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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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  • $\begingroup$ Have you thought using Arzela-Ascoli's theorem? It follows (in a not-so-hard way) from this theorem. $\endgroup$
    – Patrick Da Silva
    Mar 14, 2012 at 1:02

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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  • $\begingroup$ I thought of the solution that uses Beronulli's inequality to show that it is eventually monotone increasing, so that we can conclude the uniform convergence by Dini's theorem. But it seems redundant compared to your elegant answer... $\endgroup$
    – Sangchul Lee
    Mar 14, 2012 at 1:13
  • $\begingroup$ Just one thing : you say "expanding log in a Taylor series", a thing that you can't do over an arbitary interval of the form $[-A/n,A/n]$. What you're actually doing is supposing $n$ large enough so that $x/n \in ]-1,1[$ for $x \in [-A,A]$ (in other words, $n > A$), and then using Taylor's theorem (that only requires the fact that $\log$ is a $C^2$ function over some open interval containing $1$ and contained in $]0,\infty[$). But nevertheless this is a great idea and I like your answer. +1 $\endgroup$
    – Patrick Da Silva
    Mar 14, 2012 at 1:17
  • $\begingroup$ Do you guys know where I can find documentation regarding this particular Taylor expansion? I would like to know how you reached that equality. $\endgroup$
    – wjmolina
    Mar 14, 2012 at 1:22
  • $\begingroup$ I edited the answer to clarify the issue raised by Patrick Da Silva. The Taylor series is using the so-called Lagrange remainder: $$f(x+h)=f(x)+f'(x)h+\frac12f''(x)h^2+\cdots+\frac{1}{n!} f^{(n)}(x) h^n + \frac{1}{(n+1)!} f^{(n+1)}(y) h^{n+1},$$ where $y$ is in the interval between $x$ and $x+h$. $\endgroup$
    – David Moews
    Mar 14, 2012 at 1:29
  • $\begingroup$ To simplify it a bit, I suppose you can just use the inequality $x - x^2/2 \le \log(1+x) \le x$ $\endgroup$
    – Aryabhata
    Mar 14, 2012 at 1:40
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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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  • $\begingroup$ Thank you very much! But is it really possible to show that rigorously? The first thing that came to my mind was to rewrite the sum using sigma notation, apply the triangle inequality to the expressions in $n$ and somehow manage to show that they can be made arbitrarily smaller than any chosen $\epsilon$ for a sufficiently large $n$. $\endgroup$
    – wjmolina
    Mar 14, 2012 at 3:28
  • $\begingroup$ @JosuéMolina Indeed, that would work. $\endgroup$
    – Pedro ♦
    Mar 14, 2012 at 15:19

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