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[2011-May-18] Mathness of 1

PostPosted: Wed May 18, 2011 11:13 am
by Astrogirl
http://www.smbc-comics.com/index.php?db=comics&id=2249

Is this last thing really 1? It surely somehow draws from e^(i*pi) being -1, but how?

I can't even read properly where is the base and where is the exponent. Either it says e^(i*pi/2) * i^i or it says e^(i*pi/2 * i^i).

i^i has infinitely many results: e^(-Pi/2 - 2*Pi*n) with n an integer. This is where I get lost.

Re: Mathness of 1 [2011-May-18]

PostPosted: Wed May 18, 2011 11:21 am
by zomgmouse
(e^[pi*i]) +2 = 1

Re: Mathness of 1 [2011-May-18]

PostPosted: Wed May 18, 2011 11:28 am
by Astrogirl
Is this ... an answer to my question?

Re: Mathness of 1 [2011-May-18]

PostPosted: Wed May 18, 2011 11:39 am
by Felstaff
HE MATHED AT YOU DIDN'T HE?!?

Re: Mathness of 1 [2011-May-18]

PostPosted: Wed May 18, 2011 12:00 pm
by K^2
Astrogirl wrote:Is this last thing really 1? It surely somehow draws from e^(i*pi) being -1, but how?

e^(pi/2) * i^i = e^(pi/2) * e^(i*ln(i)) = e^(pi/2) * e^(i * i * pi/2) = e^(pi/2 - pi/2) = 1.

And yes, there are multiple roots for e^x = i, but only i*pi/2 is used here. I can't quite think of a good argument for that just now.

Re: Mathness of 1 [2011-May-18]

PostPosted: Wed May 18, 2011 12:37 pm
by shining2k1
Hey Astrogirl,

you smuggled an extra i into your first expression (force of habit, I guess). With e^(pi/2) *i^i instead of e^(i*pi/2) *i^i, the whole thing resolves to
e^(Pi/2-Pi/2 - 2*Pi*n)=e^(-2Pi*n)=1 for all whole numbers n

Re: Mathness of 1 [2011-May-18]

PostPosted: Wed May 18, 2011 1:36 pm
by Astrogirl
Ah, that suddenly makes a lot of sense. Indeed an extra i crept in on me.

Re: Mathness of 1 [2011-May-18]

PostPosted: Wed May 18, 2011 1:39 pm
by K^2
shining2k1 wrote:e^(Pi/2-Pi/2 - 2*Pi*n)=e^(-2Pi*n)=1 for all whole numbers n

e^(-2i*Pi*n) = 1 for all integer n, but what you actually get is only 1 for n=0. That doesn't go away on its own, and has to be part of definition of imaginary exponents.

Re: Mathness of 1 [2011-May-18]

PostPosted: Wed May 18, 2011 1:49 pm
by orionsbelt
No one is going to point out that the caption says "Mathematicians are no longer allowed to sporting events"?
Sorry guys. I was a math major but I was also a writing minor. Awkward wording pops out to me. At me. Whatever you prefer.

Re: Mathness of 1 [2011-May-18]

PostPosted: Wed May 18, 2011 2:09 pm
by JS
Oh, no, Zach; do you know what kind of can of worms you've opened? http://tinyurl.com/3tnvqh5

Re: Mathness of 1 [2011-May-18]

PostPosted: Wed May 18, 2011 3:27 pm
by Kaharz
In the pop up he did note that that 0^0=1 depends on your definition. Always read the footnotes...

Re: Mathness of 1 [2011-May-18]

PostPosted: Wed May 18, 2011 4:56 pm
by Mr. Pedantic
It is multi-valued.
e^(Pi/2) i^i = e^(Pi/2) e^(i ln i) = e^(Pi/2) e^(i (ln |i| + i arg(i))) = e^(Pi/2) e^(- Pi/2 - 2 n Pi) = e^(- 2 n Pi)
for any integer n. There is no 'i' term in the exponent, so this is not a single value. For n=0, we get 1 (and indeed, this would be the value associated with the principal branch of the complex logarithm), but it's just as reasonable to claim that this sign says "We're #e^(16 Pi)", (approximately 6.7 x 10^21)

Re: Mathness of 1 [2011-May-18]

PostPosted: Wed May 18, 2011 6:01 pm
by Oldrac the Chitinous
You know, if you just glance at the comic quickly, it looks kind of like the last sign says "We're Hentai."

Or maybe that's just me.

Re: Mathness of 1 [2011-May-18]

PostPosted: Wed May 18, 2011 6:16 pm
by Edminster
Oldrac the Chitinous wrote:You know, if you just glance at the comic quickly, it looks kind of like the last sign says "We're Hentai."

Or maybe that's just me.

You being hentai does bad things to my peace of mind.

Re: Mathness of 1 [2011-May-18]

PostPosted: Wed May 18, 2011 6:19 pm
by Tom Swift
Although it's a reasonable gripe that i^i has infinitely many possible values, I'm more concerned that the guy in the middle thinks 0^0 is 1. I think that maybe the mathematicians had a buddy from the physics department come along.

The value of 0^0 is "undefined". It's a classic trick question about double limits (limits of limits). It isn't 1: think about the limit of the limit of f(u,v) = u^v as both u and v go to 0. If you take the limit as u goes to 0 first (with v not 0) then that limit is 0. Do it the other way around and take the limit as v goes to 0 first and you get 1. So you can approach 0^0 two different ways and get two different answers for what it should be. If 0^0 really had a value, you couldn't get two different limits for it.