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.