Christian Bower wrote: << Gene wrote: << Here's a nice challenge problem. Let f(z) = sum( z^(2^n) / n!, n=0..infinity). We know that f(z) is analytic on the open unit disk and that f and all its derivatives are continuous and bounded on the closed unit disk. Find a function g(z) analytic on the open complement of the unit disk (possibly including infinity), with g and its derivatives continuous and bounded on the closed complement, and that agrees with f and its derivatives on the unit circle. The obvious extension method, Schwartz reflection, does not work, since f does not map the unit circle into itself.

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Can't we just use f(1/z)?

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Trouble is, for z = i, say, we have f(1/z) = f(-i) = f(z) - 2i != f(z). --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele