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Great xkcd today

  1. Apr 25, 2012 #1
    Great xkcd today

    http://xkcd.com/

    At the bottom, he claims some identities expressing some roots in terms of pi.

    Is he right? What's the easiest way to prove these identities?

    Sqrt(2) = 3/5 - pi/(7-pi)
     
    Last edited by a moderator: Apr 27, 2012
  2. jcsd
  3. Apr 25, 2012 #2

    mathwonk

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    Re: xkcd

    wouldn't that identity imply that pi is an algebraic number?
     
  4. Apr 25, 2012 #3

    Hurkyl

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    Re: xkcd

    Pro-tip: the following two phrases are not synonymous:
    • not all of these are wrong
    • all of these are right
     
  5. Apr 25, 2012 #4

    Char. Limit

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    Re: xkcd

    Pro-tip: Only one of those "identities" is true. Bonus points if you figure out which it is.

    I'm still trying to figure out HOW said identity is true. It's really kind of cool.
     
  6. Apr 25, 2012 #5

    Hurkyl

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    Re: xkcd

    It's a geometric series, if you can figure out what the complex part of the numbers are supposed to be.
     
  7. Apr 25, 2012 #6

    Char. Limit

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    Re: xkcd

    Huh? I was talking about the cosine one.
     
  8. Apr 25, 2012 #7

    Hurkyl

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    Re: xkcd

    So was I. :biggrin: IIRC you can also work it out with induction and the sum-of-cosines formula, but the trick to turn it into a geometric series is faster. Of course, it's a bit annoying extracting the answer after you use the trick. There's another trick you can do that I think works out, but I haven't bothered fleshing it out.
    cos x = Re{ exp(i x) }
     
  9. Apr 26, 2012 #8
    Re: xkcd

    Squaring both sides results in the following equation which is false:

    this is not even close
    364 π+ 14 π^2 = 2009
     
  10. Apr 26, 2012 #9
    Re: xkcd



    Yes, and mathwonk first noted it without the operations: it'd imply [itex]\pi[/itex] is rational, which is false, of course.

    DonAntonio
     
  11. Apr 26, 2012 #10
    Re: xkcd

    So is there a simple identity for [itex]\sum_{n}n^{-n}[/itex]?
     
  12. Apr 26, 2012 #11
    Re: xkcd

    could I suggest linking to http://xkcd.com/1047/ instead, so that people on friday and so on don't become very confused reading this thread ;)
     
  13. Apr 27, 2012 #12

    Borek

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    Re: xkcd

    Done.
     
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