Wednesday, 26 September 2018

complex analysis - Do the polynomials (1+z/n)n converge compactly to ez on mathbbC?




The question is




Do the polynomials pn(x)=(1+z/n)n converge compactly (or uniformly on compact subsets) to ez on C?




I thought about expanding
pn(z)=nk=0a(n)kzk


where

a(n)k=(nk)1nk=1k!k1j=0(1jn)

and trying to show that 1k!a(n)k decreases sufficiently fast on any closed ball. That is, I tried to show
limnmaxz¯B0(A)|nk=0zkk!pn(z)|=0

for any fixed A>0, but I had difficulty with this approach.



Any help is appreciated.


Answer



You can use following steps.





  1. For a,bC and kN you have |akbk|=|ab||ak1+bak2++bk1||ab|kmk1
    where m=max(|a|,|b|)

  2. For uC you have |eu(1+u)|+k=2|u|kk!|u|2+k=0|u|kk!=|u|2e|u|

  3. Now taking a=eu,b=1+u, we get m=max(|eu|,|1+u|)max(e|u|,1+|u|)e|u|. For k1 applying (1) and (2) successively, we get |eku(1+u)k||ku|2e|ku|k

  4. Finally for zC and denoting u=zn and k=n, we obtain using (3) |ez(1+zn)n||z|2e|z|n

  5. For KC compact, one can find M>0 such that MsupzK|z| which implies supzK|ez(1+zn)n|M2eMn
    proving that (pn) converges uniformly to ez on every compact subset of C.


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