Wikipedia:Reference desk/Mathematics

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October 20

How do I prove this?

How do I prove that d/dx (cosx)=-sinx?

Thank you!

-EggYolk

If you're working from first principles, you need to use the formula $\cos(x+y)=\cos x\cos y-\sin x\sin y$, as well as the small angle approximation (that $\lim _{x\rightarrow 0}{(\sin x)/x}=1$), and then the only tricky part is a term where you have to prove that $lim_{x\rightarrow 0}{\frac {\cos x-1}{x}}=0$, and I can't recall the proof for that offhand. Of course, if you already have your proof that (sin x)' = cos x, you can use the chain rule: ${\frac {d\cos x}{dx}}={\frac {d\sin(\pi /2-x)}{dx}}=-\cos(\pi /2-x)=-\sin x$. Confusing Manifestation 05:15, 20 October 2006 (UTC)

$\lim _{x\rightarrow 0}{\frac {\cos x-1}{x}}=\lim _{x\rightarrow 0}{\frac {\cos x-1}{x}}{\frac {\cos x+1}{\cos x+1}}=\lim _{x\rightarrow 0}{\frac {\cos ^{2}x-1}{x(\cos x+1)}}=\lim _{x\rightarrow 0}{\frac {-\sin ^{2}x}{x(\cos x+1)}}$

I'll let you finish the proof that $lim_{x\rightarrow 0}{\frac {\cos x-1}{x}}=0$ from there. Yes, it does need the small angle approximation. --Ķĩřβȳ♥ŤįɱéØ 05:57, 20 October 2006 (UTC)

Also, you can check out this newly created article, lol. --Ķĩřβȳ♥ŤįɱéØ 08:53, 20 October 2006 (UTC)

A more geometric argument begins with a unit circle. For any parametric curve, the instantaneous movement at any point is a vector tangent to the curve. If the curve is parameterized by arclength, the tangent vector will have unit length. On the circle, that tangent is the radial vector to the point, rotated counterclockwise 90°. For the radial vector (cos ϑ,sin ϑ), the tangent is (−sin ϑ,cos ϑ). This simultaneously gives the derivative of cosine and of sine.

This argument uses concepts and methods not available in first-year calculus, but the insights can still be of use. --KSmrq^T 09:00, 20 October 2006 (UTC)

Not that your point isn't valid, but the idea that geometry is inherently more advanced than formal calculus seems to be a property of the Western system of elementary math education, rather than a reflection of some purely mathematical stratification. Tesseran 15:24, 20 October 2006 (UTC)

Perhaps I did not make myself clear. It is not the geometry that is learned later, but the use of vectors and differential geometry. --KSmrq^T 19:38, 20 October 2006 (UTC)

October 21

Aesthetically pleasing maths

Hi. I'm developing an idea for a peice of art which basically consists of a sheet of glass held up in a frame. On the glass in thick marker pen, I want some mathematical calculations written on one side, much like a teacher would write on a blackboard. The peice will be visible from both sides.

Although I know very little maths, the notation and symbols used are very pleasing and beautiful to look at (the integration sign is a particular favorite) and I was wondering if you maths people out there could give me some "workings" that utilise a lot of these glyphs. One simple example to give you a picture is perhaps a step by step working of the differentation of f(x) = x^2; although maybe something a bit more beautiful, perhaps involving Euler's Identity or lots of weird glyphs and what-not would be great. Thanks! 164.11.204.56 15:37, 21 October 2006 (UTC)

Well, Maxwell's equations look very pretty to me. ☢ Ҡi∊ff⌇↯ 17:14, 21 October 2006 (UTC)

Ah, another one you might be interested is the series expansion for the trigonometric functions:

$\sin x=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}+\cdots =\sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{2n+1}}{(2n+1)!}}$

$\cos x=1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-{\frac {x^{6}}{6!}}+\cdots =\sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{2n}}{(2n)!}}$

Which I think have an amazing simplicity and importance. The sigma notation is not an integral, but I think it is also very aesthetically pleasing. ☢ Ҡi∊ff⌇↯ 17:41, 21 October 2006 (UTC)

Cool! I'll try to come up with suggestions. I immediately thought about the integration sign, but apparently, you already had it in mind. :-) —Bromskloss 17:23, 21 October 2006 (UTC)

So, here I show off a variety of mathematical symbols that might be of interest. You'll have to judge by yourself which of them, if any, are good looking. I have chosen examples that actually mean something useful in mathematics – I suppose that would make it more interesting. However, I'm afraid I sometimes mix up graphical beauty with matematical beauty, so mabye they are only attractive for their meaning and not for their typography! If you like a particular symbol and want more of it, just ask. But then again, you seem to know quite a bit of mathematics yourself, don't you?

• $\zeta (s)=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}}$
• $\aleph _{\lambda }=\bigcup _{\beta <\lambda }\aleph _{\beta }$ Hebrew!
• ${\begin{pmatrix}\cos {\theta }&-\sin {\theta }\\\sin {\theta }&\cos {\theta }\end{pmatrix}}{\begin{pmatrix}\cos {-\theta }&-\sin {-\theta }\\\sin {-\theta }&\cos {-\theta }\end{pmatrix}}={\begin{pmatrix}1&0\\0&1\end{pmatrix}}$
• $\langle x,y\rangle ={\frac {1}{2}}\left(\|x+y\|^{2}-\|x\|^{2}-\|y\|^{2}\right)$
• $\gamma ={\frac {1}{\sqrt {1-v^{2}/c^{2}}}}$ Lorentz factor, important in special relativity
• $\operatorname {div\ curl\ } \mathbf {A} =\nabla \cdot (\nabla \times \mathbf {A} )=0$
• $\Delta f=\nabla ^{2}f={\partial f \over \partial r}\mathbf {\hat {r}} +{1 \over r}{\partial f \over \partial \theta }{\boldsymbol {\hat {\theta }}}+{1 \over r\sin \theta }{\partial f \over \partial \phi }{\boldsymbol {\hat {\phi }}}$
• $n\in \mathbb {N} \Rightarrow {\sqrt {n}}\in \mathbb {N} \cup (\mathbb {R} \setminus \mathbb {Q} )$ I believe I proved this some time ago, but it should probably be checked before you use it. It has a number of funny symbols in any case. :-)
• $\ell ^{2}=\left\{\left(x:\mathbb {N} \rightarrow \mathbb {C} \right)\;:\;\sum _{n\,\in \,\mathbb {N} }\left|x(n)\right|^{2}<\infty \right\}$
• $\int \limits _{\varphi =0}^{2\pi }\mathrm {e} ^{\mathrm {i} \varphi }\,\mathrm {d} \varphi =0$
• ${\frac {\mathrm {d} ^{4}}{\mathrm {d} \varphi ^{4}}}\mathrm {e} ^{\mathrm {i} \varphi }=\mathrm {e} ^{\mathrm {i} \varphi }$
• $\iiint \limits _{V}\left(\nabla \cdot \mathbf {F} \right)\mathrm {d} V=\iint \limits _{\partial V}\mathbf {F} \cdot \mathrm {d} \mathbf {S}$
• $\int \limits _{\Sigma }\nabla \times \mathbf {F} \cdot \mathrm {d} \mathbf {\Sigma } =\oint \limits _{\partial \Sigma }\mathbf {F} \cdot \mathrm {d} \mathbf {r}$

—Bromskloss 20:36, 21 October 2006 (UTC)

(edit conflict) For weird symbols, you can't beat physics equations. Maxwell's equations are neat looking, as is the Schrödinger equation. Depending on how meaningful the result is supposed to be and how much space you have you could include all sorts of things. Combining, for instance, Taylor series and the Euler identity, you could prove e^ix = isin(x)+cos(x).

To someone who has not had a course in diff. eq (where, in my experience, the result is first seen), the asthetic of symbols is all that is there. The result, however, would be meaninful to people who understand it. The outline of the proof (omitting questions of convergence and such) is

• $e^{ix}=\sum _{k=0}^{\infty }{\frac {(xi)^{k}}{k!}}$
• $=\sum _{k\mod 2=0}^{\infty }{\frac {(xi)^{k}}{k!}}+\sum _{k\mod 2=1}^{\infty }{\frac {(xi)^{k}}{k!}}$
• $=\sum _{k=0}^{\infty }{\frac {(xi)^{2k}}{(2k)!}}+\sum _{k=0}^{\infty }{\frac {(xi)^{2k+1}}{(2k+1)!}}$
• $=\sum _{k=0}^{\infty }\left(-1\right)^{k}{\frac {(x)^{2k}}{(2k)!}}+i\sum _{k=0}^{\infty }\left(-1\right)^{k}{\frac {(x)^{2k+1}}{(2k+1)!}}$
• $=\cos(x)+i\sin(x)$

(Doing this on the fly, I hope I didn't make a mistake.) Many similar short proofs of interesting results could be found (espeically since you could leave out some steps). --TeaDrinker 20:47, 21 October 2006 (UTC)

Here are some of my favorites from pure analysis:

$f(a)={1 \over 2\pi i}\oint _{C}{f(z) \over z-a}\,dz$

$\zeta (z)\;\Gamma (z)=\int _{0}^{\infty }{\frac {t^{z-1}}{e^{t}-1}}\;dt$

$\Gamma \left({\frac {s}{2}}\right)\zeta (s)\pi ^{-s/2}=\Gamma \left({\frac {1-s}{2}}\right)\zeta (1-s)\pi ^{(1-s)/2}$

$\pi =\sum _{k=0}^{\infty }{\frac {1}{16^{k}}}\left({\frac {4}{8k+1}}-{\frac {2}{8k+4}}-{\frac {1}{8k+5}}-{\frac {1}{8k+6}}\right).$

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

It's hard to compete with many of the equations from vector calculus and mathematical physics, though, simply because they so compactly encode properties of extremely complex systems (at least to a mathematician, depth makes formulas visually appealing :-). I'd give the Navier-Stokes equations a look. Fredrik Johansson 21:42, 21 October 2006 (UTC)

My field is computer science and discrete mathematics, and there are some fine ones. I've always had a soft spot for the binomial theorem

$(x+y)^{n}=\sum _{k=0}^{n}{n \choose k}x^{k}y^{n-k}$

It's not terribly advanced but I think it's beutiful as hell. If we're going for pure awesomeness, the hypergeometric function is unbeatable. Unfortunatly, the notation that's used in that article is not the coolest one, but the one in Donald Knuths works is mindblowing. Also, anything involving Stirling numbers of the second kind. Like

$\left\{{\begin{matrix}n\\k\end{matrix}}\right\}=\left\{{\begin{matrix}n-1\\k-1\end{matrix}}\right\}+k\left\{{\begin{matrix}n-1\\k\end{matrix}}\right\}$

or

$\left\{{\begin{matrix}n\\k\end{matrix}}\right\}={\frac {1}{k!}}\sum _{j=1}^{k}(-1)^{k-j}{k \choose j}j^{n}.$

Curly brackets are cool! Also, as you are an artist, it might be a nice idea to have your equations mean something deeper. The axioms of Zermelo–Fraenkel set theory are pretty neat, and they are the most commonly used way to define mathematics itself! Personally, I find something mezmerising about that, what those formulas stand for. How much comes from them. Oskar 03:36, 22 October 2006 (UTC)

Wow, there's a lot of great stuff here!

e^ix = isin(x) + cos(x) is a bloody good one actually since it has strong ties to musical analysis for obvious reasons (Fourier Series / Transform)! I play guitar. The Schrödinger equation also looks wonderful although I am not familiar with what it describes - hopefully something profound about the nature of the physical world?

I was also wondering if anyone recognises the beast below. Something to do with digital sampling, by the look of those braces? I don't even know if it's actually a valid statement or an arbitrary collection of stuff, but it's exactly the kind of thing I'm looking for...

$\Delta M_{i}^{-1}=-a\sum _{n=1}^{N}D_{i}\left[n\right]\left[\sum _{j\in \mathbb {C} \{i\}}F_{ij}\left[n-1\right]+F\operatorname {ext} _{i}\left[n^{-1}\right]\right]$

164.11.204.56 02:17, 23 October 2006 (UTC)

Why not Euler's Identity?

$e^{i\pi }+1=0\!\$

The five most important constants in all of mathematics (e, i, pi, 1, and 0), and the three most important operators (addition, multiplication, and exponentiation). Most beautiful --Ķĩřβȳ♥ŤįɱéØ 06:28, 24 October 2006 (UTC)

I recommend the expansion in chord diagrams of the Kontsevich integral of the trefoil knot. It uses such exotic glyphs, I can't even show it to you using Wikipedia's software! It occurs in the last page of [1]. Go download it; you won't be disappointed. Melchoir 06:55, 24 October 2006 (UTC)

Working out the point at which lines of longnitude intercept the equator on an orthographic projection

I have seen the equations at Orthographic projection (cartography)#Mathematics, although I am told there is an easier way to create these arc by simply finding the equatorial intercept. I would be very obliged if anyone could tell me how this can be worked out. Thankyou Lofty 15:51, 21 October 2006 (UTC)

Is the equation Radius * sin (angle)? Lofty 16:17, 21 October 2006 (UTC)

The equations from the article are:

$x=R\,\cos(\phi )\sin(\lambda -\lambda _{0})$

$y=R\,[\cos(\phi _{1})\sin(\phi )-\sin(\phi _{1})\cos(\phi )\cos(\lambda -\lambda _{0})]$

in which $\lambda \,$ stands for the latitude and $\phi \,$ for the longitude of the point being projected, and $(\lambda _{0},\,\phi _{1})$ is the point on the sphere to be mapped to the origin. (No idea why one has a subscript 0 and the other a subscript 1.) Indeed, if we substitute $(\lambda ,\,\phi ):=(\lambda _{0},\,\phi _{1})$, we obtain $(x,\,y)=(0,\,0)$. The equator corresponds to $\phi =0\,$. Applying the substitution $\phi =0\,$, and using the fact that sin 0 = 0 and cos 0 = 1, we can simplify the equations, obtaining this for the intercept of the line of longitude for $\lambda \!\,$ with the equator:

$x=~~R\,\sin(\lambda -\lambda _{0})$

$y=-R\,\sin(\phi _{1})\cos(\lambda -\lambda _{0})$

As on this image, due to Snyder himself, the equator becomes in general an ellipse. If we furthermore assume that the origin (on the sphere) is at (0,0), the intersection of the equator and the main line of longitude, this can be further simplified to:

$x=R\,\sin \,\lambda$

$y=\,0\,$

in which the projected equator is a straight line segment, just like here. The formula for x corresponds to what you wrote.  --Lambiam^Talk 03:51, 22 October 2006 (UTC)

Solution for a proof?

Hi guys, I need a bit of help. I do not know exactly how to finish this proof. I need to show that the set of all infinite subsets of the natural numbers is uncountable. I need to do this directly, as opposed to showing that it must be uncountable since the set of all finite subsets of the natural numbers is itself countable and the union of two countable sets is countable (and the powerset of the natural numbers just is the set of all infinite and finite subsets of the natural numbers). Thanks. Also I do not think Cantor's diagonal argument for the uncountability of the reals does not work in this case. Thanks again.--69.171.125.20 16:58, 21 October 2006 (UTC)

The diagonal argument does work, you just have to be a little creative. Suppose you have a list of all the sets. Construct a new set in the following way: If the first set in the list contains 1, the new set does not contain 1, and vice versa. I'll let you take it from there. —Keenan Pepper 18:42, 21 October 2006 (UTC)

The diagonal argument will not work unmodified: for example, if set n is all natural numbers except n+1 (which admittedly is not at all a complete list), then the diagonal set you construct will be the empty set, which is decidedly not infinite. It's possible to modify this argument so that it works, though -- think about all the sets whose complements are finite and convince yourself that the set of such finite-complement sets is countable. What does this suggest our proof will need to take into account? Tesseran 03:20, 23 October 2006 (UTC)

Goodness of fit for a rodent Population

I will try to be as brief as possible, but please forgive me if I do not include needed information. I am totally lost. I have calculated ratios of sexes for 7 rodent species I have captured in the field. My professor told me that he would like to have a calculation for goodness of fit for these populations. I have been reading all over the internet, wikipedia and various books, and I have no idea how to calculate goodness of fit. I have seen the equation $\chi ^{2}=\sum {\frac {(O-E)^{2}}{E}}$ but I'm not sure what numbers to use, or if this is even the correct equation to use. Am I totally missing something here? Can I even do a goodness of fit test for rodent sex, as there are only 2 sexes (it is my understanding that chi square tests for 3 or more events)? If I forget about sex alltogether, is there a way to calculate goodness of fit? I can provide data if needed, but any help would be greatly appreciated! Thank you so much! —The preceding unsigned comment was added by ChibiChibiChan (talk • contribs) .

The equation you have seen gives the statistic for Pearson's chi-square test, and yes it can be used to see how well your data resembles your model. The O in the equation refers to what you've Observed, and the E is for what you Expect to observe. You can use the test when you have 2 categories; there should be a section in the lower part of the article about that :) Yesitsapril 00:28, 22 October 2006 (UTC)

So The $64,000 Question is: What is your model? The goodness of fit is the goodness of the fit between two things: between (a) the actually observed frequencies and (b) the expected values according to the distribution given by your model (after you tweaked the parameters to get the best fit you could). For each possible category (for instance "Transsexual Gerbil", or "Male Capybara"), you should have a value O for the observed number (e.g. 20) and a value E for the expected number (e.g. 22.4), which you find from your model. Then you compute (20−22.4)²/22.4 = 0.2571, which is the contribution of this category towards the total χ². If all observations are close to the expectations, you get a low value, which indicates a good fit.  --Lambiam^Talk 04:10, 22 October 2006 (UTC)

Make sure you calculate frequencies in O and E and not ratios and percentages. Basically, imagine you have a set of data, and you want to see if the data resembles a Normal distribution, for example. You would find the mean and variance of the data and use this to make a model for a Normal distribution, assuming it did represent one. Then you would calculate the expected frequencies using a formula for your Normal distribution (total frequency x probability for that frequency) and then use the chi-squared test on each of these values of data, then add them up. You would then use tables or manual calculations to determine whether the values fall under your required significance level. You don't need to have a formula to calculate E, either. If you have a set of values a test should've had (such as pH level), and you did an experiment and found that your values look fairly stupid, you could run the same test using these values as O. If you are still confused, I'll give you an example. x42bn6 Talk 13:09, 22 October 2006 (UTC)

October 22

calculus

Role of trignometic results in differenciation —The preceding unsigned comment was added by 61.17.223.27 (talk • contribs) .

Do you mean things like Trigonometric substitution in integration? x42bn6 Talk 13:03, 22 October 2006 (UTC)

Do you mean the discussion at Wikipedia:Reference desk/Archives/Mathematics/2006 October 20#How do I prove this??  --Lambiam^Talk 15:24, 23 October 2006 (UTC)

Powers of ten using first-order logic and simple arithmetic

I've been re-reading Gödel, Escher, Bach and I'm puzzled by one of the problems offered to exercise the reader's grasp of TNT (a number theory system). Basically, I'm trying to work out how to represent the statement, a is a power of ten in one statement using only first order logic, with operations only of addition and multiplication. I've managed to represent a is a power of two in the following way (using equivalent mainstream notation rather than that developed for TNT, which is isomorphic):
$\forall x:(((\exists y:y\times x=a)\land \neg (\exists y:(\exists z:z\times y=x)))\rightarrow (x=2))$
which can be interpreted as "for every x, if x divides a and no y divides x, then x is two," or, more succinctly, "the only prime that divides a is two." but I can't work out how to do it for powers of ten. Every time I try to do it, I either say something not powerful enough, or too powerful, or just wrong. I can say things like "the only primes dividing a are 2 and 5" and "The only numbers that divide a that are less than (some fixed limit) are (some powers of ten)" but neither is what I want. I still want to solve this problem for myself but I would like a gentle nudge in the right direction. Can anybody help me out? Maelin 13:55, 22 October 2006 (UTC)

I should probably add that the book does warn not to try doing the power of ten one unless you have several hours to spare for it and also know quite a bit of number theory. Maelin 14:34, 22 October 2006 (UTC)

This is not a direct answer, but have a look at Matiyasevich's theorem. It implies that the problem posed by GEB has a solution. The proof of this theorem is constructive, so it allows you to find a solution. Most likely there are much simpler solutions than the one you will obtain this way.  --Lambiam^Talk 18:09, 22 October 2006 (UTC)

sphere clusters

Yesterday someone mistook me for an expert on sphere packing ;) and asked about what I think is called clustering: the shapes formed when hard spheres seek to minimize the sum of distances between them (or their mutual gravitational potential energy), without constraint on the convex hull. I know I've seen something on this subject on the web, but couldn't find anything; perhaps I misremembered the keyword. —Tamfang 16:57, 22 October 2006 (UTC)

Try Kissing number, maybe? - Rainwarrior 17:36, 22 October 2006 (UTC)

No, because there's no requirement that some central sphere touch all the others. —Tamfang 18:52, 22 October 2006 (UTC)

Oh, there is a name for that? Cool. I remember, as a kid, realising that six circles (probably coins, glasses or boccia balls) seemed to fit around a seventh, but I didn't know much maths then, so I couldn't prove it. —Bromskloss 20:35, 22 October 2006 (UTC)

Though I am also reminded of image processing techniques opening and closing which (our articles on them aren't great) take an image and pass a shape around it (usually some sort of circle), using the area covered by that shape as the new image. It could be done in 3 or more dimensions too. I can't quite figure out what you're looking for by your description though. - Rainwarrior 17:47, 22 October 2006 (UTC)

What do you mean by "without constraint on the convex hull"? —Bromskloss 20:35, 22 October 2006 (UTC)

I think the problem can be formulated in one form as: find n points in Euclidean space maximizing $\sum _{1\leq i<j\leq n}d_{ij}^{-1}$ subject to the constraint $d_{ij}\geq 1\,$, where $d_{ij}$ is the distance between point i and point j. This is related to the question how many unit spheres you can pack inside a sphere of (large) radius R. For both problems the asymptotic density should be that given by Kepler's conjecture. I don't know a name for this version or similar versions. I'd call it huddling spheres.  --Lambiam^Talk 22:13, 22 October 2006 (UTC)

Yes, that's exactly the problem I had in mind (or, alternatively, minimize $\sum d_{ij}$). One important difference from packing is that in this problem you're less likely to get "rattlers", balls which have some freedom of movement even in the optimal solution. —Tamfang 01:55, 23 October 2006 (UTC)

In mail, one Henry Cohn pointed me to [2] which has lists of solutions and a few pix. —Tamfang 07:05, 24 October 2006 (UTC)

This might not actually answer your question, but it's an interesting read nonetheless. Rob Cockerham from Cockeyed.com has a page about filling a car with small plastic balls for a competition. You can read about it here: http://cockeyed.com/inside/trailblazer/trailblazer.html -Maelin 23:41, 22 October 2006 (UTC)

Crystal systems formed from identical atoms follow several spherical packing patterns (crystal structures). Among the most common are hexagonal close pack, body-centered cubic, and face-centered cubic. StuRat 00:51, 23 October 2006 (UTC)

Yeah, so? —Tamfang 07:05, 24 October 2006 (UTC)

This is an answer to your question. Look at those articles at the links for more info on sphere packing. StuRat 17:35, 25 October 2006 (UTC)

No, it isn't an answer to my question, because packing and clustering are distinct problems (though the solution of one is often a solution to the other). Why is it that the more I say "I'm interested in X rather than Y" the more people tell me about Y? —Tamfang 06:19, 30 October 2006 (UTC)

October 23

Two degrees off course

If I were two degrees off course, flying from Los Angeles to New York City, where would I land (north and or south) . In addition, if I were two degrees off course while flying to the moon, by how much would I miss my objective.

~~Thomas A. Coss (email removed) —Preceding unsigned comment added by 68.4.244.159 (talk • contribs)

This depends entirely of the distance you are going to travel and which direction the two degrees are relative to (example, two degrees CW or CCW from north). ☢ Ҡi∊ff⌇↯ 03:05, 23 October 2006 (UTC)

In the simplest aproximate way of presenting the answer it would be (sin 2º = 0.035) * 2500 miles = 87 miles for NY to LA. Just change the distance for the error for any other 2º deviation --hydnjo talk 05:20, 23 October 2006 (UTC)

Problem with this page

I just noticed that the first eight topics on this page have been duplicated. This seems to have been caused by RefDeskBot. --KSmrq^T 14:09, 23 October 2006 (UTC)

Eccentricity question

Is the eccentricity of a straight line 1 or infinity? I realize the article claimed it is 1 (I changed it recently, before realizing I should have asked for some sources), but I can't find another source via a brief Googling that corroborates or denies it. Based on my experience (mostly dealing with orbital mechanics, and thus a different interpretation from traditional mathematics), the defining equations, for any value of k, all give a value of 1 for an ellipse with a minor axis of 0 (hence a line). It may be possible it's just defined as such (presumably to differentiate it from a parabola), but again, I couldn't find a source that didn't apparently mirror Wikipedia.

I'd say you're right – it's ∞. I just replaced the first image in the article with a vector version (which I didn't make, it was already lying around) and the new one agrees with you. —Bromskloss 18:25, 23 October 2006 (UTC)

Oh, another thing. In what way would mathematics and orbital mechanics treat the subject differently? —Bromskloss 18:28, 23 October 2006 (UTC)

I'd say it is somewhat indeterminate, mathematically. On the one hand, considering the family of parabolas y = λx², all have e = 1 if λ ≠ 0. It would be reasonable then to extend this to the case λ = 0, a straight line. On the other hand, considering the family of hyperbolas μx²−y² = μ², as the parameter μ tends to 0 the eccentricity e = √(1+1/μ) tends to infinity, while the hyperbola does what it can to degenerate to a straight line.  --Lambiam^Talk 19:14, 23 October 2006 (UTC)

Sorry, I accidentally wrote that the article used to say 1, when I meant infinity. So everything Bromskloss wrote should say "dis"agree =O). In another interpretation I thought of, if one uses a conic section visualization, a line is really just the parabolic section passing through the (conic) origin, in which case the eccentricity would be one (assuming the eccentricity is proportional to the normalized slope of the plane). In this case an infinite eccentricity still refers to a hyperbola.

Re: orbital mechanics versus traditional interpretations, in OM we're mostly concerned with velocities and extrapolating positions, so the equations used to define the orbits (usually ellipses) tend to focus on time derivatives of position rather than position itself - also, the semi-minor axis is rarely mentioned. Ultimately everything is the same, but the biggest difference is the vocabulary and concepts used to describe the objects.

There is some confusion here. The article, at this moment, says a line is an ellipse with vanishing minor axis. Wrong; a line goes to infinity, an ellipse does not. Flattening a parabola by "spreading its legs" does produce a line, in the limit, but not a proper conic. A hyperbola loops through infinity twice, so flattens to a double line. The geometric answer is that a line is not a conic, and its eccentricity is not defined.

Take a circular cone around a vertical axis, with two halves meeting at a vertex. Slice it with a plane to get a conic section. Vertical planes produce hyperbolae. Sliding the plane to pass through the vertex produces a pair of lines intersecting at the vertex. Rotate the plane around the vertex until it is tangent to the cone; the two lines become one. Sliding the plane off the vertex produces a parabola. Hmm.

As for orbits, it is one thing to restrict attention to satellites and the like; those orbits are ellipses. But suppose some perturbation sends an object in the Oort cloud inward to loop around the Sun. When we spot it with telescopes, we begin to accumulate data to construct an orbit. Perhaps it will reveal itself as a new periodic comet, whose orbit is elliptical. Perhaps it will be a one-time visitor to be flung back into space and lost forever, whose orbit is hyperbolic. Our methods for fitting Keplerian orbital elements must be prepared to cope. --KSmrq^T 12:26, 24 October 2006 (UTC)

I'm not really sure what you're getting at here - orbital mechanics also includes parabolas and hyperbolas, not just ellipses (though the last are obviously the most common). OM typically addresses the type and characteristics of an orbit through its "energy" (KE and PE, zeroed on a parabolic trajectory), which applies to all conics. Energy is calculated via Keplerian elements.

Consider a line L, a point P not on the line, and the locus of points Q in their plane such that |PQ| = b |LQ| (where |LQ| is the perpendicular distance). Amazingly enough, that locus is a conic with eccentricity b. As b increases toward infinity, the conic converges to L. —Tamfang 06:54, 24 October 2006 (UTC)

Duh, as the picture cited above illustrates. —Tamfang 06:55, 24 October 2006 (UTC)

Comment: This particular derivation also has the limit of eccentricity 0 (a circle) vanishing to a point focused at P, so I don't really know if it should also apply to the limit of an infinite eccentricity either (which, in my understanding, generates two lines, one at L and the other at infinity). 192.31.106.34 17:30, 27 October 2006 (UTC)

So ultimately the varied definitions of eccentricity (including the picture, which is just one user's interpretation) that seem to be used here imply (to me, anyways) that there is no set value of e for a line, unless someone has a definitive text that says so. —The preceding unsigned comment was added by 192.31.106.34 (talk • contribs) 15:23, 2006 October 24.

(In future, please, as requested at the top of this page:

• Sign your question. Type --~~~~ at its end.

Actually, sign all your posts here and on talk pages. Among other things, it helps us know who is asking or stating what.) --KSmrq^T 19:06, 24 October 2006 (UTC)

In this specific case, it wouldn't, since I'm on a floating IP which seems to change every few hours =O) 192.35.35.34 19:09, 24 October 2006 (UTC)

System of DifEqs

It's been a couple years since I took difeq, and then I stumble across this in lab.

${\frac {dy_{1}}{dt}}=-Ay_{1}$

${\frac {dy_{2}}{dt}}=Ay_{1}-By_{2}$

${\frac {dy_{3}}{dt}}=By_{2}$

Any bright ideas? I thought I could solve the first eq and get y1 in terms of t, put that into the second to get an equation of just y2 and t, but that came out to be something like

${\frac {dy_{2}}{dt}}=ACe^{-At}-By_{2}$

which I'm not sure how to solve. --Registrar

Found it. If you're interested, Steady state (chemistry). --Registrar

Should you run into a situation with 17 reactants in your lab, a general way of solving this kind of linear ODEs can be found in our article Matrix exponential.  --Lambiam^Talk 19:27, 23 October 2006 (UTC)

Möbius Space Station

I was reading about the Möbius strip and I was wondering what would happen if a space station was designed like one. I was thinking a space station that is shaped like a ring (possible spinning to create artificial gravity) but had a twist in it. Would there be a part where the ceiling and floor met?Ed Dehm 22:18, 23 October 2006 (UTC)

You wouldn't be able to do that. A Möbius strip doesn't have an interior, the reason it works is that you're taking a 2 dimensional object and using the 3rd dimension to do the twist. (A space station has to have an interior in which people can live.) The nearest analogy in 3 dimensions is the Klein bottle (take a look at the Klein Bottle discussion above), which actually requires 4th dimension to pull off properly... and furthermore, a Klein Bottle has no inside or outside, so you wouldn't be able to protect astronauts from space. - Rainwarrior 23:27, 23 October 2006 (UTC)

I disagree. A paper Möbius strip has some thickness, and the thickness could be increased to be sufficient to accommodate a crew. However, it would be far more expensive to construct a space station shaped like that, and, if rotated about it's center, the apparent gravitational force would be toward the "floor", then one wall, then the "ceiling", then the other wall, then the "floor" again, as you ran around it. StuRat 23:55, 23 October 2006 (UTC)

Topologically speaking, the Möbius strip has no thickness, and if you think of it as a flattened-tube, it would be homeomorphic to just a regular tube, twisted or not (this has a lot to do with needing the extra dimension to accomplish the twist). So... you could have a Möbius strip painted on the inside of a toroidal tube, sure, but there'd be no real reason to call it the "floor", it would appear to go around the tube twice (crossing over once). - Rainwarrior 04:29, 24 October 2006 (UTC)

I don't see why you can't have a rectangular cross section with a twist in it (as it rotates around the center). Are you just saying that this isn't called a Möbius strip ? StuRat 04:47, 24 October 2006 (UTC)

A Möbius strip doesn't have a cross section. That negates its uniqueness. If you give it thickness it is homeomorphic to a torus, which is the regular space station shape we were starting with. A paper model of a Möbius strip is only a Möbius as much as you ignore its edges. If you include the edgies, then it has a continuous surface that goes all the way around (around its cross section), making it the same as any other ring. It is only with the 2-dimensional strip that the notion of "top" and "bottom" surfaces apply, and the peculiar property of the Möbius strip having only a single surface that is neither.

So, yes, you could construct a space station like you would a paper Möbius model, and ignore the "edges", the non-edge surface would be a Möbius strip (either the inside one or the outside one). But, since gravity is going to be radial, I don't think you can really say that it's the floor. If you built a space station with a twist in it, it's just going to be an unusual twisty hump in the floor, you won't continue onto the ceiling.

Maybe if you had a flexible station that continually moves the twisted portion along the length, then as long as you walked along it at the same speed that the twist is moving along, you could remain on the floor, which would eventually cover both the floor and ceiling. (Or, alternatively if you could generate gravity some other way that is local to the Möbius floor?)

The shape of the station though wouldn't have to have a rectangular cross section. It doesn't really matter what the cross section is shaped like, as long as the cross section is a continous curve that loops back on itself (homeomorphic to a circle), you can put a Möbius loop on its surface (inside or out). If you chose a circular cross section, the twist isn't going to look different from any other part in terms of cross section, but where you are assigning the "floor" would proceed in a twisted fasion. - Rainwarrior 05:12, 24 October 2006 (UTC)

So is there a name for a geometric figure like a Möbius strip, but with thickness added ? StuRat 05:23, 24 October 2006 (UTC)

A twisted prism would probably suffice. One can probably find such a shape (though not as flat) in a number of different failure modes related to buckling under torsion. I don't think there's an actual name for it, since it's not a very useful (or geometrically interesting) shape. Virogtheconq 06:04, 24 October 2006 (UTC)

Well, not really. As I said, with the thickness, topologically speaking you've just got some sort of deformed torus. You could call your shape the product of a Möbius loop and a short line segment, but I don't think there's a particular name for this. There's a Klein bottle, but that's a bit different (it's more like two Möbius loops grafted together). Also, the buckling suggestion is unlikely, as if a rectangular torus (or your rectangular torus) with no twist is going to be crushed and begin to buckle, under no circumstances would it be allowed to form a half twist in itself. The half twist has to be part of its construction, you can't add it later without tearing the object and reattaching it. - Rainwarrior 15:45, 24 October 2006 (UTC)

Yes, that was the question, what if the space station was originally designed with a twist. StuRat 06:39, 25 October 2006 (UTC)

Perhaps I should have said such a twisted shape was then bent into a circle - I realize there's no way to get a failure quite like that in reality (aside from a sudden cold weld, perhaps?). And I withdraw my comment on not being a geometrically interesting shape, because it's a three dimensional object that only has four sides, which I think is the minimum required for any 3d shape. That being said, I don't think there's a name for it, and it's still not a good shape for a space station, since there's no inherent gain on volume or surface area in the design over a traditional torus (within orders of magnitude, anyway), and it's technically more challenging (though not impossible) to build. Virogtheconq 14:46, 26 October 2006 (UTC)

I'd never heard of cold welding before. That's pretty interesting. - Rainwarrior 16:04, 26 October 2006 (UTC)

At a much higher level, for those who are interested, we're saying that the real projective line bundle over the circle (that is, the Möbius strip) has no nonzero section and thus is a nontrivial vector bundle (and is nonorientable as a manifold and so on). Conversely, you can get a rank-2 vector bundle by taking the trivial rank-2 real bundle on the interval [0,1], then identifying the endpoints 0 and 1 of the base space while identifying their fibers by the antipodal map (which is the same as a half-twist). This vector bundle is equivalent to a "thickened" Möbius strip, but thanks to the added dimension this bundle has a nonzero section. In fact, it has 2 orthogonal nonzero sections and thus is the trivial bundle $S^{1}\times \mathbb {R} ^{2}$. Tesseran 07:13, 26 October 2006 (UTC)

Think outside the box. How about a Möbius Ringworld? :-D --KSmrq^T 12:32, 24 October 2006 (UTC)

Terry Pratchett says: "I'll start thinking outside the box when I see any evidence that there's any thought going on inside the box." :) JackofOz 10:20, 25 October 2006 (UTC)

I thought that was what he was talking about - a ring-like world/station/whatever, with a half-twist in it somewhere along its length. The problem is, as they said (at some point, if you wade back through the rest of it) if you base the gravity off rotation, you've essentially just made a normal ringworld with an inconvenient sort of hump thing partway along, where coming along the inside, the land turns over and exposes you (fatally) to space. Imagine hiking down into a valley with no floor. If the twist were drawn out farther along the ring, more of the world would be sideways and it would be pretty well impossible to live on that section, unless you like living on the side of a vertical mountain face. Sadly, this marvel of topology rather loses its luster when you try to actually use it for something. One thing that might be interesting, though, going with the half-turn idea, would be to build a donut-shaped station, made up of a bunch of donut-shaped tubes, with a half-twist that attaches each tube to a different one. You could be walking down that hallway for a very long time, just going round and round the station, without realizing how many loops you'd completed. Not that you'd have any way of judging that in the first place, though. Black Carrot 22:30, 27 October 2006 (UTC)

bases

teacher says that [101(base b)]^5 = (b^2 + 1)^5

but didn't explain where this came from or what the general case was. anyone seen this strange identity before? some kind of binomial theorem?

It's true that 101(base b) = b^2 + 1. I wouldn't worry about the ^5. Melchoir 00:20, 24 October 2006 (UTC)

Yeah, indeed it's true.

  1    0    1
× b^2  b^1  b^0
= b^2 + 1

The 5 is completely irrelevant here. ☢ Ҡi∊ff⌇↯ 03:30, 24 October 2006 (UTC)

• It might help to think about this in the case of base 10 -- in that case, all you're saying is that 101 is 10*10 + 1. The biggest digit is the 100's column, which is the 10-squared column. The middle digit is the 10's column, and the smallest digit is the 1's column (10-to-the-zero-power). -- Creidieki 21:13, 27 October 2006 (UTC)

October 24

Complete list of Memoirs of the American Mathematical Society

As a member of our LOCAL Library committee, I am supposed to find out the complete list of the Memoirs of the American Mathematical Society, including the volume number, the author(s) and the title. It is less than 190 entries. From (MathSciNet, Math Reviews on the Web), I got the following message:

Mem. Amer. Math. Soc.

Memoirs of the American Mathematical Society.

Amer. Math. Soc. , P.O. Box 6248, Providence RI 02940.

Status: No longer indexed

I filled in the form at

http://www.ams.org/mathscinet/support_mail.html

asking for instructions to use their site for this specific purpose but so far nobody paid any attention to my call for about the last 5 days. This is why I come to here. I am sorry that this is NOT a mathematical question but a mathematical ADMINISTRATIVE question. If not appropriate for this site, please delete me and I have no hard feeling. However, I would be very grateful if somebody could help. Thanks in advance. Twma 03:24, 24 October 2006 (UTC)

Maybe you have already seen this, but the overview page about the Memoirs says "This journal is indexed in Mathematical Reviews, Zentralblatt MATH, Science Citation Index®, SciSearch, Research Alert, CompuMath Citation Index®, and Current Contents®, Physical, Chemical & Earth Sciences" - so that gives you some possible sources. Alternatively, if a printed list would meet your requirements, you could perhaps request one by snail mail. Gandalf61 11:57, 24 October 2006 (UTC)

I cannot find a complete list of their *** the volume#, author(s), title *** from the above overview page. My problem is not yet solved. Thanks. Twma 03:47, 25 October 2006 (UTC)

Hi: "MELVYL" catalog has 883 entries going back to 1954.I don't know when they started or if that is complete. My way was to google "uc davis",click on "library" on ucdavis main page then click on melvyl and search under titles. I hope this is helpful. Let us know how it's going & don't hesitate to ask again and again here. Good luck,Rich 04:36, 25 October 2006 (UTC)

Following your instructions faithfully, I got 21 lines probably because I am an outsider. Extending your idea, I tried a few other university libraries including our own Australian National University. No luck. Perhaps the best solutions is to accept the statement of my colleague that it is POSSIBLE but not a SIMPLE way to get the complete listing of merely three items: volume number, author(s), title although we are in an information technological era. Originally I would expect that clicking the name of the Journal from mathscinet would give me the answer with 100 records per page. Once that failed, I should have stopped immediately. Mission abort. Instead, I came here. Cannot afford the time to work any further. Thanks to both Gandalf61 and Rich. Twma 09:28, 26 October 2006 (UTC)

Google gives me this... Does it meet your needs? --TeaDrinker 20:40, 26 October 2006 (UTC)

This is EXACTLY what I am looking for. I did try Google but I missed it. Thank TeaDrinker for his/her help with appreciation. My problem is solved. My colleague said, POSSIBLE but NOT SIMPLE. From TeaDrinker, it is very simple. This case is close. Thanks again. Twma 01:07, 30 October 2006 (UTC)

Difference between Diminishing Rate of interest and Flat Rate of Interest

What is the difference between Diminishing Rate of interest and Flat Rate of Interest? Which is more advantageous?

Diminishing rate of interest is diminishing i.e. getting lower over time and flat rate of interest stays the same. If you take out a loan you'd want diminishing interest more but if you were earning interest then you'd prefer flat rate of interest. --WikiSlasher 08:11, 24 October 2006 (UTC)

Complex number dimensions

So we can have 1 dimension (line), 2 dimensions (square), 3 dimensions (cube)... but can we have i dimensions? I mean in the sense that if you "square" the line you add 1 dimension, but if you square the i you get -1 dimension. But if you square -1 dimension you get 1 dimension. So then i would be the 4th root of the 1st dimension... would that serve any useful purpose? Sorry it's 4:30am here this might be a garbled message. Thanks --Ķĩřβȳ♥ŤįɱéØ 11:38, 24 October 2006 (UTC)

When you square the imaginary number i you get the number −1, not some dimension –1, which is not a particularly meaningful notion. I would say, No, we cannot have i dimensions. The number of dimensions is the number of characterizing values you need to supply in some context, such as height and width. There can't be i of such any more than you can have 2-3i grandchildren.  --Lambiam^Talk 12:58, 24 October 2006 (UTC)

Imaginary numbers are neither ordinal numbers nor cardinal numbers, so it wouldn't make sense to describe a dimension (which are described either ordinally or cardinally, depending on the semantics used) using an imaginary number.

Well ... there are definitions of dimension which extend the concept to fractional dimensions (e.g. box-counting dimension, Hausdorff dimension, correlation dimension). But they all very roughly depend on calculating or measuring how some quantity related to the "size" of a set varies as a power of some other quantity, which can be interpreted as a "length". So it is difficult to see how any extension of these ideas could give a meaning to an imaginary number of dimensions. I think you would need to start by finding a complex-valued equivalent of the concept of measure. Gandalf61

It is almost meaningful to raise a (real or complex) number to the power i: the "almost" because it's multivalued. See Exponential_function#On the complex plane. This means that you can determine (up to multiple values) the volume of an i-dimensional cube of any given side length. Similarly, you can compute the Hypersphere#Hyperspherical volume of an i-dimensional sphere. —Blotwell 02:57, 26 October 2006 (UTC)

what is progressive matrix that are asked in aptitude test

what is progressive matrix that are asked in aptitude test

Raven's Progressive Matrices is the page your looking for. --Salix alba (talk) 14:22, 24 October 2006 (UTC)

Probability question

I am a maths teacher - in class this question came up - but I can't get my head around it - but I'm sure that someone here can help:

If we have two identical lines A and B next to each other

________ A 

________ B

if I randomly choose any point on each line then there is a probability of 1/2 that the point from line B will be further right than the point from line A (symmetry)

If the two line look like this

________ A
                    ________ B  

now there is a probability of 1 that the point from line B will be further right than the point from line A

My question:

What if the two line look like this:

________ A
    ________ B

This time line B starts 1/2 way along from A (or any other fraction) What is the probabilty that if I choose a random point from each line then the point from line B will be further right than the point on line A? --yakov Korer 18:21, 24 October 2006 (UTC)

How about we draw a square on a grid, with corners at (0,0.5), (0,1.5), (1,0.5) and (1,1.5). The x-values represent the possible values of point A. The y-values represent possible values of point B. As we assume that they are both distributed uniformly - any unit area in that square has the same probability of occurring as any other unit area. Now we need to look at what it means for point B to be further right than point A - it's the same as y - x > 0. Now draw the line y - x = 0 on the grid. Anything above that line has y-x > 0, so point B is further right than point A. Anything below the line has y-x<0, so point A is further right than point B. It's straightforward to see what the required probability is - 7/8. To satisfy you that this is right, you could try this method with the other examples you posted - to see that you get the correct answers i.e. 1/2 and 1. For a more general look, see our article on convolutions, in particular, the bit about half-way down that page. Richard B 20:10, 24 October 2006 (UTC)

gear ratio selection

If i have a block that fits on a ball screw with a pitch of 6 mm and a diameter of 6 mm, how can I determine a gear box, with an appropriate gear ratio, that will allow the block to move along the ball screw with an accuracy of +-1 micron?

Spherical Pythagorean theorem

I had read once about the spherical Pythagorean theorem and discovered some, which appeared to be complex, explanations of the theorem. I can not find any reference to this anywhere on Wikipedia, and I have searched the Internet to no avail. Could someone please enlighten me on the spherical Pythagorean theorem? Thank you. —The preceding unsigned comment was added by 72.160.189.183 (talk • contribs) .

Look at the law of cosines for the case that C is a right angle.  --Lambiam^Talk 20:51, 24 October 2006 (UTC)

Radius/semi-axis vs. diameter/axis

With a circle, ellipse and ellipsoid, the radius/semi-axis is half the diameter/axis. When finding an elliptic parameter (eccentricity, flattening, etc.), "a" and "b" are used to calculate these parameters (ultimately trig functions of the angular eccentricity, $o\!\varepsilon \,\!$). Are "a" and "b" technically radii or diameters in calculating $o\!\varepsilon \,\!$? While this may seem just semantical, consider an ellipsoid where "b_north" ≠ "b_south":

Since the polar radii aren't equal, the four semi-axes don't intersect in the middle. Does this mean the calculation of $o\!\varepsilon \,\!$ (and the associated elliptic paramters) isn't as simple as $\arccos \!\left({\frac {b_{n}+b_{s}}{2a}}\right)\,\!$ (since $\arccos \!\left({\frac {b}{a}}\right)=\arccos \!\left({\frac {2b}{2a}}\right)\,\!$), or is it?  ~Kaimbridge~23:04, 24 October 2006 (UTC)

The image does not look symetric about any horizontal line, hence its not an ellipse. This means that talking about the usual description of an ellipse do not hold. I don't know much about angular eccentricity as it is a rather non standard presentation. --Salix alba (talk) 12:53, 25 October 2006 (UTC)

It might not be an ellipse, but it is an ellipsoid——consider Mars: (Mean) equatorial radius = 3396.200, north polar radius = 3376.189 (typo corrected) and south polar radius = 3382.580. Is there an equation for an oval's eccentricity?  ~Kaimbridge~14:24, 25 October 2006 (UTC)

An ellipsoid is a three-dimensional shape. The curve in the figure is neither an ellipse nor an ellipsoid, and it has no eccentricity because it's not a conic section. —Keenan Pepper 18:38, 25 October 2006 (UTC)

Okay then, forget that figure. Keeping in mind the above Mars example (so this example is certainly possible), let's say there is an oblate spheroid, where a = 10000, b_north = 7000 and b_south = 8000. The obvious, single value for b is 7500, but would that be technically correct for finding elliptic parameters (such as eccentricity or flattening), since (e.g.) ${\frac {(2a)^{2}-(2b)^{2}}{(2a)^{2}}}={\frac {a^{2}-b^{2}}{a^{2}}}\,\!$, or——like the radius of an ellipse——would the parameters vary from latitude to latitude, perhaps utilizing some integrand?  ~Kaimbridge~22:04, 25 October 2006 (UTC)

The keyword is ovoid (or, if you wish, oval). As a class, they are not defined precisely enough to make the original question meaningful. One could study a specific class of ovoids, for example those satisfying an equation like ax²+kx³+by² = 1, and try to define notions similar to minor axis and such. Technically, definitions developped for ellipses do not apply, because these shapes are not ellipses.  --Lambiam^Talk 19:35, 25 October 2006 (UTC)

Abstraction of Boolean Rings

I am wondering whether there is an abstraction of the concept of a Boolean ring. This would be some ring defined by the n-th power of each element being the element itself, for some fixed n. Are there any interesting results that can be found from these rings?

I have nothing useful to say about this, other than to suggest you think about associative algebras over finite fields. (Other articles that might interest you are modular arithmetic and roots of unity.) –Joke 05:09, 25 October 2006 (UTC)

October 25

Math

What is the factors of 24

Half of me is saying that I shouldn't answer this because it looks like a homework question, while the other half says I should... But I guess I won't. I'd like to point you to the Divisor article if you don't know what a factor is.

Anyway, the factor of a number (X) are all whole numbers other than 0 which can be multiplied by another whole number (which is also a factor of X) to get X, such as y (8) and z (3) which, multiplied together, result in yz = x (8 x 3 = 24). There isn't a "formula" to find factors of numbers aside from dividing the number by all whole numbers from 1 to whatever until you get one that works - that's what supercomputers are for (see RSA). The number ends in 4, so you can assume that it's divisible by 2, possibly by 3, 4 or 6 but not by 5 or 7, etc. But that's not needed for a small number like this. Also, the smallest factor is always 1 (assuming it's a positive nonzero integer) and the largest factor is always the number itself (1 x 24 = 24).

Just out of curiosity, how old are you and what grade are you in? I'm just asking because this is middle-school stuff. Corporal 02:10, 25 October 2006 (UTC)

To answer your question, I suggest that you read these articles:

• Fundamental theorem of arithmetic
• Prime factorization algorithm

The fundamental theorem of arithmetic establishes the importance of prime numbers. Prime numbers are the basic building blocks of any positive integer, in the sense that each positive integer can be constructed from the product of primes with one unique construction. Finding the prime factorization of an integer allows derivation of all its divisors, both prime and non-prime.

202.168.50.40 05:07, 25 October 2006 (UTC)

Yo, any whole number you can divide it by to make a whole number is a factor. End of story.

1, 2, 3, 4, 6, 8, 12, 24 --Proficient 02:55, 26 October 2006 (UTC)

Economic Production function to profit function:

I would like to make sure I am accurate in my technique, am I right?

Once you have the cost function, formulated from a production function, there are two ways to find maximum profit:

You can either plug what optimal Q is equal to w.r.t. the constraint, and then maximize with respect to inputs X and Y, and then replug these optimals into the constraint for the optimal Q...

Or, you can plug in the cost function, and merely derive w.r.t. Q, then solve for the optimal Q?

After both of these are solved for Q, just plug them into: pi = Q(PQ) - X(pX) - Y(pY).

Correct? Thanks so much, ChowderInopa 02:49, 25 October 2006 (UTC)

You are not making much sense to me. What is X , Y , P and Q ? What is pi (I don't think you mean 3.1416)? Is lower case p different from upper case P? What is a production function? Is it a function of quantity? 202.168.50.40 05:01, 25 October 2006 (UTC)

(after edit conflict) The method you propose is not entirely clear. What is the relationship between the cost function and the production function? What is the role of the constraint? Is it a constraint on the inputs (for example X+2Y < 100)?

In general you cannot do optimization unless you actually know the objective function. If it is given analytically, by an equation, you may hope to use methods such as that of Lagrange multipliers to find an optimum. Or perhaps you can use quadratic programming. Otherwise, you could resort to numerical techniques such as hill climbing, modified to obey the constraint.  --Lambiam^Talk 05:10, 25 October 2006 (UTC)

Are all real numbers with multiple decimal representations rational?

Today's featured article, besides drawing out an unusually broad array of cranks, got me to thinking. Obviously, 0.999... = 1, 0.34 = 0.3399999..., 330 = 329.999..., etc. But all of these numbers with multiple decimal represenations are constructed by taking a simple terminating decimal number and replacing a final digit n with the digit sequence (n-1)99999... Are there any other types of numbers that also have multiple representations? --Pyroclastic 07:32, 25 October 2006 (UTC)

e (number) = 1 + 1/1 + 1/2! + 1/3! + ... Which is a basis for Euler's formula, and actually quite elegant. Also look at pi. 169.229.89.53 09:25, 25 October 2006 (UTC)

To make it even more elegant, why not write it as e (number) = 1/0! + 1/1! + 1/2! + 1/3! + ... StuRat 17:29, 25 October 2006 (UTC)

Um, but that's not a definition in terms of an infinite decimal sequence?--140.180.157.26 07:18, 26 October 2006 (UTC)

Well, there's 0.000… = -0.000…, although in some authors' development of decimals, this is written 0+.000… = -1+.999…, so it may or may not fit the pattern. Anyway, every number with multiple decimal representations is rational, yes: in particular, a rational which can be represented by a fraction with a power of 10 in the denominator. Melchoir 09:31, 25 October 2006 (UTC)

Well, I've no proof but a strong hunch that there are an infinite number of series representations of any rational number, since there's an infinite number of series that converge. But the number is probably finite for the irrationals? --BluePlatypus 12:37, 25 October 2006 (UTC)

It draws cranks for two reasons. One, there will always be some for such a topic. Two, the article threw away the instructional care of earlier versions to throw up a façade of being “encyclopedic”. Far from representing the best of our mathematical writing, it is an object lesson in what not to do. (Pardon me while I dismount my soapbox.)

Do you mean to restrict attention to decimal expansions? If so, let's review the options. A number like 0.84375, the finite decimal expansion of ²⁷⁄₃₂, can be normalized to an infinite expansion by adding trailing zeros. Such numbers are always rational, and — if nonzero — their denominator contains only prime factors of two and five. Exclude zero, and we have the only real numbers that admit two decimal expansions, the second one having trailing nines.

If we do not restrict ourselves to decimal expansions, every real number has numerous representations. We have alternatives like expansions in a radix of our choosing, continued fractions, and so on. We can represent a real algebraic number as a select root of a polynomial with integer coefficients. Expanding our scope, we can write down the digits using a different alphabet, like Devanāgarī or Hangul. The idea of a representation is broad!

As far as the series question, no, every real number is the limit of an infinite number of different series. A proof is trivial: Take any given series and preface it with two new terms that cancel each other; the result is a new series with the same limit. --KSmrq^T 14:13, 25 October 2006 (UTC)

A trivial proof but for a trivial case. Not a very interesting or enlightening answer. --BluePlatypus 20:23, 25 October 2006 (UTC)

Nonsense. You speculated that any given irrational number was the limit of a finite number of series. I showed that is false. More elaborate variations could be given, but there is no need; false is false. If you claim every mammal has a placenta, and I show you a platypus (!), must I also point out a Virginia Opossum, and a Tasmanian Devil, and a Monito del Monte? --KSmrq^T 01:59, 26 October 2006 (UTC)

No one seems to have directly answered the original poster's question. The answer is yes, all numbers with two distinct decimal representations are rational, and in fact are of the form ${\frac {n}{10^{m}}}$, where n and m are integers. --Trovatore 17:33, 25 October 2006 (UTC)

Oh, actually I guess Melchoir did answer it; sorry about that. --Trovatore 17:41, 25 October 2006 (UTC)

Is there some reason you ignore my answer? Also, your form of the fraction is misleading, because n cannot always be relatively prime to 10, which we would expect from the presentation of a rational number. The very example I gave, 0.84375 = ²⁷⁄₃₂, illustrates that. Sheesh. --KSmrq^T 18:39, 25 October 2006 (UTC)

Well, you sort of went on a bit, KSmrq, and your answer to the original question was in the middle of it all. The OP's question had a direct answer—"yes"—which no one bothered to give directly. --Trovatore 18:45, 25 October 2006 (UTC)

My answer mirrored the question sequence: cranks, decimal expansions, other representations, series. And if you think the answer to the poster's question is "yes", maybe you should go back and decide which question you were answering, the one in the title or the one in the post itself. Oy vey! --KSmrq^T 01:24, 26 October 2006 (UTC)

Help with diff eq

Hey,

I'm looking for a solution for next differential equation:

(b'(t))²/2 = A*b - B*b^4/4 + U²/2 , with A,B en U positive constants.

I actually need an explicit formula for b(t)! I would be very happy if someone could help me with this problem.

Solutions can be send to: #@#@#@#@#

Yours sincerely,

Sam

Please read the guidelines at the top of this page, and follow them. --KSmrq^T 17:06, 25 October 2006 (UTC)

On the look of it, this has no analytic solution. For that, we need an antiderivative of the function f(x) = (1 + λx − μx⁴)^−1/2, and that antiderivative has to have an explicit inverse to boot. But I'm already stuck on the first part. You could perhaps try looking at expansions of b (for example as a sum of exponentials near the roots of the polynomial Ax - Bx^4/4 + U²/2), using the method of equating coefficients (in a more general sense than treated in our article under that name).  --Lambiam^Talk 19:13, 25 October 2006 (UTC)

Basically the equation is ((b')²)/2 = Ab - (Bb^4)/4 + U²/2 (I'm just confused about b^4/4 because that could be (b^4)/4, or b^1), right? --AstoVidatu 22:32, 25 October 2006 (UTC)

I'm not done, but try rewriting the equation as b' = (k)^(1/2), and set k equal to all that junk in the middle. Then take the integral of that equation (easy as pie) and sub in k and k' using k = 2*(Ab - (Bb^4)/4 + U²/2)... And remember U is just a number, so it goes to 0 when you take the derivative. --AstoVidatu 22:37, 25 October 2006 (UTC)

Infinitesimal

Is an infinitesimal the recipricol of infinity? — Preceding unsigned comment added by 86.142.195.245 (talk • contribs)

No, I believe the reciprocal of infinity is 0. It is somewhat debated, however, because infinity can have more than one value, while 0 is only 0. PullToOpen ^talk 20:10, 25 October 2006 (UTC)

But couldn't a case be made for 0 to be equal to an infintesimal? If you define 1/0 as infinity, the reciprocal is 0. However, that's just the ramblings of a 15-year-old, so don't set too much store by it. —Daniel (‽) 20:30, 25 October 2006 (UTC)

No it is not debated within professional mathematicians. The reciprocal of infinity is undefined, because infinity is not a real number. There are any number of Math FAQs out there that explain this, as well as the article on Infinity. --BluePlatypus 20:33, 25 October 2006 (UTC)

Let's clarify things a bit. The answer depends on the context. In the real numbers, there is no infinity or infinite quantity. In the real projective line, the reciprocal of ∞ is 0. In the surreal numbers, there is no ∞, but there are infinite quantities like ω, the reciprocal of which is 1/ω, an infinitesimal. -- Meni Rosenfeld (talk) 20:55, 25 October 2006 (UTC)

For "infinitesimal" to be a useful concept, you want to retain some conventional algebraic properties, such as, letting ε stands for some infinitesimal, (2ε)/ε = (ε+ε)/ε = ε/ε + ε/ε = 1 + 1 = 2. If ε was equal to 0, this would not be possible; we would get the useless form 0/0.  --Lambiam^Talk 21:45, 25 October 2006 (UTC)

Lambiam, isn't there a problem there? Your example says that 2ε/ε = (ε+ε)/ε = ε/ε + ε/ε = 1 + 1 = 2, but if ε is infintesimal, surely adding multiplying it by any finite number will leave it unchanged, thus 2ε/ε = ε/ε = 1? Have I missed something?

Yes, you have missed something. The idea is related to "differential quotients" suggesting the ratio of two infinitesimal quantities, as in dy/dx = 2. Mathematicians find it useful to manipulate this like dy = 2dx, as if dy and dx are actual quantities. Actually that practice preceded the invention of the differential calculus. Think of dy and dx as very very (as in VERY) small changes. It took quite some time before someone invented a way to put that on a safe footing.  --Lambiam^Talk 22:45, 25 October 2006 (UTC)

0 would be the recipricol of infinity. Infinity is defined as (infinity/1). And the recipricol of that would be (1/infinity), which is 0. --AstoVidatu 22:28, 25 October 2006 (UTC)

Yes, the typical infinitesimal is the reciprocal of an infinity. By definition, a positive infinitesimal number is smaller than the reciprocal of any positive integer: ε < ¹⁄_n. (Formally, we would add "for all n ∈ Z⁺".) Inverting, ¹⁄_ε > n; this says the reciprocal of an infinitesimal is greater than any integer.

However, if we want to do mathematics, not just wave our hands, we must be careful and more precise. The ordered systems of arithmetic we use most often — the integers, the fractions, the real numbers — exclude both infinities and infinitesimals. There is no integer greater than every other integer; there is no positive fraction smaller than every other positive fraction. So, if we are to do mathematics with infinities and/or infinitesimals, we must define a different system of arithmetic.

We have choices. In many of the new systems what I said is true. (An example is hyperreal numbers.) In some, we have no infinities, and no infinitesimal has a reciprocal. (An example is dual numbers.) Or we can extend the real numbers with no infinitesimals, but with a single infinite value, ∞, whose reciprocal is zero. (An example is the real projective line.) We also have exotic possibilities like surreal numbers.

All of these systems are valid choices and allow us to do interesting mathematics. However, they can have strange new properties and lack familiar old properties. Choose wisely. --KSmrq^T 00:06, 26 October 2006 (UTC)

If .999... = 1in a base 10 number system what about others?

Would it also be true that .777... = 1 in a base 8 system? If so, would .000... = 1 in base 2? I'm just curious and hope this is not an inappropriate post. Downhill.geezer 21:36, 25 October 2006 (UTC)

No worries! Yes, in base 8, .777... = 1. In base 2, however, .111... = 1, not .000.... There's some more information at 0.999...#Generalizations. Melchoir 21:44, 25 October 2006 (UTC)

Your post is welcome.

If d is an integer greater than 1, then we can use it as a radix (base) just like 10 or 8 or 2. When we do, the equivalent of 9 repeating is d−1 repeating. So 0.777…₈ = 1 for radix 8, as you propose; but 0.111…₂ = 1 for radix 2, contrary to your guess. Independent of which d we choose, 0.000… is always exactly zero.

Mathematicians have a habit of seeing how far an idea can go. Donald Knuth, in volume 2 of his series The Art of Computer Programming (ISBN 978-0-201-89683-1), presents a wide assortment of positional notation systems. For example, in balanced ternary, with digits 0, 1, and −1 (written 1), the notation 0.111… represents ¹⁄₂. With a radix of −10, digits 0 through 9, every real number (positive, zero, and negative) can be written without a sign. With a radix of i−1 (where i² = −1), digits 0 and 1, every complex number can be written. It might be fun and educational to explore some of these, seeking the equivalent of repeating 9s. --KSmrq^T 00:58, 26 October 2006 (UTC)

October 26

Terms in polynomial expansion

How can you determine the number of terms in a simplified (like terms have been combined) polynomial expansion? I am particularly interested in ${(a+b+c+d)}^{17}$, as that was on my math contest today, but any general help would be appreciated. Thanks. —Mets501 (talk) 01:49, 26 October 2006 (UTC)

Well for starters the powers of any of the terms should add up to 17. So the question is how many different ways can you make $a^{x}b^{y}c^{z}d^{w}$ where $x+y+z+w=17$? I'd have to think about this for a moment to give you an answer to that question though... - Rainwarrior 02:50, 26 October 2006 (UTC)

For just three variables ${(a+b+c)^{n}}$ should have $\sum _{i=1}^{n+1}{i}$, which is an easily computed arithmetic series. To do a fourth as well, it would compound, I think, as something like $\sum _{j=0}^{n}{\sum _{i=1}^{j+1}{i}}$? Since 17 is a fairly small number you could calculate the 18 values (for j=0 to n) and sum them yourself, I suppose, but I don't know of an easier way to compute the sum of this series. - Rainwarrior 03:07, 26 October 2006 (UTC)

Ah! It's a Tetrahedral number! (The 3 term polynomial problem was a Triangular number.) So... according to that formula, I'd say, 1140? - Rainwarrior 03:35, 26 October 2006 (UTC)

Tetrahedral numbers also turn up as a diagonal of Pascal's triangle. So, ${{n+3} \choose {3}}$ or ${{17+3} \choose {3}}=1140$. (I suspected there would be a solution related to the choose operation.) - Rainwarrior 04:08, 26 October 2006 (UTC)

The general question asks for the number of terms in $(x_{1}+x_{2}+\cdots +x_{k})^{n}$. A good way to tackle this type of question is to find the answers for specific cases and look for a pattern. For instance, if k = 2 then you're looking at $(a+b)^{n}$ which has $n+1$ terms according to the binomial theorem. The trivial case k = 1 is also interesting here; there is only one term if k = 1. Look at some more cases: n = 0, n = 1, perhaps n = 2, etc., and put all the results in a table. I think you will recognize the pattern in the table once you have enough entries. -- Jitse Niesen (talk) 03:06, 26 October 2006 (UTC)

OK, I'll try it out tomorrow. Thanks for the help! —Mets501 (talk) 03:10, 26 October 2006 (UTC)

See the multinomial theorem, which deals with this in generality, and may be useful to the problem you are considering. Dysprosia 08:42, 26 October 2006 (UTC)