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suppose that $f$ is an entire function such that $f'(0)=0$ and $f''(1+1/n)=7-3/n$ for all $n$ natural number. we have to find all $f$ that satisfies these properties.

What I have done is: define $g(z)=f''(z)-10+3z$, so $g(1+1/n)=0$ so uniqueness theorem implies that $f''(z)=10-3z$ and therefore $f'(z)=10z-3z^2/2 +a$, As $f'(0)=0$ so $f(z)=5z^2-z^3/2+b(constant)$ is my solution is correct?

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What is the uniqueness theorem? – William Jun 10 '12 at 5:47
if zero set has limit point in the domain of the analytic function then it is identically zero there. – Un Chien Andalou Jun 10 '12 at 5:50
I think you meant $f''(z)=10-3z$. – Gerry Myerson Jun 10 '12 at 6:38
ah yes u r right... – Un Chien Andalou Jun 10 '12 at 6:41
OK, I have edited in the correction for you. – Gerry Myerson Jun 10 '12 at 12:34
up vote 2 down vote accepted

The solution is correct: since the equality $f''(z)=10-3z$ holds on a set with a limit point, it holds for all $z\in\mathbb C$. By integration, using the condition $f'(0)=0$ it follows that $f(z)=5z^2-z^3/2+\text{const}$.

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