## [2011-May-18] Mathness of 1

Blame Quintushalls for this.

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Astrogirl
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### [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.
Microaggression? Microaggression!

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

(e^[pi*i]) +2 = 1
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Astrogirl
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### Re: Mathness of 1 [2011-May-18]

Is this ... an answer to my question?
Microaggression? Microaggression!

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

HE MATHED AT YOU DIDN'T HE?!?
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K^2
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### Re: Mathness of 1 [2011-May-18]

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.

shining2k1

### 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

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

Ah, that suddenly makes a lot of sense. Indeed an extra i crept in on me.
Microaggression? Microaggression!

K^2
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### Re: Mathness of 1 [2011-May-18]

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.

orionsbelt

### 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.

JS

### 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
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### 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...
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Mr. Pedantic

### 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)

Oldrac the Chitinous
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### 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.
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### Re: Mathness of 1 [2011-May-18]

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.
ol qwerty bastard wrote:bitcoin is backed by math, and math is intrinsically perfect and logically consistent always