# Analytic function $f$ agrees with $\tan x$ on $0 \leq x \leq 1$—is $f$ entire?

Suppose an analytic function $f$ agrees with $\tan x$, $0 \leq x \leq 1$. Could $f$ be entire?

Since $f$ and $\sin z/\cos z$ agree at a set of points and both are analytic in an open neighborhood of $(0,1)$, $f(z)=\sin z/\cos z$ in that open neighborhood. Now the problem for $f$ being entire is that $\sin z/\cos z$ is not differentiable at $z=(2n-1)\pi/2$, which aren't in $(0,1)$. I'm not sure if this is problem or if we can let $f$ be a different analytic function at the points where $\sin z/\cos z$ isn't defined.

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No that is not possible. $\tan(x) - f(x) = 0$ for $x \in (0,1)$ and that implies that $f = \tan$ since the zeroes of a holomorphic function can have no limit points in its (connected) domain unless it is identically zero.
@caligurl11: It shows that $f$ cannot be entire since you cannot extend it over the points where $\tan$ (and hence $f$) has poles. That is, $f$ can be extended exactly to the domain of $\tan$ and both are equal on that domain. –  WimC Feb 28 '12 at 18:20
@caligurl11: how would you even extend it to a continuous function, given that $|f| \rightarrow \infty$ near such a pole? –  WimC Feb 28 '12 at 19:47