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)=n∑k=0a(n)kzk
where
a(n)k=(nk)1nk=1k!k−1∏j=0(1−jn)
and trying to show that 1k!−a(n)k decreases sufficiently fast on any closed ball. That is, I tried to show
limn→∞maxz∈¯B0(A)|n∑k=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.
- For a,b∈C and k∈N you have |ak−bk|=|a−b||ak−1+bak−2+⋯+bk−1|≤|a−b|kmk−1where m=max(|a|,|b|)
- For u∈C you have |eu−(1+u)|≤+∞∑k=2|u|kk!≤|u|2+∞∑k=0|u|kk!=|u|2e|u|
- Now taking a=eu,b=1+u, we get m=max(|eu|,|1+u|)≤max(e|u|,1+|u|)≤e|u|. For k≥1 applying (1) and (2) successively, we get |eku−(1+u)k|≤|ku|2e|ku|k
- Finally for z∈C and denoting u=zn and k=n, we obtain using (3) |ez−(1+zn)n|≤|z|2e|z|n
- For K⊂C compact, one can find M>0 such that M≥supz∈K|z| which implies supz∈K|ez−(1+zn)n|≤M2eMnproving that (pn) converges uniformly to ez on every compact subset of C.
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