until I asked my 12U students to make a conjecture about whether an odd or even function taken to a natural number power becomes odd, even, or neither. I didn't ask for a proof, it was just a little 2-mark thinking question.
In case you need reminding: x^2 is even because (-x)^2 = x^2. sin(x) is odd because sin(-x)=-sin*(x).
Anyway, it got me thinking: what does a real number exponent do to the symmetry? What does a real number exponent MEAN anyway? I thought I had a handle on rational number exponents. But even then, the domain (valid x values) gets messed up. And now my real problem (though not real enough for me to look up the "answer....")
What about irrational (or better yet, transcendental) exponents? I'd say, "Who cares? just approximate to a rational," but it comes up in polar coordinates all the time. e^ipi anyone? My graphing calculator seems to tell me that
x^2pi
(x^pi)^2
(x^2)^pi
are all different. Try a table of values. What does -0.7^2pi even MEAN? I don't know how to calculate that by hand. Excel should work for those lacking graphing calculators.
Actually: I was putting x^2pi into my calculator wrong. It's graphed the same as (x^2)^pi.
How's my calculator doing it anyway? Taking an approximation of pi, converting it to a rational, then doing the exponent? Then the different curves (domain aside) are simply artefacts of rounding errors that blow up due to exponentiation? Or Dion suggests that it might be interpolating between two nearby rationals. Rob thinks it probably depends on the calculator but likely it just chooses one nearby rational.
Maybe it all makes sense if exponents are seen as the inverse of logarithms (which were invented first, BTW). But then I would need to understand logarithms.
Discuss.
Uh yeah, whatever, Karen.
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2 comments:
Normally a^b is interpreted as exp(b*log(a)).
The exp() function is nicely behaved, analytic everywhere with Taylor series converging for all complex arguments. For log() you can get a series for your preferred subset of the positive reals, then extend your domain to any complex argument using a polar representation: a = r*exp(i*t) where r and t are positive reals.
Then log(a) = log(r)+(i*(t+2*pi*k)) for integer k, which, in general, you have to choose.
So since -0.7 = 0.7*exp(i*pi), then -0.7^(2*pi) = exp(2*pi*(log(0.7)+i*pi*(2*k+1)) = 0.7^(2*pi)*exp(i*(4k+2)*pi^2). If your exponent b is rational, you end up with a finite number of choices, if irrational then an infinite number.
I'm not sure exactly how it works, but there are people at places like Intel that work on fast ways to compute elementary functions for their specialized floating-point domains. I think it's like quick approximations and interpolation, maybe even table-lookup. More about numeric approximation of a known function than about the meaning of rational exponents.
Alex,
1. Thanks. Believe it or not, that did clarify a couple of things for me.
2. Nice to hear from you.
3. Dion says, "Nerd." Which, coming from Dion....
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