Can you prove the identities for sqrt(2) and pi from the xkcd comic?

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The xkcd comic presents identities relating square roots to pi, specifically questioning the validity of the identity sqrt(2) = 3/5 - pi/(7-pi). It is discussed that if this identity were true, it would imply that pi is an algebraic number, which is false. Only one of the identities presented in the comic is correct, and participants are exploring methods to prove its validity, including geometric series and induction. Squaring the proposed identity leads to a false equation, further supporting its inaccuracy. The conversation highlights the complexities of these mathematical identities and encourages further exploration.
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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)
 
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wouldn't that identity imply that pi is an algebraic number?
 


flatmaster said:
At the bottom, he claims some identities expressing some roots in terms of pi.
Pro-tip: the following two phrases are not synonymous:
  • not all of these are wrong
  • all of these are right
 


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.
 


Char. Limit said:
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.
It's a geometric series, if you can figure out what the complex part of the numbers are supposed to be.
 


Hurkyl said:
It's a geometric series, if you can figure out what the complex part of the numbers are supposed to be.

Huh? I was talking about the cosine one.
 


Char. Limit said:
Huh? I was talking about the cosine one.
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) }
 


flatmaster said:
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)

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

this is not even close
364 π+ 14 π^2 = 2009
 


coolul007 said:
Squaring both sides results in the following equation which is false:

this is not even close
364 π+ 14 π^2 = 2009



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

DonAntonio
 
  • #10


So is there a simple identity for \sum_{n}n^{-n}?
 
  • #11


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 ;)
 
  • #12


SHISHKABOB said:
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 ;)

Done.
 
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