
trekkiee
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How to get latex to display when browsing
trekkiee replied to trekkiee's topic in Feedback to Hypography
How do I change my IE6 browser running on windows XP to display the proper theme for the website, which I guess is the same as allowing the website to choose its own colors? Thx in advance -
How to get latex to display when browsing
trekkiee replied to trekkiee's topic in Feedback to Hypography
I am still using IE6 'cus of other unrelated isues, but I think this problems effects most browsers 'cus I've seen others complain of the same or a similar thing. On physicsforums.com some1 complained of a problem very similar to what I described, latex code displaying as nearly unreadable or unreadable white marks against a black background. That person described it as an anit-aliasing problem, which it might be, I don't know. But a forum mentor said that these images were stored that way on the server, implying that they were, perhaps, incorrectly formated when initially posted, and that one's browser can do nothing to properly display the image. Or not. I'm still not sure. -
Thx all for the replies. I should have known the problem was mathematica's [wolfram alpha] behavior when doing certain calculations.
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Hi, I'm sure this has been asked many times before, and I tried to find a sticky for this, but can some1 plz tell me the easiest way to get latex code to display properly when I'm browsing the math & physics forums? The code usually displays for me as unreadable white marks against a black gackground. I usually click on the code and then try and piece together what it's saying from the written LaTex code. Thx in advance.
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I noticed some unexpected behavior in the real-valued f(x)=(1+x)^1/x, as a function of real numbers, when plotting it on wolfram alpha. I inputed: plot (1+x)^1/x from x=-0.0000001 to x=0.0000001 and saw that it unexpectedly seemed to oscillate near zero. I took a closer look with: plot (1+x)^1/x from x=-0.00000000001 to x=0.00000000001 and saw that it definitely seems to oscillate near zero. My original rough graph on paper using a hand calculator suggested the curve was smooth near zero, and even windows calculator's 32 decimal places were unable to reveal the oscillation when I manually calculated many different values near zero. I don't think f(x) is smooth near zero, though, since d/dx[f(x)]=f(x)d/dx[1/xln(1+x)] so all of f(x)'s derivatives have a discontinuity at zero. Since Leonhard Euler showed limf(x)=e as x approaches zero, and this limit is the same as you approach zero from the left or from the right, then the discontinuity of f(x) at zero is removable and the function f(x)=(1+x)^1/x, x not = 0 f(0)=0 x=0 is continous. But the discontinuities in the derivatives don't appear to be removable. Each derivative will have a term containing the factor 1/x (not the 1/x in the exponent of f(x)) which cannot be removed. E.g. d/dx[f(x)]=f(x)[(1/x)(1/(1+x)+(-1/x^2)(ln(x+1))] the 1/x in the 1st term is not removable. I'm not saying it doesn't oscillate near zero, I'm just saying that this is unexpected. I need confirming opinions and confirming logic to explain it so I can believe it. So the questions are: 1. Does f(x) oscillate near zero? 2. Why does f(x) oscillate near zero? 3. Why does this behavior not appear until |x|<~10^-7? 4. Is it a glitch in Mathimatica? [doubt it] 5. If it doesn't oscillate [which it probably does] then how is the behavior of the derivatives explained near zero? 6. If it oscillates, then why is it so hard to replicate this oscillation manually on a calculator? Note#1: This behavior concerns the real-valued f(x). I don't think Re[f(z)] or Im[f(z)], z complex, are of interest here. Note#2: This behavior concerns (1+x)^1/x. I dont think either (1+1/x)^x or lim(1+1/x)^x as x goes to infinty are of interest here. Thx in advance and Kudos to the person with the explanation!
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Mean nearest neighbor distance in 3d
trekkiee replied to trekkiee's topic in Physics and Mathematics
aswoods at S.O.S. Mathematics CyberBoard :: View topic - mean nearest neighbor in 3d helpfully pointed out that Gradshteyn and Ryzhik 3.381.10, and Wolfram Alpha agree that the integral from 0 to infinity of x^3 e^(-ax^3)dx = 1/3 gamma(4/3) a^(-4/3) so the mean nearest neighbor distance in 3d is: <r> = [4 pi rho] integral from 0 to infinity r^3 e^(-[4/3] pi rho r^3) dr <r> = 1/3 gamma(1/3) ([4/3] pi)^(-1/3) rho^(-1/3) <r> = 0.55396 rho^(-1/3) <r> = 3.93 light-years for the 23 star systems within 12.5 ly :) -
Mean nearest neighbor distance in 3d
trekkiee replied to trekkiee's topic in Physics and Mathematics
since what we really want is the average distance to the nearest star system, and atlasoftheuniverse.com reports 23 star systems within 12.5 ly (35 stars but 3 trinaries, 6 binaries, and 14 singles), we have: rho = 23/([4/3]pi 12.5^3) rho = 0.0028113 star systems/cubic light-year rho is only an estimate since some of the star systems near the outer edge of the volume (of 12.5 ly radius) might have nearest neighbors outside the volume and some star systems just outside the volume might have nearest neighbors inside. since the integral of x^3 exp(-a x^3)dx doesn't seem to be integrable, I used people.hofstra.edu/stefan_waner/RealWorld/integral/integral.html to numerically integrate: the integral from 0 to infinity of 0.035328(x^3)(e^(-0.011776(x^3)))dx = 3.93 light years as the average distance from an arbitrary star system in the solar neighborhood to its nearest nighbor :) -
Mean nearest neighbor distance in 3d
trekkiee replied to trekkiee's topic in Physics and Mathematics
Modest at http://hypography.com/forums/physics-mathematics/21509-mean-nearest-neighbor-distance-3d.html was helpful in pointing me to this link: Probability, statistical optics, and ... - Google Books And now I need to integrate: integral of x^3 exp(-a x^3) dx, with a = constant, but I couldn't. Hopefully, it's an easy integral and and someone will figure it out. In a 3-dimensional random distribution, the basic idea for finding the average distance from any given particle to its nearest neighbor begins with: P®dr = [1 - integral from 0 to r of P®dr][4 pi r^2 pho dr] (1) where pho = average number of particles/unit volume P® dr = the probability of a particle's nearest neighbor occurring in the interval [r,r+dr] integral from 0 to r of P® dr = probablity that an arbitrary particle's nearest neighbor lies within a distance r of the particle 1 - integral from 0 to r of P® dr = the probability that the nearest neighbor is no closer than r. differentiating & separating eq. 1: dP/P = [2/r - 4 pi rho r^2] dr integrating: P = [constant] r^2 exp(- [4/3] pi rho r^3) normalizing: 1 = integral from 0 to infinity of P® dr 1 = [constant] integral from 0 to infinity of r^2 exp(-[4/3] pi rho r^3) dr 1 = [constant] [-[1/(4 pi rho)] exp(-[4/3] pi rho r^3)] evlauated from 0 to infinity gives the constant = 4 pi rho and P® = 4 pi rho r^2 exp(-[4/3] pi rho r^3) The average distance from any given particle to its nearest neighbor in 3 dimensions is then the expectation value of r: <r> = integral from 0 to infinity of r P® dr <r> = [4 pi rho] [integral from 0 to infinity of r^3 exp(-[4/3] pi rho r^3) dr] I was unable to do the last integral, but I'm sure someone can :) -
Hi. In 3 dimensional Euclidean space with the usual metric, d=[(delta x)^2+(delta y)^2+(delta z)^2]^1/2, I'm trying to figure out the average distance between nearest neighbors in a randomly distributed sample of particles. My best initial guess for the average distance from any given particle to its nearest neighbor is d_nearest neighbor_mean=(volume/n)^1/3 where n particles are randomly distributed in a 3 dimensional volume. The question originated when I wondered what was the average distance between stars in the solar neighborhood. atlasoftheuniverse.com gives 35 stars (including the Sun) within 12.5 light-years, and the above formula yields 6.16 ly as the avg distance from any given star to its closest neighbor. This seemed a little high to me, since the distance from the Sun to its nearest neighbor (Proxima Centauri) is 4.4 ly. But perhaps the Sun has a closer-than-avg nearest neighbor, since, after all, the distribution should be very close to random. Let us assume that the stars are randomly distributed. I originally thought it would be easy to figure this out, but after trying unsuccessfully for an hour to work out a better formula, then another hour trying to google one, I gave up. Thanks in advance :(
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i'm confused about how COBE (the Cosmic Background Explorer) measured the frequency peak of the Cosmic Microwave Background (CMB). the following seems clear to me and relatively straightforward: for black body radiation, i.e., for an ideal photon gas in local thermodynamic equilibrium with matter, e.g., the surface of last scattering of the CMB, the spectral radiance, I_nu (T) = [2h/c^2][nu^3/(exp(h nu/[k T])-1)] or I'_lamba (T) = [2hc^2][1/lamba^5(exp(h c/[lambda k T])-1)] <forgive my clumsy notation, i don't know how to do LaTeX notation>, peaks at nu_max or lambda_max, respectively, such that c/lambda_max = 1.76*nu_max. although the lambda_max peak is the actual energy peak at which a black body radiates the maximum energy [photons at lambda_max are 1.76 times as energetic as photons at nu_max], radio astronomers seem to prefer the frequency peak. in fact, doing a google image search of "cosmic microwave background," then picking out spectral graphs [spectral radiance vs frequency or wavelength], will find almost exclusively graphs depicting the frequency peak. examples include: http://map.gsfc.nasa.gov/media/ContentMedia/990015b.jpg http://www.phy.duke.edu/~kolena/cmbspectrum1.gif which show the frequency peak of 384 MJy/sr = 384 megaJanskys per steradian = 3.84e-18 W/(m^2-Hz-sr) = I_nu (2.725 K) at 1.87mm (160 GHz). File:Firas spectrum.jpg - Wikipedia, the free encyclopedia which shows the frequency peak of 1.15e-4 erg/(s-cm^2-cm^-1-sr) = 3.84e-18 W/(m^2-Hz-sr) = I_nu (2.725 K) at 1.87mm = 5.34 cm^-1 (160 GHz). my understanding is that the FIRAS interferometer on board COBE compared the spectral radiance of the CMB to an on-board black body. what is NOT clear to me is exactly how these measurements were made. i would think that if u measured the CMB's [or any black body's] spectral radiance, u would measure the wavelength peak, not the frequency peak, since the wavelength peak is the physical maximum, that is, the point on the EM spectrum where the black body radiates the maximum energy. the frequency peak seems to me to have no physical significance and to only be a mathematical tool. i don't see how physical measurements can measure any energy peak other than the wavelength peak. i would very much appreciate it if some1 could clear this up for me. thx in advance. ö¿ö¬ E=mc² ~