Question

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

Answer 1

`(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`

`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!) + ...`