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How can I show that for every $\epsilon>0$, there exists an $N\in\mathbb{N}$ such that $\left|f_n(x)-f(x)\right|=\left|\left(\frac{x}{n}+1\right)^n-e^x\right|<\epsilon$ whenever $n\geq N$ and $x\in\left[-A,A\right]$? By the way, $n\in\mathbb{N}$.

In a previous exercise, I was able to show that $f_n$ does indeed converge pointwise to $f$. However, I have been stuck for hours trying to prove uniform convergence. Would anyone lend me a hand? Thanks in advance.

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    Have you thought using Arzela-Ascoli's theorem? It follows (in a not-so-hard way) from this theorem.2012-03-14

2 Answers 2

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Write $(\frac{x}{n}+1)^n=\exp\left( n \log(1 + \frac{x}{n})\right).$ Assuming that $N>2A$ and hence $|x/n|<\frac12$, $\log (1+\frac{x}{n})$ can be expanded in a Taylor series with remainder, giving $\log(1 + \frac{x}{n})=\frac{x}{n}-\frac{1}{(1+(y/n))^2} \frac{x^2}{2n^2},\qquad |y|\le |x|$ $=\frac{x}{n}-\theta \frac{x^2}{2n^2},\qquad \theta\in [0,4].$ Substituting this into the first equation gives$(\frac{x}{n}+1)^n = \exp\left(x-\theta \frac{x^2}{2n}\right)$ so $\left|(\frac{x}{n}+1)^n-e^x\right|=e^x (1-\exp -\theta \frac{x^2}{2n}).$ The right-hand side of this is no bigger than $e^A (1-e^{-4A^2/2n})$, so uniform convergence follows.

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    Yes, I apologize for this mistake: you can't take $\theta\in[0,1]$ if x<0. However, it's enough to find a bounded interval which must contain $\theta$, which is possible if $|x/n|$ is bounded away from 1. Hopefully, this bug is now fixed.2012-03-14
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You (might) know that

$e^x=1+x+\frac{x^2}{2!}+\cdots$

and that

$\left(1+\frac{x}{n} \right)^n=1+x+\frac{n(n-1)}{n^2}\frac{x^2}{2!}+\frac{n(n-1)(n-2)}{n^3}\frac{x^3}{3!}+\cdots$

So you get

$\left|\left(1+\frac{x}{n} \right)^n-e^x \right|\leqslant \left| \frac{n(n-1)}{n^2}-1 \right|\frac{x^2}{2!}+ \left| \frac{n(n-1)(n-2)}{n^3}-1 \right|\frac{x^3}{3!}+\cdots$

All the expressions in $n$ will be rational and of the form $\dfrac{P(n)}{Q(n)}$, where the degree of $P$ will be one less than $Q$. For example, the first two are:

$ \left| \frac{1}{n} \right|$

$ \left| \frac{3n^2-2n}{n^3} \right|$

They will all behave like $\dfrac{C}{n}$ for some constant $C$ for large $n$. It shouln't be too hard for you to show that this difference can be made arbitrarely small for sufficiently large $n$.

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    @JosuéMolina Indeed, that would work.2012-03-14