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I know that $$e:=\lim_{n\to \infty }\left(1+\frac{1}{n}\right)^n,$$ by definition. Knowing that, I proved successively that $$e^{k}=\lim_{n\to \infty }\left(1+\frac{k}{n}\right)^n,$$ when $k\in \mathbb N$, $k\in \mathbb Z$ and $k\in\mathbb Q$. Now, I was wondering : how can I extend this result over $\mathbb R$ ? I tried to prove that $f_n(x):=(1+\frac{x}{n})^n$ converge uniformly on $\mathbb R$ but unfortunately it failed (I'm not sure that it's even true). Any idea ?


My idea was to define the function $x\longmapsto e^x$ as $$e^x=\begin{cases}e^x& x\in \mathbb Q\\ \lim_{n\to \infty }e^{k_n}&\text{if }k_n\to x \text{ and }(k_n)\subset \mathbb Q\end{cases}.$$ But to conclude that $$e^x=\lim_{n\to \infty }\left(1+\frac{x}{n}\right)^n,$$ I need to prove that $f_n(x)=\left(1+\frac{x}{n}\right)^n$ converge uniformly on a neighborhood of $x$, but I can't do it. I set $$g_n(x)=f_n(x)-e^x,$$ but I can't find the maximum on a compact that contain $x$, and thus can't conclude.

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  • $\begingroup$ How do you define $\exp(x)$ for an arbitrary $x$, by the way? Anyhow, there exists a sequence $\left(x_n\right)_{n\in\mathbb{Z}_{>0}}$ of rational numbers such that $x_n\to x$ as $n\to\infty$. $\endgroup$ – Batominovski Jul 6 '18 at 16:16
  • $\begingroup$ as $\sum_{k=0}^\infty \frac{x^k}{k!}$ $\endgroup$ – MathBeginner Jul 6 '18 at 16:20
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    $\begingroup$ See this. Not sure it's an exact enough duplicate? $\endgroup$ – Jyrki Lahtonen Jul 6 '18 at 16:34
  • $\begingroup$ math.stackexchange.com/questions/1557074/… $\endgroup$ – Robert Wolfe Jul 6 '18 at 16:38
  • $\begingroup$ You may also try to prove this for complex $x=a+ib$ given the definition $e^{a+ib} =e^a(\cos b+i\sin b) $. $\endgroup$ – Paramanand Singh Jul 6 '18 at 19:18
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We can use that exists $p_n, q_n \in \mathbb{Q}$ such that $p_n,q_n \to x$ and $p_n\le x\le q_n$, therefore

$$\left(1+\frac{p_n}{n}\right)^n\le \left(1+\frac{x}{n}\right)^n\le \left(1+\frac{q_n}{n}\right)^n$$

and

$$\left(1+\frac{p_n}{n}\right)^n=\left[\left(1+\frac{p_n}{n}\right)^\frac{n}{p_n}\right]^{p_n}\to e^x$$

$$\left(1+\frac{q_n}{n}\right)^n=\left[\left(1+\frac{q_n}{n}\right)^\frac{n}{q_n}\right]^{q_n}\to e^x$$

indeed for $\frac{n}{p_n}\in (m,m+1)$ with $m\in \mathbb{N}$ we have

$$\left(1+\frac1{m+1}\right)^m\le \left(1+\frac{p_n}{n}\right)^\frac{n}{p_n}\le \left(1+\frac1m\right)^{m+1}$$

and therefore $\left(1+\frac{p_n}{n}\right)^\frac{n}{p_n}\to e$.

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  • $\begingroup$ I think this is by far the simplest approach, and it does not rely on other properties of the exponential (e.g. its derivative or the properties of the logarithm as its inverse). The key observation is basically that $x$ can be moved from the inside of the parentheses to the exponent by substituting the variable used in the limit. $\endgroup$ – sasquires Jul 6 '18 at 16:35
  • $\begingroup$ Sorry but why $\left(\left(1+\frac{p_n}{n}\right)^{\frac{p_n}{n}}\right)^{p_n}\to e^x$ ? $\endgroup$ – MathBeginner Jul 6 '18 at 16:37
  • $\begingroup$ @MathBeginner since $p_n/n \to 0$ we have $$\left(1+\frac{p_n}{n}\right)^{\color{red}{\frac{n}{p_n}}}\to e$$ and $p_n \to x$. $\endgroup$ – gimusi Jul 6 '18 at 16:39
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    $\begingroup$ Since the exponent depend on $n$, we can't do that, no ? $\endgroup$ – MathBeginner Jul 6 '18 at 16:43
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    $\begingroup$ @MathBeginner Yes we can since $p_n/n \to 0$. $\endgroup$ – gimusi Jul 6 '18 at 17:12
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It is not difficult to prove the result for real irrational $x$ if you have already proved the case for rational $x$. The only idea you need to establish first as a part of your definition of $e^x$ is that $f(x) =e^x$ is continuous everywhere. I leave this as an exercise for you (hint: show that $\lim_{x\to 0}e^x=1$ using your definition).

Now let $x$ be any irrational number. Given any $\epsilon>0$ there is $\delta>0$ such that $$e^x-\epsilon<e^t<e^x+\epsilon$$ whenever $|t-x|<\delta$. Consider two rationals $r, s$ with $x-\delta<r<x<s<x+\delta$ and then we have $$e^x-\epsilon <e^r<e^s<e^x+\epsilon$$ Now we have $$\left(1+\frac{r}{n}\right)^n<\left(1+\frac{x}{n}\right)^n<\left(1+\frac{s}{n}\right)^n$$ and taking limits as $n\to\infty$ we get $$e^x-\epsilon<e^r\leq \lim_{n\to\infty} \left(1+\frac{x}{n}\right)^n\leq e^s<e^x+\epsilon $$ (the above assumes that the limit in question exists for irrational $x$ also and you can prove it using the fact that a bounded monotone sequence is convergent, or better apply liminf/limsup to the above inequalities). Since $\epsilon$ is arbitrary it follows that $$e^x=\lim_{n\to\infty} \left(1+\frac{x}{n}\right)^n$$

Based on feedback from Mark Viola via comments I am giving a link to my blog posts which discuss various routes to the theory of exponential and logarithmic functions :

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  • $\begingroup$ (+1) Well done, as always! $\endgroup$ – Mark Viola Jul 6 '18 at 17:24
  • $\begingroup$ @MarkViola: thanks. Btw this exponential and logarithmic functions of a real variable is one of my favorite topics. $\endgroup$ – Paramanand Singh Jul 6 '18 at 17:26
  • $\begingroup$ Yes, I know. I have read your "blog" and its three-part series. It is well-written. I suggest that you consider embedding a link to it as a solid reference on this topic. $\endgroup$ – Mark Viola Jul 6 '18 at 17:27
  • $\begingroup$ @mathbeginner This is the answer that you covet. $\endgroup$ – Mark Viola Jul 6 '18 at 17:31
  • $\begingroup$ @MarkViola: added the links. $\endgroup$ – Paramanand Singh Jul 6 '18 at 17:33
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To prove $$e^x=\lim_{n\to \infty }\left(1+\frac{x}{n}\right)^n$$

Let $$ y=\left(1+\frac{x}{n}\right)^n$$

$$ \ln y=n \ln(1+x/n)$$

$$= \frac {\ln(1+x/n)}{(1/n)}$$

$$\lim_{n\to \infty }\ln y=\lim_{n\to \infty }\frac {\ln(1+x/n)}{(1/n)}=x$$

Thus $$\lim_{n\to \infty } y= e^x$$

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    $\begingroup$ This doesn't address the question, which is "How does one show that $$\left(\lim_{n\to\infty}\left(1+\frac1n\right)^n\right)^x =\lim_{n\to \infty}\left(1+\frac xn\right)^n?"$$ $\endgroup$ – Mark Viola Jul 6 '18 at 16:29
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$$\frac xn(\frac n{n+x})\le\int_1^{1+\frac xn}\frac1t dt\le\frac xn(1)\implies \frac x{n+x}\le\ln (1+\frac xn)\le\frac xn\implies e^{\frac x{n+x}}\le1+\frac xn\le e^{\frac xn}\implies e^{\frac{xn}{n+x}}\le(1+\frac xn)^n\le e^x\implies e^x\le\lim_{n\to\infty}(1+\frac xn)^n\le e^x$$, by the squeeze or sandwich theorem...

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Left side:

The exponential function may be written as a Taylor series:

$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$

Right side:

$(1+\frac{x}{n})^n$ is a binomial expansion like:

$(1+y)^n=\binom{n}{0}y^0+\binom{n}{1}y^1+\binom{n}{2}y^2+...+\binom{n}{n-1}y^{n-1}+\binom{n}{n}y^n$

Where $\binom{n}{k}$ is the Binomial coefficient given by the formula : $\binom{n}{k}=\frac{n!}{k!(n-k)!}$

Some basic properties of $\binom{n}{k}$:

a)$\binom{n}{0}=1$ because $\frac{n!}{0!(n-0)!}=\frac{n!}{1*n!}$

b)$\binom{n}{1}=n$ because $\frac{n!}{1!(n-1)!}=\frac{(n-1)!*n}{(n-1)!}$

c)$\binom{n}{n-1}=n$ because $\frac{n!}{(n-1)!(n-(n-1))!}=\frac{(n-1)!*n}{(n-1)!*1!}$

d)$\binom{n}{n}=1$ because $\frac{n!}{n!(n-n)!}=\frac{1}{1!}$

e) The formula does exhibit a symmetry that is less evident from the multiplicative formula: $\binom{n}{k}=\binom{n}{n-k}$

Returning:

$(1+\frac{x}{n})^n=1+n*\frac{x}{n}+\frac{n!}{2!(n-2)!}\frac{x^2}{n^2}+\frac{n!}{3!(n-3)!}\frac{x^3}{n^3}+...+\frac{n!}{3!(n-3)!}\frac{x^{n-3}}{n^{n-3}}+\frac{n!}{2!(n-2)!}\frac{x^{n-2}}{n^{n-2}}+n*\frac{x^{n-1}}{n^{n-1}}+\frac{x^n}{n^n}$

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

$(1+\frac{x}{n})^n=1+x+\frac{n-1}{n}\frac{x^2}{2!}+\frac{(n-2)(n-1)}{n^2}\frac{x^3}{3!}+...+\frac{(n-2)(n-1)}{n^{n-4}}\frac{x^{n-3}}{3!}+\frac{n-1}{n^{n-3}}\frac{x^{n-2}}{2!}+\frac{x^{n-1}}{n^{n-2}}+\frac{x^n}{n^n}$

Let's analyze what happens for $n\rightarrow\infty$-here we have three types of limits:

-First type:

$\displaystyle\lim_{n \to \infty}\frac{n-1}{n}=\displaystyle\lim_{n \to \infty}[1+\frac{1}{n}]=1+0=1$ $\displaystyle\lim_{n \to \infty}\frac{(n-2)(n-1)}{n^2}=\displaystyle\lim_{n \to \infty}\frac{n^2-3n+2}{n^2}=\displaystyle\lim_{n \to \infty}[1-\frac{3}{n}+\frac{2}{n^2}]=1-0+0=1$

Hense $\displaystyle\lim_{n \to \infty}\Bigg(\frac{\displaystyle\prod_{i=1}^{k} (n-i)}{n^k}\Bigg)=1$

-Second type is $\displaystyle\lim_{n \to \infty} \frac{x^{n-\alpha}}{n^{n-\beta}}$-Because ${n^{n-\beta}}$ grows much fasten than $x^{n-\alpha}$ hense: $\displaystyle\lim_{n \to \infty} \frac{x^{n-\alpha}}{n^{n-\beta}}=0$

-Third type:

$\displaystyle\lim_{n \to \infty}\Bigg(\frac{\displaystyle\prod_{i=1}^{k} (n-i)}{n^{n-k-1}}\frac{x^{n-k}}{k!}\Bigg)$

We have to show on the biggest power (similar to the first type) as the most relevant:

$\frac{\displaystyle\prod_{i=1}^{k} (n-i)}{n^{n-k-1}}\frac{x^{n-k}}{k!}\sim\frac{n^{k-1} }{n^{n-k-1}}\frac{x^{n-k}}{k!}=n^{k-1-(n-k-1)}\frac{x^{n-k}}{k!}=n^{2k-n}*\frac{x^{n-k}}{k!}=\frac{1}{k!}*\frac{x^{n-k}}{n^{n-2k}}$

Again: ${n^{n-\beta}}$ grows much faster than $x^{n-\alpha}$

Hense: $\displaystyle\lim_{n \to \infty}\Bigg(\frac{\displaystyle\prod_{i=1}^{k} (n-i)}{n^{n-k-1}}\frac{x^{n-k}}{k!}\Bigg)=0$

Our right side equals:

$\displaystyle\lim_{n \to \infty}(1+\frac{x}{n})^n=1+x+1*\frac{x^2}{2!}+1*\frac{x^3}{3!}+...+0+0+0+0$

$\displaystyle\lim_{n \to \infty}(1+\frac{x}{n})^n=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$

We got the same elements like in the Taylor series of $e^x$. Q.E.D.

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