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I need to find all entire functions $f$ such that $f(x) = e^x$ on $\mathbb{R}$.

At first it seems that, since the function $f$ will be real analytic on $\mathbb{R}$ and will have a power series expansion with radius of convergence = $\infty$, only possible function is $f(z) = e^z$. But now I understand that my logic has many faults.

Can anybody please help ?

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What are the faults of your logic? I cannot see them. –  rom Feb 14 '13 at 4:11
    
@rom f only takes the real part on $\mathbb{R}$, i.e, one may define f to be $f(x+iy) = e^{x+ig(y)}$ for some entire function $g$. –  user44349 Feb 14 '13 at 4:22
    
Don't forget the Cauchy-Riemann equations. –  rom Feb 14 '13 at 4:30
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Any two holomorphic functions on a connected domain that agree on a set with a limit point are equal. –  JSchlather Feb 14 '13 at 5:03
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2 Answers 2

up vote 2 down vote accepted

Its clear that one of the possibilities is $e^z$. Now let $f$ be analytic such that $f(x)=e^x$ consider the set $A=\{z\in \mathbb C\mid f(z)=e^z$}, but then we have $\mathbb R\subset A$, which implies that $A$ has a limit point, Thus, $f(z)=e^z$.

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  1. Your method is basically correct. Because $f$ is entire, it has an everywhere convergent Maclaurin series $f(z)=\sum_n a_n z^n$. Consider the restriction of $f$ to $\mathbb R$, $g=f|_\mathbb R$. Then we have $g(x)=e^x$, and $g(x)=\sum_n a_nx^n$. But power series coefficients are unique: $a_n=\dfrac{1}{n!}g^{(n)}(0)=\dfrac{1}{n!}$. Thus $f(z)=e^z$.
  2. For an alternative approach, see the identity theorem. (This approach was also mentioned in i.a.m's answer and Jacob Schlather's comment.)
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Thanks a lot to all for the help ! :) –  user44349 Feb 14 '13 at 5:20
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