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May 6

Spherical Triangle

How many degrees are there in a spherical triangle? I was given the example - using the earth for ease of communication - of drawing the equator, followed by two circles perpendicular to this going through both poles. To me it seems that - in this case at least - there are 270 degrees in a spherical circle. Is this right and does it always hold? Algebra man 16:55, 6 May 2007 (UTC)[reply]

That is correct, but it varies. In the limit of small triangles, the sum of angles goes to the familiar 180 degrees. See non-Euclidian geometry. —Ben FrantzDale 17:03, 6 May 2007 (UTC)[reply]

In particular, see spherical geometry. —Ben FrantzDale 17:07, 6 May 2007 (UTC)[reply]

The formula is π + area/r². So for very small triangles it's close to the usual sum of angles = π formula. —David Eppstein 17:06, 6 May 2007 (UTC)[reply]

Does the same follow, ie that in spehercial geometry the sum of interior angles is always greater than in Euclidean gemoetry, for any 2D shape and that in small shapes the sum approaches the number of degrees in a Euclidean shape? Algebra man 17:20, 6 May 2007 (UTC)[reply]

David Eppstein has stated the formula for angle excess of a triangle, which may answer your question for simple spherical polygons. Here's why. Dissect the polygon into triangles, say by slicing off ears. Any triangle that has a finite area will have an angle sum exceeding the Euclidean sum. The dissection introduces no new vertices, so the sum of all the triangle excesses is the excess of the sum of the polygon angles. Formally, the only challenge is to show that a suitable triangulation exists. --KSmrq^T 19:17, 6 May 2007 (UTC)[reply]

If you really want to know what's going on (for general non-polygonal shapes), you probably want the Gauss-Bonnet theorem, but our article is not as accessible as it might be. Algebraist 09:58, 7 May 2007 (UTC)[reply]

Large equations in Latex

Hi! I have a really big equation which spans a several lines. I am trying to typeset it using Latex but I am unable to. I have tried using \begin{equation}, eqnarray, align, multline, split etc. but all of the them seem to work for multi-euqations and not multi-line equation. Please help! The basic problem is that there is a \left( very early in the equation and it is closed on the last line. Using any template, whenever I change the line using "\\", latex gives the error that \left( is not closed.

Here is the tex that is giving the error:

\documentclass [a4paper,10pt]{article}
\usepackage{amsmath}
\begin{document}
\begin{equation*}
\ln\dfrac{f(T,\rho)}{RT\rho}=\dfrac{1}{RT}\left\lbrace2B\rho+\dfrac{3}{2}C\rho^2+\dfrac{4}{3}D\rho^3+\dfrac{5}{4}E\rho^4+
\dfrac{6}{5}F\rho^5+\dfrac{1}{2a_{20}}\left( G+\dfrac{H}{a_{20}}\right) +\left[G\left(\rho^2-\dfrac{1}{2a_{20}}\right) +
H\left(\rho^4-\dfrac{\rho^2}{2a_{20}}-\dfrac{1}{2a_{20}^2}\right)\right] \exp\left(-a_{20}\rho^2\right)\right\rbrace \tag{A2}
\end{equation*} 
\end{document}

Here is how it should look like: [1] It is basic Bender equation.--129.69.36.89 19:42, 6 May 2007 (UTC)[reply]

From memory, I think you can use \right. to close the first \left(, and \left. to match the now unmatched \right). The sizes may not match though then, which perhaps can be fixed – if necessary – by using struts. I can't run Latex on this computer to check this out. --Lambiam ^Talk 19:56, 6 May 2007 (UTC)[reply]

You mean something like this?

\documentclass [a4paper,10pt]{article}
\usepackage{amsmath}
\begin{document}
\begin{align*}
\left(land alknfdo hgsod sigf bfic yvej uvco ouewof b uogoi a oiajs sohgposi23908975 230920375023 20174023 \right \\
\left 52 59265092 5ß0285ß23 5023975  0ß57 230 52905620 745250236502 5025z02562035 2 5 \right)
\end{align*} 
\end{document}

It is not working :(--129.69.36.89 20:26, 6 May 2007 (UTC)[reply]

No, you omitted the stop, "\right.", instead of a parenthesis, "\right)"; likewise, "\left.". This can be made to work, but it may be easier to use explicit sizes, such as "\bigg(", instead of automatic ones. The problem with the automation is that the left and right markers must balance within a table entry. --KSmrq^T 20:31, 6 May 2007 (UTC)[reply]

Ahh! Thanks a lot! I was missing the stop. Here is the final version that works:

\begin{align*}
\ln\dfrac{f(T,\rho)}{RT\rho}=\dfrac{1}{RT}\left\lbrace2B\rho+\dfrac{3}{2}C\rho^2+\dfrac{4}{3}D\rho^3+\dfrac{5}{4}E\rho^4+ \right. \\
\dfrac{6}{5}F\rho^5+\dfrac{1}{2a_{20}}\left( G+\dfrac{H}{a_{20}}\right) +\left[G\left(\rho^2-\dfrac{1}{2a_{20}}\right) + \right. \\
\left. \left. H\left(\rho^4-\dfrac{\rho^2}{2a_{20}}-\dfrac{1}{2a_{20}^2}\right)\right] \exp\left(-a_{20}\rho^2\right)\right\rbrace \tag{A2}
\end{align*}

Thanks again.--129.69.36.89 20:38, 6 May 2007 (UTC)[reply]

Note for Lambiam: there are various online LaTeX compilers for when you don't have it handy. --TotoBaggins 21:02, 6 May 2007 (UTC)[reply]

Thanks. This one, which produces pdf, allows you to upload images, styles, and classes, and even understands amsspeak! --Lambiam ^Talk 22:10, 6 May 2007 (UTC)[reply]

1/3 != 0.(3) ?

I dont have any degree at math, but after reading http://polymathematics.typepad.com/polymath/2006/06/no_im_sorry_it_.html I got a question about $0.(9)=1$ . Isnt $1/3!=0.(3)$ but actually $1/3=0.(3)+(0.(0)1)/3$ and $2/3=0.(6)+(0.(0)2)/3$ . Im thinking here that the reason 1/3 is recurring is because the last number is always 1/3 higher then it was written so you always have to add a 3 to the end which again is 1/3 higher then is actually written... so 0.(3) is actually 1/3 higher then the last digit if you could get to it, thus 0.(3) is not exactly equal to 1/3 but (0.(0)1)/3 less. Since $1-0.(9)=0.(0)1=(0.(0)1)/3+(0.(0)2)/3$ then $0.(9)!=1$ ? If you cant use $1/3+2/3=3/3$ to justify $0.(3)+0.(6)=0.(9)=1$ then $0.(9)!=1$ . This question is really bugging my mind right now, couldnt find it answered anywhere so I hope you can help. —The preceding unsigned comment was added by 80.202.226.35 (talk) 22:28, 6 May 2007 (UTC).[reply]

Wikipedia:Reference_desk/Mathematics#Way_to_prove_.999.2C_or_theories_against_it.3F. Scrolling up is your friend.--Kirby♥time 22:53, 6 May 2007 (UTC)[reply]

If the arguments given at the page you linked to did not convince you, then probably nothing will. Any mathematician can prove that (using your notation)

1/3=0.(3)

. The fraction on the left-hand side and the infinite decimal on the right-hand side represent the same number. The reason that the decimal expansion of the fraction is recurring is simple: 1/3 cannot be expressed in the form P/Q, in which Q is a power of 10. In mathematics, you must be able to define the meaning of the expressions you use; you cannot step up to a mathematician and ask: "prove or disprove that !@#$% = ^&*|\". That is meaningless. The expression you used, 0.(0)1, likewise has no known meaning in mathematics. You cannot append digits after the end of an infinite sequence of digits in a decimal representation in any known meaningful way. If you insist on using the notation 0.(0)1, it is upon you to supply a definition of what this is supposed to mean. --Lambiam ^Talk 23:50, 6 May 2007 (UTC)[reply]

Ohanian saids "Eat this!"

According to you 1 = 0.(9) + 0.(0)1

Let U = 1

thus

U = 0.(9) + 0.(0)1

thus

10 U = 9.(9) + 10 * 0.(0)1

thus

10U - U = 9.(9) + 10 * 0.(0)1 - 0.(9) - 0.(0)1

thus

9 U = 9 + 9 * 0.(0)1

thus

U = 1 + 1 * 0.(0)1

And now we have both " 1 = 0.(9) + 0.(0)1 " and " U = 1 + 1 * 0.(0)1 "

But U = 1

so

0.(9) + 0.(0)1 = 1 + 1 * 0.(0)1

Which means

0.(9) = 1

Case Closed. Ohanian 08:02, 7 May 2007 (UTC)[reply]

Lambiam said that "you cannot append digits after the end of an infinite sequence of digits in a decimal representation in any known meaningful way." Has anyone bothered to ponder the consequences of decimal expansions indexed by ordinals? The def. of decimals would then have to use nets instead of sequence of partial sums, I suppose. Weird idea. nadav 08:22, 7 May 2007 (UTC) And the ω 'th partial sum would be an infinite limit...But I suppose you get a negative ordinal exponent, which doesn't make sense. Alternatively, I suppose you could define the new number system to consist of mappings from ordinals to {0,...,9}, along with an appropriate equivalence relation if necessary. nadav 08:35, 7 May 2007 (UTC)[reply]

You could consider these mappings, but can they be made to constitute a "number system"? The challenge is to turn this into an ordered field (where the order is the easy part) extending the usual real numbers. If you succeed in doing that, we know that the result can be embedded in the surreal numbers. --Lambiam ^Talk 12:29, 7 May 2007 (UTC)[reply]

The seemingly impossible problem when defining addition, for example, is how to carry the 1 from limit ordinal decimal places. If these "hyperdecimals" could somehow be made to work then it might be an alternative approach to hyperreals, but I doubt it's possible. nadav 18:54, 7 May 2007 (UTC)[reply]

I think perhaps the questioner is trying to understand the difference between an infinite sequence of decimal digits and a real number. Clearly as infinite sequences of digits, 0.(9) is different to 1.(0). But we have some ideas about what real numbers are, i.e. perhaps we consider them to be a complete ordered field or something. One way to talk about these numbers is by using infinite sequences of symbols, but we don't put these in one-to-one correspondence with the real numbers, in fact all rational numbers have two sequences of symbols that correspond to them. The reason is, that if we don't set up the correspondence in this way then we don't end up with something that corresponds with our idea of what real numbers should be. Trious 12:53, 11 May 2007 (UTC)[reply]