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Find the flaw: Picard's Theorem

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In complex analysis, an entire function is defined as a function which is infinitely differentiable at every point in C (for example: constants, polynomials, e^x, etc.). Picard's Theorem says that every nonconstant entire function f misses at most one point (i.e. f© = C or C-{x0}). For example, every nonconstant polynomial hits every point, and e^x misses only 0.
Now consider the function f(x) = e^(e^x). Since e^x is entire, f is also entire by the chain rule. But it misses 0 since the base e^y misses 0, and it misses 1 since the top e^x misses 0 so that e^(e^x) misses e^0 = 1. But by Picard's Theorem there can be only one missing point, so the two missing points must be the same. Therefore, 0 = 1.
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Posted · Report post

e

z=1 is true not only for z=0, but for z=n*tau*i (where tau = 2*pi and n is any integer). So whenever ez=n*tau*i, we have f(z)=1.

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I'm not sure I understand Picard's Theorem. e^x misses all of the negative points as well as 0; even-degree polynomials also miss infinitely many points.



e^(e^x) simply misses all of the points below, and including, 1.

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I'm not sure I understand Picard's Theorem. e^x misses all of the negative points as well as 0; even-degree polynomials also miss infinitely many points.

e^(e^x) simply misses all of the points below, and including, 1.

It's a theorem in complex analysis, not in real analysis. Do you know about complex numbers?

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Little Picard Theorem: If a function f : CC is entire and non-constant, then the set of values that f(z) assumes is either the whole complex plane or the plane minus a single point.

from wikipedia: http://en.wikipedia.org/wiki/Picard_theorem

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Posted · Report post

I'm not sure I understand Picard's Theorem. e^x misses all of the negative points as well as 0; even-degree polynomials also miss infinitely many points.

e^(e^x) simply misses all of the points below, and including, 1.

It's a theorem in complex analysis, not in real analysis. Do you know about complex numbers?

Ah, that was the part I missed. Thanks. :blush:

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