# Exponential Theorem

## Exponential Theorem

If x ϵ R then $${{e}^{-x}}=1-\frac{x}{1!}+\frac{{{x}^{2}}}{2!}-\frac{{{x}^{3}}}{3!}+…+\frac{{{x}^{n}}}{n!}$$.

Key points:

(1) The series $${{e}^{-x}}=1-\frac{x}{1!}+\frac{{{x}^{2}}}{2!}-\frac{{{x}^{3}}}{3!}+…$$ is called exponential series. Thus $${{e}^{x}}=\sum\limits_{n=0}^{\infty }{\frac{{{x}^{n}}}{n!}}$$.

In this series (n + 1)th term is called general term. It is denoted by Tn ₊ ₁.

Now Tn ₊ ₁ = xⁿ/n!

(2) $${{e}^{-x}}=1-\frac{x}{1!}+\frac{{{x}^{2}}}{2!}-\frac{{{x}^{3}}}{3!}+…=\sum\limits_{n=0}^{\infty }{{{(-1)}^{n}}\frac{{{x}^{n}}}{n!}}$$. In this expansion Tn₊₁ = (-1)ⁿ xⁿ/n!

(3) $${{e}^{-1}}=1-\frac{1}{1!}+\frac{{{1}^{2}}}{2!}-\frac{{{1}^{3}}}{3!}+…=\sum\limits_{n=0}^{\infty }{{{(-1)}^{n}}\frac{{{x}^{n}}}{n!}}$$. In this expansion Tn ₊ ₁ = (-1)ⁿ 1ⁿ/n!

$${{e}^{-1}}=\frac{1}{2!}-\frac{1}{3!}+…=\sum\limits_{n=0}^{\infty }{{{(-1)}^{n}}\frac{1}{n!}}$$. In this expansion Tn ₊ ₁ = (-1)ⁿ/n!.

(4)  e⁻x > 0 for all x ϵ R

(5) $$\sinh x=\frac{{{e}^{x}}-{{e}^{x}}}{2}=x+\frac{{{x}^{3}}}{3!}+\frac{{{x}^{5}}}{5!}+….$$.

(6) $$\cosh x=\frac{{{e}^{x}}+{{e}^{x}}}{2}=1+\frac{{{x}^{2}}}{2!}+….$$.

(7) If x, a ϵ R and a > 0 then $${{a}^{x}}=1+x.{{\log }_{e}}a+\frac{{{x}^{2}}}{2!}{{({{\log }_{e}}a)}^{2}}+…….+\frac{{{x}^{n}}}{n!}{{({{\log }_{e}}a)}^{n}}$$.

Example: Find 1 + 4/2! + 7/3! + 10/4! + . . . is equal to

Solution: Given that 1 + 4/2! + 7/3! + 10/4! + … = ?

1, 4, 7, 10 are in Ap

Tn = a + (n – 1) d

= 1 + (n – 1) 3

= 1 + 3n – 3

$$=\sum\limits_{n=1}^{\infty }{\frac{3n-2}{n!}}$$.

$$=\sum\limits_{n=1}^{\infty }{\frac{3n}{n!}}-\sum\limits_{n=1}^{\infty }{\frac{2}{n!}}$$.

$$=3\sum\limits_{n=1}^{\infty }{\frac{1}{(n-1)!}}-2\sum\limits_{n=1}^{\infty }{\frac{1}{n!}}$$.

= 3e – 2(e – 1)

= e + 2

1 + 4/2! + 7/3! + 10/4! + … = e + 2.