05, 9 February 2012 (UTC)

Something else to consider is that trains only move cargo from one train depot to another. Unless your factory is at one depot and customer at another, you will therefore incur additional shipping costs for delivery. StuRat (talk) 10:45, 7 February 2012 (UTC)

We have an article Fuel efficiency in transportation which may be of some use.--Salix (talk): 17:43, 7 February 2012 (UTC)

Note that the 185.363 km/l (per short ton) quoted there is 436 mi/gal. -- ToE 01:39, 9 February 2012 (UTC)

Not too bad but not quite 500 but they must have had a reason for that, I wonder if having better locomotives in shunting yards could have made the difference. I'm not sure if I read it right but it looks to me that for passenger trains something like half the power is used in the passenger cars rather than for driving the train along! Dmcq (talk) 02:05, 9 February 2012 (UTC)

They might also be talking about movement on straight, level ground, with no accel or decel (so excluding starts and stops). StuRat (talk) 02:19, 9 February 2012 (UTC)

I'm pretty certain they mean delivered cargo and total fuel for the freight trains over the whole operation. Association of American Railroads: Freight Railroads Help Reduce Greenhouse Gas Emission in 2010 describes what they are doing. Dmcq (talk) 10:31, 9 February 2012 (UTC)

Angle bisector problem

Can someone help me how to do this? Triangle ABC has AB = 27, AC= 26, BC = 25. Let say I is the incenter point, where all three angle bisectors met. Find BI. Thanks!Pendragon5 (talk) 23:51, 7 February 2012 (UTC)

ABI is another triangle. Use the cosine rule to find the angles in ABC, divide the angles at points A and B by 2 to get the angles of this new triangle (ABI), then use the sine rule to find the length of BI. Widener (talk) 00:43, 8 February 2012 (UTC)

Ops forgot to say. No calculator is allowed.Pendragon5 (talk) 01:12, 8 February 2012 (UTC)

Is this homework? WP:DYOH RudolfRed (talk) 03:19, 8 February 2012 (UTC)

Nah?? If it is homework then i'm pretty sure i can do it by myself. Why would i not know how to do it unless i was not paying attention in class, which is not my type. If you wonder where this problem comes from then let me tell you it is part of the AMC 12 test, which i just took it today. And they are not easy. Don't try to advise me on whether or not i do my homework. Even if this is my homework, which it is not, so what? You're not helping me that's your choice. I have the right to ask.Pendragon5 (talk) 03:52, 8 February 2012 (UTC)

Chill out man. The policy of the reference desk is that we don't do people's homework for them. It's not obvious from your question that it's not homework, so RudolfRed asked. How can we know what "your type" is?

As for the question, drop the perpendiculars from the incenter, and you get 3 pairs of congruent right triangles. You can find the inradius from the side lengths with the formula in the article (it's √56). You can solve for the lengths of the remaining legs of the right triangles x,y,z by setting up the three equations x+y = 27, x+z = 26, y+z = 25 and find that y is 13. Then by Pythagorean theorem, BI is 15. Rckrone (talk) 06:02, 8 February 2012 (UTC)

If you don't know the formula for calculating the inradius, then you can also find it in the following way: The triangles ABI, AIC, and IBC all have height equal to the inradius r and base 27, 26, and 25 respectively. Therefore, their areas are 13.5r, 13r, and 12.5r respectively. The sum of these areas is equal to the area of the whole triangle ABC and is equal to 39r. From Heron's formula (which you should probably memorize) you also know that the area of ABC is $\sqrt{39\cdot 12 \cdot 13 \cdot 14}$ so we have $39r = \sqrt{39\cdot 12 \cdot 13 \cdot 14} \implies r=\sqrt{\frac{12 \cdot 13 \cdot 14}{39}}=\sqrt{\frac{12 \cdot 14}{3}}=\sqrt{56}$ which fortunately is all quite easy to do by hand. So clearly the method suggested by Rckrone is the way you're supposed to do it in this case. Widener (talk) 06:42, 8 February 2012 (UTC)

The no homework policy in my opinion is pretty stupid. What is the different between people who don't know something and people who don't know how to do their homework? This should be the place where everyone gets help regardless of whatever. I don't think people can really exploit this to do homework for them. First of all, homework is not just 1 problem but usually ranging from 10 up to 50 problems. Plus when you asked it here, you're not going to get the answer right the way. You probably have to wait few hours to just get the answer for ONE problem, it would takes forever (i meant they won't have enough time to finish it) for someone to try to do homework base on this. This desk reference can not be exploited as a tool to do someone else's homework anyway even if someone wants to. And I see no problem with helping someone on their homework with 1 or 2 problems. At least give them some direction to head to. So i don't think people here should even bother to ask is this homework because it doesn't do anything good. Plus people can always lie if they want to. The no homework policy here is just useless to be honest.

"You can solve for the lengths of the remaining legs of the right triangles x,y,z by setting up the three equations x+y = 27, x+z = 26, y+z = 25" I don't really understand this part. Can you explain in more details? Which right triangles are you talking about? What segments are you talking about? Name all of them please. And what is x, y, z stand for. By the way, is this base off from angle bisector theorem Pendragon5 (talk) 21:59, 8 February 2012 (UTC)

Hopefully this diagram makes it clearer: [3] Widener (talk) 00:53, 9 February 2012 (UTC)

IT IS! Thanks! Anyway i wonder what theorem is that property come from?Pendragon5 (talk) 01:19, 9 February 2012 (UTC)

What property? Widener (talk) 01:39, 9 February 2012 (UTC)

The property that enable us to set up three equations like this: x+y = 27, x+z = 26, y+z = 25. Pendragon5 (talk) 03:48, 9 February 2012 (UTC)

The reason we know that the two segments labelled as having length x both have the same length is that they are the corresponding sides of congruent triangles. We know those two triangles are congruent because they share the same hypotenuse and they have the same angle at vertex A (because the hypotenuse is the angle bisector). Similarly for the segments labelled y and z. Rckrone (talk) 04:22, 9 February 2012 (UTC)

Oh wow, pretty much common sense.Pendragon5 (talk) 19:20, 9 February 2012 (UTC)

About the homework policy: First is that people are allowed to ask for help with their homework questions. Reference deskers will guide people in the right direction and help them understand the material. What we won't do is hand out answers so they can get around grasping the material. Second, you are wrong about people being able to exploit the reference desk if we did give out answers. You're assuming that everyone's homework is structured like yours, which is not true. For instance in a college or graduate class, problem sets often consist of far fewer but more difficult proof questions. In some ways the question you asked is much more like these problems than the homework you're used to (although the subject matter is not something you'd typically see a class on). Rckrone (talk) 04:34, 9 February 2012 (UTC)

I didn't say people can. I meant they can't exploit it even if they want to because of the waiting time needed to received the answer. And of course beside that i agree with your points. That's also my point of view too anyway.Pendragon5 (talk) 19:18, 9 February 2012 (UTC)

February 8

Problem solving : work backward

How do I solve the problem? In June 2005, a sixth grade class planted a tree in the schoolyard. The tree grew about 3 inched a year. If the tree was 38 inches high in June 2010, about how high was the tree when it was planted? — Preceding unsigned comment added by 67.8.185.89 (talk) 04:16, 8 February 2012 (UTC)

Start with:

PRESENT HEIGHT = INITIAL HEIGHT + GROWTH

Now subtract GROWTH from both sides:

PRESENT HEIGHT - GROWTH = INITIAL HEIGHT + GROWTH - GROWTH

PRESENT HEIGHT - GROWTH = INITIAL HEIGHT

Now expand GROWTH:

PRESENT HEIGHT - (YEARS OF GROWTH)×(GROWTH PER YEAR) = INITIAL HEIGHT

Now just plug in your values to get the answer. StuRat (talk) 04:50, 8 February 2012 (UTC)

Time lag in second-order positive feedback

In a second-order positive feedback loop that would produce hyperbolic growth with no time lag, what's the shape of the curve if there's a constant time lag? And what about a time lag that's inversely proportional to the quantity -- does the latter still produce a singularity? NeonMerlin 04:46, 8 February 2012 (UTC)

Value of infinite limit

I'm having trouble with this, the equation of motion from the free fall article (position as a function of time) for a body falling under gravity but subject to air resistance proportional to the square of velocity:

$y = y_0 - \frac{v_{\infty}^2}{g} \ln \cosh\left(\frac{gt}{v_\infty}\right)$

As t becomes infinite, the equation should become

$y = y_0 - v_{\infty}{t}$

but I get a -ln2 term appearing. What's wrong?86.174.199.35 (talk) 17:19, 8 February 2012 (UTC)

It doesn't start going at rerminal velocity immediately so you need y0 adjusted by some constant amount. Dmcq (talk) 17:45, 8 February 2012 (UTC)

Of course. 86.174.199.35 (talk) 08:28, 9 February 2012 (UTC)

Actually thinking about it one can have an amount that grew to infinity if it did it slowly enough but that's not what's happening in this case thankfully :) Dmcq (talk) 13:06, 9 February 2012 (UTC)

February 9

Unique Factorization Domain

Are Z[sqrt(2)], Z[cuberoot(2)], and Z[sqrt(2),cuberoot(2)] unique factorization domains? Black Carrot (talk) 06:43, 9 February 2012 (UTC)

No. For instance, $119=7\times 17=(11-\sqrt{2})(11+\sqrt{2})$ and $123=(5-\sqrt[3]{2})(5^2+5\sqrt[3]{2}+(\sqrt[3]{2})^2)=3\times 41$. Sławomir Biały (talk) 12:23, 9 February 2012 (UTC)

But don't $7=(3-\sqrt 2)(3+\sqrt 2)$ and $3=(1+\sqrt[3]{2})(1-\sqrt[3]{2}+(\sqrt[3]{2})^2)$?

Plus fixed a minus to a + in your bit above hope okay Sławomir. Dmcq (talk) 12:42, 9 February 2012 (UTC)

Sorry, I was just being stupid. The class number of $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt[2]{2})$ is one. So both of rings of integers are UFDs. Sławomir Biały (talk) 13:17, 9 February 2012 (UTC)

How do you calculate the class number? Black Carrot (talk) 00:03, 10 February 2012 (UTC)

Well there's articles Ideal class group and class number formula here but if you really want to get into all that I can warn you that entire books have been written on the subject. Dmcq (talk) 09:33, 10 February 2012 (UTC)

The first, Z[sqrt(2)] (see Quadratic integer) is an Integrally closed domain (necessary for it to be a UFD, you have to remember to check the Z [?] guy is the whole ring of integers of the number field, as indicated above), and is even a Euclidean domain, using the norm, and hence a UFD, so everything is as nice & tangible as you could want there.John Z (talk) 11:08, 12 February 2012 (UTC)

Offsetting an ellipse

It's been a long time since I thought much about geometry, so bear with me, I'm sure this is rather elementary.

I am drawing an ellipse using a rather straightforward function that takes x,y as the center, two radius lengths (a and b), and a rotational angle as its arguments. The ellipse draws perfectly fine under such conditions. For a rotational angle of 0, a ends up being the horizontal radius and b ends up being the vertical radius, like this:

The function spits out a bunch of coordinates (the number depends on how many steps there are, but that isn't very important for this question) that I can then plot on a grid.

Here's the thing. I need to adjust the plotting of the ellipse so that x,y isn't the center, but coincides with one of the edges of the ellipse. So in the above example, I would shift the ellipse over so that what is now -a would be at point C. No problem for a single axis, of course (just offset every x coordinate by a), but how do I do this for arbitrary angles, keeping the -a point in the above ellipse centered on x,y? It smells like trigonometry to me and I'm very rusty on that.

Another way to do this would be to plot the ellipse coordinates at 0 degrees rotation, then rotate the ellipse with the rotational point being one edge of it. But I've no idea how to do those sorts of transformations on a bunch of x,y coordinates.

Any suggestions? --Mr.98 (talk) 17:06, 9 February 2012 (UTC)

As to rotating points, the general (computer guy, at least) thing is to apply a rotation matrix to each point. Slightly more generally, affine transformations (shears, translations, dilatations, and rotations) can be composed ad-nauseum by taking their corresponding transformation matrices and multiplying them together (in order of application). When applied to a point, the resulting matrix applies all those transformations in one go. So for example if you wanted to rotate something by 0.1 radians about (3,-10). First you'd translate to move the coordinate system so the centre of rotation is at the origin) - so you'd apply the translation matrix for x-3, y+10. Then you'd apply the rotation matrix. And then you'd translate back to where you came from (x+3, y-10). -- Finlay McWalterჷTalk 20:11, 9 February 2012 (UTC)

To be a bit more specific, let's say your basic ellipse is given by

$x=a \cos t,\,y=b \sin t,$

then the rotation counterclockwise by θ is

$x=a \cos \theta \cos t - b \sin \theta \sin t,\,y=a \sin \theta \cos t+ b \cos \theta \sin t$

and the translation so the center is at (h, k) is

$x=h+a \cos \theta \cos t - b \sin \theta \sin t,\,y=k+a \sin \theta \cos t+ b \cos \theta \sin t.$

--RDBury (talk) 21:57, 9 February 2012 (UTC)

(ec) I remember you do VB, so this may be helpful. Here's R (programming language) code that does what you want:

function(x,y,a,b,theta)
{
  # (x,y) is leftmost intersection of first axis with ellipsis
  # a and b are first and second axes
  # theta is rotation angle of first axis with x axis (in radians, of course)
 
  pp=seq(0,2*pi, by=0.01); # Note that pp now is an array
  xx= -0.5*a*cos(pp); # xx is an array (minus sign to place first point at left side of origin)
  yy=0.5*b*sin(pp); # and yy is an array
 
  # do the rotation (of the arrays xx and yy)
  xrot=cos(theta)*xx-sin(theta)*yy;
  yrot=sin(theta)*xx+cos(theta)*yy;
 
  # translate to make x,y correspond to intersection of first axis with ellipsis
  tx = xrot[1] # Arrays in R are 1-based, tx is first point in xrot
  ty = yrot[1] # Ditto for yrot
  xx = xrot + x - tx # xrot is array, x and tx are scalars
  yy = yrot + y - ty # yrot is array, y and ty are scalars
 
  # return result
  data.frame(x=xx,y=yy);
}

If you want to try it out in R, create the function using "fix(elip)", and substitute the empty function with the code above. To see the result, after defining the function type

dd=elip(5,90,7,2,0.3)
plot(dd,asp=1) # asp is the aspect ratio
abline(h=90) # horizontal line to highlight the leftmost intersection of first axis with ellipsis
abline(v=5) # ditto vertical line

--NorwegianBlue ^talk 22:48, 9 February 2012 (UTC)

Thanks! I will puzzle over this a bit further. --Mr.98 (talk) 20:55, 10 February 2012 (UTC)

Stationary functions of a functional

What kind of techniques would a mathematician employ when looking for the functions which minimise or maximise the functional given by $\mathcal{F}[f(x)]= \int_0^1 x^2 f(x) \text{d}x - \left( \int_0^1 x f(x) \text{d}x \right)^2$ subject to a constraint of the form $\int_0^1 f(x) \text{d}x = \text{constant}$ Thanks for your help — Preceding unsigned comment added by 129.67.37.224 (talk) 23:57, 9 February 2012 (UTC)

In general, the techniques for functional optimization problems are those of the calculus of variations. Here you have a quadratic functional on an affine hyperspace, which is unbounded both above and below. Consider f = c + th where $c\in\R$ is the given constant for the constraint, $\scriptstyle t\in\R$ and $\scriptstyle \int_I h= 0$. Then $t\mapsto \mathcal{F}[c+th]$ is polynomial in t:

$\mathcal{F}[c+th]= \big(c/3 - c^2/4\big) + \bigg( \int_Ix^2 h -c\int_I xh\bigg)t - \bigg(\int_I xh\bigg)^2 t^2 .$

Now, if for instance we fix an h with $\scriptstyle \int_I h = \int_I x h= 0$ and $\scriptstyle \int_I x^2 h \neq0$, e.g. h = x² − x + 1 / 6, the above quantity is a non-constant affine function of t, hence unbounded from above and from below (homework?)--pma 08:50, 11 February 2012 (UTC)

February 10

Minimum area of a quadrilateral

The maximum area has been resolve above at Wikipedia:Reference desk/Mathematics#Max Quad Area which made me curious as to the resolution of the minimum area of a quadrilateral.

Assume a convex quadrilateral with its longest side = a with successive and arbitrary sides = b, c, d and with a < (b + c + d). What would be a method to find the minimum area quadrilateral. hydnjo (talk) 01:41, 10 February 2012 (UTC)

I don't think there is a minimum. Imagine a really flat trapezoid. Widener (talk) 02:05, 10 February 2012 (UTC)

If you count degenerate quadrilaterals (and you imply that you can with the condition a ≤ (b + c + d) rather than a < (b + c + d)), then the minimum area is obviously zero. Widener (talk) 02:10, 10 February 2012 (UTC)

Hmm. Of course, notice that a > (b + c + d) is never possible anyway, so the condition a ≤ (b + c + d) is redundant actually. Widener (talk) 02:12, 10 February 2012 (UTC)

Yeah, I just added a < (b+c+d) so as to limit discussion about the degenerates (a ≥ (a + b + c) quads and it almost worked ;-) hydnjo (talk) 02:42, 10 February 2012 (UTC)

Oh, you changed the condition to from a ≤ (b + c + d) to a < (b + c + d). Be aware that it is still possible to create a degenerate quadrilateral which satisfies a < (b + c + d) Widener (talk) 09:15, 10 February 2012 (UTC)

There's the problem of do you allow self intersecting quadrilaterals, and if so do you count one of the areas as positive and the other negative which would be quite usual in maths so the minimum absolute are would always be zero. Otherwise it looks like the minimum would be give by one of the triangles like a-b,c,d where two adjacent sides were made to coincide. Dmcq (talk) 09:41, 10 February 2012 (UTC)

If you do use the interpretation which allows negative areas, then a self intersecting quadrilateral will not be convex. Widener (talk) 10:53, 10 February 2012 (UTC)

Am I misreading your question? Are the lengths a,b,c,d allowed to vary (subject to the conditions you mentioned)? Widener (talk) 10:28, 10 February 2012 (UTC)

I had thought at first that a was fixed but that b,c,d were allowed to vary (since you used the word "arbitrary" to describe them). Widener (talk) 11:56, 10 February 2012 (UTC)

I think we need more information. Precisely what is allowed to vary? Widener (talk) 11:00, 10 February 2012 (UTC)

If you keep all lengths fixed then you get a Four-bar linkage, a little off topic but a fascinating study.--Salix (talk): 11:20, 10 February 2012 (UTC)

That's assuming that the arrangement of the four sides is also fixed i.e. the only thing which varies is that one angle, in which case you can find the area of the quadrilateral as a function of the angle and then use optimization techniques such as or including basic calculus to find the angle which generates the minimum area, and then use that angle to find the area of the smallest quadrilateral. Widener (talk) 11:47, 10 February 2012 (UTC)

To review, the sides of the convex quadrilateral are successively a, b, c, and d of arbitrary length where the longest side is side a. Also, a < (b + c + d). In a previous question I learned that the maximum area is when the angles are such as to form a cyclic quadrilateral and then the area and the interior angles can be calculated and expressed in terms of the sides. After that discussion (which satisfied my need to know) I became curious as to the minimum area quadrilateral with the assumption that all of the interior angles are variable so long as the quadrilateral remains convex (no crossovers). Sorry for any confusion about the givens. hydnjo (talk) 22:40, 10 February 2012 (UTC)

Your use of the phrase "arbitrary length" may be causing some confusion. Are you asking, "Given positive values a, b, c, & d such that a is the largest value and a < (b + c + d), what is the minimum area of a convex quadrilateral whose sides are, in order, a, b, c, & d?"? -- 182.232.144.48 (talk) 03:28, 11 February 2012 (UTC)

Yes, a much more concise statement than my version - thanks. hydnjo (talk) 13:33, 11 February 2012 (UTC)

Then Salix's mention of Four-bar linkage is entirely on topic, as is Widener's optimization suggestion. While you have only one degree of freedom here, you will likely have to consider several cases, as illustrated in the "Types of four-bar linkages" diagram. -- ToE 05:38, 12 February 2012 (UTC)

This KMODDL page does not address minimum area, but it works through the different types of four-bar linkages. -- ToE 05:57, 12 February 2012 (UTC)

modulo arithmetic

Suppose n is an integer in the range 0-1000 inclusive. Let p, q and r be qual to n reduced modulo 7, 11 and 13 respectively (e.g. if n=20, then p=6, q=9, r=7). I believe that as 7, 11 and 13 are all prime, and 7*11*13=1001, then the combination of (p,q,r) will be unique. Is there a proof of this, other than by trial and error? Also, is there a formula for establishing n given p, q and r?  An optimist on the run! 15:07, 10 February 2012 (UTC)

Chinese remainder theorem. Gandalf61 (talk) 15:12, 10 February 2012 (UTC)

Sorry, but that article's way over my head. Is there a simpler answer to this?  An optimist on the run! 18:03, 10 February 2012 (UTC)

Have you read the section Chinese_remainder_theorem#Finding_the_solution_with_basic_algebra_and_modular_arithmetic? If you can work through that example, then you can solve your own question, and also determine whether it's unique. If you cannot work through that example, we can go from there. SemanticMantis (talk) 21:24, 10 February 2012 (UTC)

... and specifically for the 7, 11, 13 case, you can find n from p, q and r by calculating

$715p + 364q - 77r \ \text{mod} \ 1001$

So, for example,

$715 \times 6 + 364 \times 9 - 77 \times 7 = 7027 = 20 \ \text{mod} \ 1001$

Gandalf61 (talk) 09:15, 11 February 2012 (UTC)

I've tried working through the example, but floundered at the post where it said "(1/3) (mod 4) = 3 (mod 4)". Can I ask Gandalf how he derived the numbers 715, 364 and -77 in the above example? Thanks.  An optimist on the run! 13:32, 11 February 2012 (UTC)

For the first coefficient, 715, you want a number that is a multiple of 11 and 13, and leaves a remainder of 1 when divided by 7. To be a multiple of 11 and 13 it has to be a multiple of 11 x 13 = 143. So you can check multiples of 143 in turn:

$143 = 3 \ \text{mod} \ 7$

$2 \times 143 = 286 = 6 \ \text{mod} \ 7$

$3 \times 143 = 429 = 2 \ \text{mod} \ 7$

$4 \times 143 = 572 = 5 \ \text{mod} \ 7$

$5 \times 143 = 715 = 1 \ \text{mod} \ 7$

and that's how you get 715 (you can short-cut the last part a little if you know how to find multiplicative inverses in modular arithmetic). In the same way, the second coefficient has to be a multiple of 7 and 13 and leave a remainder of 1 when divided by 11, so you check the multiples of 91 until you find

$4 \times 91 = 364 = 1 \ \text{mod} \ 11$

and -77 is a multiple of 7 and 11 and leaves a remainder of 1 when divided by 13. Gandalf61 (talk) 13:56, 11 February 2012 (UTC)

Checking multiples mod 7 of 11×13 in turn, even easier:

11×13 = 4×6 = 24 = 3 mod 7

2×3 = 6 mod 7

3×3 = 9 = 2 mod 7

4×3 = 12 = 5 mod 7

5×3 = 15 = 1 mod 7

Bo Jacoby (talk) 16:16, 11 February 2012 (UTC).

Thanks for the explanation - it seems to make sense!  An optimist on the run! 23:45, 11 February 2012 (UTC)

Greatest Common Divisor of Two Polynomials

What is the difference between finding the greatest common divisor of two polynomials and finding their GCD to some (prime) modulus p? Could I be provided with any references for where to read around the issue? Thanks. 131.111.216.115 (talk) 17:01, 10 February 2012 (UTC)

Perhaps this is a question that lies in territory unfamiliar to the members of the Reference Desk. Could someone at least voice their thoughts on the issue, even if they don't have an answer as such? I would appreciate any help with which you could provide me. Thanks. 131.111.216.115 (talk) 12:01, 12 February 2012 (UTC)

The greatest common divisor between two polynomials is a polynomial. The greatest commom divisor between two polynomials modulo a prime is a polynomial modulo a prime. Different beasts. Do I understand you question correctly? Bo Jacoby (talk) 16:19, 12 February 2012 (UTC).

Solving the 3-body problem if two of the objects are stationary

The simulation here is of a gravitational system with two fixed sun and one to four movable planets. The values that can be changed are the mass of the first sun, the x position that the first planet starts at, and the number of planets. The starting x value of additional planets is an offset of the starting x value of the first planet. Additional planets do not interact with each other and instead are provided for comparing the differences in paths. I want to know if a 3-body problem with two fixed objects and one movable object can be solved exactly. --Melab±1 ☎ 20:49, 10 February 2012 (UTC)

You can't have two fixed objects. What is going to hold them still? They'll fall towards each other. What you can do, is have two objects in circular orbits around their centre of gravity (so they'll look fixed in a rotating reference frame) and then have a third object small enough for the effect of its gravity on the first two objects to be negligible. Then you can solve it reasonably easily. --Tango (talk) 22:37, 10 February 2012 (UTC)

There's no reason mathematically you can't have two fixed objects, the force on a satellite would just be the sum of the forces from the two objects individually. It may be that there could never be two real fixed objects in space, but that's something for physicists to determine. It appears from the simulation that the solution is chaotic, generally an indication that there is no closed form solution. That could be due to inaccuracies in the simulation though.--RDBury (talk) 23:18, 10 February 2012 (UTC)

It doesn't make much sense to solve a problem about gravity without taking into account the gravitational attraction between your two largest objects. --Tango (talk) 14:33, 11 February 2012 (UTC)

I think it is solvable in bipolar cylindrical coordinates. The problem is closely related to that of the electron in a diatomic molecule, and unless I am mistaken this is reasonably well understood. Note that by fixing two of the bodies, it is actually a two-body problem, i.e. there are only two frames relevant to the problem, albeit with an awkward force relating their motion. — Preceding unsigned comment added by 129.67.37.224 (talk) 23:58, 10 February 2012 (UTC)

If you have the two massive objects (e.g. stars) fixed, is that equivalent to having them in a circular orbit around each other and having the coordinate system rotate? Would that give the correct results? Bubba73 ^{You talkin' to me?} 06:29, 11 February 2012 (UTC)

I don't know what you mean by "the correct results". You can get results that way. The fact that you have a rotating reference frame would be very significant for the results, though. --Tango (talk) 14:33, 11 February 2012 (UTC)

What I'm saying is, suppose you have two stars of equal mass and they are at a distance and velocity to be in a circular orbit around each other. If you have a coordinate system rotating at that rate, the two stars would be fixed in that coordinate system. Now suppose you add an object to the system that has very low mass compared to the stars. It is easy to calculate the trajectory of the small body in that rotating coordinate system, with the two fixed stars. My question is: does that correspond to what would happen in the real world? Bubba73 ^{You talkin' to me?} 15:38, 11 February 2012 (UTC)

I would expect so. Intuitively, I'd expect the small object could either be far away from the two large objects, orbiting their barycenter in a roughly elliptical path, or close in, orbiting each object to form a figure eight. I would predict a range of instability between the distant orbit and close orbit. Do simulations match these predictions ? StuRat (talk) 20:18, 12 February 2012 (UTC)

Mercator latitude scaling

Tissot's indicatrix

Let's imagine I have a Mercator projection map centered on the equator and I've drawn a perfect square on it. Let's say that because of the scale of the map, the square is 1" by 1" which corresponds to 1 mile by 1 mile. Now if I get a map of a more northern latitude, and I apply the same 1" by 1" square, but while the 1" of width will still correspond to a mile, the 1" of height now corresponds with a greater value, like 1.25 miles.

What's the general formula to figure out how that 1" square will scale as I go to different latitudes? I've gone over the Mercator page, but math is not my strongest suit, and applying those equations to this specific problem is causing me some difficulty.

I'm using this for Javascript, and I'm not terribly good at translating mathematical formulae into Javascript, so if you could express this in a way that doesn't involve fancy Greek letters, I'd be very grateful... thanks. --Mr.98 (talk) 20:54, 10 February 2012 (UTC)

It looks like the Gudermannian function article might be of some use to you. See the applications section. — Fly by Night (talk) 21:36, 10 February 2012 (UTC)

Doesn't the fact that the Mercator projection is a conformal map suggest that, for relatively small squares, the distances represented by the sides are equal? -- 182.232.144.48 (talk) 03:43, 11 February 2012 (UTC)

When I plot a 20mi x 20mi square in Google Maps, the deformation is really palpable, even at straightforward (e.g. non-polar) latitudes. --Mr.98 (talk) 15:14, 11 February 2012 (UTC)

[4] shows no deformation for a ~20mi x ~20mi square at 60°N. Note that the tip of the marker is at equal distance from the lines at 60°N and 102°W, as it should be since 0.29 = 0.58 * cos(60°N). What's your evidence that Google Maps has deformation for relatively small squares? 98.248.42.252 (talk) 19:36, 11 February 2012 (UTC)

I am using a Ground Overlay in the API that has a fixed height and width. To plot Ground Overlays in Google Maps you have to specify it as a pair of lat,lng coordinates corresponding to the SW and NE corners of the image. I'm dynamically playing the Ground Overlay depending on whatever the lat,lng of the center of the map is. So the formulae for where to plot the lat,lng is to take the center coordinates and subtract or add the height or width (divided by two) to them to end up with the box. If you do this, the deformation is really palpable. It's actually not even 20mi on each end; it's more like 5mi on each end. --Mr.98 (talk) 21:31, 11 February 2012 (UTC)

Do you mean that the image has a "fixed height and width" in pixels and you specify the lat/lng of the corners? If so, then you are fully in control of the scaling; if you want the height and width to scale equally, you need to choose your corners taking cos(latitude) into account. If not, then can you better explain the situation, possibly providing an image? -- ToE 01:36, 12 February 2012 (UTC)

The width of the map is equal at each latitude, but the actual distance around the sphere is proportional to the cosine of the latitude. If 1" on the map is 1 mile of actual distance at the equator, then at latitude θ, 1" will be cosθ miles. As 182.232.144.48 mentioned, height and width scale equally with the Mercator projection. Rckrone (talk) 05:26, 11 February 2012 (UTC)

February 11

Using mod (i.e. remainder after division) on complex numbers

I'm currently trying my best to solve a Project Euler problem, and I have a question about using the mod function on complex numbers (i.e. the remainder after the division one, not the traditional "modulus" for complex numbers). One of the lovely things about mod is that it sort of "distributes" over addition and multiplication, right? Like this:

(a+b) mod m = ((a mod m) + (b mod m)) mod m

(a*b) mod m = ((a mod m) * (b mod m)) mod m

(btw, I realize that you usually don't write it like that, like an operator, but I only really have experience in computer science, so forgive my poor math-notation). First off all: this obviously applies when a and b are integers, but it also applies when they are arbitrary reals, right (at least assuming that m is an integer)?

But what if you defined the mod function for complex numbers, so that it applies to both the real and imaginary part? I.e., what if you defined it "complex mod" like this:

z = a + bi

z mod m = (a mod m) + i*(b mod m)

Does those "distributive" properties still hold? I.e., if z and v are complex numbers with the mod-function defined just like I did, is this true?

(z+v) mod m = ((z mod m) + (v mod m)) mod m

(z*v) mod m = ((z mod m) * (v mod m)) mod m

In all these examples, a and b can be arbitrary reals, but m is always a positive integer.

PS. The problem I'm working on is Problem 258, and I think I've almost cracked it. I figured out a way that would work but would take a few hours or so, but since Project Euler has an official "should run in under a minute" policy, I'm trying to figure out a better way to do it rather than just leaving the computer on over night. If anyone is curious, I'd be happy to explain why I need to take mods of complex numbers in order to solve it (or at least why I think I do, I may be completely barking up the wrong tree). 80.216.1.161 (talk) 18:52, 11 February 2012 (UTC)

Yeah, it will "distribute" in the way you want. To see this, consider how you calculate addition and multiplication of complex numbers in terms of the real and imaginary components:

(a + bi) + (c + di) = (a+c) + (b+d)i

(a + bi) * (c + di) = (ac - bd) + (ad + bc)i

Both operations can be built up with addition, subtraction and multiplication, which all behave nicely mod m as you mentioned. Rckrone (talk) 03:14, 12 February 2012 (UTC)

If you know something about rings, this setting that you are working in is the quotient ring (Z/mZ)[x]/(x² + 1). So basically we are taking the integers mod m, and then adding a new element x, which satisfies the property that x² = -1 (so x is the imaginary unit). Rckrone (talk) 03:24, 12 February 2012 (UTC) Edit: Maybe it's clearer to write the ring as Z[x]/(m, x² + 1).

BTW, Project Euler, per se, doesn't appear to have a one minute requirement, if you read all the about and problem pages then time is not mentioned anywhere. I agreed that many problem discussion threads mention folk endeavouring to achieve under 1 minute and it is probably mentioned in the discussion forums, it is too heavily dependent upon the technology used. FWIW, I solve the problems in interpreted QBASIC and am happy with solutions that run in under an hour. -- SGBailey (talk) 20:33, 12 February 2012 (UTC)

February 12

Preliminary tests for regression analysis in SPSS; ANCOVA in SPSS

How do I check the validity of the assumptions of normality and homoscedasticity for a simple linear regression, in SPSS (please give detailed instructions if possible)?

Another issue - when I run an ANCOVA in SPSS, why is it that in the output table "Test of between-subjects effects" sometimes the "corrected model" is mentioned in the first line, but sometimes the table does not mention this parameter?

Thank you in advance. Gidip (talk) 08:57, 12 February 2012 (UTC)

Division by zero

Is it right to say that division by zero is infinite sometimes, meaning it's undefined? 212.170.181.95 (talk) 13:43, 12 February 2012 (UTC)

See division by zero. Infinite is different from undefined. It is sometimes okay to say division by zero gives infinity, for instance when dealing with the real projective line. Also in floating point computing they have signed zero which enables them to give either a positive infinity or negative infinity. Normally though division by zero is considered as giving an undefined result. Dmcq (talk) 14:20, 12 February 2012 (UTC)

My theory is this.

The infinitesimal (which is the theoretically smallest possible number greater than 0) is 1/infinity. If you consider the following:

$\begin{align} &V &\frac{1}{\tfrac{1}{V}+V}\\ &.1 &.099009900990099\dots\\ &.01 &.009999000099990000999900\dots\\ &.001 &.000999999000000999999000\dots\\ &.00001 &.0000099999999990000000000999999999900000\cdots\\ &\quad\dots{\color{white}_\Big|}&\quad\dots\end{align}\,\!$

and, if the infinitesimal (“iota”), ${\color{white}|}\iota\,\!$, is the smallest possible number, then

$\frac{1}{\tfrac{1}{\iota}+\iota}=\frac{1}{\infty+\iota}=0\,\!$

but now you have to consider $-\iota\,\!$: As $\sqrt{V{\cdot}V}=V\,\!$, then $\sqrt{V\cdot-V}=\sqrt{-1}{\cdot}V.\,\!$ Thus,

$\frac{1}{0}:=\lim_{V \to \iota}\frac{1}{\sqrt{-1}(\tfrac{1}{V}+V)}=\frac{1}{i(\infty+\iota)}\,\!$

So, since it involves $\sqrt{-1}\,\!$, it is still not a “real” quantity!

Of course, this is just my personal, unsubstantiated, unprovable theory, so don’t even think of using it anywhere in Wikipedia! P=)  ~Kaimbridge~ (talk) 15:24, 12 February 2012 (UTC)

We already have infinitessimal plus articles on non-standard arithmetic. Dmcq (talk) 16:18, 12 February 2012 (UTC)