[2011-May-18] Mathness of 1

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

Post by K^2 »

Hey. We, Physicists, know perfectly well that 0^0 is undefined. We only use it as a short-hand in certain general formulae. It's shorter than writing "x^y, unless both are zero, in which case, read it as 1." And it just happens to work out this way more often than not.

WK1
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Re: Mathness of 1 [2011-May-18]

Post by WK1 »

I also came to say that 0^0 is undefined, but I have an simpler explanation. 0^(anything else) is 0. (anything else)^0 is 1. 0^0 can't be both 0 and 1, it is undefined.

Zach Weiner should take a pre-algebra class.

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

Post by Edminster »

So do none of you guys know that if you hover over the red button you get a further comic? Because Zach already knows about this stuff and it makes you look kinda retarded to go over the same material over and over.
ol qwerty bastard wrote:bitcoin is backed by math, and math is intrinsically perfect and logically consistent always

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Ben G

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

Post by Ben G »

Although Tom Swift is right in stating that basic calculus says that 0^0 should not have a value (since no value would make the function x^y continuouss), the fact remains that the accepted mathematical convention is to let 0^0 be 1.

http://www.askamathematician.com/?p=4524

While the definition of 0^0=1 doesn't necessarily square (that is, hypocube) to make x^y continuous, it does implicitly create a contradiction and apparently makes several other mathematical definitions/formulae simpler.

Ben G

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

Post by Ben G »

edit:

"...doesn't necessarily square (that is, hypocube) WITH ATTEMPTS to make x^y continuous, it does NOT implicitly create a contradiction...."

hooray for proofreading! I think we're good here.

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

Post by Astrogirl »

K^2 wrote:
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.
How does this have to be part of the definition of imaginary exponents? How would you decide which n you use? Don't say 0, I could just shift the index.
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Roman Cilicia
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Re: Mathness of 1 [2011-May-18]

Post by Roman Cilicia »

this is so ghey

what are we, xkcd?

okay, regarding the comic (get it? I was talking about this stupid layout):

this is so gay

what is this, xkcd?

Ben G

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

Post by Ben G »

Astrogirl wrote:
>
K^2 wrote:
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.
> How does this have to be part of the definition of imaginary exponents? How would
> you decide which n you use? Don't say 0, I could just shift the index.
>

Okay, so a clarification of the ambiguity involved (for anyone who cares):

Euler's formula is that e^(ix) = cosx + isinx (e is Euler's constant, that is 2.71282..., i is the square root of -1 and pi is the ratio of circumference to diameter, that is 3.14159...). This means that e^(2pi*i) = cos(2pi) + isin(2pi) = 1. In fact, we can equivalently state that 1^n = [e^(2*pi*i)]^n = e^(2*pi*i*n).
So, now we have infinitely many ways of expressing any number by multiplying or dividing by 1 as many times as one pleases.
So, for example, "i" can be expressed as e^(pi*i/2), but it can also be expressed as e^(pi*i/2) * 1^n = e^(pi*i/2 + 2*pi*i*n), where n is any integer.
So far so good?
While the exponent of the expression for i has infinitely many potential values (corresponding to the choice of "n") there is only 1 value this expression represents, since we've been multiplying it by powers of 1. However, things change when we take the "i"th power of this expression:

i^i = [e^(pi*i/2 + 2*pi*i*n)] = e^(-pi/2 - 2*pi*n); n is any integer

this expression has infinitely many (non-equal) values, each of which is equally an answer to "what is i^i?" In fact, this is generally true when you take the complex or irrational power of a complex number.
If we were to stick with this convention, however, we would be stuck with a multivalued function, which in the context of practical computation can be unwieldy. So, often a single member from this solution set is chosen, and the one chosen is usually the one for which "n=0," which can mean different things in different contexts. In this context, we state that i^i = e^(-pi/2) is a representative, useful, selection from our solution set.

Guest

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

Post by Guest »

You could say e^{\pi/2}i^i is a set valued mapping, so maybe the rightmost mathematician is saying "We're the cardinality of the natural numbers!"

Guest

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

Post by Guest »

Edminster wrote:So do none of you guys know that if you hover over the red button you get a further comic? Because Zach already knows about this stuff and it makes you look kinda retarded to go over the same material over and over.
Thank you for pointing that out, Edminister. However, that only works if you have javascript enabled. And it is still hidden. Plus, it doesn't really depend on definition, it is undefined, there is no other reasonable mathematical explanation.

The fact that Weiner even has a little thing that requires javascript, without any note specifying so, shows that he sucks at HTML too. Although the 84 errors say the same thing.

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

Post by Kimra »

So, apparently when I look at this tread it just looks like a whole lot of people drawing incomprehensible smiley faces.
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Re: Mathness of 1 [2011-May-18]

Post by Rileyc »

Mathematicians are also not allowed to not wear glasses, apparently!

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

Post by Felstaff »

It's against the law for any scientist to be out of their labcoat at any time.
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Edminister
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Re: Mathness of 1 [2011-May-18]

Post by Edminister »

Guest wrote:
Edminster wrote:So do none of you guys know that if you hover over the red button you get a further comic? Because Zach already knows about this stuff and it makes you look kinda retarded to go over the same material over and over.
Thank you for pointing that out, Edminister. However, that only works if you have javascript enabled. And it is still hidden. Plus, it doesn't really depend on definition, it is undefined, there is no other reasonable mathematical explanation.

The fact that Weiner even has a little thing that requires javascript, without any note specifying so, shows that he sucks at HTML too. Although the 84 errors say the same thing.
I wasn't the one that pointed it out, but thanks for giving me credit?

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

Post by Astrogirl »

The clones, the clones are a-coming!
Microaggression? Microaggression!

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