Great xkcd today

1. Apr 25, 2012

flatmaster

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. Apr 25, 2012

mathwonk

Re: xkcd

wouldn't that identity imply that pi is an algebraic number?

3. Apr 25, 2012

Hurkyl

Staff Emeritus
Re: xkcd

Pro-tip: the following two phrases are not synonymous:
• not all of these are wrong
• all of these are right

4. Apr 25, 2012

Char. Limit

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.

5. Apr 25, 2012

Hurkyl

Staff Emeritus
Re: xkcd

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

6. Apr 25, 2012

Char. Limit

Re: xkcd

Huh? I was talking about the cosine one.

7. Apr 25, 2012

Hurkyl

Staff Emeritus
Re: xkcd

So was I. 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) }

8. Apr 26, 2012

coolul007

Re: xkcd

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

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

9. Apr 26, 2012

DonAntonio

Re: xkcd

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

DonAntonio

10. Apr 26, 2012

jacobrhcp

Re: xkcd

So is there a simple identity for $\sum_{n}n^{-n}$?

11. Apr 26, 2012

SHISHKABOB

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 ;)

12. Apr 27, 2012

Re: xkcd

Done.