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Define $p_n (x) = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \dots + \frac{x^n}{n!}$. Show that $\{ p_n (x) \}_{n=0}^{\infty}$ is a Cauchy sequence.

I'd just like to know the facts required to show this. We know that if a function $f: A \mapsto \mathbb{R}$, $A \subseteq \mathbb{R}$, is uniformly continuous on $A$ then, if $(x_n)$ is a Cauchy sequence in $A$, then $(f(x_n))$ is also a Cauchy sequence. Is this a relevant fact? Do I need to find some sequence, say $y_n$, that is Cauchy, and then find an explicit function that maps the terms of $y_n$ to the terms of $p_n(x)$?

Is there an easier way to do this?

Edit: I need to expand on this... the norm is $||p|| =$max$_{[0,1]} | p(x)|$. So for $p_n$ and $p_m$ the difference is $\sum _{ i = \min (n,m)} ^{\max (n,m)} \frac {x^i}{i!}$. The maximum term is the first term, because I've shown this to be a decreasing sequence. Now, is it still bounded by the geometric series with the factor 1/2?

Jared
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  • How easy or otherwise the problem is kind of depends what you are allowed o assume to start with. Usually, there are no shortcuts to prove that sequence is Cauchy from first principles. – Geoff Robinson May 11 '13 at 19:52

2 Answers2

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HINT: For sufficiently big $n$ (say $n>2|x|$) you can compare with a geometric series with a suitable factor (say $\frac12$).

Hagen von Eitzen
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  • By compare, do you mean I can observe that the geometric series will be smaller, so I have a bound? – Jared May 11 '13 at 20:27
  • I need to expand on this... the norm is $||p|| =$max$_{[0,1]} | p(x)|$. So for $p_n$ and $p_m$ the difference is $\sum _{ i = \min (n,m)} ^{\max (n,m)} \frac {x^i}{i!}$. The maximum term is the first term, because I've shown this to be a decreasing sequence. Now, is it still bounded by the geometric series with the factor 1/2? – Jared May 11 '13 at 20:48
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A sequence $f_n(x)$ is uniformly Cauchy if $\lim_{n\to \infty} ||f_{n+q}(x)-f_n(x)||_{\infty}=0,\quad \forall q\geq 1 $.

Now, applying this to our problem, we have

$$| p_{n+q} (x)-p_n(x) | = \Big| \sum_{k=0}^{n+q}\frac{x^k}{k!}-\sum_{k=0}^{n}\frac{x^k}{k!}\Big|= \Big| \sum_{k=n+1}^{n+q}\frac{x^k}{k!} \Big|\leq \sum_{k=n+1}^{n+q}\frac{x^k}{k!}\leq \sum_{k=n+1}^{\infty}\frac{x^k}{k!}<\epsilon \rightarrow (1) $$

$$\implies \sup_{0\leq x\leq 1} |p_{n+q} (x)-p_n(x)|=||p_{n+q} (x)-p_n(x)||_{\infty}<\epsilon.$$

The last inequality in $(1)$ follows from the fact that $ \sum_{k=0}^{\infty}\frac{x^k}{k!} $ is a convergent series.

Mhenni Benghorbal
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