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
lim
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 \in \mathbb C and k \in \mathbb N you have \vert a^k -b^k \vert =\vert a-b \vert \vert a^{k-1}+b a^{k-2}+\dots+b^{k-1}\vert\le \vert a - b \vert k m^{k-1} \tag{1} where m = \max (\vert a \vert, \vert b \vert)
- For u \in \mathbb C you have \left\vert e^u-(1+u) \right\vert \le \sum_{k=2}^{+\infty} \frac{\vert u \vert^k}{k!} \le \vert u \vert^2 \sum_{k=0}^{+\infty} \frac{\vert u \vert^k}{k!}=\vert u \vert^2 e^{\vert u \vert} \tag{2}
- Now taking a=e^u,b=1+u, we get m=\max(\vert e^u \vert,\vert 1+u \vert) \le \max(e^{\vert u \vert},1+\vert u \vert) \le e^{\vert u \vert}. For k \ge 1 applying (1) and (2) successively, we get \left\vert e^{ku} -(1+u)^k\right\vert \leq\frac{\vert k u \vert^2 e^{\vert ku \vert}}{k} \tag{3}
- Finally for z \in \mathbb{C} and denoting u=\frac{z}{n} and k=n, we obtain using (3) \left\vert e^z -\left(1+\frac{z}{n}\right)^n \right\vert \le \frac{\vert z \vert^2 e^{\vert z \vert}}{n} \tag{4}
- For K \subset \mathbb C compact, one can find M > 0 such that M \ge \sup\limits_{z \in K} \vert z \vert which implies \sup\limits_{ z \in K} \left\vert e^z -\left(1+\frac{z}{n}\right)^n \right\vert \le \frac{M^2 e^{M}}{n} \tag{5} proving that (p_n) converges uniformly to e^z on every compact subset of \mathbb C.
No comments:
Post a Comment