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August 25

Help on ACT/SAT math

Hey guys. I'm taking the ACT and SAT in the near future. The concepts tested in the math section are pretty elementary, in my opinion (but I take AP calc :|). BUT! I still get one or two wrong in easy places when doing practice tests due to arithmetic errors. SAT math is ~25 questions per section and ACT math is 60 questions, and I generally have 5-10 minutes left at the end. I don't usually get the "hard" one wrong because to me they're not really hard to me. So does anyone have tips on how to reduce these, and how to check over my work? I'm most interested in the latter; I can't go over all of the questions in the time left over, but I also can't really identify problems I might have done wrong because of an arithmetic error. 76.230.231.191 (talk) 00:11, 25 August 2010 (UTC)

The most important point is that if you are scoring above 700 (for the mathematics section of the SAT), and there are still a few weeks to go before the test, then you are doing fine. What score are you achieving on an average? Unfortunately, the mathematics section of the SAT is all too easy these days and so if you make just one mistake in the whole test, you might get 720. I know you probably do not want to be hearing this, but I think that these points are important to keep in mind. (And for some universities such as Caltech, they might reject your application if you have scored below 760 on the mathematics section without even bothering to read the rest of your application - this is also unfortunate.)

As for your question, the best advice I can give you is to write all your computations without ommitting a single detail - not one. It does not matter if you are moving from $(5+4)-(2+3)$ to $4$ or something more complex - expand the brackets carefully in at least 3 steps! I know this sounds boring, but it is the only way I know to avoid silly errors. There is ample space to work out your questions so you should use this as much as you can. And once you have worked out a question in many steps, it is very likely that you will not make a mistake. However, always double check at the end of the test.

Also confirm that you are bubbling everything correctly - double, no triple check this! One mark is "gold" on the higher end of the scale and so you should not be losing marks because of incorrect bubbling.

You are probably thinking that if you do all these things, then you will run out of time. I think that taking tests at home is the key here. Try different strategies, decide how to manage your time on the questions, and you should be fine. The first 15 questions should take 5 minutes at the most especially if you have done AP calculus. The last 10 questions should take 10 minutes at the most. And this should give you ample time to check. But I strongly believe that you should try to do everything carefully the first time around - then, there will be no need to check! (But still check!) In particular, while this strategy is possible, you should also try other strategies that enable you to spend more time on the easier questions. If you have 10 minutes left over, then you are doing the test too fast. Try to divide that extra 10 minutes among the easier questions. Hope that helps! (Disclaimer: I am no expert on these matters.) PST 02:37, 25 August 2010 (UTC)

Caltech is not going to throw out an application just because of one or two missed questions on the SAT math section. SAT scores are not a particularly good predictor of how good a student will be, so the admissions committee cares far more about the rest of the application.

My advice to the OP is to just stay calm at the test and be SURE to get enough sleep the night before. The others have good advice, too. But don't worry if you miss a few questions. Eric. 82.139.80.104 (talk) 16:27, 29 August 2010 (UTC)

When you check an answer, it's useful to try to solve the question in a different way. If you have an error you will be less likely to repeat it. -- Meni Rosenfeld (talk) 09:01, 25 August 2010 (UTC)

I have one possibly weird piece of advice. A lot of the multiple-choice questions on those tests are of the form "what number x satisfies [some condition]". These can be further split into two types. It's usually obvious at a glance what type a question is:

• "Which number gives 15 if you double it and then add 3?" (type 1)
• "Which number is 2 less than the square of a prime number?" (type 2)

Type 1 means you can set up an equation with a unique solution (2x+3=15 in the example), then solve for x, then look at the multiple choices and find the one equal to x and fill in the bubble. Type 2 doesn't have a unique answer so you have to examine the 5 choices and see which one satisfies the condition. Type 1 is more common than type 2, or at least it was on the SAT that I took, back in the day.

With a type 1 question, it's usually faster and less error-prone to solve the equation first and look at the choices afterwards. It's faster because you do just one calculation instead of trying out several guesses, and less error-prone because if you make a mistake, you'll probably get an x that doesn't match any of the choices, so you know something is wrong and can go back and fix it. With type 2, you just check the choices one by one.

It is possible to hit a type 2 question that looks like it could be a type 1 question, causing you to burn a lot of time if you're trying to stick to the above strategy rigorously (I did that and probably lost a few points because of it). Just remember that all the questions are basically easy, so if you're spending more than a few moments figuring out a type 1 approach, it's possible that there isn't one or that you're missing it somehow, so just switch to type 2 and get on with it. 67.117.146.38 (talk) 07:34, 26 August 2010 (UTC)

here's the non-answer, what is the question?

Resolved

There is a certain unsolved problem in combinatorics (Ramsey theory I guess) that seeks to determine the value of a certain number n. The situation is something like this: just about everyone believes n=6. It's obvious that n>4, and it can be proven with a little work that n>5. There is a ton of experimental evidence pointing to n=6, though it might not come as a complete shock if it turns out that n=7 or even n=8.

However, the best lower upper bound anyone has been able to rigorously prove on n is something like n = the 10 millionth Ackermann number.

Anyone know what I'm referring to? Thanks.

67.117.146.38 (talk) 21:48, 25 August 2010 (UTC)

You're thinking of Graham's problem. The upper bound you're thinking of is probably Graham's number (which article also states the problem), though that's not the best bound known. Even the best known upper bound is absurdly large, however: much bigger than the 10 millionth Ackermann number. For a long time the solution to the problem was believed to be 6, but it is now known to be at least 11. Algebraist 22:09, 25 August 2010 (UTC)

Thanks! I guess the deal with the number is less clear than I thought. I remembered there was a general belief that the actual value of the number was quite small, not that it was somewhere unknown within a vast interval. 67.117.146.38 (talk) 22:34, 25 August 2010 (UTC)

Line integrals

Hey guys. We're studying line integrals in class, and I've got three homework assignments about them. The trouble is we don't have our books yet and I left my notes on campus :( The problems are pretty basic (two involve linear bottom function things--forgot the word, sorry!--and the third is an ellipse). Since it would be simple to solve these with basic calculus or even geometry, we're getting graded on the work more so than the answer. So i guess what I'm asking is, can someone demonstrate how you would do the line integral (being sure not to skip steps please), with any function you think is best or just a circle centered at the origin and with radius 1 (That's not one of the problems, I just want to see how it's done)? thanks. 68.249.1.8 (talk) 23:39, 25 August 2010 (UTC)

All right, let's do the easy thing and integrate $f(x,y)=1$ along the unit circle, using the natural parametrization ($x=\cos t,y=\sin t$). By the first definition in Line integral, we then have:

$\int _{C}1\,ds=\int _{0}^{2\pi }1\,|(-\sin t,\cos t)|\,dt=\int _{0}^{2\pi }{\sqrt {(-\sin t)^{2}+(\cos t)^{2}}}\,dt=\int _{0}^{2\pi }{\sqrt {1}}\,dt=\int _{0}^{2\pi }dt=2\pi .$ 85.226.206.72 (talk) 06:29, 26 August 2010 (UTC)

The crucial step was deciding the best parameterization for a circle. In the case of a circle, it is easy to map two-dimensions (x,y) to a single parameter (t) (distance along the circle). From basic geometry/trigonometry, we know that the relationship {x=sin(r), y=cos(r)} is a suitable description of a circle. In general, you need to find the "simplest" parameterized representation of your line path - preferably as a one-dimensional function in the form of easily-integrable parts. Nimur (talk) 21:46, 26 August 2010 (UTC)

August 26

Volume of revolution

How do you find the volume of revolution generated by the area enclosed between $y=x^{n}$, $x=0$, $y=a^{n}$, $y=b^{n}$ rotated around the x axis?--115.178.29.142 (talk) 03:24, 26 August 2010 (UTC)

It would be easier to help if you show us how much you've already done. The obvious first step here is to draw a sketch, however crude, of the curves and find out what species the region in question belongs to. Have you done that? 85.226.206.72 (talk) 06:18, 26 August 2010 (UTC)

... and tell us if you don't understand the methods given in the article: Solid of revolution. Dbfirs 06:50, 26 August 2010 (UTC)

Inclined sit-up

OK, tonight I did 3 sets of 5 inclined sit-ups. The incline was at approximately 25-30 degrees. I placed a 25 lb weigh on my chest, and sat straight up. I also tried to maintain myself parallel to the ground for a few seconds, which was surprisingly difficult.

Can you help me come up with a formula, based on my body weight of 185 lbs, for how many calories I actually burned tonight. I do not know the breakdown of how weigh is distributed in the body (sorry). I would like to have an accurate representation, so if we have to add just a little bit for calories wasted, that's fine with me (I imagine most exercises have this type of fudge factor, though I could be wrong). Magog the Ogre (talk) 04:14, 26 August 2010 (UTC)

The amount of calories burned is pretty small, but you are building muscle tone. If you are trying to lose weight through mathematics, you might like the downloadable book The Hacker's Diet. 67.117.146.38 (talk) 08:41, 26 August 2010 (UTC)

I did a very rough calculation and it turned out about 5 kcal (it could easily be 10 or something).

Assuming that by "formula" you meant just that rather than a number, then I'd say the energy burned per sit-up is

${\frac {mgh(\sin(\alpha _{2})-\sin(\alpha _{1}))}{\eta }}$

Where

• m is the moving mass (not including, say, your legs)
• g is the acceleration due to gravity
• h is the average (weighted by mass) distance of the moving mass from the fulcrum
• $\alpha _{1},\ \alpha _{2}$ are the angles at the start and finish (negative angle for the mass being below the fulcrum)
• $\eta$ is the efficiency of the muscles involved.

If you're using metric units you'll get a result in Joules; you'll need to convert them using $1kcal=4184J$.

-- Meni Rosenfeld (talk) 15:10, 27 August 2010 (UTC)

Optimal decision making under uncertainty.

100 cups are placed on a desk. Underneath one of them is $100, and the rest contain nothing. At any time, you may pick a cup and claim the contents of it. You may only do this once. You may also pick a cup and identify the contents of it, but doing so costs $1. You may do this as many times as you like. What is the optimal strategy to maximise the expected profit?--Alphador (talk) 09:04, 26 August 2010 (UTC)

Keep spending $1 until you find the $100. On average you'll have looked at ~50 cups before you find it and expect to earn ~$50. Dragons flight (talk) 09:26, 26 August 2010 (UTC)

With the obvious, very minor caveat that if you investigate 99 and don't find it, take the 100th without investigating. —Preceding unsigned comment added by 203.97.79.114 (talk) 10:55, 26 August 2010 (UTC)

This would be much harder if, as you wait idly trying to make up your mind, other players are allowed to participate in the same game with the same set of cups. The mathematics for deciding when to act are described in Nash equilibrium (to maximize your own profit), and the mathematics to maximize the total profit for all participants (i.e., somebody gets the $100, and the least amount of money is spent) is described in Pareto optimality. Nimur (talk) 21:41, 26 August 2010 (UTC)

It's customary to say "thank you" when people answer one's question, along with any followup questions. I don't know about others, but for me, knowing that my answer was helpful gives me closure which allows me to stop thinking about the question. -- Meni Rosenfeld (talk) 14:50, 27 August 2010 (UTC)

Assigning probabilities to curve fittings

Let's suppose you're given the following data points: (0,1) (1,2) (2,4) (3,8) (4,16) (5,32) (6,64) (7,128) (8,256) (9,512)
There are infinitely many curves that will fit these data points, however if you saw this as a natural phenomenon (for instance, the population of bacteria with respect to time), intuition would tell you that the most likely equation that would fit these is $y=2^{x}$. Is there any mathematical justification for believing that $y=2^{x}$ is the equation which most probably fits these data points out of the infinitely many that could potentially fit them?--Alphador (talk) 11:10, 25 August 2010 (UTC)

The general principle for this is Occam's razor. One concretization for it is that your prior probability for a sequence (or anything else) should be smaller the higher its Kolmogorov complexity. Because the sequence $y=2^{x}$ has such a short description, it will have a relatively high prior probability, and its posterior probability will be overwhelming after observing these data points. -- Meni Rosenfeld (talk) 11:01, 26 August 2010 (UTC)

The exponential function assumes fractional values while the population at any time is an integer. So the population is not an exponential function.

The exponential function grows without limit while the population is limited. So the population is not an exponential function.

So the best you can get is an approximate fit in a finite interval of time.

Consider the function g(t)=f(t)−2^t. Then g(t)=0 for t=0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. Consider only a finite number, N, of points of time in the finite interval, because you cannot measure with unlimited frequency. So this is like a lottery with N tickets, and K of these tickets are marked 'zero', and N−K of these tickets are marked 'nonzero', and you have opened n=10 tickets, out of which k=10 are all marked 'zero' and n−k=0 are marked 'nonzero'. The likelihood function L(K)=$\scriptstyle {\binom {K}{k}}{\binom {N-K}{n-k}}$=$\scriptstyle {\binom {K}{10}}$ for 0≤K≤N. The maximum likelihood estimate is K=N. So YES, there is a mathematical justification for believing that y = 2^x is the equation which most probably fits these data points out of the infinitely many that could potentially fit them. Bo Jacoby (talk) 18:30, 25 August 2010 (UTC). The question and the answer has been removed, and the question reinstalled without the answer. Don't know why. Bo Jacoby (talk) 11:07, 26 August 2010 (UTC).

Sorry, the question was removed several times, I restored the question as it appeared the first time while I should have looked for the last time. -- Meni Rosenfeld (talk) 11:22, 26 August 2010 (UTC)

how to integrate 1/(1+cosx)?

tried making denominator (1^2+(sqrt(cosx)^2)) but stuck after that. the answer supposedly has a tan(0.5x). —Preceding unsigned comment added by 218.186.8.247 (talk) 18:34, 26 August 2010 (UTC)

Divide and multiple by 1-cos x -Shahab (talk) 19:11, 26 August 2010 (UTC)

You're right that the answer is tan(½x). Why not differentiate tan(½x) and then try to retrace the steps to prove the integration? For example:

${\frac {\text{d}}{{\text{d}}x}}\tan \!\left({\frac {x}{2}}\right)={\frac {\text{d}}{{\text{d}}x}}\left({\frac {\sin \left({\frac {x}{2}}\right)}{\cos \left({\frac {x}{2}}\right)}}\right)={\frac {{\frac {1}{2}}\cos ^{2}\!\left({\frac {x}{2}}\right)+{\frac {1}{2}}\sin ^{2}\!\left({\frac {x}{2}}\right)}{\cos ^{2}\!\left({\frac {x}{2}}\right)}}={\frac {1}{2}}\left({\frac {\cos ^{2}\!\left({\frac {x}{2}}\right)+\sin ^{2}\!\left({\frac {x}{2}}\right)}{\cos ^{2}\!\left({\frac {x}{2}}\right)}}\right)={\frac {1}{2\cos ^{2}\!\left({\frac {x}{2}}\right)}}.$

If you use the cosine angle formulas, which say that cos(α ± β) = cos(α)cos(β) ∓ sin(α)sin(β), then you will be able to show that 2⋅cos²(½x) = 1 + cos(x). Notice that cos(x) = cos(½x + ½x) = cos²(½x) − sin²(½x), and so 1 + cos(x) = 1 + cos(½x + ½x) = 1 + cos²(½x) − sin²(½x). Since cos²(½x) + sin²(½x) = 1 it follows that 1 − sin²(½x) = cos²(½x), and so 1 + cos(x) = 2⋅cos²(½x) as required. Thus:

${\frac {\text{d}}{{\text{d}}x}}\tan \!\left({\frac {x}{2}}\right)={\frac {1}{1+\cos x}}\ .$

If you run these trigonometric identities backwards then you'll have the proof. Don't forget the constant of integration! — Fly by Night (talk) 19:38, 26 August 2010 (UTC)

There's some material on this in the article titled tangent half-angle formula. Michael Hardy (talk) 20:43, 26 August 2010 (UTC)

differential equation

80(dx/dt) = 1 + k/(x+1) where k is a constant.

i can see that if k/(x+1) is combined with the dx/dt term, can integrate by ln. also tried multiplying whole equation by x+1 but then the 1 will become an x+1 and I can't transfer the x over. also tried making RHS (x+1+k)/(x+1) but after transferring reciprocal over to LHS, cant think of a formula for (x+1)/(x+1+k). —Preceding unsigned comment added by 218.186.8.247 (talk) 18:40, 26 August 2010 (UTC)

I'm going out on a limb, but you might have to solve this approximately or numerically - I'm pretty sure it's nonlinear. So: taylor series! k/(x+1) is exactly k(x+1)^-1, which can be expanded and approximated as k*(1-x). This linearizes the equation for small x. For large x, dx/dt approximately equals 1/80. So, you have a suitable first-order approximation. The function diverges, so if this were a real-world problem, it would be of little value to solve it exactly or approximate it very accurately anyway. Maybe a pure mathematician can come through and clean up my messy engineer's work. Nimur (talk) 21:34, 26 August 2010 (UTC)

Separation of variables is the way to go, try writing (x+1)/(x+1+k)=1-k/(x+1+k).--RDBury (talk) 23:25, 26 August 2010 (UTC)

Not nonlinear! Just ugly. Nimur (talk) 23:51, 26 August 2010 (UTC)

Separate variables:

$80{dx \over dt}={x+1+k \over x+1}$

and then

${80(x+1)\,dx \over x+1+k}=dt.$

Then integrate both sides. Michael Hardy (talk) 00:58, 27 August 2010 (UTC)

Substitute t=80s, dt=80ds, and get rid of the common factor 80:

${x+1 \over x+1+k}dx-ds=0.$

Substitute x=y-1-k, dx=dy:

$dy-k{dy \over y}-ds=0.$

Integrate:

$y-{k\log y}-s\,$ is constant.

Substitute back, y=x+1+k, s=t/80:

$x+1+k-k\log(x+1+k)-t/80\,$ is constant.

Discard the constants:

$x-k\log(x+1+k)-t/80\,$ is constant.

Exponentiate:

$(x+1+k)^{-k}e^{x-t/80}\,$ is constant.

Bo Jacoby (talk) 20:34, 27 August 2010 (UTC).

August 27

Semicircles problem

This was a question in math league a while ago, and I did not know how to do the problem.

Givens: There are 3 semi circles, one with radius 3, one with radius 2, and one with radius 1. They are inlaid as shown in the diagram. There is a circle tangent to the sides of each of the semicircles. What is $x$, the radius of that circle?

What I attempted to do before getting confused: I made a triangle (not drawn in the diagram), connecting the center of the cicle with the center of the semicircle radius=1, connecting the circle center with the center of semicircle radius=2, and the final line connecting the two centers of the semicircles r=1 and r=2. So now I got a triangle with sides of 3, 1+x, and 2+x.

Was making a triangle helpful in solving the problem? What is the next step I should take? Spencer^T♦C 01:36, 27 August 2010 (UTC)

I seem to recall this very same question being asked here maybe between six and twelve monts ago. I'm not sure if I can find it though. Michael Hardy (talk) 01:42, 27 August 2010 (UTC)

That's when it was the math league problem, so it's quite possible someone else asked how to do it. Spencer^T♦C 01:45, 27 August 2010 (UTC)

Found: Wikipedia:Reference_desk/Archives/Mathematics/2010_March_25#Circle. Can someone explain how the Descartes' theorem applies? Spencer^T♦C 01:46, 27 August 2010 (UTC)

Wikipedia's article on Descartes' theorem says it's a relationship among the four curvatures. Three of the curvatures you've given are 1, 1/2, and 1/3. The problem is to find the fourth. So plug them in. Michael Hardy (talk) 02:07, 27 August 2010 (UTC)

...and it turns out the quadratic equation you get works out very neatly, since its solutions are rational. Michael Hardy (talk) 02:13, 27 August 2010 (UTC)

...but remember that the largest circle surrounds the other three circles, so you have to take its curvature as -1/3, which leads to only one solution (there are two kissing circles, but they are the same size). In general, given circles with radii a, b and a+b, the term under the square root is 0 because

${\frac {1}{ab}}-{\frac {1}{a(a+b)}}-{\frac {1}{b(a+b)}}=0$

and the radius of the kissing circles is

$\left({\frac {1}{a}}+{\frac {1}{b}}-{\frac {1}{a+b}}\right)^{-1}={\frac {ab(a+b)}{a^{2}+ab+b^{2}}}$

Gandalf61 (talk) 10:47, 27 August 2010 (UTC)

Break the side of length 3 into two parts: from the center of the circle of radius 2 to the center of the circle of radius 3, of length 1, and from there to the center of the circle of radius 1, of length 2. Draw a line from the center of the big circle (of radius 3) to the center of the circle of unknown radius, and keep going until that line hits the point where the big circle and the unkown circle touch each other. (Notice that if a line passes through the centers of two circles in the same plane that touch each other, then it must pass through the point where they touch each other.) The portion of that line that lies within your triangle has length 3 − x. Keep going from there. Michael Hardy (talk) 02:01, 27 August 2010 (UTC)

Apollonian gasket has some relevance here.--RDBury (talk) 14:50, 27 August 2010 (UTC)

This problem is discussed in H. S. M. Coxeter's "Introduction to Geometry" (1961), where he quotes Frederick Soddy's poem "The Kiss Precise", which ends with the couplet The sum of the squares of all four bends/Is half the square of their sum (where the "bend" of a circle is its curvature, i.e. the reciprocal of its radius). AndrewWTaylor (talk) 10:43, 28 August 2010 (UTC)

Oldest open problem

What is the oldest open problem in mathematics? --84.61.172.89 (talk) 07:53, 27 August 2010 (UTC)

There must be countless old forgotten open problems in mathematics, thus there is no telling what is the oldest. However, among well-known problems, I believe that the problem of the existence of odd perfect numbers goes back to Euclid, making it fairly old.—Emil J. 10:33, 27 August 2010 (UTC)

sin(ln(z)) in annulus around z=0

Although ln(z) is multivalued and discontinuous near 0, i'm thinking sin(ln(z)) collapses the multivalues into a single valued function and is continuous. Am i correct? If so, what's the Laurent series? Thanks, 24.7.28.186 (talk) 09:57, 27 August 2010 (UTC)

No, you are not correct. For example, the values of ln 1 are 2kπi for any integer k, and plugging it into the definition $\sin z={\tfrac {1}{2i}}(e^{iz}-e^{-iz})$ gives $\sin \ln 1={\tfrac {i}{2}}(e^{2k\pi }-e^{-2k\pi })$. However, sin(i ln z) is single-valued, it is a meromorphic function with a simple pole at 0, and it equals ${\tfrac {i}{2}}(z-z^{-1})$.—Emil J. 10:27, 27 August 2010 (UTC)

(To provide more context: this question appears to be a followup of Wikipedia:Reference desk/Archives/Mathematics/2010 August 16#Laurent Series for ln(z) in annulus around z=0?.—Emil J. 10:54, 27 August 2010 (UTC))

Right you are, thanks.-Rich Peterson199.33.32.40 (talk) 22:40, 27 August 2010 (UTC)

Roots of unity for non-integer exponents

Does $x^{n}=1$ have multiple solutions for $x$ when $n$ is not an integer, and if so, how many?--Alphador (talk) 11:23, 27 August 2010 (UTC)

First, how do you interpret the equation? If n is not an integer, then xⁿ is multi-valued. Do you call x a solution if some value of xⁿ is 1, or if all values of xⁿ are 1 (i.e., if 1 is the only value)?—Emil J. 11:30, 27 August 2010 (UTC)

Have you studied root of unity? Bo Jacoby (talk) 13:14, 27 August 2010 (UTC).

Why do you ask me? Anyway, I know the article, I even contributed to it, and as far as I am aware it does not contain a single word on the topic of roots with non-integer exponents.—Emil J. 13:25, 27 August 2010 (UTC)

Bo was probably just misusing indentation. -- Meni Rosenfeld (talk) 14:43, 27 August 2010 (UTC)

Yes. Sorry. Bo Jacoby (talk) 22:12, 27 August 2010 (UTC).

I think the answer depends on the exact nature of your non-integer n; specifically if it's rational, and if so whether the numerator is odd or even in simplest form. If we extend the general technique we use for the roots of unity then we can rewrite the equation as:

$\textstyle x^{n}=e^{2\pi ki}$ (where k is an integer) which rearranges to:

$\textstyle x=e^{2\pi i{k \over n}}$

giving us $\textstyle x=1$ whenever $\textstyle k=0$ OR $\textstyle k/n$ is a non-zero whole number (which can happen easily enough for rational-but-non-integer n)

but we can also get $\textstyle x=-1$ when $\textstyle k/n=m+1/2$ for m an arbitray integer. Unless I'm missing something, this can happen when n is rational and has a factor of 2 in the numerator in its simplest form (trivial example: $\textstyle n=2/3$ gives $\textstyle k/n=3/2$ when $\textstyle k=1$, and $\textstyle e^{2\pi i{2 \over 3}}=e^{3\pi i}=-1$).

For irrational n, though, we only have the $\textstyle k=0$ case giving us $\textstyle x=1$, and then an infinite number of complex roots. (I think it's safe to say that we're sort of precessing around the Argand diagram in steps of $\textstyle k/n$, and I'm sort of tempted to say that $\textstyle x=-1$ when $\textstyle m=\infty$, but that seems like wishful thinking.)

Summarising the conclusions:

If n is rational with an even numerator, there are two real roots, -1 and 1, and some number of complex roots..

If n is rational with an odd numerator, there is one real root, 1, and some number of complex roots.

If n is irrational, we have 1 and an infinite number of complex roots.

I've probably missed a subtlety somewhere in all that, though. (Using n for an explicitly non-integer number is messing with my head; too much FORTRAN in my youth, perhaps.) --217.41.233.67 (talk) 15:52, 27 August 2010 (UTC)

(Replying to myself, sorry) An easier way to think about the requirement that n has an even numerator to get -1 as a root is to write $\textstyle n=2a/b$, giving $\textstyle x^{2a/b}=1$ which gives $x^{2}=1^{b/a}$. For $\textstyle x=-1$ this gives the obviously valid $\textstyle (-1)^{2}=1^{b/a}$ which works for any a and b.

Also, I think the 2/3 example I gave is something of a special case, as AFAICS there are no complex roots there at all -- each time k increases by 1, we're back to either an odd or even multiple of $\textstyle \pi$ in the exponential form. --81.158.2.129 (talk) 17:04, 27 August 2010 (UTC)

Non atomic poset

Resolved

I have the following definition in my book:

Definition: If a poset T has a smallest element 0, then any cover of 0 is called an atom or point of T. A poset with 0 is atomic if every nonzero element contains an atom.

Now can someone give me an example of a non-atomic poset. Since 0 is the smallest element all nonzero elements contain it, and so either are atoms or contain atoms. So isnt any poset with 0 an atomic poset?-Shahab (talk) 11:46, 27 August 2010 (UTC)

The rational or real interval [0,1] with the usual ordering has no atoms.—Emil J. 12:29, 27 August 2010 (UTC)

Oh, of course. Dont know why I was thinking only finite. Thanks-Shahab (talk) 16:29, 27 August 2010 (UTC)

partial derivative

find the wxtreme values of function $f(x,y)=x^{3}y^{2}(1-x-y)$ —Preceding unsigned comment added by Himanshu.napster (talk • contribs) 19:23, 27 August 2010 (UTC)

I have written the definition of your function in LaTeX so it is easily intelligible by volunteers at this reference desk. I hope you do not mind. PST 06:19, 28 August 2010 (UTC)

Partial derivative should help you with your homework. But then I guess you knew that since you made it the title of your question. --81.158.2.129 (talk) 20:50, 27 August 2010 (UTC)

On what region do you wish to work out the extreme values of the function f? If you wish to work out its extreme values on the entire plane, do the standard routine. Work out the critical points, compute the determinants of the corresponding Hessians, and use the relevant extrema tests. (All of this should be possible to find in whatever textbook you are using.) PST 06:19, 28 August 2010 (UTC)

For instance, we can write $f(x,y)=x^{3}y^{2}-x^{4}y^{2}-x^{3}y^{3}$ and thus $f_{x}(x,y)=3x^{2}y^{2}-4x^{3}y^{2}-3x^{2}y^{3}$ and $f_{y}(x,y)=2x^{3}y-2x^{4}y-3x^{3}y^{2}$. (Where f_x and f_y denote the partial derivatives of f with respect to x and y, respectively.) Determine the ordered pairs (x,y) for which both f_x(x,y) and f_y(x,y) vanish. Once you have done this, compute the determinant of the 2x2 Hessian matrix of f:

${\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}}.$

(The above partial derivatives are all "second order" partial derivatives of f.) Then, evaulate this determinant at each of the critical points to test whether they are extrema or not. I will leave you to work out how to do this as an exercise. PST 06:31, 28 August 2010 (UTC)

Calculus on science desk: atmospheric carbon scenario difference projections

There is a calc problem that I don't understand at Wikipedia:Reference desk/Science#What is the wind-water-solar climate change mitigation scenario atmospheric carbon projection? Why Other (talk) 22:22, 27 August 2010 (UTC)

From what I can tell, since there seems to be a lot of jargon there that would be better understood by a atmospheric scientist than a mathematician, the model being used is that the atmosphere is a giant tank of some fluid where some impurity is being added at a given rate while at the same time it's being removed at a rate proportional to its concentration. The tank is assumed to be well mixed, meaning you don't have to worry about the concentration not being the same in different parts of the tank. This is a fairly standard problem in ODE's and the solution is a straightforward use of separation of variables. I hope that helps with the mathematical aspect of the model at least.--68.40.56.142 (talk) 15:04, 28 August 2010 (UTC)

Thanks! Why Other (talk) 03:24, 30 August 2010 (UTC)

mgf of the square of a standard normal random variable

How can I get the moment generating function of the square of a standard normal random variable? Do I just replace $x$ with $x^{2}$ in the integral? I know it's meant to come out the same as a gamma mgf but I can't get it to work. —Preceding unsigned comment added by 118.208.51.232 (talk) 23:08, 27 August 2010 (UTC)

The law of the unconscious statistician applies. (I finished my Ph.D. without ever hearing that name for it, although I used it all the time. However, apparently some respectable authors call it that.) The moment-generating function of a continuous random variable with density ƒ is given by

$m_{X}(t)=E\left(e^{tX}\right)=\int _{-\infty }^{\infty }e^{tx}f(x)\,dx.$

The moment-generating function of the square is

$m_{X^{2}}(t)=E\left(e^{tX^{2}}\right)=\int _{-\infty }^{\infty }e^{tx^{2}}f(x)\,dx.$

To be continued..... Michael Hardy (talk) 02:01, 28 August 2010 (UTC)

Michael, I have known that rule/law/theorem since I began high school but to this day I never knew it had a name! Thank you for noting that! This is exactly the reason Wikipedia is so great. (I do not (really) work in probability theory by the way, so I probably have not heard of many results in the field and I guess I can be excused for that ;), but I feel awkward knowing that one of the very first (and easiest) results I learnt in the field has a name, and I never knew it for so many years ...) Or perhaps the name was invented recently. Was it? If not, do you know when it was invented? PST 06:11, 28 August 2010 (UTC)

Let me try doing this...

$\int _{-\infty }^{\infty }e^{tx^{2}}f(x)\,dx.$

$={\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{\infty }e^{tx^{2}}e^{\frac {-x^{2}}{2}}\,dx.$

$={\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{\infty }e^{{\frac {-1}{2}}(x^{2}-2tx^{2})}\,dx.$

$={\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{\infty }e^{{\frac {-x^{2}}{2}}(1-2t)}\,dx.$

$={\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{\infty }e^{\frac {-x^{2}}{2}}e^{(1-2t)}\,dx.$

$=e^{(1-2t)}\int _{-\infty }^{\infty }{\frac {1}{\sqrt {2\pi }}}e^{\frac {-x^{2}}{2}}\,dx.$

$=e^{(1-2t)}$

Alas! I'm sure the answer is supposed to be $(1-2t)^{\frac {1}{2}}$. What am I doing wrong? —Preceding unsigned comment added by 130.102.158.15 (talk) 06:02, 28 August 2010 (UTC)

$e^{{\frac {-x^{2}}{2}}(1-2t)}\neq e^{\frac {-x^{2}}{2}}e^{(1-2t)}$. -- Meni Rosenfeld (talk) 17:11, 28 August 2010 (UTC)

Wrong!

$e^{{\frac {-x^{2}}{2}}(1-2t)}$

is not the same as

$e^{\frac {-x^{2}}{2}}e^{(1-2t)}$

You were OK until that point.

More later.... Michael Hardy (talk) 18:50, 28 August 2010 (UTC)

I see Meni Rosenfeld already noted the error.

Now remember that once you've got

${\frac {1}{\sqrt {2\pi \,{}}}}\int _{-\infty }^{\infty }e^{-x^{2}/2}\,dx=1,$

then via the substitution of ax for x you get

${\frac {1}{\sqrt {2\pi \,{}}}}\int _{-\infty }^{\infty }e^{-(ax)^{2}/2}\left(a\,dx\right)=1,$

and hence

${\frac {1}{\sqrt {2\pi \,{}}}}\int _{-\infty }^{\infty }e^{-(ax)^{2}/2}\,dx={\frac {1}{a}},$

provided a > 0. (If a is negative, then the bounds of integration get reversed, and you go on from there....)

Now apply this in the case where

$a=(1-2t)^{1/2}.\,$

Michael Hardy (talk) 19:00, 28 August 2010 (UTC)

Thanks so much, I would never have figured out that substitution step! —Preceding unsigned comment added by 130.102.158.15 (talk) 01:24, 29 August 2010 (UTC)

A geometric sequence

Can anyone help me with this question? A geometric sequence is given. If one replaces the signs of the items with even index, the sum of the new sequence will be 3 times smaller then the original one. What is the sum of the original sequence? I get from the equation I write that q, the quotient of the sequence, is 1\2. But what next? —Preceding unsigned comment added by 87.68.249.109 (talk) 07:37, 28 August 2010 (UTC)

Surely if you multiply any such sequence by an arbitrary nonzero constant, you get something else with the same property, and so the only answer is "whatever you like provided it's nonzero"? 95.150.22.219 (talk) 12:05, 28 August 2010 (UTC)

You should double check your calculations for q = 1/2. That's correct if you replace swap the signs on the odd terms. If you replace those of the even terms, you get a different ratio. There is a unique answer. —Preceding unsigned comment added by 203.97.79.114 (talk) 14:24, 28 August 2010 (UTC)

If q is the ratio between term in the original series, then the the original sum is

${\frac {a}{1-q}}$

and the new sum is

${\frac {a}{1+q}}$

so the problem reduces to solving

${\frac {a}{1-q}}={\frac {3a}{1+q}}$

for q.--RDBury (talk) 15:14, 28 August 2010 (UTC)

You've made the same mistake as the OP. That's if you replace the 'odd terms. —Preceding unsigned comment added by 203.97.79.114 (talk) 15:19, 28 August 2010 (UTC)

Surely we can treat this as the sum of two independent geometic sequences. That is if the original sequence is n + a.n + a^2.n + a^3.n + ... then we have two geometric sequences (using b=a^2) which sum as n + b.n + b^2.n + ... and also a.n + b.a.n + b^2.a.n + ... If these sums are Se (sum of even terms) and So (sum of odd terms) and the original total was T then So + Se = T. Then if we change the sign of the even terms we get So - Se = T/3. Solving gives So = 2Se. We can pair the terms of the two sequences: n <-> a.n; a^2.n with b.a.n = a^3.n; a^4.n with a^5.n; ... The tricky bit is do the indices start at 0 or at 1? If they start at 1, then we can pair terms of So is n, a^2.n, a^4.n and Se is a.n, a^3.n, a^5.n so a=2. If the indices start at 0 then think Se = So/2 and a=0.5. If a=2 the the sum of the original sequence is infinite. If a=0.5 then the sum of the sequence = 2.n and the actual value depends upon the chosen 'n'. -- SGBailey (talk) 16:24, 28 August 2010 (UTC)

(ec) The first geometric series is Σqⁱ=1/(1−q). The second geometric series is Σ−(−q)ⁱ=−1/(1+q). The equation −1/(1+q)=(1/3)(1/(1−q)) has the unique solution q=2. The series 1+2+4+8+... is not convergent but by analytic continuation the value is −1. Bo Jacoby (talk) 17:26, 28 August 2010 (UTC).

([[wp:edit conflict|edit conflict]) Assume, for | r | < 1, that you have a geometric series given by S = a + ar + ar² + ar³ … then the standard formula tells us that

$\sum _{k=0}^{\infty }ar^{k}=a+ar+ar^{2}+ar^{3}+\cdots ={\frac {a}{1-r}}.$

I think the OP wants us to reverse the signs of the even-powered terms. If we replace r with −r then we chance the sign of the odd-powered terms. Taking the negative of the whole series will give us what we want:

$-\sum _{k=0}^{\infty }a(-r)^{k}=-a+ar-ar^{2}+ar^{3}-ar^{4}\pm \cdots ={\frac {-a}{1+r}}.$

If the second series is three times smaller than the first then we are left with the following:

$\sum _{k=0}^{\infty }ar^{k}=-{\frac {1}{3}}\sum _{k=0}^{\infty }a(-r)^{k}\ \iff \ {\frac {a}{1-r}}={\frac {-a}{3(1+r)}}\ \iff \ a(2+r)=0.$

Clearly either a = 0 or r = −2. If r = −2 then | r | > 1, and the series cannot converge! So there is no such series. I guess the only answer is the trivial answer where a = 0 and the sum of the series is also zero! — Fly by Night (talk) 19:11, 28 August 2010 (UTC)

Fly by Night, you made the second series three times bigger than the first. If the second series is three times smaller than the first then:

$\sum _{k=0}^{\infty }ar^{k}=-{3}\sum _{k=0}^{\infty }a(-r)^{k}\ \iff \ {\frac {a}{1-r}}={\frac {-3a}{1+r}}\ \iff \ a(2-r)=0.$

Bo Jacoby (talk) 06:25, 29 August 2010 (UTC).

...which gives a=0 or r=2, and thus |r| > 1 still, so Fly by Night's objection that it doesn't converge still applies. --81.158.2.129 (talk) 08:43, 29 August 2010 (UTC)

The function f(x)=(1−x)⁻¹ is an analytic function for x≠1. The series Σxⁱ is convergent for |x|<1 with value f(x). So f(x) is the analytic continuation of the series, and f(2)=−1. Convergence is not the only way to assign a value to a series. Bo Jacoby (talk) 11:32, 29 August 2010 (UTC).

Thanks for pointing out my error. I don't think the OP was set the question to test their knowledge of analytic continuation. It seems quite a simple question put in simple language. The OP didn't mention complex numbers, let alone analytic continuation. My guess is that the question was set to see how s/he could manipulate simple series. And I think the correct answer would have been "No such non-zero, convergent series exists." A series does not have a well defined value outside of its radius of convergence. Finding an analytic function that agrees with the series in the region of convergence and then defining the value of the series outside of that region as equal to the value of the function is subtly different. It gives a value to the series where the series does not have a value of its own. (Do you not feel slightly dishonest assigning the value of −1 to the sum of positive numbers given by 1 + 2 + 2² + 2³ + 2³ + 2⁴ + …?) — Fly by Night (talk) 20:00, 30 August 2010 (UTC)

The series s=1+2+4+8+... satisfies the equation s−1=2s, (because s−1=2+4+8+.. and 2s=2+4+8+... too) and s=−1 is the unique solution to this equation. The series t=−1+2−4+8... satisfies the equation t+1=−2t, and t=−1/3 is the unique solution. That the sum of positive numbers is positive is not necessarily the case for infinite series. That the sum of integers is integer is not necessarily the case for infinite series either. The OP did not require the series to be convergent. No, I don't feel dishonest. Bo Jacoby (talk) 22:06, 30 August 2010 (UTC).

But the infinite sums you mention in you last post; the ones you then perform arithmetic upon, aren't defined unless they converge. It doesn't make sense to say that ∞ − 1 = 2×∞, and then conclude that ∞ = −1 is the unique solution to the "equation". — Fly by Night (talk) 23:10, 30 August 2010 (UTC)

I didn't say that ∞ = −1. If the value of the series s is not defined, then it is free to be defined. You define s = ∞ and I define s = −1. These are different definitions, and we can investigate which definition makes better sense. Note that the binary number 1111111111111111 represents −1 in two's complement, so I am not the only one to define s = −1. The Riemann zeta function is also defined as the analytic continuation of a convergent series. Bo Jacoby (talk) 07:17, 31 August 2010 (UTC).

August 28

Proper Form for Factoring Polynomial

I'm doing this summer work for high school and the instructions are "factor completely". The problem is -15x²-5x and I used [1] to factor and it came out as -15(x+1⁄3)*x. But, if I were to factor it such that it has only integers it comes out as -5(3x+1)x. I want to know if factoring it properly would be with only integers (-5(3x+1)x) or such that the monomial portions of the result are in the form of x+a and not nx+a (eg. -15(x+1⁄3)*x). --Melab±1 ☎ 00:50, 28 August 2010 (UTC)

That really depends on what your teacher means by the phrase. It doesn't have a standard mathematical meaning, at least at this level of detail. --Trovatore (talk) 01:04, 28 August 2010 (UTC)

For some purposes you definitely want −15x(x + 1/3). In particular, if you want to know that the whole thing is equal to 0 when x is either −1/3 or 0, but not otherwise, then that's the form you need. But for some other purposes using only integers is approrpriate. But as "Trovatore" notes, the meaning of the phrase "factor completely" is context-dependent. Michael Hardy (talk) 01:24, 28 August 2010 (UTC)

In the high school context, though, 99 times out of 100, factorisation is all about integers (lowest common factor, prime factorisation, etc. etc.). I'll bet you dollars to donuts Melab's teacher is expecting the answer $-5x(3x+1)$ (and depending on how competent they are, would or wouldn't accept the arguments being made here for the context dependence of the instructions). If I'm right, the teacher would argue that you obtain the x=-1/3 solution by setting $3x+1=0$ and doing trivial algebra (that's certainly the way I was taught it). --81.158.2.129 (talk) 08:40, 29 August 2010 (UTC)

... and that's the way I taught it, though I wouldn't necessarily mark other complete factorisations as wrong, I'd just comment that this way is usually considered the simplest and neatest. The "standard" order (given in all text-books) is integer; powers of x; other linear factors without fractions. Dbfirs 14:28, 29 August 2010 (UTC)

August 29

linear transformation of standard normal random variables

${\mathbf {Y}}=A{\mathbf {X}}$ where ${\mathbf {X}}$ is a vector of iid $N(0,1)$ random variables.

I need to show that ${\mathbf {Y}}$ is also iid $N(0,1)$ if $A$ is an orthogonal matrix.

I have the result that for ${\mathbf {Y}}=A{\mathbf {X}}$ we have $f_{\mathbf {Y}}({\mathbf {y}})={\frac {f_{\mathbf {X}}(A^{-1}{\mathbf {y}})}{|A|}}$

With an orthogonal matrix substituted into the multivariate standard normal formula things cancel and I get a numerator that works out to be multivariate standard normal which is what I want.

But I'm left with that determinant on the bottom, which for an orthogonal matrix is 1 or -1. If it's 1, no problem. But if it's -1, does that mean my distribution is upside down? —Preceding unsigned comment added by 130.102.158.15 (talk) 02:40, 29 August 2010 (UTC)

Consider first the one-dimensional case. The transformation Y=−X swaps the positive and negative values of X. It does not turn the distribution upside down. Bo Jacoby (talk) 06:41, 29 August 2010 (UTC).

...which means that the denominator should actually be the absolute value of the determinant. -- Meni Rosenfeld (talk) 08:43, 29 August 2010 (UTC)

In this case, if the determinant of the matrix $A$ is 1, I have $f_{y}(y)=f_{x}(y)$ where $f_{x}$ is a normal pdf...but if the determinant is -1, I get $f_{y}(y)=-f_{x}(y)$ and that's not a normal pdf unless $-f_{x}(y)=f_{x}(-y)$, which I'm pretty sure it's not —Preceding unsigned comment added by 130.102.158.15 (talk) 08:44, 29 August 2010 (UTC)

Where did the minus sign come from? -- Meni Rosenfeld (talk) 09:46, 29 August 2010 (UTC)

aaaargh, sorry I had my notation mixed up and didn't realize that the denominator is the ABSOLUTE value of the determinant, I thought it was just the determinant! —Preceding unsigned comment added by 130.102.158.15 (talk) 21:55, 29 August 2010 (UTC)

Euler-Maruyama and Milstein schemes

Can anybody suggest where I might be able to find proofs of the order of convergence for the Euler-Maruyama and/or Milstein schemes used for stochastic differential equations? —Preceding unsigned comment added by Damian Eldridge (talk • contribs) 06:24, 29 August 2010 (UTC)

The Euler-Maruyama method article has a reference that you might look at. 67.119.3.248 (talk) 09:03, 30 August 2010 (UTC)

Trig Question

My days of battling with trig identities are unfortunately a little far behind me. Is it possible to express ${\frac {d}{b^{2}c^{2}-(a^{2}-b^{2})^{2}}}[(a^{2}-b^{2})(\cos(bt))+bc\sin(bt)]$ as $Acos(bt-\theta )$? Thanks asyndeton talk 16:33, 29 August 2010 (UTC)

Actually, maybe they're closer than I think. Is it just $A={\frac {d}{b^{2}c^{2}-(a^{2}-b^{2})^{2}}}$, $tan(\theta )={\frac {a^{2}-b^{2}}{bc}}$? asyndeton talk 16:45, 29 August 2010 (UTC)

Certainly a linear combination of sine and cosine functions of the same period can be expressed as a single cosine function with that same period and with a phase shift. More later...... Michael Hardy (talk) 16:48, 29 August 2010 (UTC)

More specifically, see List_of_trigonometric_identities#Linear_combinations. Michael Hardy (talk) 17:09, 30 August 2010 (UTC)

To quote:

In the case of a linear combination of a sine and cosine wave (which is just a sine wave with a phase shift of π/2), we have

$a\sin x+b\cos x={\sqrt {a^{2}+b^{2}}}\cdot \sin(x+\varphi )\,$

where

$\varphi ={\begin{cases}\arcsin \left({\frac {b}{\sqrt {a^{2}+b^{2}}}}\right)&{\text{if }}a\geq 0,\\\pi -\arcsin \left({\frac {b}{\sqrt {a^{2}+b^{2}}}}\right)&{\text{if }}a<0,\end{cases}}$

or equivalently

$\varphi =\arctan \left({\frac {b}{a}}\right)+{\begin{cases}0&{\text{if }}a\geq 0,\\\pi &{\text{if }}a<0.\end{cases}}$

Michael Hardy (talk) 17:11, 30 August 2010 (UTC)

http://en.wikipedia.org/wiki/Wikipedia:RDMA#Semicircles_problem

here Michael Hardy said that "if a line passes through the centers of two circles in the same plane that touch each other, then it must pass through the point where they touch each other". I can see why this is for externally tangent circles, but why is it true for internally tangent circles as well? in other words, how do you know that the line from the center of the big circle passes through the point the passes through the center and the point of tangency? —Preceding unsigned comment added by 68.248.229.115 (talk) 18:17, 29 August 2010 (UTC)

The circles have a common tangent at the point where they touch. The lines from the centre of each circle to this point meet the tangent line at right angles, hence they coincide. Or just appeal to the symmetry of the layout. →86.132.161.214 (talk) 22:45, 29 August 2010 (UTC)

Or just reflect the inner circle in the common tangent, producing the case of externally tangent circles.→86.132.161.214 (talk) 19:50, 30 August 2010 (UTC)

August 30

Graph question

What is the function for a graph that starts at x=a, y=k, which begins to rise slowly then much more steeply until nearly perpendicular to the x axis, then cuts off at x', and resumes in a mirror image at x'', descending until it reaches y=k again? (Apologies for using non-standard notation, I'm afraid I don't know any better. (The shape I'm envisioning is sort of like the elevation of a cooling tower at a power plant.) I suppose it's a hyperbola, but I looked at that article and it's too much information for me to wade through.--82.113.121.52 (talk) 18:39, 30 August 2010 (UTC)

Just cutting off $y=x^{-2}$ at suitable $\pm a,\pm b$ (that is, defining it only when $|x|\in [a,b]$) would do something like that — choose a and b to get the right amount of "slowly/nearly vertical" and the right amount of separation between the two branches. --Tardis (talk) 21:43, 30 August 2010 (UTC)

Taking $a={\frac {1}{2}},b=2$ seems to be a nice choice. Cooling tower says that they are hyperboloid, so the cross section is a hyperbola: in this case, it's $y=-{\sqrt {x^{2}-1}}$, which becomes completely vertical at $x=\pm 1$. Replace x and y with $a_{x}x-b_{x}$ and $a_{y}y-b_{y}$ to scale/shift to taste. --Tardis (talk) 21:56, 30 August 2010 (UTC)

Dear Dr. Who, thank you! Reason I was asking was, I read an article about a speculative theory that attempts to explain why gravity is so much weaker a force than, say, electromagnetism. According to the theory, if our universe is envisioned as a two-dimensional sheet lying flat then other universes are stacked on both sides like pancakes and gravity dissipates by becoming spread out among universes. I was trying to picture in my mind how universes could interact gravity-wise without them slamming into each other and hit on the idea of a graph that rises from nothingness asymptotically and, after an interval, reverses back to nothing. The graph is made by one function so the two halves are in fact a single entity but there is a break in between them. Hmm. On second thought, maybe part 1 and part 2 should switch places. A few more, if I may: (1) Why are so few of us mathematically talented? (2) I am still heartbroken about Rose Tyler, she should still be your companion! (3) Do you ever tire of stupid questions from earthlings?--82.113.106.31 (talk) 01:34, 31 August 2010 (UTC)

Magic Numbers

Can someone explain why there are "magic numbers", and particularly why they work.

I was thinking of the number: 142857. When multiplied by 7 and multipiles thereof produce unusual results, or indeed of any non-multipiles-of-seven.

When multiplied by 55 and then 7, look at the result! Why? MacOfJesus (talk) 19:21, 30 August 2010 (UTC)

There's an article on this at Cyclic number. 85.226.205.150 (talk) 20:02, 30 August 2010 (UTC)

1/7 has the decimal value 0.142857142857142..., so 142857 is just below 1000000/7. Multiplying by 7 gives a whole number just below 1000000. Similarly, 1/13 is 0.076923076..., so 76923 shows the same result when multiplied by 13.→86.132.161.214 (talk) 20:13, 30 August 2010 (UTC)

Thank you. MacOfJesus (talk) 21:11, 30 August 2010 (UTC)

The Cyclic number article mentions a number of restrictions:

This restriction also excludes such trivial cases as:

1. repeated digits, i.e.: 555
2. repeated cyclic numbers, i.e.: 142857142857
3. single digits preceded by zeros, i.e.: "005"

I understand the first two, but how is #3 problematic? 005 * 2 = 010, so it doesn't seem to be cyclic, trivially or not. -- ToE^T 00:14, 31 August 2010 (UTC)

Is it because "Zero" "0" is used by us so differently? "Zero" can mean "Zilch" or that there is no value in that space, such as the difference between "ten" and "one", a zero!? MacOfJesus (talk) 11:37, 31 August 2010 (UTC)

I don't buy that. The article explicitly addresses the way leading zeros are handled. -- ToE^T 13:05, 31 August 2010 (UTC)

Anything like "00005" is trivially cyclic because the cyclic permutations are "50000", "05000", "00500", "00050", and "00005". All of these are multiples of 5. (The fact that 005 * 2 = 10 doesn't mean it's non-cyclic. Cyclic means the permutations are all multiples, not that all multiples are permutations.) Staecker (talk) 13:24, 31 August 2010 (UTC)

But according to the article, cyclic numbers specifically deal with "successive multiples". -- ToE^T 13:30, 31 August 2010 (UTC)

FWIW, the three restrictions were in place when the article burst forth fully formed from the prolific keyboard of 198.99.123.63. -- ToE^T 13:38, 31 August 2010 (UTC)

August 31

conditional probabilities

Given $P(A)=0.5$
and $P(A|B)=0.1$
and $P(A|C)=0.9$
is it possible to find $P(A|B,C)$ ?115.178.29.142 (talk) 01:20, 31 August 2010 (UTC)

No. Here is one extreme case consistent with the information given above:

The sample space consists of five points: v, w, x, y, z, with respective probabilities 1/20, 1/20, 8/20, 1/20, 9/20, and the events are

A = {w, z}

B = {w, x, y}

C = {y, z}

Then P(A) = 1/20 + 9/20 = 0.5,

P(A | B) = (1/20)/(1/20 + 8/20 + 1/20) = 0.1

P(A | C) = (9/20)/(9/20 + 1/20) = 0.9

and P(A | B & C) = 0.

Here is the opposite extreme case, also consistent with the information given above:

The sample space consists of five points: v, w, x, y, z, with respective probabilities 1/20, 1/20, 9/20, 8/20, 1/20, and the events are

A = {v, w, y}

B = {w, x}

C = {w, y, z}

Then P(A) = 1/20 + 1/20 + 8/20 = 0.5

P(A | B) = (1/20)/(1/20 + 9/20) = 0.1

P(A | C) = (1/20 + 8/20)/(1/20 + 8/20 + 1/20) = 0.9

and P(A | B & C) = 1.

Michael Hardy (talk) 02:29, 31 August 2010 (UTC)

Which experiment is the best to perform?

You have a hypothesis that you wish to test. Based on your current knowledge, you evaluate the probability that the hypothesis is true as 50%. You can choose to perform one of three experiments. You have already previously evaluated the expected probability that the hypothesis is true after performing the experiment. For experiment 1, this is 30%. For experiment 2, 60%, and experiment 3, 90%. Which experiment is expected to best perform in getting the most accurate reading for the truth/falsity of the hypothesis?--Alphador (talk) 07:13, 31 August 2010 (UTC)

I'm not sure this question makes sense. Shouldn't the a priori expected probability that the hypothesis is true after the experiment also be 50%? An experiment will have multiple outcomes, and we could evaluate the probability that the hypothesis is true given each of the outcomes, and also evaluate the probabilities of each outcome based on our current knowledge. But if the weighted average of these probabilities is other than 50%, wouldn't that give us a different probability for the hypothesis based on our current knowledge? —Preceding unsigned comment added by 203.97.79.114 (talk) 09:45, 31 August 2010 (UTC)

May I ask a silly question; regarding the Calandar and how we work-out rates to pay ?

I am finding myself trying to explain to well educated people on this subject, but are not Maths. / Accountancy perhaps. I have been trying to explain that a rate of rent per calandar month is a payment of twelve per year and so is to 12 * 4 = 48 weeks. A rate per year is to 52 weeks. Yet a Company is trying to fit one into soft-ware that does not fit. Am I wrong? MacOfJesus (talk) 11:13, 31 August 2010 (UTC)

If you pay monthly then there are twelve payments a year, but this doesn't mean there are 48 weeks in a year. If you pay yearly then there is one payment a year, regardless of whether there are exactly 52 weeks in a year (which there aren't). 86.173.36.196 (talk) 14:06, 31 August 2010 (UTC).

Random permutation

Initially:

a(1) = 1
a(2) = 2
...
a(n) = n

Then:

for i = 1 to n
   r = random number between 1 and n (inclusive)
   swap the values of a(i) and a(r)
next

Question: will this produce a uniformly random permutation of 1,2..n in a() (i.e. all permutations equally likely)? If not, is there a simple tweak that will fix it so it does? —Preceding unsigned comment added by 86.173.36.196 (talk) 14:01, 31 August 2010 (UTC)