# What is the answer?



## sossego (Jan 15, 2014)

∞√(ω)/{R}


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## throAU (Jan 15, 2014)

42.


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## fonz (Jan 15, 2014)

throAU said:
			
		

> 42.


For sufficiently small values of ω, that is :OOO


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## JanJurkus (Jan 15, 2014)

fonz said:
			
		

> For sufficiently small values of ω, that is :OOO



Or for small values of infinity...  :beergrin 

ω is a bit misleading, isn't it? I had to think about angles, and expected ω to be within 0 and 2*pi. That's probably wrong anyway, it's been a while. Anyway, I think the answer is 1.


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## fonz (Jan 16, 2014)

JanJurkus said:
			
		

> ω is a bit misleading, isn't it? I had to think about angles, and expected ω to be within 0 and 2*pi.


Actually, I initially thought of the (principal) cube root of unity: -1/2 + (√3)/2 i  :h


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## igorino (Jan 17, 2014)

Sometimes it is more useful to know the right question to the/an answer.


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## SirDice (Jan 17, 2014)

Is it something to do with angular momentum? I'm not sure what {R} would represent though.


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## phoenix (Jan 17, 2014)

It took me a long time staring at that before I realised it was an actual mathematical equation and not pictographs talking about the infinite beauty of boobs and their relation to whatever {R} was supposed to represent.


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## JanJurkus (Jan 17, 2014)

Uh, I think {R} is the set of real numbers. Thus, ω can take any value, as long as it is a real number. It's not very clear whether it's (sqrt (ω))*inf or (sqrt (ω))^(1/inf) either. (You can write sqrt (2) also as 2^(1/2).)


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## fonz (Jan 18, 2014)

phoenix said:
			
		

> It took me a long time staring at that before I realised it was an actual mathematical equation and not pictographs talking about the infinite beauty of boobs and their relation to whatever {R} was supposed to represent.


Actually, it's not an equation: there's no = sign. I'm feeling increasingly inclined to think that your initial hunch was correct: it's probably some kind of visual joke.


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## sossego (Jan 18, 2014)

Having an equals/equal to sign "=" is not always necessary for an equation. Such as ³√27 has the answer of 3. We understand that the problem did not require "=" to be answered.
Now, the question is "{ What is the infinite root}∞√ [of the value {of the least infinite ordinal}(ω)/{divided by any number from the set of real numbers}{R}]?"


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## fonz (Jan 18, 2014)

sossego said:
			
		

> Having an equals/equal to sign "=" is not always necessary for an equation. Such as ³√27 has the answer of 3. We understand that the problem did not require "=" to be answered.


An equals sign is not necessary for a _question_, but it sure is for an _equation_, which was the word @phoenix used  :stud 



			
				sossego said:
			
		

> Now, the question is "{ What is the infinite root}∞√ [of the value {of the least infinite ordinal}(ω)/{divided by any number from the set of real numbers}{R}]?"


When infinity is involved one usually needs to speak in terms of limits. In any case, I think there isn't one definite answer:

The infinite root of a positive (real) number is/approaches 1.
The infinite root of 0 is still 0.
The infinite root of a negative (real) number probably isn't even defined because it doesn't converge.

P.S. I'm somewhat disappointed that it wasn't a joke  P


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## sossego (Jan 18, 2014)

You could always make it into a joke. 

I was thinking along the same lines with the answers being (+1, -1, 1, ±1) and (+0, -0, 0, ±0). Even roots with negative values weren't taken into consideration outside of the " Does  _4 √i_ have a  value of (+, -, ±, unsigned) or not?"

The above: If the root value is even, then a negative value would be hard to determine; yet, if the root is odd then the value would be dependent upon the symbol. Yes, I corrected my mistake.

Hmmm, I was trying to find a formula to describe a qubit.


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## fonz (Jan 18, 2014)

sossego said:
			
		

> Does  _4 √i_ have a  value of (+, -, ±, unsigned) or not?"


As you probably know, any complex number _z_ (except zero) has exactly _n_ (all distinct) complex _n_-th roots. In the complex plane these lie on a circle centred at the origin with radius _n_√|_z_| (i.e. the _n_-th root of the absolute value of _z_, which is obviously a positive real number), spaced at angles that are 2π/_n_ apart, with the argument (angle in polar representation) of the principal root being the argument of _z_ itself divided by _n_. So the four quartic roots of _i_ are e^_i_φ = cos(φ) + _i_ sin(φ) where φ is in the set {π/8, 5π/8, 9π/8,13π/8}. Note that none of those quartic roots are either real or purely imaginary, they really are properly complex (i.e. both real and imaginary parts are non-zero).

However, this only applies when _n_ is a finite natural number (and for convenience let's disregard 0 and 1). Once you start throwing infinity into the mix things get screwed up royally. The modulus of the roots converges to 1 (except when _z_ equals zero) but the arguments of the roots diverge. They could essentially be anything (i.e. the entire circle), but raising such a root to infinite power means multiplying the argument by ∞, which means the argument becomes ∞, which is indeterminate modulo 2π  :q When _z_ is real and positive, the *principal* root will always be likewise (the argument remains zero after all), but beyond that it escalates.



			
				sossego said:
			
		

> Hmmm, I was trying to find a formula to describe a qubit.


That sounds interesting, so feel free to elaborate. You might want to have a look at Bloch spheres if you hadn't already.

P.S. I can't blame Freddie (@phoenix) for thinking of certain squishy lady parts when you wrote (ω)  :e


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