Paul Rubin wrote:
> Steve Holden <[EMAIL PROTECTED]> writes:
>> but the reals aren't. Clearly you *can* take the square root of all
>> real numbers, since a real number *is* also a complex number with a
>> zero imaginary component. They are mathematically equal and equivalent.
>
> Ehhh, I let it slide before but since the above has been said a few
> times I thought I better mention that it's mathematically a bit bogus.
> We could say there is an embedding of the real numbers in the complex
> numbers (i.e. the set of complex numbers with Im z = 0). But the
> usual mathematical definition of the reals (as a set in set theory) is
> a different set from the complex numbers, not a subset. Also, for
> example, the derivative of a complex valued function means something
> considerably stronger than the derivative of a real valued function.
> The real valued function
>
> f(x) = { exp(-1/x**2, if x != 0,
> { 0, if x = 0
>
> for real x is infinitely differentiable at x=0 and all the derivatives
> are 0, which makes it sound like there's a Taylor series that
> converges to 0 everywhere in some neighborhood of x=0, which is
> obviously wrong since the function itself is nonzero when x!=0. The
> discrepancy is because viewed as a complex valued function f(z), f is
> not differentiable at z=0 even once.
>
> It's pretty normal for a real function f to have a first derivative at
> x, but no second derivative at x. That can't happen with complex
> functions. If f'(z) exists for some z, then f is analytic at z which
> means that all of f's derivatives exist at z and there is some
> neighborhood of z in which the Taylor series centered at z converges.
Much as I'd like to argue with that I can't, dammit :-)
regards
Steve
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