[2011-May-18] Mathness of 1
- Astrogirl
- so close, yet so far
- Posts: 2114
- Joined: Wed Aug 11, 2010 10:51 am
[2011-May-18] Mathness of 1
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.
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.
- zomgmouse
- ss No More!
- Posts: 862
- Joined: Tue Sep 28, 2010 12:38 pm
- Location: Melbourne, Australia
- Contact:
- Astrogirl
- so close, yet so far
- Posts: 2114
- Joined: Wed Aug 11, 2010 10:51 am
Re: Mathness of 1 [2011-May-18]
Is this ... an answer to my question?
- Felstaff
- XKCD spy
- Posts: 787
- Joined: Sun Jul 12, 2009 2:37 pm
Re: Mathness of 1 [2011-May-18]
HE MATHED AT YOU DIDN'T HE?!?
255 characters of free advertising space? I'm selling these line feather jackets...
- K^2
- Posts: 13
- Joined: Tue Mar 23, 2010 11:30 pm
Re: Mathness of 1 [2011-May-18]
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.Astrogirl wrote:Is this last thing really 1? It surely somehow draws from e^(i*pi) being -1, but how?
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]
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
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
- Astrogirl
- so close, yet so far
- Posts: 2114
- Joined: Wed Aug 11, 2010 10:51 am
Re: Mathness of 1 [2011-May-18]
Ah, that suddenly makes a lot of sense. Indeed an extra i crept in on me.
- K^2
- Posts: 13
- Joined: Tue Mar 23, 2010 11:30 pm
Re: Mathness of 1 [2011-May-18]
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.shining2k1 wrote:e^(Pi/2-Pi/2 - 2*Pi*n)=e^(-2Pi*n)=1 for all whole numbers n
Re: Mathness of 1 [2011-May-18]
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.
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]
Oh, no, Zach; do you know what kind of can of worms you've opened? http://tinyurl.com/3tnvqh5
- Kaharz
- This Intentionally Left Blank
- Posts: 1572
- Joined: Fri Feb 25, 2011 12:17 pm
Re: Mathness of 1 [2011-May-18]
In the pop up he did note that that 0^0=1 depends on your definition. Always read the footnotes...
Kaharz wrote:I don't need a title. I have no avatar or tagline either. I am unique in my lack of personal identifiers.
Re: Mathness of 1 [2011-May-18]
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)
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)
- Oldrac the Chitinous
- Chicken O' the Sea
- Posts: 3477
- Joined: Fri Dec 12, 2008 11:41 pm
- Location: The Perfect Stillness of the Deep
- Contact:
Re: Mathness of 1 [2011-May-18]
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.
Or maybe that's just me.
Police said they spent some time working out if they could charge the man with being armed with a weapon, as technically he was armed with part of a fish.
- Edminster
- Tested positive for Space-AIDS
- Posts: 8835
- Joined: Fri Sep 09, 2005 9:53 pm
- Location: Internet
- Contact:
Re: Mathness of 1 [2011-May-18]
You being hentai does bad things to my peace of mind.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.
ol qwerty bastard wrote:bitcoin is backed by math, and math is intrinsically perfect and logically consistent always
gödel stop spreading fud
Re: Mathness of 1 [2011-May-18]
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.
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.