\(e = 2.718\cdots\): Napier's constant

-

We have defined the constant known as Napier's constant \(e\) in a previous post, as \(e = \exp(1)\) where \(\exp(x)\) is the inverse of \(\log(x)\). 

See also: \(\log\) and \(e\)

Here we provide an alternative definition of this constant. That is,

\[e = \lim_{n\to\infty}\left(1 + \frac{1}{n}\right)^n.\]

For this definition to be valid, we need to show that the sequence \(\{a_n\}\) defined by

\[a_n = \left(1 + \frac{1}{n}\right)^n\]

converges.



Example. The first several terms of the above sequence are:

\[\begin{align*}a_1 &= 2,\\ a_2 &= 2.25,\\ a_3 &= 2.370\cdots,\\ a_4 &= 2.441\cdots,\\ a_5 &=2.488\cdots,\\ &\vdots\\a_{10}&=2.593\cdots,\\a_{100} &= 2.704\cdots,\\a_{1000} &= 2.716\cdots,\\&\vdots\end{align*}\]

□

Lemma

The sequence \(\{a_n\}\) defined above is monotone increasing.

Proof. We show that \(a_n < a_{n+1}\) for all \(n\in\mathbb{N}\).

By the Binomial Theorem, we can write

\[a_n = \sum_{k=0}^{n}\frac{s_k}{k!}\]

where

\[ \begin{eqnarray*} s_k &=& \frac{n(n-1)\cdots(n-k+1)}{n^k}\\ &=& \frac{n}{n}\cdot\frac{n-1}{n}\cdot\cdots\cdot\frac{n-k+1}{n}\\ &=&1\cdot\left(1 - \frac{1}{n}\right)\cdot\cdots\cdot\left(1 - \frac{k-1}{n}\right). \end{eqnarray*} \]

Similarly,

\[a_{n+1} = \sum_{k=0}^{n+1}\frac{t_k}{k!}\]

where

\[t_k = 1\cdot\left(1 - \frac{1}{n+1}\right)\cdot\cdots\cdot\left(1 - \frac{k-1}{n+1}\right). \]

For each \(i = 1, 2, \cdots, k-1\), we have

\[1 - \frac{i}{n} < 1 - \frac{i}{n+1}\]

so that

\[s_k < t_k ~ (k = 1, 2, \cdots, n).\]

Therefore, for \(n\geq 1\),

\[a_n = \sum_{k=0}^{n}\frac{s_k}{k!} < \sum_{k=0}^{n}\frac{t_k}{k!} < \sum_{k=0}^{n}\frac{t_k}{k!} + \frac{t_{n+1}}{(n+1)!} = a_{n+1}.\]

■

Lemma

For all \(n\in\mathbb{N}\), the following holds:

\[2 \leq \left(1 + \frac{1}{n}\right)^n < 3.\]

In particular, the sequence \(\{a_n\}\) defined by \(a_n = \left(1 + \frac{1}{n}\right)^n\) is bounded.

Proof. \(a_1 = 2\) and \(\{a_n\}\) is strictly monotone increasing. Therefore \(a_n \geq 2\) for all \(n \in\mathbb{N}\). Using \(s_k\) defined in the previous lemma,

\[s_k = 1\cdot\left(1 - \frac{1}{n}\right)\cdot\left(1 - \frac{2}{n}\right)\cdot\cdots\cdot\left(1 - \frac{k-1}{n}\right) \leq 1\]

for \(k = 1, \cdots, n-1\). Therefore

\[ \begin{eqnarray*} a_n &=& 1 + 1 + \frac{s_2}{2!} + \frac{s_3}{3!} + \cdots + \frac{s_n}{n!}\\ &\leq& 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots + \frac{1}{n!}. \end{eqnarray*} \]

Noting that \(n! = 1\cdot 2\cdot 3\cdot\cdots\cdot n > 1\cdot 2\cdot 2\cdot\cdots\cdot 2 = 2^{n-1}\) for \(n\geq 3\),

\[ \begin{eqnarray*} a_n &\leq& 1 + \left(1 + \frac{1}{2} + \frac{1}{2^2} + \cdots + \frac{1}{2^{n-1}}\right)\\ &=& 1 + \left(2 - \frac{1}{2^{n-1}}\right) < 3 \end{eqnarray*} \]

■

The next theorem follows immediately from these lemmas.

Theorem

The sequence \(\{a_n\}\) defined by \(a_n = \left(1 + \frac{1}{n}\right)^n\) converges.

Now we can define

\[e = \lim_{n\to\infty}\left(1 + \frac{1}{n}\right)^n = 2.71828\cdots.\]


Comments

Post a Comment

Popular posts from this blog

Open sets and closed sets in \(\mathbb{R}^n\)

-
Image
Open sets In \(\mathbb{R}\), we have the notion of an open interval such as \((a, b) = \{x \in \mathbb{R} | a < x < b\}\). We want to extend this idea to apply to \(\mathbb{R}^n\). We also introduce the notions of bounded sets and closed sets in \(\mathbb{R}^n\). Recall that the \(\varepsilon\)-neighbor of a point \(x\in\mathbb{R}^n\) is defined as \(N_{\varepsilon}(x) = \{y \in \mathbb{R}^n | d(x, y) < \varepsilon \}\) where \(d(x,y)\) is the distance between \(x\) and \(y\). Definition (Open set) A subset \(U\) of \(\mathbb{R}^n\) is said to be an  open set  if the following holds: \[\forall x \in U ~ \exists \delta > 0 ~ (N_{\delta}(x) \subset U).\tag{Eq:OpenSet}\] That is, for every point in an open set \(U\), we can always find an open ball centered at that point, that is included in \(U\). See the following figure. Perhaps, it is instructive to see what is  not  an open set. Negating (Eq:OpenSet), we have \[\exists x \in U ~ \forall \delta > 0 ~ (N_{\delta}(x) \not
Read more

Euclidean spaces

-
Image
We would like to study multivariate functions (i.e., functions of many variables), continuous multivariate functions in particular. To define continuity, we need a measure of "closeness" between points. One measure of closeness is the Euclidean distance. The set \(\mathbb{R}^n\) (with \(n \in \mathbb{N}\)) with the Euclidean distance function is called a Euclidean space. This is the space where our functions of interest live. The real line is a geometric representation of \(\mathbb{R}\), the set of all real numbers. That is, each \(a \in \mathbb{R}\) is represented as the point \(a\) on the real line. The coordinate plane , or the \(x\)-\(y\) plane , is a geometric representation of \(\mathbb{R}^2\), the set of all pairs of real numbers. Each pair of real numbers \((a, b)\) is visualized as the point \((a, b)\) in the plane. Remark . Recall that \(\mathbb{R}^2 = \mathbb{R}\times\mathbb{R} = \{(x, y) | x, y \in \mathbb{R}\}\) is the Cartesian product of \(\mathbb{R}\) with i
Read more

Applications of multiple integrals

-
Image
We can use multiple integrals to compute areas and volumes of various shapes. Area of a planar region Definition (Area) Let \(D\) be a bounded closed region in \(\mathbb{R}^2\). \(D\) is said to have an area if the multiple integral of the constant function 1 over \(D\), \(\iint_Ddxdy\), exists. Its value is denoted by \(\mu(D)\): \[\mu(D) = \iint_Ddxdy.\] Example . Let us calculate the area of the disk \(D = \{(x,y)\mid x^2 + y^2 \leq a^2\}\). Using the polar coordinates, \(x = r\cos\theta, y = r\sin\theta\), \(dxdy = rdrd\theta\), and the ranges of \(r\) and \(\theta\) are \([0, a]\) and \([0, 2\pi]\), respectively. Thus, \[\begin{eqnarray*} \mu(D) &=& \iint_Ddxdy\\ &=&\int_0^a\left(\int_0^{2\pi}rd\theta\right)dr\\ &=&2\pi\int_0^a rdr\\ &=&2\pi\left[\frac{r^2}{2}\right]_0^a = \pi a^2. \end{eqnarray*}\] □ Volume of a solid figure Definition (Volume) Let \(V\) be a solid figure in the \((x,y,z)\) space \(\mathbb{R}^3\). \(V\) is sai
Read more