I'm kind of curious.. Shouldn't (-2)^(-2)^(-2)= (-2)^0,25..?
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I'm kind of curious.. Shouldn't (-2)^(-2)^(-2)= (-2)^0,25..?
It is. At least, that's how a calculator interprets it. And (-2)^(1/4) = 0.841 + 0.841i (correct to three decimal places).Quote:
Originally Posted by narko
But another interprettation might well be [(-2)^(-2)]^(-2) = (1/4)^(-2), which will give a different answer.
hmm, so negative numbers cannot be exponentiated by fractions (like -10^0,5=imaginary number).. well OK, thanks. We haven't learned yet how to calculate things like 10^0,5 without calculators or programs yet. Thank you, again.
PS. Heh, these are interesting numbers. So basically (-2)^(-2)^(-2) have multiple answers depending on how u interpret it (I always thought you had to solve these issues in "(-2)^(-2)^(-2)= (-2)^0,25" this way, and not this; "0,25^-2" way)..
Wrong.Quote:
Originally Posted by narko
If n is odd, then (-ve)^1/n = -ve. eg. (-8)^1/3 = -2. (There are two other answers that are both non-real).
If n is even, then (-ve)^1/n is not real. eg. (-4)^1/2 = -2i. (There are two other answers that are also non-real).
The standard interpretation is (-2)^[(-2)^(-2)].Quote:
Originally Posted by mr fantastic
Reference: Wikipedia - Tetration
Is this a typo? I thought the only solutions were 2i and -2i.Quote:
Originally Posted by mr fantastic
hmm, OK. I think I understand this allot better. for example:
10^1/5 solving this would need:
x^5=10
and then u use the 5√ 10 to solve it.
Thinking of it, we HAVE went through this. heh I just forgot it. 10^2/6= 3/1√10 and so on.
And yeah, now that I can see this it is only logical that even exponent fractions are imaginary (square root of -1?) while odd ones are not (3√(-1)=-1*-1*-1=-1).
Hello, narko!
Quote:
^{(-2)^{(-2)}} \:=\:(-2)^{0.25}\;?)
You are correct . . .
Now how do you take the 4th root of a negative number?