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.
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