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

[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)

 

[mathwonk]

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

 

[Hurkyl]

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

 

[Char. Limit]

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.

 

[Hurkyl]

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.

Spoiler

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

 

[Char. Limit]

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.

 

[Hurkyl]

Huh? I was talking about the cosine one.

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.

Spoiler

cos x = Re{ exp(i x) }

 

[coolul007]

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

 

[DonAntonio]

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

 

[jacobrhcp]

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

 

[SHISHKABOB]

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 ;)

 

[Borek]

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.