# How do you find the taylor series series for #((e^x)-1)/x# at c=0?

##### 1 Answer

Apr 13, 2018

# (e^x-1)/x = 1 + x/(2!) + x^2/(3!) + x^3/(4!) + x^4/(5!) + ... #

#### Explanation:

Whilst we could start from first principles and derive using the MacLaurin formula:

# f(x) = f(0) + (f'(0))/(1!)x + (f''(0))/(2!)x^2 + (f'''(0))/(3!)x^3 + ... (f^((n))(0))/(n!)x^n + ...#

This would seem to be to a significant amount of work when we can utilise the well known series for

# e^x = 1 + x + x^2/(2!) + x^3/(3!) + x^4/(4!) + ...#

From which we get:

# (e^x-1)/x = ({1 + x + x^2/(2!) + x^3/(3!) + x^4/(4!) + ...}-1)/x #

# \ \ \ \ \ \ \ \ \ \ \ = (x + x^2/(2!) + x^3/(3!) + x^4/(4!) + ...)/x #

# \ \ \ \ \ \ \ \ \ \ \ = 1 + x/(2!) + x^2/(3!) + x^3/(4!) + x^4/(5!) + ... #