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The Problem :

We state the following two definitions of the real exponential function from the Pr$\infty$fWiki page. We're interested in showing that the two definitions are valid $($i.e. the defining sequence/series does converge to a unique real number$)$ and that the two definitions are equivalent.

I'm stuck at a couple of points $($which are described in highlighted lines$)$. Any help would be much appreciated. Thank you!

Definition $1$. The exponential function can be defined as the following limit of a sequence $$\exp x := \lim_{n \to \infty} \left({1 + \frac x n}\right)^n$$

Definition $2$. The exponential function can be defined as a power series $$\exp x := \sum_{n = 0}^\infty \frac {x^n} {n!}$$

My Progress and two places where I'm stuck : Essentially the solution consists of three parts, namely validity of definition $1$, validity of definition $2$ and equivalence of the two definitions.

Validity of Definition $1$. I'm stuck here! Can we show that the sequence $(a_n)$ given by $a_n = \left(1+\frac{x}{n}\right)^n$ converges for every $x \in \mathbb{R} ??$ Is it eventually monotone and bounded $??$

Validity of Definition $2$. The radius of convergence of the power series is $$r=\lim_{n \to \infty}\left| \frac{\frac{1}{n!}}{\frac{1}{(n+1)!}} \right|=\lim_{n \to \infty}\left| \frac{(n+1)!}{n!} \right|=\lim_{n \to \infty}(n+1)=+\infty$$ Thus the infinite series in the right-hand side of Definition $2$ converges to a unique real number for all $x \in \mathbb{R}$. Hence the definition is well-defined.

Equivalence of Definition $1$ and Definition $2$.

There's this proof of Definition $1$ $\implies$ Definition $2$ in the Pr$\infty$fWiki page, but it's kind of under construction and I'm not really convinced by it. So I decided to try to take my own shot at it.

For all $n \in \mathbb{N} \cup \{0\},$ let $$T_n=\left(1+\frac{x}{n} \right)^n, ~S_n=\sum_{k=0}^n \frac{x^k}{k!}$$ We have to show that $\lim_{n \to \infty} T_n = \lim_{n \to \infty} S_n$

Now, \begin{align} T_n &= \left(1+\frac{x}{n}\right)^n\\ &= 1+n\cdot\frac{x}{n}+\frac{n(n-1)}{2!}\cdot\frac{x^2}{n^2}+\cdots +\frac{n(n-1)\cdots 1}{n!}\cdot\frac{x^n}{n^n}\\ &= 1+x+\left(1-\frac{1}{n}\right)\cdot \frac{x^2}{2!}+\cdots +\left(1-\frac{1}{n}\right)\cdots \left(1-\frac{n-1}{n}\right)\cdot \frac{x^n}{n!} \end{align}

Clearly, $$S_n-T_n=\left\{1-\left(1-\frac{1}{n}\right)\right\}\frac{x^2}{2!}+\cdots +\left\{1-\left(1-\frac{1}{n}\right)\cdots \left(1-\frac{n-1}{n}\right)\right\}\cdot \frac{x^n}{n!}\geq 0$$

I'm stuck at this point. Can we show that $S_n-T_n \leq B_n$ such that $B_n \to 0$ as $n \to \infty ??$

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  • $\begingroup$ I have presented this in great detail in my blog post paramanands.blogspot.com/2014/05/… $\endgroup$ – Paramanand Singh Jul 6 '17 at 17:33
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    $\begingroup$ It is much easier to show that $(1+x/n)^n$ converges to $e^x$ after developing some more properties of $e^x.$ But I have posted an answer that doesn't use them $\endgroup$ – DanielWainfleet Jul 6 '17 at 23:17
  • $\begingroup$ I would consider trying to prove, in sufficient generality to apply to the case at hand, a theorem about limits of sequences of power series in terms of limits of the coefficients. Sometimes general theorems are easier to prove than special cases, because the special case is complicated by irrelevant details of the special case. $\endgroup$ – Hurkyl Jul 7 '17 at 0:39
  • $\begingroup$ Also see math.stackexchange.com/q/637255/72031 $\endgroup$ – Paramanand Singh Jul 7 '17 at 5:27
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Starting point: $E(x) = \sum_{n = 0}^\infty \frac {x^n} {n!}$ converges.

1st Question: Show that $a_n = \left(1+\frac{x}{n}\right)^n$ converges.

Solution: Show that $(a_n)$ converges to $E(x)$.

Look at Konrad Königsberger's analysis, just tone it down a bit by replacing the $w$ stuff with $x$.

Lemma: ${n \choose k} \frac 1 {n^k} = \frac{1}{k!} \prod_{i = 1}^{k-1} \Bigl(1 - \frac i n \Bigr) \le \frac 1 {k!}$
Moreover, the sequence $t_n = {n \choose k} \frac 1 {n^k}$ converges to $\frac {1}{k{!}}$ as $n \to +\infty$.

The OP should be familiar with the triangle inequality for infinite sums.

If $K$ and $n$ are positive integers and $n \ge K$ we have a RHS side broken down into three sums:

$\left|\Bigl(1 + \frac{x}{n}\Big)^n - E(x) \right| \le \sum_{k=0}^{K-1} \left|{n \choose k}\frac{x^k}{n^k} - \frac{x^k}{k!}\right| + \sum_{k=K}^n{n\choose k} \frac{|x|^k}{n^k} + \sum_{k=K}^\infty \frac{|x|^k}{k!}$

So far this is just expressive notation showing the binomial expansion of $a_n$ combined with $E(x)$ (in a useful way) while using the triangle inequality.

Let $\varepsilon \gt 0$ be given. We choose $K$ so that the last term is less than $\varepsilon/3$. By the Lemma, the middle term will also be less than $\varepsilon/3$ (true for any selected $n$). By the Lemma, we can choose $N$ so that for all $n \gt N$ the first term is also less than $\varepsilon/3$.

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    $\begingroup$ What a pain! I tried to highlight stuff and the math symbols would get garbled! Who ya gonna call - no idea. $\endgroup$ – CopyPasteIt Jul 7 '17 at 1:07
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Let $E(x)=\sum_{j=0}^{\infty}x^j/j!.$

Given any $x\in \mathbb R$ and given any $\epsilon >0,$ take $k_1\geq 1$ such that $$(I).\quad \sum_{j=1+k_1}^{\infty }|x|^j/j!<\epsilon.$$

For brevity let $1-\prod_{i=0}^{j-1}(1-i/n)= A(n,j)$ for $1\leq j\leq n.$

Take $k_2>k_1$ such that $$(II). \quad n\geq k_2\implies \max \{\;|(1-A(n,j)\cdot x^j/j!|\;: 1\leq j\leq k_1\}<\epsilon /k_1.$$

For $n\in \mathbb N$ we have $$E(x)-(1+x/n)^n=\sum_{j=0}^{\infty}B_jx^j$$ where $B_0=0$ and $B_j=(1-A(n,j))/j!$ for $1\leq j\leq n$ and $B_j=1/j!$ for $j>n.$

Observe that $0\leq B_j\leq 1/j!$ for all $j.$

Now when $n\geq k_2$ we have $$|\sum_{j=0}^{k_1}B_jx^j|= |\sum_{j=1}^{k_1}B_jx^j|\leq \sum_{j=1}^{k_1}|B_jx^j|=$$ $$=\sum_{j=1}^{k_1}|(1-A(n,j)\cdot x^j/j!|<\epsilon \quad \text {by }\; (II),$$ and also we have $$|\sum_{j=1+k_1}^{\infty}B_jx^j |\leq \sum_{j=1+k_1}^{\infty}|B_jx^j|\leq$$ $$ \leq \sum_{j=1+k_1}^{\infty}|x^j/j!|<\epsilon \quad \text { by }\; (I).$$ So $|E(x)-(1+x/n)^n|<2\epsilon$ for all $n\geq k_2.$

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  • $\begingroup$ Wow, Within minutes we both answer (first ones) a 13 hour question I give the hint and you give the solution! And we both used $E(x)$ since we don't know it is $exp(x)$. $\endgroup$ – CopyPasteIt Jul 6 '17 at 23:23
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    $\begingroup$ I know it is $\exp (x).$ Just gave it another name, to make it clear that I'm showing that the power series is the limit of $(1+x/n)^n$ without using any of the other properties of $\exp (x).$ Kudos. $\endgroup$ – DanielWainfleet Jul 7 '17 at 2:19

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