Successive derivative of analytic function

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SUMMARY

The discussion centers on proving that if an analytic function f on the disk |z| <= R has its nth derivatives converging uniformly to f(z), then f(z) must be of the form c * exp(z). The proof utilizes the properties of uniform convergence of analytic functions and Bach's differentiation formula. By defining a new function h(z) = f(z) * exp(-z), it is shown that h' = 0, leading to the conclusion that f(z) is indeed proportional to the exponential function.

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sbashrawi
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Homework Statement


Let f be analytic function on |z|<= R and lim fn(z)( the nth derivative) = f(z) uniformly in |z| <=R. Show that f(z) = c* exp(z)


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The Attempt at a Solution



I tried the power series expansion but I couldn't prove it
 
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sbashrawi said:
Let f be analytic function on |z|<= R and lim fn(z)( the nth derivative) = f(z) uniformly in |z| <=R. Show that f(z) = c* exp(z)
Let {Gn(z)} be a sequence of analytic functions converging to G(z), an analytic function. If the convergence is uniform in |z| <=R, then Gn'(z) converges to G'(z). (This can be proved either by an elementary limit argument or by Bach's differentiation formula).
Here, Gn = fn -> f. As Gn' = fn'(z) = f_(n+1)(z), the same sequence converges to f' as well. Hence, f=f'. Taking h(z) =f(z) exp(-z), we find that h' =0 .QED.
 
Thank you
 

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