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For best results in downloading the Destiny Beta to your Xbox One, don't launch the game until the download is complete. The Destiny Beta FAQ has been updated with info on how to troubleshoot your Xbox One download (scroll down). http://www.bungie.net/en/View/Bungie/DestinyBetaFAQ If encountering Party Stabilization errors on Xbox One: try turning off your console, fully unplugging, then plugging in and trying again.
2014-07-23T06:50:45Z

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2/13/2012 5:27:40 AM
Whence one might presume that the integral, from the negative of infinity towards the infinite, of the simple function exp(-x^2) might equate to the root of pi? I shall set forth to prove to you this assertion, that you should be struck by its audacity as much as I on this Sunday's eve. We begin by considering the square, such that the integral under examination would be the product of an integral of exp(-x^2) and of an integral of exp(-y^2), for {x:-r,r} and {y:-r,r}. As we should desire r to tend toward an infinite quantity, it necessarily must be that we have, indeed, a product of two limits. As one must recall from elementary calculus, Lim(r->oo) int[-r,r]exp(-x^2)dx * Lim(r->oo) int[-r,r]e^(-y^2)dy is equivalent to Lim(r->oo) { int[-r,r]exp(-x^2) int[-r,r]exp(-y^2)dy } One may then be struck with an occurrence: a theorem of Fubini may be applied to conjoin the integration. The result follows as Lim[r->oo] int[(-r <= x <= r), (-r <= y <= r)]exp(-(x^2+y^2))dxdy This is, of course, the same as int[R^2]exp(-(x^2+y^2))dxdy; we are therefore not constrained by a square modicum of integration, and indeed we should choose a circular region to evaluate this integral. Thus, we have the result Lim(q->oo) int[x^2+y^2<=q^2]exp(-(x^2+y^2))dxdy When considering circular regions it is most convenient to use a polar co-ordinate system, thus we apply the angular transformation to result in Lim(q->oo) int[0,q]int[0,2pi]q*exp(-q^2)dtdq Which can be evaluated by elementary calculus techniques to yield Lim(q->oo) (2pi)*(1 - exp(-q^2))/2 = pi. We recall that we have calculated the square of the integral of interest. Thus, it follows by taking the principal root that the infinite-integral of exp(-x^2) is the root of pi. [url=http://www.wolframalpha.com/input/?i=integrate+e%5E(-x%5E2)+from+x+%3D+-infinity+to+infinity]tl; dr[/url] [Edited on 02.12.2012 11:49 PM PST]
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