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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)
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)
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)
\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}
\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)
Note for Lambiam: there are various online LaTeX compilers for when you don't have it handy. --TotoBaggins 21:02, 6 May 2007 (UTC)
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 . Isnt but actually and . 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 then ? If you cant use to justify then . 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).
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)
Ohanian saids "Eat this!"
According to you 1 = 0.(9) + 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
Which means
0.(9) = 1
Case Closed. Ohanian 08:02, 7 May 2007 (UTC)
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)
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)