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Old 2007-12-24, 01:16   #1
nibble4bits
 
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Default rational powers of negative one

Let's say I have y=f(x)=(-1)^x as a function.

If x=...,-1,0,1,... then y alternates between opposite signs.

If x=...,-1.5,0.5,2.5,... then y is the square root of -1 which is of course i.

f(5.3)=f(1.3) * f(2)^2=f(1.3)
(-1)^1.3=(-1) * 10th root of (-1)^3 = (-1)* 10th root of (-1) = -i

I think that there could be rational numbers with more than one possible value since A) Those are just the primary roots in my example. B) Other ways to rewrite the fraction which might give different signs?

I see that this is close to how e^xi = cis(x) = cos(x)+ i sin(x) works for some values. My reasoning for this statement is that the angle (aka argument) of y is the same as the value of x for right angles. Too bad this property doesn't hold for all values of x. (After scaling)

My goal was to try to make a derivative of (-1)^x for nonintegers.
The closest I've gotten is a function that looks like this:
(x ln -1) * (-1)^x = xi pi * (-1)^x
Is this remotely correct? If so, there's a follow up topic I'd like to discuss involving derivatives. I need this derivative for another.

BTW Rational powers meaning negative one taken to a rational power, not that the result is rational or even real.

Last fiddled with by nibble4bits on 2007-12-24 at 01:17 Reason: Forgot to mention scaling
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Old 2007-12-24, 15:51   #2
fivemack
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The obvious function equal to (-1)^x at integers is exp(pi * I * x), which does have the derivative you want.

-1 has ten tenth roots, amongst which is I; the cube of any tenth root of -1 is another tenth root of -1, but that doesn't mean that they're either equal or equal to I.
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Old 2007-12-24, 23:04   #3
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Actually, what you have is a "multivalued function" (which is not technically a function). To actually get a function, you need to choose a branch cut.

(-1)^x=(e^{i\pi+2ni\pi})^x=e^{i\pi(1+2n)x}

where n can be any fixed integer.

Last fiddled with by jinydu on 2007-12-24 at 23:05
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Old 2007-12-26, 05:21   #4
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jinydu: I think I mentioned that - the comment about primary roots. It means there's non-primary roots such as 1 at the angles 2k*Pi/10 where k is an integer. Of course for k>9 or k<0, there are 'mirrors' that are equivilent. This is because k mod 10 is always in the set {0,1,2,3,4,5,6,7,8,9}.

I was just wondering if there's something that makes the derivative useless for my purposes. It seems OK for integers but it would be nice to find out why it doesn't work if it is wrong. For one thing, finding the derivative of a discontinous function...
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Old 2007-12-26, 09:47   #5
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Quote:
Originally Posted by nibble4bits View Post
For one thing, finding the derivative of a discontinous function...
A function that is discontinuous at a point is never differentiable at that point.

e^{i\pi x} is definitely a continuous (in fact infinitely differentiable) function of x. So is e^{i\pi(1+2n)x} for any fixed integer n.

Last fiddled with by jinydu on 2007-12-26 at 09:49
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Old 2008-01-08, 04:58   #6
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Quote:
A function that is discontinuous at a point is never differentiable at that point.
That is the only thing I thought was likely to go wrong.

Last fiddled with by nibble4bits on 2008-01-08 at 04:58
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