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        riddles >> general problem-solving / chatting / whatever >> An Exponential Curiosity
        
(Message started by: Sir Col on Jan 29th, 2005, 10:19am)
Title: An Exponential Curiosity
Post by Sir Col on Jan 29th, 2005, 10:19am
Clearly x2/6 = x1/3, or does it?

As x2/6 = (x1/6)2 = (x2)1/6...

(-1)1/3 = -1, (1[pm][smiley=i.gif][surd]3)/2 [three values]
((-1)1/6)2 = -1, (1[pm][smiley=i.gif][surd]3)/2
However, ((-1)2)1/6 = -1, (1[pm][smiley=i.gif][surd]3)/2, but also 1, -(1[pm][smiley=i.gif][surd]3)/2 [six values]

Consider the two graphs below: (i) y = x1/3, (ii) y = (x2)1/6.

Spot the difference?  ;)



[e]Thanks for the correction, rmsgrey.[/e]
Title: Re: An Exponential Curiosity
Post by rmsgrey on Jan 29th, 2005, 2:12pm

on 01/29/05 at 10:19:29, Sir Col wrote:
((-1)1/6)2 (has six complex values)

I get [pm]i6=-1 and that:
((-1)1/6)2=-1 (and 2 complex values)=(-1)1/3
Title: Re: An Exponential Curiosity
Post by Icarus on Jan 29th, 2005, 9:02pm
Impudent monkey indeed. You know very well that exponentiation cannot be extended to domains completely encircling 0 in the complex plane. Yet that is exactly what you attempt here!
Title: Re: An Exponential Curiosity
Post by Sir Col on Jan 30th, 2005, 2:15am
Cheeky by name, cheeky by nature...

But what about the real graphs?

Why is the graph of y = (x2)1/6 different to y = x1/3, when (x2)1/6 = x2/6 = x1/3?

Does that mean that 2/6 is not equal to 1/3? ::)
Title: Re: An Exponential Curiosity
Post by Barukh on Jan 30th, 2005, 6:26am

on 01/30/05 at 02:15:15, Sir Col wrote:
But what about the real graphs?

Why is the graph of y = (x2)1/6 different to y = x1/3, when (x2)1/6 = x2/6 = x1/3?

Does that mean that 2/6 is not equal to 1/3? ::)

Why complicate the question? Consider two functions: (i) y = x; (ii) y = [sqrt](x2)
Title: Re: An Exponential Curiosity
Post by Icarus on Jan 30th, 2005, 11:24am
More to the point, that (xa)b = xab rule comes with an "x [ge] 0" qualifier.


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