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Singularity Functions

ThemThem
edited 2011-11-24 18:51:05 in IJAM
Good people don't end up here.
How come we didn't learn this until now?!

Comments

  • MadassAlexMadassAlex
    One foot in front of the other, every day.
    If this has anything to do with Tang, I'll be extremely upset.
  • ThemThem
    Good people don't end up here.
    ...I'm confused. Tang?

    No, a singularity function is a high-class party in a black hole a mathematical equation with a special set of rules so that you can construct discontinuous functions while still being able to integrate and differentiate them!
  • MadassAlexMadassAlex
    One foot in front of the other, every day.
    NGE joke.

    Continue unmolested.
  • MillerCrossMillerCross
    You can change. You can.
    Does this involve people turning into ta--

    >Alex's post

    welp

    fuck
  • ForzareForzare
    Till shade is gone, till water is gone, into the Shadow with teeth bared, screaming defiance with the last breath, to spit in Sightblinder’s eye on the last Day.
    I didn't understand a word of what this topic is actually about, so insert your favorite overused tang joke here.
  • MillerCrossMillerCross
    You can change. You can.
    Poor Them. He doesn't know this place is full of inbred hicks who don't understand math. 
  • ThemThem
    Good people don't end up here.
    I've determined the optimal way to express my displeasure. With science. You monsters.

    :Þ
  • StormtroperStormtroper
    > Singularity functions

    I know, right? They make working with beams so much fun.

    Also, the table in that page is wrong, isn't it? Where'd those denominators come from?
  • IcalasariIcalasari
    No rainbow star
    I don't understand the math or the tang jokes
  • StormtroperStormtroper
  • ThemThem
    Good people don't end up here.
    Win!

    ...I think the table is only wrong in the n >= 0 row, actually. The denominators are the result of integrating x^n --> (x^(n+1))/(n+1), but in the last row the denominator and the exponent don't match.
  • StormtroperStormtroper
    No, I mean, unless it's different in solid mechanics than elsewhere, \langle x-a\rangle ^n=  (x - a)^n for x >= a, and 0 for x < a, with no denominator anywhere (or anything resembling the < 0 rows, but Dirac deltas work in mysterious ways so whatevs) and is not the result of integrating anything. I'd guess someone mistakenly inserted a list of integrals/derivatives as the definition (and got at least the last one wrong).
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