Does there exists a non-constant entire function with the properties $f^n(0)=3^n$ for n is even and $f^n(0)=(n-1)!$ for n is odd ?
I have no idea,can i use coefficient formula of power series(Taylor series) ?please someone help.Thanks.
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2What would be the radius of convergence of the Taylor series about $0$? – Daniel Fischer Nov 16 '15 at 14:23
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sir Is it 1/3 ? – user290641 Nov 16 '15 at 14:31
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Use the hint of @Daniel Fisher and please look here http://math.stackexchange.com/questions/125201/every-power-series-expansion-for-an-entire-function-converges-everywhere – Nov 16 '15 at 14:32
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No, it's larger than that. The coefficients of the Taylor series are not $f^{(n)}(0)$, but …? – Daniel Fischer Nov 16 '15 at 14:33
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I doubt that you have no idea. For example, you know the root test. Report back when you try that test. – zhw. Nov 17 '15 at 01:52
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Hint. If such function existed then its expansion at $z=0$ would be $$ f(z)=\sum_{n=0}^\infty \frac{3^{2n}}{(2n)!}z^{2n}+\sum_{n=1}^{\infty}\frac{z^{2n-1}}{2n-1}. $$ Then, what would the radius of convergence be?
Yiorgos S. Smyrlis
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