Euler’s Number

$e=\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^n$

Consider $\left(1+\frac{1}{n}\right)^n$

The $k^{\text {th }}$ term for $k \leqslant n$ can be obtained from Binomial Theorem.

The $k^{\text {th }}$ term for $k=$ $0,1,2, \ldots n$ is

${}^n C_k \left( \frac{1}{n} \right)^k$

$= \frac{n!}{k !(n-k) !} \frac{1}{n^k}$

$= \frac{n(n-1) \cdots(n-\overline{k-1)}}{k !} \frac{1}{n^k}$

$=\frac{1\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right) \cdots\left(1-\frac{k-1}{n}\right)}{k !}$

$\text { As } n \rightarrow \infty$ $\frac{1}{k!}$

$\therefore \text { The limit of }\left(1+\frac{1}{n}\right)^n$

$\rightarrow \frac{1}{0 !}+\frac{1}{1 !}+\cdots+\frac{1}{k !}+ \cdots$

The infinite series can be seen as

$1+1+\frac{1}{2 !}+\frac{1}{3 !}+\cdots \cdot \cdot$

This is called Euler’s Number.

This is an irrational number

$2 \cdot {71828}1828459045 230365$

$2 < e$

Check if $e$ is greater than 3 or not.

Solution:

Consider $\frac{1}{2 !}+\frac{1}{3 !}+\frac{1}{4 !}+ \cdots$

$=\frac{1}{2}+\frac{1}{2 \cdot 3}+\frac{1}{2 \cdot 3 \cdot 4}+\frac{1}{2 \cdot 3 \cdot 4 \cdot 5} +\cdots$

$=\frac{1}{2}+\frac{1}{2 \cdot 2}+\frac{1}{2 \cdot 2 \cdot 2}+\frac{1}{2^4} +\cdots$

$=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\cdots$

$=\frac{1}{2}\left(\frac{1}{1-\frac{1}{2}}\right)=1$

$2 < e < 2+1=3$

Let us consider $e^x$ for some real or complex $x.$ What is the series for $e^x$ ?

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

Intuitive idea
Consider $e^2 = e \times e$

$\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{2 n}$

$= \lim _{n \rightarrow \infty}\left(\left(1+\frac{1}{n}\right)^2\right)^n$

$= \lim _{n \rightarrow \infty}\left(1+\frac{2}{n}+\frac{1}{n^2}\right)^n$

The $k^{\text {th }}$ term for $0 \leq k \leq n$

is ${}^n C_k\left(\frac{2}{n}+\frac{1}{n^2}\right)^k$

$= \frac{n !}{k !(n-k) !} \frac{1}{n^k}\left(2+\frac{1}{n}\right)^k$

$= \frac{1\left(1-\frac{1}{n}\right) \cdots\left(1-\frac{k-1}{n}\right)}{k !}\left(2+\frac{1}{n}\right)^k$

$\lim _{n \rightarrow \infty} \frac{1}{k !} 2^k=\frac{2^k}{k !}$
(for a fixed $k$ )

$\therefore e^2= \frac{2^0}{0 !}+\frac{2^1}{1 !}+\cdots+\frac{2^k}{k !}+\cdots$

$\text { or } 1+\frac{2}{1 !}+\frac{2^2}{2 !}+\cdots+\frac{2^k}{k !}+\cdots$

This gives the idea for

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

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

Considers the power $e^{i x}$

$=1+i x+\frac{(i x)^2}{2 !}+\frac{(i x)^3}{3 !}+\frac{(i x)^4}{4 !} +\frac{(i x)^5}{5 !}+\frac{(i x)^6}{6 !}+\cdots$

$=1+i x-\frac{x^2}{2 !}-i \frac{x^3}{3 !}+\frac{x^4}{4 !}+i \frac{x^5}{5 !} \cdots$

$= \underbrace {\left(1-\frac{x^2}{2 !}+\frac{x^4}{4 !}-\frac{x^6}{6 !} \cdots \right)}_ {=\cos x} + i \underbrace {\left(x-\frac{x^3}{3 !}+\frac{x^5}{5 !} \cdots \right)}_{=\sin x}$

$e^{i x} = \cos x+i \sin x$

Find the value of

$\frac{1}{1 !}+\frac{2}{2 !}+\frac{3}{3 !}+\cdots .$

Solution:

$\frac{1}{1 !}+\frac{2}{2 !}+\frac{3}{3 !}+\cdots = \frac{1}{1 !}+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !} \cdots .$

This is also $=e$.

1) $\frac{x}{1 !}+\frac{2 x}{2 !}+\frac{3 x}{3 !}+ \cdots$

$=x\left(\frac{1}{1 !}+\frac{2}{2 !}+\frac{3}{3 !}+\cdots\right)=x e$

2) $\sum_{n=0}^{\infty} \frac{1}{(n-1) !}=\sum_{n=1}^{\infty} \frac{1}{(n-1) !}$

$=\sum_{m=0}^{\infty} \frac{1}{m !}$ where $m=x-1$

$=e$.

$\sum_{n=0}^{\infty} \frac{1}{(n-2) !}=\sum_{n=2}^{\infty} \frac{1}{(n-2) !}$

$=\frac{1}{0 !}+\frac{1}{1 !}+\frac{1}{2 !}+\cdots$

$=e$

$\sum_{i=0}^{\infty} \frac{i^2}{i !}=\sum_{i=1}^{\infty} \frac{i^2}{i !}$

The $k^{\text {th }}$ term $=\frac{k^2}{k !}=\frac{k}{(k-1) !}$

$=\left(\frac{k-1+1}{(k-1) !}\right)=\frac{k-1}{(k-1) !}+\frac{1}{(k-1) !}$

$=\sum_{k=2}^{\infty} \frac{1}{(k-2) !}+\sum_{1}^{\infty} \frac{1}{(k-1) !}$

$=e+e=2 e$

$\sum_{n=0}^{\infty} \frac{n^3}{n !}=\sum_{n=1}^{\infty} \frac{n^2}{(n-1) !}$

$=\sum_{n=1}^{\infty} \frac{(n-1)^2+2 n-1}{(n-1) !}$

$=\sum_{n=1}^{\infty} \frac{(n-1)}{(n-2) !}+2 \sum^{\infty} \frac{n}{(n-1) !}-\sum_{n=1}^{\infty} \frac{1}{(n-1)}$

$=\sum_n \frac{(n-2)+1}{(n-2) !}+2 \sum \frac{n-1+1}{(n-1) !}-\sum \frac{1}{(n-1)}$

$=\sum \frac{1}{(n-3) !}+\sum \frac{1}{(n-2) !}+2 \sum \frac{1}{(n-2) !}$

$+2 \sum \frac{1}{(n-1) !}-\frac{1}{(n-1) !}$

$= e + e + 2e + 2e -e = 5e$

Consider $\frac{1}{1 !}+\frac{1+2}{2 !}+\frac{1+2+3}{3 !}+\cdots$

We can see the $k^{\text {th }}$ term is $\frac{\sum_{1}^{k} i}{k !}=\frac{\frac{k(k+1)}{2}}{k !}$

$= \frac{1}{2}\left(\frac{k}{k !}+\frac{k+1}{k !}\right)$

$=\frac{1}{2}(\frac{1}{(k-1) !}+ \frac{1}{(k-1) !}+\frac{1}{k !})$

$\therefore$ Taking sum, $\sum_{i=1}^{\infty} \frac{1+2+3+\cdots=i}{i!}$

$=\frac{1}{2}\left(\sum \frac{1}{(k-1) !}+\sum \frac{1}{(k-1) !}+\sum \frac{1}{k !}\right)$
$=\frac{3}{2} e$

Find the coefficient of $x^4$ in $\left(1+2 x+3 x^2\right) e^{-x}$.

Solution:

$\left(1+2 x+3 x^2\right)(1-x+\frac{x^2}{2 !}-\frac{x^3}{3 !}+\frac{x^4}{4 !}\cdots)$

$=\frac{1}{4 !}-\frac{2}{3 !}+\frac{3}{2 !}$

$=\frac{1}{24}-\frac{2}{6}+\frac{3}{2}$

$=\frac{1-8+36}{24}=\frac{29}{24}$

Find the value of

$1+5 ln 3+\frac{(5 ln 3)^2}{2 !}+ \cdots$

$= e^{5 ln 3}$

$= e^{ln 3^5}$

$= 3^5$

Find the expansion for $\frac{e^x}{\cos x}.$

Solution:

Let the corresponding series be

$C_0+C_1 x+C_2 x^2+C_3 x^3+ …$
$e^x=\cos x(C_0+C_1 x +C_2 x^2+C_3 x^3 +\ldots)$ $\left(1+x+\frac{x^2}{2 !}+\frac{x^3}{3 !}+\cdots\right)$

$=\left(1-\frac{x^2}{2 !}+\frac{x^4}{4 !}\cdots \right)$

$\left(C_0+C_1 x + +C_2 x^2+ C_3 x^3+\cdots\right)$

$\therefore C_2 =1$

$\therefore C_3 =\frac{2}{3}$

The coefficient & $\frac{e^x}{\cos x}$ are

$C_0=1$

$C_1=1$

$C_2=1$

$C_3=\frac{2}{3}$

$C_4=\frac{1}{2}$

$C_5=\frac{3}{10}$
$\vdots$