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September 22

rational number is group under addition

the set of rational numbers (gcd(a,b)=1) whose denominator are odd is group under addition?(include zero).

Yes (as long as negatives are also included, of course). It is left as an easy exercise for the reader to prove it. -- Meni Rosenfeld (talk) 07:31, 22 September 2006 (UTC)

Original poster, ignore this: we can also look at that as $\mathbb {Z}$ localized at the prime ideal (2), right? Tesseran 08:48, 22 September 2006 (UTC)

To be honest, this is the first time I've heard about localization of a ring. If I understand that article correctly, then no - rather, it would be Z localized at the set of odd integers. -- Meni Rosenfeld (talk) 20:22, 22 September 2006 (UTC)

You can localize in multiplicative set, for example odd integers (here You are right), but if one sais "to localize in prime ideal" he means to localize in a multiplicative set consisting of all elements not in a given prime ideal. So "to localize at the prime ideal (2)" means exactly "to localize using multiplicative set of odd numbers".

Oh, okay. -- Meni Rosenfeld (talk) 06:02, 23 September 2006 (UTC)

Furthermore, it's even a ring, isn't it? – b_jonas 18:01, 23 September 2006 (UTC)

"difference between average mode mean?"

In statistics, what is exactly the difference between mode, mean and average?"

Usually "mean" and "average" mean the same thing. See our articles on Mode, Mean, and Average. You may also be interested in Median. --Lambiam^Talk 15:20, 22 September 2006 (UTC)

"Arithmetic mean" is the same as "average", but "geometric mean" is something else. StuRat 15:23, 22 September 2006 (UTC)

A distinction should be drawn between the precise meaning of these terms and the popular meaning. In the latter case, "average" nearly always refers to the arithmetic mean, but more precisely it refers to a "typical value", or with more jargon, a "measure of central tendency". Thus any mean, or mode, or median, is an average. (Though the mode, in particular, can be anything but central.)

Again, "mean" without qualification usually refers to the AM, but there are any number of different means which can be calculated from a set of figures, the commonenest are maybe the geometric and harmonic varieties.

It's a great pity that a subject concerned with the careful evaluation of data is cursed with such sloppy usage.--86.132.238.249 18:01, 22 September 2006 (UTC)

chi-square in statistics

how do I calculate the chi-square in very clear and simple steps?Mariesaintmichel 15:12, 22 September 2006 (UTC)thank you Marie Saint Michel

What is the nature of your data? Is it a 2 by 2 contingency table? --Lambiam^Talk 15:22, 22 September 2006 (UTC)

If you can calculate the expected values, then Pearson's chi-square test tells you the calculation ($\chi ^{2}=\sum _{i=1}^{N}{\frac {({O_{i}-E_{i}})^{2}}{E_{i}}}$) to use.

For a congintency table, here's a hint:

${\mbox{expected frequency of cell}}={\frac {{\mbox{row total where cell is}}\times {\mbox{column total where cell is}}}{\mbox{sum of all cell values in contingency table}}}$

x42bn6 Talk 01:39, 23 September 2006 (UTC)

September 23

Identities

What are identities involving nested hyperbolic functions and inverse trigonometric functions called? --HappyCamper 04:43, 23 September 2006 (UTC)

Say, something of this sort: $\tanh ^{-1}(\sin 2x))\equiv 2\tanh ^{-1}(\tan x).$

This is also equal to ${\mbox{gd}}^{-1}(2x),$ which uses the inverse Gudermannian function. I don't think the species has been named; perhaps call them "inverse-Gudermannian identities"? --Lambiam^Talk 15:57, 23 September 2006 (UTC)

Ah, I did not spot that quite so quickly. Actually, this came from a foreign language textbook that I found, there were other interesting identities invovling inverse sines and hyperbolic cosines, which was what prompted this question. --HappyCamper 20:32, 23 September 2006 (UTC)

Questions about the shape of the earth and distances

I was thinking, what is the relationship to a torus and a sphere? That is, let's assume that we have a torus and a sphere with equal surface area and equal volume. So let's say that the earth was a torus, with two equators, and the north pole and south pole coalesce into each other, in the middle. And let's say that the continents and oceans of the earth were proportionally mapped onto the torus. So my question would be, let's say someone wanted to get from Alaska to South America. In a spherical earth, a person would have to fly a plane south from Alaska to reach to South America. However, in a toroidal earth, a person could theoretically travel north, through the pole, and arrive at South America, approaching it from the south. Let's ignore the affects of weather. So would it be a shorter distance? And is there a formula to decide whether or not a distance from random points A and B on a torus would be longer or shorter than the proportional points A' and B' on a sphere? Thanks. --Ķĩřβȳ♥ŤįɱéØ 06:05, 23 September 2006 (UTC)

Let's begin by saying that there is not torus with equal surface area and volume as a sphere. Even if we just stick to "same surface area" (and lesser volume), I do not understand what your transformation from the sphere to the torus is, but I am inclined to believe that distances will typically be longer. -- Meni Rosenfeld (talk) 06:13, 23 September 2006 (UTC)

Ok, they have the same surface area then. I mean it the earth to look like this: [1]. --Ķĩřβȳ♥ŤįɱéØ 06:54, 23 September 2006 (UTC)

Here are some thoughts:

• The image is misleading, because the "second equator", the line where the poles are stitched together, has a discontinuity, where in the image it looks continuous.
• Suppose the earth has a radius of R, and the torus has an inner radius of a and an outer radius of b. Then:
  • The distance around the equator is 2πR on earth, (b+a)π on the torus.

What was I thinking? It's 2bπ. -- Meni Rosenfeld (talk) 08:01, 25 September 2006 (UTC)

• • The distance from the south pole, through the equator, to the north pole is πR on earth, π(b-a) on the torus (and of course, just the distance from the equator to a pole is half of that).
  • The surface area of earth is 4πR², of the torus is (b²-a²)π.
  • The distance around latitude θ is 2πRcos(θ) on earth, ((b-a)cos(2θ)+a+b)π on the torus. (a nicer formula: 2π(asin²θ + bcos²θ)
• We would want the first three of these to be equal for earth and the torus, but they are easily seen to be incompatible. If the area is to be equal, then at least one of the first two distances must be greater on the torus.
• The fourth distance, for θ = 90°, is 0 for earth but (b-a) for the torus. Not nice. This also means that whatever our choice for a and b, there will be near the poles points which are very close on earth but distant on the torus (and the ratio can be as large as we please).
• All that aside, regarding your original question: Yes, since we can "magically" leap from the south pole to the north pole, there will necessary be points which on earth are distant (one near the north pole, one near the south pole) but will be close on the torus.
• About a general formula, it could be complicated, but I have given above examples of some specific cases. Of course, since they are all circles, the corresponding arcs will have proportional lengths.

I hope this answers your question. -- Meni Rosenfeld (talk) 08:24, 23 September 2006 (UTC)

Yes, it answers many of my questions, but I don't understand one part. How is it the second equator discontinuous? --Ķĩřβȳ♥ŤįɱéØ 08:38, 23 September 2006 (UTC)

It is not continuous in the same sense as a sphere because you're mapping a sphere (the map of Earth) on a torus. The north pole is connected to the south pole in the torus, so if you did morphed Earth as a torus, you'd have a discontinuity in the landmasses. Here's a visualization:

That horizontal middle line is how it would look like in the second equator, and that's what I think he meant with discontinuity. Hope that helps. ☢ Ҡi∊ff⌇↯ 09:33, 23 September 2006 (UTC)

Yes, suppose the north pole would be painted blue and the south pole painted red. Then in the stitch, you'll see a discontinuous jump from red to blue. -- Meni Rosenfeld (talk) 10:03, 23 September 2006 (UTC)

In the most general terms, it's not possible to go from a surface without a hole, to one with a hole, without tearing the surface. StuRat 10:53, 23 September 2006 (UTC)

Iterated sine function

Is there a name and a documented use for an iterated sine function? Example (my notation): $\sin _{5}x=\sin \sin \sin \sin \sin x$

I've been playing with it and it's been kinda interesting: when i tends to infinity, $\sin _{i}x\times \left({\frac {1}{\sin _{i}\pi /2}}\right)$ approaches a smooth square wave $f(x)=\operatorname {sgn} \sin x$ (using the sign function), which looks better than one made by additive synthesis of sine waves. I was also wondering if there would be a way to get the triangle and sawtooth waves with a similar method. $\tan \sin x$ is pretty close from a triangle wave, but I can't find a way to make it converge to a true triangle wave. $\sin \tan x$ resembles a sawtooth wave, but it gets chaotic near the jumps... Still, pretty interesting behaviors. ☢ Ҡi∊ff⌇↯ 08:30, 23 September 2006 (UTC)

Well, the infinite iteration could simply reprsented as $\sin x=x$

sawtooth

And that could be represented as an infinite series: $\sin x,\sin(\sin x),\sin(\sin(\sin x))),\sin(\sin(...))$

So the iterated sine would be the nth term of this series, where n is the number of iterations. I don't think this has a name though.

I'm not sure about your second question though. My guess would be a piecewise function involving absolute value. I hope that helps --Ķĩřβȳ♥ŤįɱéØ 09:03, 23 September 2006 (UTC)

I'm not sure if I get your point. The infinite iteration doesn't fall back to simply $\sin x=x$. Just plot sin(sin(sin(...x))) yourself and take a look. About the second question, I really don't want to involve piecewise functions here. ☢ Ҡi∊ff⌇↯ 09:26, 23 September 2006 (UTC)

The iterated sine function converges pointwise (i.e., for each fixed x) to zero. I would not like to express this exactly as the infinite iteration could simply reprsented as $\sin x=x$; but there is a connection with the fact that that equation has no other (real) solution than zero. That makes the idea of Kieff to multiply with a 'normalising factor' rather interesting. (Note, that there is nothing magical connected with $\pi /2$ in this context. If you pick any number c strictly between 0 and $\pi$, and use $(\sin _{i}c)^{-1}\,$ as normalising factor, you'll get the same limit function.) The 'normalisation' counteracts a very slow convergence close to the origin.

I've no idea whether or not this is done - this is not exactly my branch of mathematics - but if it isn't, I think it would be worth doing. JoergenB 15:43, 23 September 2006 (UTC)

Experimentally, the normalizing factor appears to be asymptotically equal to ${\sqrt {\frac {i}{3}}}$. --Lambiam^Talk 16:33, 23 September 2006 (UTC)

How come you say any number c strictly between 0 and π will get the same limit function? I used π/2 because it is a point when the function is at its highest value, so the normalization would make its max be exactly 1 at any iteration. Does that detail become completely irrelevant when dealing with infinite iterations? ☢ Ҡi∊ff⌇↯ 17:17, 23 September 2006 (UTC)

OK, I'll answer, but I fear I can't without getting slightly technical.

Starting with c strictly between 0 and $\pi /2$ yields a positive number not greater than 1 in the next step, and continues with a strictly decreasing sequence of positive numbers:

$\sin c>\sin _{2}c>\sin _{3}c>\sin _{4}c>\ldots >0$

(since, by an exercise in elementary calculus, indeed $0<\sin x<x\,$ for $0<x<\pi /2\,$). Thus, this sequence has an infimum $L\geq 0\,$. Now, if L were strictly greater than 0, then for some i we would have $\sin _{i}c<\arcsin L\,$, yielding the contradiction $\sin _{i+1}c<L\leq \sin _{i+1}c\,$. Thus, instead, $L=0$, for any start value c in the open interval $(0,\pi /2)\,$.

In particular, for any two given (fixed) c and d in that interval; there is a natural number j, such that $\sin _{1+j}c<\sin d\,$ and $\sin _{1+j}d<\sin c\,$. Moreover, the sine function is strictly increasing in the interval from 0 to 1 (check its derivative!), whence these properties carry over to all higher iterations as well:

$(1)\ \ \sin _{i+2j}c<\sin _{i+j}d<\sin _{i}c,\ i=1,2,3,4,\ldots$.

If you've followed the argument this far, then you're ready for the main point: Every time we do the iteration, we are replacing the value of some x with this x times a factor ${\frac {\sin x}{x}}$, and these factors converge to 1 as x approaches 0. In fact, in this interval, $x-{\frac {x^{3}}{6}}<\sin x<x$, whence the factor is a number between $1-{\frac {x^{2}}{6}}$ and 1. Thus, also the j'th power of this factor tends to one. Summing up, if we compare the quotients of the terms in formula (1) with $\sin _{i}c\,$, then we find that the first quotient converges to 1, while the third one indeed is (the constant) 1; whence the squeezed middle quotient also converges to 1.

I hope I glossed over an appropriate amount of details (giving you just enough to fill in yourself) :-) JoergenB 18:59, 23 September 2006 (UTC)

For a triangle wave, try $\arcsin ~\sin x$ or $|x-\lfloor x\rfloor -1/2|$. For sawtooth, try $\arctan ~\tan x$ or $x-\lfloor x\rfloor$. – b_jonas 17:46, 23 September 2006 (UTC)

Thanks for those but they're not what I'm looking for right here. I'm trying to find smooth functions that approximate these ideal waveforms as some value increases, like the iterated sine with the square wave. Using the floor function is kinda bad in this case too, since it's a bit of a "hack". ☢ Ҡi∊ff⌇↯ 06:13, 24 September 2006 (UTC)

Well, the triangle one might be possible by modifying the square wave one you have... if you are transforming a sin into a sqaure gradually via some process, it seems that the triangle should be available, since the square is, I believe, $\sum _{i=1}^{\infty }{\frac {\sin {((2i-1)x})}{(2i-1)}}$, and the triangle the milder $\sum _{i=1}^{\infty }{\frac {\sin {((2i-1)x})}{(2i-1)^{2}}}$, a less severe transformation of your initial sin might get you there (or something with an equivalent spectrum, maybe). Saw waves might be harder, since you will need to find some way to generate the even harmonics as well. Do you know about Frequency modulation synthesis? It is similar to what you are doing, though it is not usually taken to infinity like what you are doing here. - Rainwarrior 18:20, 24 September 2006 (UTC)

But by the way, if you are trying to find a smooth function that approximates one of these, why not just use the Fourier series (the sums I just mentioned above) of them and stop after a certain number of harmonics? - Rainwarrior 18:30, 24 September 2006 (UTC)

Oh, I don't have a direct use for these stuff, it's just something I've been toying with, really. But the problem with Fourier series is that they have artifacts. The iterated sine gives a much better result than a square wave Fourier series, and it has no artifacts at all, so I thought "hey, maybe there are similar functions for the other waves, and they could be pretty handy"... That's pretty much it.

I already tried getting some info from the Fourier series, btw, but that didn't really help. :( ☢ Ҡi∊ff⌇↯ 20:53, 24 September 2006 (UTC)

What do you mean "artifacts"? A summation of sine waves is continuous... But, if you're interested in finding other curves which approximate various waveforms, why not construct a polynomial curve with suitable guidelines between 0 and 2pi? (For starters, make it pass through (0,0) and (2pi,0), and make both tangents at those points equal as well, then keep going, add point or tangent constraints until you have the curve you like.) - Rainwarrior 03:57, 25 September 2006 (UTC)

The artifacts I'm talking about is the Gibbs phenomenon. And no, a polynomial would take away all the fun of it (too easy), plus it would not be cyclic. ☢ Ҡi∊ff⌇↯ 07:49, 25 September 2006 (UTC)

Sorry for my stupidity, but what does an X with the L and the mirrored L around it mean? ChowderInopa 00:09, 24 September 2006 (UTC)

It denotes application of the floor function to the argument. For example, ⌊3.14⌋ = 3, because 3 is the largest integer that does not exceed 3.14. If we had a skyscraper with a floors at the height of every integer, and real numbers had to live at their own heights, then 3.14 would be living on the 3rd floor. --Lambiam^Talk 00:32, 24 September 2006 (UTC)

September 24

Random Variable X and its Mean Cancelling

If I have a random variable X with a mean mu(X), and both are in the same equation, one being negative, can the two simply cancel out?

Basically, how do the two relate?

Like in X – mu(X)? No, they won't simply cancel out. This is a new random variable, say Z. Have you read the articles Random variables and Expected value? Take for example that X is: the result of throwing two dice and adding the number of eyes. The outcomes of X may range from 2 to 12, with an expected value (arithmetic population mean) of 7. Suppose you throw the dice 10 times and observe for X this sample: [3, 7, 10, 9, 9, 7, 2, 7, 2, 7]. (I actually threw two dice ten times here.) That means for Z = X − 7 this sample: [−4, 0, 3, 2, 2, 0, −5, 0, −5, 0]. Occasionally the outcome of Z is 0, but the random variable Z itself is clearly not the constant 0. However, mu(Z) = 0. To see this, we need three facts: mu(V) = E(V) for a random variable V – which is true by definition of mu(.), E(X − Y) = E(X) − E(Y), and E(C) = C for constant C. Then mu(Z) = E(Z) = E(X – mu(X)) = E(X) – E(mu(X)) = mu(X) – mu(X) = 0. Indeed, the sample mean for our sample of 10 outcomes for Z is −0.7 – not quite 0 due to the random fluctuations, but fairly close. --Lambiam^Talk 01:13, 24 September 2006 (UTC)

Okay, but in the long run, they will cancel? Like if I have a function Y equal to the equation you said, Y = X - mu(X), the E(Y) will be 0, simply due to the rule that X can be transformed into mu (X)?

Yes, E(Y)=0, but I'm not sure what you meant by "X can be transformed into mu (X)". -- Meni Rosenfeld (talk) 16:38, 24 September 2006 (UTC)

Well in order for E(Y)=0, X must be equal to mu (X) to cancel...how does this come about? How do you show that E(Y)=0?

That's what I showed above, except that I called it Z instead of Y. --Lambiam^Talk 16:59, 24 September 2006 (UTC)

Wow Sorry you are completely correct, that makes perfect sense, thank you. Just as a final clarification, the E(muX) is the part that pertains to E(C) = C for constant C?

Exactly. E(X) (aka μ(X)) is a constant, therefore E(E(X)) = E(X). -- Meni Rosenfeld (talk) 17:11, 24 September 2006 (UTC)

the other delta function?

OK, so I am trying to understand how the fourier transform of a pulse train in the time domain gives also a pulse train in the frequency domain. Actually i'm stuck on a little detail. I get that: $F[\delta (t)]=\int \delta (t)\exp ^{-jwt}dt={1}$, that $F^{-1}{[1]}=\int \exp ^{jwt}dw={\frac {1}{jt}}\exp ^{jwt}$, and that $F^{-1}F[\delta (t)]=\int (\int \delta (t)\exp ^{-jwt}dt)\exp ^{jwt}dw={\frac {1}{jt}}\exp ^{jwt}$. And that in the end, you make the jump and say that $\delta (t)={\frac {1}{jt}}\exp ^{jwt}$...

I don't see it. Did I miss something? Anybody has some intuition (hopefully on physical grounds) on how the expression ${\frac {1}{jt}}\exp ^{jwt}$ is equivalent to a delta function? please? --crj

What makes you think a pulse train in the time domain should give a pulse train in the frequency domain? On the contrary, it should be all over the place in the frequency domain (see "Localization property" in the article Continuous Fourier transform). This should be intuitively obvious: you can't assign any one frequency to a single spike. It also fits with your result: 1, which is hardly a spike. In the expression you give for $F^{-1}{[1]}$, there is an occurrence of the variable $w$ (the traditional notation is $\omega$). That can't be right; the result should only depend on $t$ . This is not an indefinite integral (antiderivative; primitive function) but a definite integral for $w$ from $-\infty$ to $+\infty$. --Lambiam^Talk 21:30, 24 September 2006 (UTC)

Ooops. I meant to type t instead of w after the integration. There are some notes on the fourier transform of pulse train in the article Dirac comb. Maybe I am doing the problem the wrong way but something smells fishy here... because what happens at t=0? oh dear. Thanks anyway. --crj 00:48, 25 September 2006 (UTC)

Whoa! Careful there. The transform of a single impulse is a constant; the transform of a periodic sequence of impulses is again a periodic sequence of impulses, whose period is inversely proportional to the original period. --KSmrq^T 22:34, 24 September 2006 (UTC)

So the Dirac delta function is a tiny bit confusing? Since it's not really a normal function at all, that's not surprising. One way we approach it is through its properties within an integral. That is,

$\int _{-\infty }^{+\infty }\delta (t-t_{0})f(t)dt\,\!$

selects the value f(t₀). We can be more formal by using a limiting process. For example, we know that a Gaussian bell curve,

${\frac {1}{\sigma {\sqrt {2\pi }}}}\exp(-{\frac {x^{2}}{2\sigma ^{2}}}),\,\!$

integrates to 1, and can be made as narrow as we like by letting σ approach zero. We also know (and can easily verify) that the Fourier transform of a Gaussian is again a Gaussian, but with width inversely proportional to σ. As we pinch the Gaussian to its delta function limit, its transform spreads and flattens to a constant.

A delta function cannot be written as you propose, as a simple exponential, which may account for your confusion.

If making a single impulse requires some "funny business", making an impulse train requires more. But perhaps we should stop here for now. --KSmrq^T 23:24, 24 September 2006 (UTC)

Let me rephrase the question:
According to the Dirac comb article the fourier transform of an impulse train is:
$\sum _{n=-\infty }^{\infty }\delta (t-nT)\quad \Longleftrightarrow \quad {1 \over T}\sum _{k=-\infty }^{\infty }\delta \left(f-{k \over T}\right)\quad =\sum _{n=-\infty }^{\infty }e^{-i2\pi fnT}$.
My question is, under what grounds are the last two terms (sum with delta function in the frequency domain, sum with complex exponential) equivalent? -- Crj 02:34, 25 September 2006 (UTC)

Let's look at the sum. $\sum _{n=-\infty }^{\infty }e^{-i2\pi fnT}={\frac {e^{-2\pi ifT(N+1)}-e^{2\pi ifTN}}{e^{-2\pi ifT}-1}}={\frac {\sin 2\pi fT(N+{1 \over 2})}{\sin 2\pi fT}}$. Now let N approach infinity. It's possible to check that this sum approaches infinity if f = k/T (with integer k), approaches zero otherwise, and that its integral over any interval containing exactly one point where f = k/T approaches 1. So it's a sum of delta functions. Conscious 11:53, 25 September 2006 (UTC)

The question is based on an elementary slip. If the sum 1+2+3 equals the sum 2+2+2, we have no grounds for assuming equality between respective terms. This is true for infinite sums as well. --KSmrq^T 14:41, 25 September 2006 (UTC)

By the way, you are wrong in saying (in your initial post) that $F^{-1}{[1]}={\frac {1}{it}}\exp ^{i\omega t}$. You have to use the definite integral, i.e. $F^{-1}{[1]}=\int _{-\infty }^{+\infty }e^{i\omega t}d\omega$, to obtain the result. Conscious 17:02, 25 September 2006 (UTC)

The question of equivalence of the sum of deltas and the sum of exponentials is discussed at the Nyqvist-Shannon sampling theorem (in the Mathematical basis for the theorem section). There, it is explained that the sum of exponentials only agrees with the sum of deltas in the sense of tempered distributions. This is slightly weaker that pointwise equality. -- Fuzzyeric 02:26, 28 September 2006 (UTC)

wow this turned out to be a funny business. my difficulty really was in recovering the delta function intact after I take the fourier transform of an impulse train: "delta functions in... delta functions out", or so I thought. Thanks for all the responses! -- Crj 14:17, 26 September 2006 (UTC)

September 25

Dice-sum probability formula

It's September 24th here on the East Coast, but anyway: suppose I roll n s-sided dice. What is the theoretical probability of getting each possible sum, irrespective of the individual die values that make up that sum? I know a little bit of combinatorics and probability, but not quite enough to figure this one out on my own. Google has failed to enlighten me, so I turn to your collective wisdom for help. —Saric (Talk) 01:20, 25 September 2006 (UTC)

I think there is no truly simple formula for that. For one die the graph of the probabilities as a function of the sum is a horizontal line: a segment from a polynomial of order 0. For two dice you get a roof shape: two segments of polynomials of order 1. I guess that continues, like N segments of polynomials of order N−1 glued together. It should be possible to find a general formula for the i-th polynomial. I haven't given the issue much thought, though. --Lambiam^Talk 02:32, 25 September 2006 (UTC)

Lambiam is right. Here are several additional suggestions:

• First, we will denote by f (n, s, k) the number of possibilities (from which the probability is easily calculated) of receiving a sum of k when throwing n s-sided dice.
• For sufficiently large n, You can approximate the function with the normal distribution as:

$f(n,s,k)\approx s^{n}\exp {\left({\frac {3(n(s+1)-2k)^{2}}{2n(1-s^{2})}}\right)}{\sqrt {\frac {6}{\pi n(s^{2}-1)}}}$

• You can define f recursively: f (1, s, k) is 1 if k is an integer between 1 and s, and 0 otherwise; And

$f(n+1,s,k)=\sum _{i=1}^{s}f(n,s,k-i)$

• If I'm not mistaken, the piecewise polynomials from which the function is made are stitched at points of the form $s(1+j)+n/2\,\!$, where 0 ≤ j ≤ n-2. You can use this and the above formula to find, for every n, the polynomials that make up the nth function (though you shouldn't hope for a closed form for the nth polynomials).

I hope this helps. -- Meni Rosenfeld (talk) 07:42, 25 September 2006 (UTC)

Ah, yes, thank you both. I implemented that function in Perl and it seems to work nicely:

sub funkshun
 {my ($n, $s, $k) = @_;
  my ($i, $sum);
  if ($n == 1)
     {if ($k >= 1 and $k <= $s)
         {return 1;}
      return 0;}
  foreach $i (1 .. $s)
     {$sum += funkshun(($n - 1), $s, ($k - $i));}
  return $sum;}

All you do is divide the result by s to the nth power, and there's your probability. There is, however, one problem: it's really slow. f(5, 20, 55), for example, takes a couple of seconds. Maybe it's just my 300-megahertz processor, but I think the recursion makes this a relatively inefficient algorithm.

Now, here's the interesting part: some searching brought me to this site. The instructions page attempts the explain the math, but the explanation eluded me, which is why I came here to the reference desk for an algorithm I could understand (and thus implement myself). However, rereading it afterwards, I realized that the words "(k = 0) to the next lowest integer of…", which had puzzled me before, were implying the use of summation, like you, Meni Rosenfield, had used in your formula. That was enough for me to figure out the rest and code it thus:

sub funkshun
 {my ($n, $s, $k) = @_;
  my ($i, $sum);
  foreach $i ( 0 .. int(($k-$n)/$s) )
     {$sum += ( ((-1)**$i) * comb($n, $i) *
                comb($k-1-($s*$i),$n-1) );}
  return $sum;}

("Comb" is the combination function; "**" is the Perl exponentiation operator.) Looks crazy, doesn't it? It does to me, at least— but it works, and a lot faster than the other way, too. Who woulda thunk? —Saric (Talk) 00:11, 26 September 2006 (UTC)

I see. Rewriting this in terms I can understand, it's:

$f(n,s,k)=\sum _{i=0}^{\left\lfloor {\frac {k-n}{s}}\right\rfloor }(-1)^{i}{n \choose i}{k-1-si \choose n-1}$

Which is simpler than I imagined. Nice! -- Meni Rosenfeld (talk) 09:21, 26 September 2006 (UTC)

By the way, the reason your first implementation was so slow is because each value of f is calculated many times, making the complexity exponential in n. If you are able to store calculated values of f in memory, and add some more fine-tuning, the complexity will only be quadratical (that is, much faster). -- Meni Rosenfeld (talk) 09:32, 26 September 2006 (UTC)

Density Curve (?)

In my basic Stat class (Algebra Based) the teacher was explaining density curves. She then gave us various examples, including one that had uniform distribution. She described the density curve as being the base times the height, where the density curve would follow the box-shaped outline exactly. However, I don't understand how that could be a "curve" because of the sharp points. The only way I could see a function like that defined is with a piece-wise equation. Help me - just try to keep it within Calc II understanding. ;) --AstoVidatu 02:48, 25 September 2006 (UTC)

If you define the function as having result 0 for arguments outside the range of the random variable, you're right, technically speaking the shape of the graph of that function is not a curve, since it's not continuous. It does not "climb up" the sides of the box, but it jumps from 0 to this constant positive value. If you pay close attention mathematicians say very sloppy things all the time, like they say "the function f(x)" when they mean "function f" and think that is fine, which is a bit amazing for a field that is supposed to be exact and precise. --Lambiam^Talk 03:41, 25 September 2006 (UTC)

If she described the graph of the density function as following the vertical edges of the box, she must have been confused; a piece-wise definition for the density function is indeed what you're looking at here.

Of course, that leaves the question of the points where the function jumps, and which function value to choose there; however, it turns out that it doesn't matter.

Using "curve" to describe this graph might indeed be a bit problematic, but the graph of the absolute value function - despite having a corner, is usually considered a curve.

RandomP 16:32, 25 September 2006 (UTC)

Parallelepipeds' volume

I know that to get the volume of a parallelepiped, when given three vectors, you use the formula ${\vec {a}}\cdot [{\vec {b}}\times {\vec {c}}]$, which gives you a 3x3 determinant. However, does it matter, when evaluating the determinant, the order in which I put the rows? My Engineering and my Calculus textbooks put the row outside the parenthesis (e.g. $a_{x}$, $a_{y}$, $a_{z}$) on the top row of the 3x3 matrix, but I don't know if this is a coincidence, a convention, or if there is some mathematical reason this happens. And the reason I'm asking is because I can't ever remember which row goes where, so it would be quite relaxing to know it doesn't matter... Titoxd^(?!?) 03:04, 25 September 2006 (UTC)

Changing the order can only cause a sign flip: see Triple product for the relevant identity. If you're looking for a positive quantity, you'll want to take the absolute value anyway, so short answer: no, it doesn't matter. Melchoir 03:26, 25 September 2006 (UTC)

If, for any reason (such as determining orientation), you wish to preserve the sign of ${\vec {a}}\cdot [{\vec {b}}\times {\vec {c}}]$, then the order does matter, but it's not a biggie - if I'm not mistaken, the order of the rows is exactly the order in which they appear in this expression. -- Meni Rosenfeld (talk) 06:30, 25 September 2006 (UTC)

Yes, it matters. Since you already know the volume is a determinant, you should also know a basic definition of a determinant:

1. The determinant of the identity matrix is 1.
2. The determinant is a linear function of each column.
3. The determinant is an alternating function of the columns.

   (Two equal columns gives a zero result, so swapping two columns negates the result.)

We can show that transposing a matrix does not change its determinant, so the above facts can use rows instead of columns. Either way, these three rules uniquely define the determinant of any n×n matrix.

The cross product is also an alternating function of its arguments, so the sign change is already visible when swapping b and c.

But why should we care about the sign, when the magnitude is the same either way? Because when we are calculating more complicated volumes with this as a portion of a larger sum, we must get the sign right. Consider a complicated polygon in 2D, but one that does not intersect itself. (Technically, it is called a "simple" polygon.) We can compute the area inside the polygon by summing triangles each of which consists of a polygon edge and the origin — but only if we get the signs right. In 3D, using faces to give tetrahedra (sliced parallelepipeds) works similarly. This convenient method can be viewed as a specialization of Stokes' theorem, which itself is just a fancy version of the fundamental theorem of calculus, telling us we can find the integral over the interior of a region from what happens on the boundary.

So resist the urge to forget the order. The investment in good habits today will pay off in easier computations tomorrow. --KSmrq^T 15:58, 25 September 2006 (UTC)

But of course, "the identity matrix" has no special parallelepiped associated to it: see right-hand rule for one convention on it, though. (To mathematicians, this rarely matters, because we're usually just as happy to see our geometry mirrored. Physicists have that whole "real world" thing to take into account, though.)

KSmrq is right, of course: it's essential to get into the habit of getting things right.

RandomP 16:23, 25 September 2006 (UTC)

Arc length

I need to find the length of a circular arc (shown). I know the length of the straight line between the two end nodes (BD), and I know the height of the arc (AC), but cannot think of any way to find the length of the curve. Is there a formula I can use in this case? smurrayinchester^{(User), (Talk)} 11:03, 25 September 2006 (UTC)

Let O be the centre of the arc. OA=OB=r is the radius of the circle. From the OBC triangle: OB²=BC²+OC², or r²=BC²+(r-AC)²; you can find r from this equation. After this, the angle BOC is equal to arcsin(BC/r), and the arc length is 2rarcsin(BC/r). Conscious 11:24, 25 September 2006 (UTC)

After simplifications, it's $l={\frac {a^{2}+4h^{2}}{4h}}\arcsin {\frac {4ah}{a^{2}+4h^{2}}}$ where a=BD and h=AC. Conscious 11:32, 25 September 2006 (UTC)

Another method: Find the center, and hence the radius, by constructing tangent lines at both ends and constructing perpendicular lines to those. The intersection of those two lines will be the center point. Then measure the angle swept by the arc. You can then apply the simple formula l = piR(theta/360), where theta = the angle swept in degrees. StuRat 11:35, 25 September 2006 (UTC)

Thanks to both of you; I've gone for Conscious's method as I don't have an accurate sketch of the arc available. The result is also close enough to my very simple approximation $l\approx 2{\sqrt {AC^{2}+CD^{2}}}$. smurrayinchester^{(User), (Talk)} 11:53, 25 September 2006 (UTC)

My preferred approximation is l ≈ BD :) Conscious 11:57, 25 September 2006 (UTC)

OK, maybe not that simple... smurrayinchester^{(User), (Talk)} 12:10, 25 September 2006 (UTC)

Factoring A Polynomial

Yes, yes, I realize this is a math HW prob. But we're kinda stuck. How does one go about factoring this polynomial: y^2-cy-dy+cd=0  ? I don't need answer, just advice on how to solve it. Thanks in advance. John 17:00, 25 September 2006 (UTC)

Just try to look what you can factor out from the first two terms, and from the two last ones. Conscious 17:04, 25 September 2006 (UTC)

Yes, I see that now. John 17:23, 25 September 2006 (UTC)

Using the angle (in radians) and length to get coordinates

Hi, if I have a line, starting at the origin (0,0), 10 units long, at a certain angle in radians, how can I work out what the coordinates of the end of the line is?

Any help would be much appreciated!

--Mary

[note: this isn't for homework: I'm writing a program where circles move around on a screen, and I want to include a line in the circle that shows a current direction.]

Take a look at our article on the trigonometric functions. Conscious 17:54, 25 September 2006 (UTC)

Better still, look at the circle article. --KSmrq^T 18:19, 25 September 2006 (UTC)

This is a conversion issue, from polar coords to 2D rectangular coords. StuRat 18:30, 25 September 2006 (UTC)

Thanks all! --Mary Bold text

Whoever adds all these "ANSWERED -> " stuff to the titles, I kindly ask that you don't do it in the future, as it will break #-links. Thanks. – b_jonas 16:15, 26 September 2006 (UTC)

Hmm, I didn't think about that. Are there many links to these questions ? StuRat 16:34, 26 September 2006 (UTC)

What's with the "m"? Are there any links to these questions? -- Meni Rosenfeld (talk) 16:56, 26 September 2006 (UTC)

Well, I occasionally link to some of them, especially when someone asks a suspicously similar question on RD again. Of course, I'm not an authority here so if the community decides for these title changes, you can do that, then I will resort to adding explicit Anchor tags instead of linking to lables. – b_jonas 18:10, 26 September 2006 (UTC)

How about if we create a subsection titled "ANSWERED" instead ? This will still show up in the TOC but won't keep your link from working. Sound good ? StuRat 22:10, 26 September 2006 (UTC)

Central Angle of a Tetrahedron

According to VSEPR theory, molecules such as methane are arranged in a tetrahedral form where the angle from one hydrogen to the carbon to another hydrogen is approximately 109 degrees. Is there any way to derive this angle mathematically? I have tried using the Pythagorean Thm in three dimensions, but it was a complete mess. Thanks, --JianLi 21:08, 25 September 2006 (UTC)

It's the top angle α of an equilateral triangle with base (2/3)×√6 and sides of length 1. Then the law of cosines gives you cos α = −1/3, as stated in the VSEPR article. No idea where the number −1/3 comes from. --Lambiam^Talk 22:30, 25 September 2006 (UTC)

Things become easier if you consider that a tetrahedron can be embedded in a cube, each face of the cube has one edge of the tetrahedron across it. If the cube has sides of length 2 and is centered at the origin then the verticies of the cube will be (+/-1,+/-1,+/-1). Four of these will be verticies of the tetrahedron, say A (1,1,1), B (1,-1,-1), C (-1,-1,1), D (-1,1,-1). Consider the triangle OAB. This has sides a=√3, b=√3, c=2√2. The law of cosines gives $c^{2}=a^{2}+b^{2}-2ab\cos(\gamma )$ so $8=3+3-2*3*\cos(\gamma )$ so $\cos(\gamma )=-2/6$ and $\gamma =109.471$ degrees. --Salix alba (talk) 23:07, 25 September 2006 (UTC)

Or you can just take the dot product of OA and OB. Melchoir 00:11, 26 September 2006 (UTC)

Wow, this problem seems so easy now. Thanks! --JianLi 01:56, 26 September 2006 (UTC)

Should some of this be mentioned in the text of VSEPR? Before cos^-1(-1/3) is mentioned there now, there is nothing constraining the shape of the tetrahedron. Is there an argument showing that the tetrahedron should be one sixth part of a cube? --Lambiam^Talk 15:34, 26 September 2006 (UTC)

September 26

uncountable set

x is uncountable set, y is subset of x, if y is countable, is that x\y uncountable.

Yes but you did not ask for proof. Twma 01:55, 26 September 2006 (UTC)

yes, the proof may be like this, assume x\y is countable, so N ~ x\y, and x\y is finite set, so every element in N, there is corresponding element in x\y, and y is subset of x, and y is countable, so x must be finite set, but this is a contradiction with x is uncountable set, so x\y is uncountable.

That doesn't make sense. An easy proof notes that x is the union of y and x\y, and the union of two countable sets is countable, so one of the two must be uncountable. You seem to have many questions-- perhaps a textbook like [2] would help? Melchoir 05:12, 26 September 2006 (UTC)

countable set

P is the set of polynomial function with rational coefficient, is P is countable set?

Yes but you did not ask for proof. Twma 01:56, 26 September 2006 (UTC)

The set of polynomials of degree 1 is clearly isomorphic to Q (the rationals). The set of polynomials of degree 2 is clearly isomorphic to QxQ. And similarly for higher degrees: Q³, Q⁴, Q⁵, ... The total number is therefore Q¹ + Q² + Q³ + ... This is clearly 1/(1-Q). Since Q isn't finite, this is -0.

Alternatively, for an actually helpful answer, you might want to start at Countable set. -- Fuzzyeric 02:43, 28 September 2006 (UTC)

The quadratic formula to solve x

For quadratic equation ax² + bx +c:

x = ( -b +- sqrt( b² - 4ac ) ) / 2a

I asked my math teacher the other day how this equation was formulated, and he told me that you must complete the square of ax² + bx + c ... he showed me, but I didn't quite understand. When I formed two squares, it was the sum of two squares to equal zero, which has no solution. Could somebody take me through it?

See quadratic equation#Derivation. Conscious 04:35, 26 September 2006 (UTC)

To have a quadratic equation, we must have an equality:

$0=ax^{2}+bx+c.\,\!$

Perhaps before studying the formal manipulations of "completing the square" we should look at simpler cases. We are looking for values of x that make the quadratic polynomial (the right-hand side of the equation) equal to zero. In more geometric terms, we wish to find where the parabola defined by

$y=ax^{2}+bx+c\,\!$

crosses the x-axis. This may happen zero, one, or two places. Suppose b is zero:

$ax^{2}+c=0.\,\!$

Then if a is not zero, we can divide both sides by a to give

$x^{2}+c/a=0.\,\!$

Now subtracting c/a from both sides, we see that the square of x must equal −c/a. So long as a and c have opposite signs, we can take a square root to get one answer and take its negation for another.

However, if b is non-zero the parabola is shifted left or right so that the zero-crossings do not occur symmetrically on each side of the y-axis. For example, suppose c is zero:

$ax^{2}+bx=0.\,\!$

Since both terms of the polynomial contain a factor of x, we can rewrite this as

$(ax+b)x=0.\,\!$

If x is zero, clearly the polynomial is zero. Otherwise we must solve

$ax+b=0.\,\!$

Divide by a and subtract to reveal the unique solution, −b/a. So in this case the parabola is centered halfway between 0 and −b/a, at −b/2a. If we shift the x values by this amount, the parabola and its zero-crossings will again be centered around zero.

Another way to describe "completing the square" is, "centering the parabola". Conveniently, we have just found the solution! For, if we add c back in to the polynomial, it bumps the parabola up or down, but does not change its center. That is, the parabola will always cross zero at −b/2a plus or minus a square root.

All that remains is to decide what should be inside the square root. We make the shifting substitution

$x=X-{\frac {b}{2a}},\,\!$

to produced the centered form

$aX^{2}-{\frac {b^{2}}{4a}}+c=0.\,\!$

This has no X term, so we can adapt the solution from above. Divide by a, subtract the constant, and simplify to produce:

$X=\pm {\frac {\sqrt {b^{2}-4ac}}{2a}}.\,\!$

Thus the solutions for x, rather than X, are

$x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}.\,\!$

To recap, the quadratic formula combines centering with square roots.

The question then naturally pops into our minds: Can we do something similar for cubic polynomials or higher? The answer is not at all obvious, took centuries to resolve, and produced some remarkable new mathematics. We can, with a little more difficulty, use shifting, square roots, and cube roots to solve cubics. With a great deal more difficulty we can similarly solve all quartics as well, though the method is so complicated it finds little use. But we can go no further; a polynomial of degree five or higher generally has no such closed-form solution.

So enjoy the quadratic formula. It's remarkably useful, and rather special. --KSmrq^T 16:51, 26 September 2006 (UTC)

equivalent class

if S is any infinite set, and x(- S, is that S is equivalent class to S\{x}.

I'm not quite sure what you're trying to say by "is that S is equivalent class to S\{x}", but if you're trying to ask whether S is in the same equivalence class (or, more succintly, "is equivalent to") S\{x}, you need to state what your equivalence relation is - for example, S and S\{x} are equivalent if the relation is "has the same cardinality" (since S is infinite), but they are not equivalent if the relation is "contains x". Confusing Manifestation 07:50, 26 September 2006 (UTC)

solve equations simultaniously

Solve equations 2x+3y=2 and 6x-y=-4 simultaneously.

What does this mean? How do I do this? I've probably been doing problems like this for about 5 years, but never got it well enough to remember anything. Thanks. 71.231.150.146 05:18, 26 September 2006 (UTC)

It means the value for X is the same in both equations and the value for Y is the same in both equations. There are many ways to solve them, the substitution method, addition method, graphing, using matrices, etc. StuRat 10:08, 26 September 2006 (UTC)

See Simultaneous equations. Conscious 05:38, 26 September 2006 (UTC)

In this case, I recommend you to read System of linear equations in the first place. JoergenB 10:00, 26 September 2006 (UTC)

Each separate equation defines a straight line in the plane; its solutions are precisely the points (x, y) lying on that line. The solution to both equations simultaneously is the coordinate where the lines intersect. Fredrik Johansson 11:26, 26 September 2006 (UTC)

It's hard to know how best to answer your question, since you do not make it clear what you already understand and what you want. The specific example you give is two equations in two variables, where each equation has at most one variable in a term. Is your question about this limited class of problems? Or are you also interested in more equations with more variables? Or perhaps more complicated equations such as quadratics? If you can "do them", what is it you'd like to remember? Are you saying you have to look up the solution technique each time, so would like help retaining the procedure? Help us help you. --KSmrq^T 17:17, 26 September 2006 (UTC)

Square Root?

What is the square root of France?

${\sqrt {(}}France)$

Thank you in advance for your assistance in solving this highly complex mathematical expression?

24.39.182.101 17:22, 26 September 2006 (UTC)

The usual definition of France is not mathematically rigorous, so we cannot use it to determine if √France is meaningful and its value. If seen as a variable which can take positive real values, then there is no way to simplify √France without knowing more about the value it represents. -- Meni Rosenfeld (talk) 17:34, 26 September 2006 (UTC)

France does not represent much of value. For most practical purposes, its square root may be considered zero. Fredrik Johansson 18:20, 26 September 2006 (UTC)

Think carefully before answering, as the verbal question would properly be notated

${\sqrt {\operatorname {France} }},\,\!$

which is not what we see. Perhaps what is meant is, "What country, if it applied itself twice over, could equal France?" But if so, it's a trick question, because France is, of course, incomparable. --KSmrq^T 19:07, 26 September 2006 (UTC)

Approximately $1.64(evil)^{\frac {1}{4}}$, of course. Factor out the root e, then apply the principle that money is the root of (all) evil. —AySz88\^-^ 22:24, 26 September 2006 (UTC)

I belive in this case, evil is the root of all money. Which, considering the widespread notion that money is the root of all evil, poses an interesting pair of questions: what is money? and what is evil? so that such an interesing pair of relations can occur.

There exist only complex numbers that, when squared, equal each other. So, economics and morality must both be complex. Black Carrot 06:44, 27 September 2006 (UTC)

Not exactly. The equations a²=b, b²=a have solutions in reals, (0,0) and (1,1). We can thus safely conclude that money = evil. -- Meni Rosenfeld (talk) 15:04, 27 September 2006 (UTC)

... or that neither money nor evil exist (or at least have no magnitude). -- Fuzzyeric 02:53, 28 September 2006 (UTC)

But we know that evil does exist. This leaves us with evil = 1, which means that evils is the fundumental unit. Therefore, everything in the world is made of evil. This explains a lot. -- Meni Rosenfeld (talk) 14:12, 28 September 2006 (UTC)

Pi

For some reason, I just can't comprehend how the Bailey-Borwein-Plouffe_formula calculates Pi. Does it really calculate individual digits? Could I easily calculate the, say, 84th digit? The article isn't too clear to me. Maybe it's not possible to explain in simple terms? --Russoc4 20:18, 26 September 2006 (UTC)

I don't really have time to answer this now, but if nobody gets back to you I'll try to explain this evening. But I have a question: is there a known, simple formula for calculating the decimal digits of π? It seems that to get an individual digit with these binary formulae you need to calculate all the preceding digits. –Joke 20:41, 26 September 2006 (UTC)

No, I know of no cheat for decimal digits. For hexadecimal digits, we can — as requested — calculate the 84th digit, say, without first accumulating all the preceeding hex digits. The reason this works is that the summation formula has two lucky properties: (1) it expands in powers of 1/16, and (2) it limits the contributions to any selected hex digit. Read carefully about the use of the modulo operator, which is key to the latter. --KSmrq^T 21:12, 26 September 2006 (UTC)

Why can't the hex values be converted to decimal? --Russoc4 23:24, 26 September 2006 (UTC)

They can, but to convert a hexadecimal expansion to decimal you need to have all its hex digits to find the last decimal digit. If I tell you the number is 3.???????????????????????DEADBABE, where each ? is a hex digit I chose not to reveal, there's nothing you can do with the deceased infant at the end. --Lambiam^Talk 23:45, 26 September 2006 (UTC)

At the mathworld article on BBP formulas, it says that it has been proven that the BBP method for PI cannot be adapted to any base that is non-binary (power of two)m, though there may still be a different method, as yet undiscovered, for finding decimal digits. - Rainwarrior 00:21, 27 September 2006 (UTC)

Shortest algorithmn for calculating infinite digits of pi (written in python programming language)

 #!/usr/bin/python
 from sys import stdout
 k, a, b, a1, b1 = 2, 4, 1, 12, 4
 while 1:
     p, q, k = k*k, 2*k+1, k+1
     a, b, a1, b1 = a1, b1, p*a+q*a1, p*b+q*b1
     d, d1 = a/b, a1/b1
     while d == d1:
         stdout.write('%d' % d)
         a, a1 = 10*(a%b), 10*(a1%b1)
         d, d1 = a/b, a1/b1

Sample run

 $ python short_pi.py 
 3141592653589793238462643383279502884197169399375105820974944592307816406286208998
 6280348253421170679821480865132823066470938446095505822317253594081284811174502841
 0270193852110555964462294895493038196442881097566593344612847564823378678316 and so on

202.168.50.40 00:05, 27 September 2006 (UTC)

It's easy to reduce this by one line. Here's a short astounding obfuscated C program by Dik T. Winter and Achim Flammenkamp:

 a[52514],b,c=52514,d,e,f=1e4,g,h;main(){for(;b=c-=14;h=printf("%04d",
 e+d/f))for(e=d%=f;g=--b*2;d/=g)d=d*b+f*(h?a[b]:f/5),a[b]=d%--g;}

However, it gives only 15000 decimals instead of an infinitude like in the Python program. --Lambiam^Talk 01:50, 27 September 2006 (UTC)

If hex digits of π are fine, I've written a Python program for calculating them with the BBP formula. The page explains how you rewrite the formula to calculate isolated digits. If a base 10 BBP-type formula existed for π, the same could be done for decimals, but as was pointed out above, there is no (arctangent-based) BBP formula for π in any non-binary base. By the way, you can calculate isolated decimal digits of ln(9/10). - Fredrik Johansson 04:41, 27 September 2006 (UTC)

That last note is interesting...can you elaborate? :-) --HappyCamper 17:09, 27 September 2006 (UTC)

Well, $\log(1-{\frac {1}{10}})\ =\ \sum _{k=1}^{\infty }{\frac {-1}{k\,10^{k}}}$. -- Fuzzyeric 03:00, 28 September 2006 (UTC)

Straight Angle's interior

My math teacher doesn't seem to know. Does a straight angle have an interior or exterior? Reywas92 21:41, 26 September 2006 (UTC)

Okay, a straight angle is 180 degrees, or π radians. A 120 degree angle has an interior angle of 120 and an exterior angle of 60. A 150 degree angle has an interior angle of 150 and an exterior angle of 30. Therefore, it would make sense for a 180 degree angle to have an interior angle of 180 degrees, and an exterior angle of 0 degrees. Makes sense to me at least. --AstoVidatu 22:42, 26 September 2006 (UTC)

Huh? Wouldn't a 150 degree angle have an exterior angle of 210 deg, not 30? Reywas92 00:20, 27 September 2006 (UTC)

Check out the Internal angle page. The interior/exterior angle diagram was the type of thing that I was thinking of. I'm pretty sure interior angles and exterior angles add up to 180 degrees. --AstoVidatu 00:51, 27 September 2006 (UTC)

September 27

Math Question

Jody paid $44 bill at the hardware store using a combination of $10, $5, and $1 bills. If she paid with 13 bills in all, how many of each bill did she use?

Let i, j and k stand for the number of $10, $5, and $1 bills, respectively. Then we have 10i + 5j + k = 44 and i + j + k = 13. Subtracting the two (corresponding to the substitution k := 13 − i − j) gives 9i + 4j = 31. Modulo 4 this is i ≡ 3, so substitute i := 4m + 3, giving 36m + 27 + 4j = 31, or 36m + 4j = 4, or 9m + j = 1. Since m and j can't be negative, this is only solved by (m, j) = (0, 1). Backsubstitution gives i = 4·0 + 3 = 3 and k = 13 − 3 − 1 = 9, so (i, j, k) = (3, 1, 9). --Lambiam^Talk 01:23, 27 September 2006 (UTC)

Yo dude, I know that Lambian basically spoonfed you the answer, but make sure you can do this. An A on the homework doesn't make up for an F on a quiz. Representin', --AstoVidatu 02:46, 27 September 2006 (UTC)

Note that there aren't very many ways to get $44 from those 3 denominations, so you could just try all the possibilities, too. Obviously, you can exclude those possibilities with 14 or more $1 bills:

10+10+10+10+ 1+ 1+ 1+ 1
10+10+10+ 5+ 5+ 1+ 1+ 1+ 1
10+10+ 5+ 5+ 5+ 5+ 1+ 1+ 1+ 1
10+ 5+ 5+ 5+ 5+ 5+ 5+ 1+ 1+ 1+ 1
 5+ 5+ 5+ 5+ 5+ 5+ 5+ 5+ 1+ 1+ 1+ 1

10+10+10+ 5+ 1+ 1+ 1+ 1+ 1+ 1+ 1+ 1+ 1
10+10+ 5+ 5+ 5+ 1+ 1+ 1+ 1+ 1+ 1+ 1+ 1+ 1
10+ 5+ 5+ 5+ 5+ 5+ 1+ 1+ 1+ 1+ 1+ 1+ 1+ 1+ 1
 5+ 5+ 5+ 5+ 5+ 5+ 5+ 1+ 1+ 1+ 1+ 1+ 1+ 1+ 1+ 1

StuRat 09:44, 27 September 2006 (UTC)

Use a reasoning like: you need 4 $1 bills to get from $40 to $44 (why?), so the problem reduces to getting $40 from 9 bills. You can have 0 or 5 $1 bills (why?), 0 is impossible (why?), so you are left with $35 and 4 $5/$10 bills. Now it should be obvious.--gwaihir 10:00, 27 September 2006 (UTC)

Plural of 'series'

On the urging of another Wikipedia member, after reversions of my ideas regarding this topic, I have decided to post here and ask your opinions.

In his Principia Mathematica, Isaac Newton, when discussion series, used the latin plural 'seriei' instead of our ambiguous english habit. While the habit of singular/plural duality is tolerated for words of english origin such as 'deer', I do not believe this ambiguity belongs in Wikipedia mathematics articles.

Quite simply, the singular/plural duality of 'series', as we somewhat haphazardly have defined it, is confusing. It would not be as confusing if the word did not end with an 's', however it does. Anyone reading these articles has to do a double take to check the verb when reading an article on a series. I have asked several collegues about this already, and they voiced support to me for a clearer, less ambiguous, more grammatically correct spelling.

I thus propose the changing of 'series' in the plural to 'seriei'. One series, several seriei; again, a distinction made by Isaac Newton and lacking the ofttimes confusing singular/plural duality we have assigned to 'series' at present. We use 'criterion/criteria', and 'nucleus/nuclei', and 'formula/formulae', and 'curriculum/curricula', many of which are not exactly common distinctions made by your average english speaker, yet which go a long way towards the goal of precision and accuracy that we strive for here on Wikipedia. I propose simply extending this customary observance to another latin borrow-word which has its own well-established plural. Dbsanfte 05:19, 27 September 2006 (UTC)

I think there must be some mistake here. In Latin, the plural of series is series, and we preserve this in English. Seriei is the genitive singular. Latin nouns ending in -es do not form their plurals in -ei. Maid Marion 07:46, 27 September 2006 (UTC)

Now I feel absolutely silly. You're correct, I misread my dictionary. Please disregard this. Dbsanfte 13:03, 27 September 2006 (UTC)

Consider it thoroughly ignored. Black Carrot 14:37, 27 September 2006 (UTC)

But just to indulge in a little bit of hypothesis, if "seriei" were the Latin plural, do you seriesly (error intended) think that people of today would catch on to either the spelling or whatever the pronunciation of "seriei" is? I rather doubt it (but than, I'm a serial doubter). JackofOz 20:50, 27 September 2006 (UTC)

In general it is not incorrect to pluralize a word with S in English. There are words with latin roots which have alternative pluralization. You should find both "nebulae" and "nebulas" in your dictionary, for instance. - Rainwarrior 04:21, 29 September 2006 (UTC)

What is this question called?

I heard this question a long time ago, but I have forgotten what it is called. Suppose you have 2 types of stamps, one with a denomination of p dollars, and another of q dollars. What is the largest amount that cannot be made with a linear combination of them? The solution is (p-1)(q-1)-1 if I recall, but I'd like to find more background about this. Thoughts? --HappyCamper 17:12, 27 September 2006 (UTC)

Coin problem. Chuck 17:28, 27 September 2006 (UTC)

Largest Number Contest.

What is the largest number you can create using 38 characters? Use any notation.- R_Lee_E (talk, contribs) 17:56, 27 September 2006 (UTC)

9!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

(an iterated factorial, exclamation points are small) –Joke 18:00, 27 September 2006 (UTC)

I interpret this as saying that the expression must have 38 chars, so you may be cheating. I'm not sure what is the best way to use Conway chained arrow notation, but I'll place my bets on:

9→9→9→9→9→9→9→9→9→9→9→9→9→9→9→9→9→99

-- Meni Rosenfeld (talk) 18:10, 27 September 2006 (UTC)

A word of caution is in order here. This is a really, really large number. For example, the factorials above are, if I'm not mistaken, much less than 2→4→3. And this grows unimaginably large for longer chains. -- Meni Rosenfeld (talk) 18:24, 27 September 2006 (UTC)

That number gives me headaches :-) --HappyCamper 18:26, 27 September 2006 (UTC)

Better B(B(B(B(B(B(B(B(B(B(B(B(B(B(B(B(B(B(9)))))))))))))))))) where B(n) is the Busy beaver function. Gives a much larger number than that with the chained arror notation. (Even only 4 or 5 or busy beavers will almost certainly surpass the chained arrow number given above. This number should give even worse headaches. Note that it is arguably not in the same category as the above numbers since it is uncomputable where one could at least in principal compute the other numbers. JoshuaZ 18:33, 27 September 2006 (UTC)

This is a pretty big cardinal number:

$2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{\aleph _{\omega }}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}$

But probably someone can write a bigger one. –Joke 18:53, 27 September 2006 (UTC)

Interesting. The rules didn't specify the number has to be finite. Here is a large ordinal number (though if I'm not mistaken, it's less than the ordinal matching the cardinal above):

$\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\omega }}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}$

Now for a question of my own. Joke's cardinal can be rewritten with Beth number notation:

$\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\aleph _{\omega }}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}$

I've heard once of a Gimel number notation, which potentially could be used to write even larger numbers, but am unable to find any information about it. Has anyone ever heard of such a notation? -- Meni Rosenfeld (talk) 19:20, 27 September 2006 (UTC)

(Edit conflict.) According to Beth number, we can rewrite the above as

$\beth _{38}(\aleph _{\omega })$.

A much larger number would be

$\beth _{9\rightarrow 9\rightarrow 9\rightarrow 9\rightarrow 9\rightarrow 9\rightarrow 9\rightarrow 9\rightarrow 9\rightarrow 9\rightarrow 9\rightarrow 9\rightarrow 9\rightarrow 9\rightarrow 9\rightarrow 9\rightarrow 9\rightarrow 9}(\aleph _{\omega })$

But we're starting to get silly. –Joke 19:40, 27 September 2006 (UTC)

Yes. That number is no joke. - R_Lee_E (talk, contribs) 21:36, 27 September 2006 (UTC)

gimel(κ) is sometimes used to mean κ raised to the power cof(κ). I don't think it necessarily blows up hugely, but it is always larger than κ. As I recall the singular cardinal hypothesis can be stated as gimel(&kappa) = κ⁺ for every singular κ. --Trovatore 19:37, 27 September 2006 (UTC)

Unfortunately, this question gets a bit silly when we realise that there is probably nothing stopping me defining a function T(n) to be B(B(B(n))) (B = Busy beaver function) and so on ... - so why do we not make it a bit more of a challenge: what is the biggest number you can create within the 38 character limit if we restrict ourselves to the symbols: 0-9, +, -, x, /, (, ) i.e. no exponentials and no factorials. Any takers? Madmath789 20:14, 27 September 2006 (UTC)

Yes, but the idea is to only use standard notation, or at least one which is defined somewhere other than this discussion. Besides, under the rules you present the answer is trivially 99999999999999999999999999999999999999. -- Meni Rosenfeld (talk) 20:32, 27 September 2006 (UTC)

Wondered how long it would take to spot that :-) Madmath789 20:40, 27 September 2006 (UTC)

Allow me to cheat: $1/0$ —Bromskloss 09:32, 28 September 2006 (UTC)

Nope, 1/0 is, at best, an unsigned infinity, which isn't larger than any of the other numbers suggested here. -- Meni Rosenfeld (talk) 09:43, 28 September 2006 (UTC)

Now the only problem is determining a winner. 207.233.9.240 20:58, 27 September 2006 (UTC)

Am I allowed to use non-notable notation? User:Salix alba/Jonathan Bowers defines some truly hugh finite numbers. --Salix alba (talk) 22:01, 27 September 2006 (UTC)

Please! No!! we had enough of a problem a few weeks ago with a large number of articles he created ... Madmath789 22:05, 27 September 2006 (UTC)

Well, if you don't require the axiom of choice, we could always go with Reinhardt cardinal, which is 18 characters and vastly larger than anything else mentioned here (or derived from the processes mentioned here). For a boggling array of very big things, see list of large cardinal properties. -- Fuzzyeric 03:07, 28 September 2006 (UTC)

If we want to stay within finite numbers, I would use Steinhaus–Moser notation in conjunction with Conway chained arrow notation:

9→9→9→9→9→9→9→9→9→9→9→9→9→9→9→9→9→9 all inside a polygon with the same amount of sides (since that polygon could be drawn as a single symbol, technically it would only count as one ^_^)

--Ķĩřβȳ♥ŤįɱéØ 06:38, 28 September 2006 (UTC)

Bah... My best guess would've been B(B(B(B(B(B(B(B(B(B(B(B(G^G)))))))))))) ☢ Ҡi∊ff⌇↯ 23:06, 28 September 2006 (UTC)

If we aren't limited need to use mathematical notation strictly, len(π) or len(${\sqrt {2}}$) or even len(1/3) in an infintely powerful computer would return the number of digits in a recurring/irrational/transcendental number (ie infinity). smurrayinchester^{(User), (Talk)} 23:20, 29 September 2006 (UTC)

That's only $\aleph _{0}$. We've already done way better than that. -- Meni Rosenfeld (talk) 08:39, 30 September 2006 (UTC)

Math Composer ThIng

Does anyone here know of a Windows program that can be used to write out math problems and such. I'm speaking of an application like MS Word only for math and numbers. Hope I'm being specific enough. Deltacom1515 20:53, 27 September 2006 (UTC)

I'm not sure, but LyX is quite good. --HappyCamper 21:02, 27 September 2006 (UTC)

MS Word will actually work for many purposes. In MS Word (I'm assuming MS Word 2003 here; I don't know whether the feature existed in earlier versions, or if so, how good it was), go to Insert - Object... - Microsoft Equation 3.0. The interface takes a bit of getting used to, but you can do quite a bit as far as writing out mathematical formulae. Chuck 21:49, 27 September 2006 (UTC)

I would definitely go with a version of LaTeX, such as the freely available MikTeX. The Word equation editor is OK for fairly trivial uses, but for anything of any complexity or of any length, you really need power and flexibility of TeX. Madmath789 21:53, 27 September 2006 (UTC)

Thanks guys. Question answered. Deltacom1515 00:20, 28 September 2006 (UTC)

Late to the party... I use OpenOffice. Much like Word... Insert|Object|Formula. -- Fuzzyeric 03:10, 28 September 2006 (UTC)

Solids

Is there a name for this solid? It is a square prism with rounded corners, but the top and botton are rounded also

File:Eclipse mints infographic.jpg

What about this one, an oval prism also with rounded top and bottom. Oblate? Prolate? Maby there is no name for these.

I actually need this info for an article. Herostratus 21:24, 27 September 2006 (UTC)

These are not shapes for which there are standard geometrical names. I've seen "pillow" used for shapes somewhat similar to the first.  --Lambiam^Talk 22:25, 27 September 2006 (UTC)

Rounded rectangular prism? Rounded cylinder? (Or whatever an oval cylinder is called.) - Rainwarrior 04:35, 28 September 2006 (UTC)

Rounded eliptical cylinder perhaps? - Rainwarrior 04:39, 28 September 2006 (UTC)

My best guess is the union between a superellipsoid and a cylinder ☢ Ҡi∊ff⌇↯ 04:51, 29 September 2006 (UTC)

Clouds can form in shapes like these, they are known as lenticular clouds. If you were defining a name for these, you could call these lenticular solids. Dysprosia 08:49, 29 September 2006 (UTC)

Thanks to all for your replies. Herostratus 07:22, 30 September 2006 (UTC)

September 28

4D "volume" ?? (length^4)

1D:length

2D:area

3D:volume

4D:??

If anyone knows this term it would be greatly appreciated. Tuvwxyz (T) (C) 02:29, 28 September 2006 (UTC)

4-volume. (Exercise for the reader: generalize to d dimensions.) –Joke 02:38, 28 September 2006 (UTC)

Actually, if I were seriously writing it I would put "four-volume" per the manual of style. –Joke 02:39, 28 September 2006 (UTC)

Volume suggests "content", which I'd never heard before, and doesn't mention "4-volume". Generally "4-volume" will do, as will "volume" if it's clear you're in four dimensions.

(If anything else finds the "content" note odd, feel free to fix it).

RandomP 02:44, 28 September 2006 (UTC)

There's also the term "hypervolume". MathWorld also has "content", and that may be where Wikipedia got it from. It may be a shortened version of Jordan content.  --Lambiam^Talk 05:06, 28 September 2006 (UTC)

4 dimensions doesnt physically exist in our world, and even if it did you'll still see it as a 3D object--RedStaR 08:48, 28 September 2006 (UTC)

• The OP did not mention the word "physical" anywhere.
• It is very common for time to be considered the 4th dimension.
• There are (prominent?) physical theories according to which the universe has as many as 26 dimensions.

-- Meni Rosenfeld (talk) 09:00, 28 September 2006 (UTC)

If we take time as the fourth dimension, it is difficult to delineate dimensions. What time do you call the "start" of your object, or the "end" of it? How does one interpret a measure of how much spacetime an object occupies meaningfully? Maelin 08:11, 29 September 2006 (UTC)

The only problem here is that objects in spacetime tend to be unbounded. If you have a cylinder in 3D with infinite length, its volume will be infinite. But if you clip it to a finite length, you'll get a "normal" cylinder with finite volume. Likewise, if you have a 3D sphere which does not move, it will have an infinite "length" in the time dimension when viewed in 4D, and thus have infinite 4-volume. However, if you take such an object and "clip" it, that is, look only at its portion in a specific time interval, you'll have a finite 4-volume. For example, a sphere of radius 2m which exists for 3s will have a 4-volume of 100.53 m³s. Another example: A ballon which starts out at 0 volume, expands and then shrinks back to 0 volume. It will have a finite 4-volume, calculated as the integral of its spatial volume with respect to time. -- Meni Rosenfeld (talk) 08:35, 29 September 2006 (UTC)

Why 1+1=2?

Can someone tell me why 1+1=2? Please...I really want to know.Thks.

It's the definition of 2. Conscious 11:26, 28 September 2006 (UTC)

In the article on natural numbers, I suggest you see the section Formal definitions for a definition of the numbers themselves and then the section Properties for a definition of addition. Return here when there is something you don't understand. (Also, please sign your comments.) —Bromskloss 12:15, 28 September 2006 (UTC)

I seem to recall that it was proven in the Principia Mathematica. Why not look there? (Expect to have to read about 300 pages to get to it though.) Myself, I just accept it as an axiom instead. - Rainwarrior 15:50, 28 September 2006 (UTC)

Metamath merely states 1+1=2 as the definition of 2. However, it has a proof that 2+2=4. If expanded fully, the proof consists of 22,607 steps. Fredrik Johansson 16:04, 28 September 2006 (UTC)

Like so much else, it depends on your foundations. Some authors define 2 to be 1+1. Others define 2 to be the successor of 1, so to prove 1+1=2 you'd need to compare that with whatever definition of "+" you're using. Given that counting underlies our intuition about naming numbers, the definition of 2 as "the next number after 1" is pretty natural. A more direct definition of 2 is this many: **. A picture proof would be good enough on that level. Melchoir 19:06, 28 September 2006 (UTC)

Or, if you prefer a proof based on number conservation as a foundation for the cardinality concept: http://www.youtube.com/watch?v=KCIHn5adOnM&NR Melchoir 19:38, 28 September 2006 (UTC)

If we use the Peano axioms for the natural numbers, we can define 1 = S(0), 2 = S(S(0)), 3 = S(S(S(0))) and 4 = S(S(S(S(0)))). Addition is defined by m + 0 = m, m + S(n) = S(m) + n. Then 2 + 2 = S(S(0)) + S(S(0)) = S(S(S(0))) + S(0) = S(S(S(S(0)))) + 0 = S(S(S(S(0)))) = 4. Only four steps needed. I wonder what Metamath is doing in the other 22,603 steps.  --Lambiam^Talk 19:10, 28 September 2006 (UTC)

I'm getting confused. How did you go from here:

S(S(0)) + S(S(0)) aka 2 + 2

to here

S(S(S(0))) + S(0) aka 3 + 1

The step above is an instance of m + S(n) = S(m) + n with m = S(S(0)) and n = S(0).  --Lambiam^Talk 01:52, 29 September 2006 (UTC)

No one has answered the actual question, which is "Why?", because it is not a proper question in mathematics (or in physics). We can show a particular set of axioms and a deduction within a system of logic and proofs, but this is mostly an exercise in definitions. Once we have defined "1", "2", "+", and "=", there is little left to say in a formal demonstration. But as for why we choose these definitions, one might as well ask, "Why pound a nail with a hammer, not pliers?" --KSmrq^T 23:17, 28 September 2006 (UTC)

Ok, so the reason why 1+1=2 is that it follows from a set of axioms that we chose to adopt because they very successfully model our experiences with collections of objects (also our experiences with magnitudes) in the world? -GTBacchus^(talk) 23:21, 28 September 2006 (UTC)

Personally I prefer to say that it follows from a set of definitions that we adopted. Yes, they are axioms, but in the sense of the axioms in the definition of a group. The validity of the conclusion 1+1=2 is independent of the motivation for choosing these definitions. As Einstein said (according to Wikiquote): As far as the laws of mathematics refer to reality, they are not certain; and as far as they are certain, they do not refer to reality. I might add that this does not say much about maths; in fact, as far as anything refers to reality it is not certain. For example, the model does not work very well for frogs in a wheelbarrow, nor for socks in the wash. You might say that these particular "axioms" proved to offer a useful mathematical model for counting objects. But, somewhat conversely, you might say that the model tells us something about us, namely what we mean by "counting" and what we consider "objects", by defining properties "objects" should have before it makes sense to "count" them.  --Lambiam^Talk 02:14, 29 September 2006 (UTC)

Good points. I'm reminded of the book Where Mathematics Comes From, by Lakoff and Nuñez, in which they discuss evidence that very young infants are capable of subitizing, which suggests that our tendency to discretize the manifold of our experience into countable objects is, to some degree, hard wired. This in turn suggests that we exist in some kind of environment that is amenable to being successfully navigated by means of discretization into "objects". Not that I'm saying we really obtain any access to "the thing in itself", but it is pretty easy to believe that there's something out there corresponding to consensual reality. I'm going to continue to assume it, anyway. ;) -GTBacchus^(talk) 18:29, 29 September 2006 (UTC)

Ha ha ha! I've actually tried pounding a nail with pliers when no hammer was available. (No, it doesn't work too well.) - Rainwarrior 04:12, 29 September 2006 (UTC)

valid range

Suppose I have a function such as f(x)=[x*(x-1)]/2 where f(0)=0 and f(1)=0 and f(2)=1 where a value of zero for the function is mathematically accurate and suggests that all values of x below two are invalid. Do I simply add a note of text to state this or is there a mathematical symbol or notation I can or should use? Adaptron 15:50, 28 September 2006 (UTC)

You should be clearer about what this function signifies. If it's just a function then there's no problem with a value of 0 for the function or with values less than 2 for x. If it represents some physical situation where x must be at least 2, you can say, "for every x ≥ 2, $f(x)={\frac {x(x-1)}{2}}$", or you can say that the domain of f is [2, ∞). -- Meni Rosenfeld (talk) 16:12, 28 September 2006 (UTC)

Would I use the same method of qualification to indicate that only integer values of x would be valid? Adaptron 17:00, 28 September 2006 (UTC)

That depends on how technical you want to be. Here are several suggestions:

• For every integer x ≥2, $f(x)={\frac {x(x-1)}{2}}$. Probably the clearest.
• For every n ≥2, $f(n)={\frac {n(n-1)}{2}}$. The letter n is usually only used for integers, and most people who read this will interpret it this way.
• $f\colon \mathbb {Z} \cap [2,\infty )\to \mathbb {Z} \quad \quad n\mapsto {\frac {n(n-1)}{2}}$. Very precise, but will probably not be legible to non-mathematicians. You can also say that $\mathbb {Z} \cap [2,\infty )$ is the domain of the function.

-- Meni Rosenfeld (talk) 17:47, 28 September 2006 (UTC)

To some extent these are matters of taste, but I might write

$f(x)={\frac {x(x-1)}{2}}\qquad ;x\geq 2\,\!$

for a simple boundary restriction, and either

$f(n)=n(n-1)/2\qquad ;x\in \mathbb {Z} \,\!$

or

$f\colon \mathbb {Z} \to \mathbb {Z} \,\!$

$n\mapsto n(n-1)/2\,\!$

to restrict to arbitrary integers. To impose both restrictions, some authors would write

$f(n)=n(n-1)/2\qquad ;2\leq x\in \mathbb {Z} .\,\!$

Personally, I find this uncomfortably compressed. Likewise, I find the intersection form needlessly baroque. I'd have to see the context to decide what works best, but generally I'd keep it simple:

For n an integer greater than 1, let f(n) = n(n−1)/2.

After all, we write mathematics for other humans to read, and the notation is supposed to help, not hinder, our efforts. --KSmrq^T 23:45, 28 September 2006 (UTC)

Venturing into physics for context and acknowledging half-life represents a transformation between two atoms where the daughter increases in direct and absolute proportion to the decrease in parent atoms where the sum total of atoms remains the same, would the above usage be adequate in your opinion to express and clarify the idea that a situation where the number of parent isotopes is less than one but greater than zero can not exist since half-life is not an infinite process (as it might be if atoms were divisible by the half-life process) but terminates when the parent isotope is less than one? 71.100.167.194 23:49, 29 September 2006 (UTC)

I assume that by "the number of parent isotopes" you mean: "the number of atoms of the parent isotope". If by "number of" you mean a (finite) cardinal number, as used for counting: 0, 1, 2, ..., and N represents that number, you can say: N is a non-negative integer. Depending on your belief you might perhaps say: N is a natural number. This applies not only to the number of atoms, but also to the number of times you've been married, or the number of pennies you have in your wallet. Saying that adequately represents the idea that N is then not equal to 0.99 or −1.  --Lambiam^Talk 00:37, 30 September 2006 (UTC)

Since I believe the likelihood of the number of atoms of the parent isotope (thanks for the correction) remaining no less than one for any reasonable multiple of the half-life after one atom is achieved is zero the statement that N is a non-negative integer as opposed to a positive integer or a natural number seems to be the right way to go. I have heard the claim that not achieving zero number of atoms of the parent isotope is possible even in a closed system subject only to decay because half-life measurement is a relative measure rather than being absolute. 71.100.167.194 01:36, 30 September 2006 (UTC)

Graphing an indifference Curve

Suppose we have a Utility function (in economics) where a certain string of baskets of goods equals the same utility, creating an indifference (preference) curve. Normally, I would find this easy to graph, creating the indifference curves eqyal to a fixed C.

However, for some reason, I can't graph U = sqrt(X) + sqrt (Y) correctly. The way I graph it in my head is like so: First, lets do the curve for U = 1. The intercepts would be 1,0 and 0,1. This would mean that the curves intersect the axis. Then, assuming equal proportions of X and Y, so both their roots equal 0.5, both X and Y are about 0.7~. Adding (0.7,0.7) gives me a graph similar to an X squared plus Y squared equals 1 graph...how is this so?

I know the answer should be a normal asymptotic curve along the axis, but I cant logically get it? Help! 152.163.100.11 17:27, 28 September 2006 (UTC)

If √X = 0.5, then X=0.25, not 0.7. -- Meni Rosenfeld (talk) 17:49, 28 September 2006 (UTC)

And it's not really an asymptotic curve. As X → 0, Y → U². In fact, it is a segment of a parabola whose axis is formed by the line x = y, touching the two coordinate axes x = 0 and y = 0.  --Lambiam^Talk 18:17, 28 September 2006 (UTC)

Oh snap thanks, I completely overlooked the possibility of a trivial mistake in my rooting. Thanks 64.12.116.11 20:35, 28 September 2006 (UTC)

Sine of a complex number?

I have looked at complex numbers but the article did not tell me wherether sin(i) is a legal mathematical operations. Or is it illegal like division by zero? 202.168.50.40 21:50, 28 September 2006 (UTC)

It's equivalent to $i\sinh 1\approx 1.1752\,i$. In general, $\cos ix=\cosh x\!$, and $\sin ix=i\,\sinh x$. You can also use Euler's formula and its corollaries. To do complex rather than just imaginary numbers, use Euler or else the sum and difference formulas. --Tardis 22:45, 28 September 2006 (UTC)

There are also more details at Sine#Relationship to exponential function and complex numbers. Melchoir 22:55, 28 September 2006 (UTC)

Once we extend the sine function to complex numbers, there is nothing illegal about feeding it a complex number. The question is, how do we so extend? In this case, we can take the series expansion,

$\sin(x)=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-\cdots ,$

and use it unchanged, thus converting a real analytic function into a complex one. Furthermore, as a complex function it is analytic over any open subset of complex numbers. Specifically, it converges at i. To find the value, it is convenient to use Euler's formula,

$e^{ix}=\cos(x)+i\sin(x).\,\!$

This equality is written in terms of complex numbers, and holds for complex x as well as real. Noting that sine is an odd function, so that sin(−x) = −sin(x), we derive

$\sin(x)={\frac {e^{ix}-e^{-ix}}{2i}}.$

Letting x equal iy for real y, this produces

$\sin(iy)\,\!$	${}=(e^{i^{2}y}-e^{-i^{2}y})/2i\,\!$
	${}=(e^{-y}-e^{y})/2i\,\!$
	${}=i(e^{y}-1/e^{y})/2.\,\!$

Thus a pure imaginary argument produces a pure imaginary result, averaging an exponential rise with an exponential decay. For large values of y the decaying term makes a negligible contribution; and the periodicity of sine for real arguments has been completely lost for imaginary arguments. --KSmrq^T 01:46, 29 September 2006 (UTC)

Please visit your new friend. Twma 04:43, 29 September 2006 (UTC)

September 29

Mixed trig / linear terms

Is there an algebraic method for finding solutions to equations of the form ax + bsin(x) + c = 0 ? Maelin 10:50, 29 September 2006 (UTC)

I don't think so. One special case would be $\cos x=x$ (almost, anyway), and there is no clean way to solve that, as far as i know. —Bromskloss 11:38, 29 September 2006 (UTC)

There's probably a way using the series definitions of the trig functions. ☢ Ҡi∊ff⌇↯ 11:59, 29 September 2006 (UTC)

No, that won't help. Replacing the trigonometric function with an infinite series just gives you an equation where one side is an infinite series. This doesn't simplify the problem. It would help if you could do some manipulation on the series and find that the result is a series for some other known function that provides an algebraic solution, but in this case the series is just the sine series with a few terms altered, and that doesn't help. Fredrik Johansson 15:35, 29 September 2006 (UTC)

I am almost certain that (except for special cases) it will be impossible to find a closed expression for the solutions in terms of a finite combination of elementary functions like exponentials, sin, cos, tan, (and inverse trig functions) log, exp, or hyperbolic functions. I suspect that the best you will get (other than a numerical answer to a certain degree of accuracy) will be an infinite series expansion by using the Lagrange inversion theorem. Madmath789 12:28, 29 September 2006 (UTC)

I think that there isn't. If there was a way, then you could construct a cos(1) angle with compass and straightedge; but http://en.wikipedia.org/wiki/Angle_trisection. Sorry, maybe in the morning my brain will work, so I can give you the exact proof.

Functor nOOb200.65.178.127 07:45, 30 September 2006 (UTC)

data

data and graphs from 1990 to 2005 of the following South African macroeconomic variables: Real GDP, Inflation, Unemployment and Balance of payments

The above is not a question, and is not about mathematics, the topic of this reference desk.  --Lambiam^Talk 12:57, 29 September 2006 (UTC)

Hello! You should be able to post your question in another ref. desk page, as the misc. one. And please remember that no mechanical nor computerized parts are tortured to answer you question, we're not googols ;-)

So the title of the question might give : South A. macroeconomics, and the question could start with "hello, where may I find ...". Now, try this link first : South Africa. Thank you. -- DLL ^{.. T} 19:16, 29 September 2006 (UTC)

Economic Bundles

I've solved an equation for a bundle that includes 4 of unit X and 2.5 of unit Y as the utility-maximizing one. How come it is okay to have non-whole numbers as part of an answer...its not like I can go out an actually buy 2 and a half bananas!

Because mathematicians care more about number than whether it possible to have 2.4 children. More seriously, there are many applications where it is useful to have fractional results. So its good practice to start using them now as it will help you a lot in the future. --Salix alba (talk) 21:36, 29 September 2006 (UTC)

Just multiply by 2, then you have 8 of unit X and 5 of unit Y. StuRat 21:45, 29 September 2006 (UTC)

There are also applications where the solution must need be in whole numbers; anything else is totally useless. That is why there is the whole area of integer programming in the field of operations research. However, this is much and much harder.  --Lambiam^Talk 22:55, 29 September 2006 (UTC)

September 30

Inequality Question

Hi, I know that this is homework, but I would be indebted to whoever can point out and explain where I have gone wrong and how I should have gone right:

The area of a rectangle is 12 cm². Find the range of possible values of the width of the rectangle if the diagonal is more than 5cm.

I got (apologies for my inability to use the Latex notation).

height=h, width=w

h^2 + w^2 > 5^2

h^2 + w^2 > 25

hw = 12

h = 12/w

(12/w)^2 + w^2 > 25

144/w^2 + w^2 > 25

144 + w^4 > 25w^2

Then the whole thing falls apart for me.

Thank you for your help, —Daniel (‽) 09:46, 30 September 2006 (UTC)

Define z = w^2. Then your last equation can be rewritten as 144 + z^2 > 25z, or equivalently z^2 − 25z + 144 > 0. This is a standard quadratic inequation. Given the range of z satisfying it, and considering that w must be non-negative, the possible range for w consists of the square roots of the non-negative part of the solution range for z. It may be easier to look at the values that violate the inequation.

A different route to the solution is found from the consideration that you can easily deduce inequations for h^2 + 2hw + w^2 = (h + w)^2 as well as for h^2 − 2hw + w^2 = (h − w)^2.  --Lambiam^Talk 10:21, 30 September 2006 (UTC)

Thanks a lot. I used your first method to come to 0 < w < 3 (or) w > 3. I don't quite understand what you mean by the second method. Thank you again! —Daniel (‽) 16:06, 30 September 2006 (UTC)

Unfortunately, I think you made an error somewhere. Take w = 2√3 = 3.4541... This satisfies w > 3. Then w^2 = 12, so h = 12/w = w. Then h^2 + w^2 = 2w^2 = 24, which is not greater than 25. Unless you made a simple copying error, to get into this situation, you must have concluded before that z > 9 satisfies the inequation in z. But clearly that is incorrect: 144 + 100 < 250.  --Lambiam^Talk 16:25, 30 September 2006 (UTC)

Venn diagrams

I was wondering, if all wigs are wags, and some wags are wogs, does that mean that all wigs are wogs? Please hurry, my exam finishes in 10 minutes!

If all students are humans, and some humans are females, does that mean that all students are females?  --Lambiam^Talk 15:51, 30 September 2006 (UTC)

If your exam finishes in 10 minutes, what are you doing writing on Wikipedia? —Daniel (‽) 16:02, 30 September 2006 (UTC)

Function question

File:Functions x2 sqrtx 0-10.JPG

I was wondering if I could find an equation to describe a curve in a Cartesian System of Axes, given that the curve I need is the graphs of 2 (or maybe more) functions combined.

i.e. for:

f(x) = x^2
g(x) = sqrt(x)

I would need an equation to describe a curve like the one at the right

Thanks, --Danielsavoiu 15:50, 30 September 2006 (UTC)

You could use the equation (x − y^2)(y − x^2) = 0, which however does not give you a curve. If you want to stay within the first quadrant, gluing these two graphs together, you could use a parametric equation, something like

     x = t^2 if t < 0, x = t  otherwise

     y = − t if t < 0, y = t^2 otherwise

 --Lambiam^Talk 15:59, 30 September 2006 (UTC)

Thanks, Lambiam. I thought about parametric equations, but I would rather have only one solid equation. Your first suggestion ( (x − y^2)(y − x^2) = 0 ) is ok, but what if I wanted to glue the graphs of:

f(x) = x^3
g(x) = 3x

--Danielsavoiu 16:17, 30 September 2006 (UTC)