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Aug 16, 2012 at 9:07 comment added Pietro Majer Indeed, $I$ has zero spectral radius, so the series for $(1-z I)^{-1}$ even converges for all $z$. Notoriously, the exponential series is a particular case of a geometric series.
May 19, 2012 at 22:51 comment added Ryan Reich As long as you verify that I is a contraction operator on continuous functions on an interval of length less than 1, this works just fine: the series converges in the max norm, i.e. uniformly. Then you can check that this particular series happens to converge everywhere. Although omitting this check is an error, it seems to me that it just exposes an error in the strategy of using a purely algebraic argument to prove an analytic statement.
May 19, 2012 at 15:39 history edited James Cranch CC BY-SA 3.0
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May 19, 2012 at 14:05 history answered James Cranch CC BY-SA 3.0