Question

How do you find the taylor series for `e^x - e^-x`?

Answer 1

Taylor series centered at a would be
`e^a {1 + (x-a) +(x-a)^2 /(2!) + ........} - e^-a {1- (x-a)+ (x-a)^2 /(2!) .....}`

Explanation:

Taylor series for f(x)= `e^x -e^-x`, centered at x=a would be
`f(x)= f(a)+ (x-a)f'(a)+(x-a)^2 (f'' (a))/(2!)` +.......

`f(a)= e^a -e^-a`
`f'(a)=e^a +e^-a`
`f''(a)= e^a -e^-a` and so on.......
`f(x)=e^a -e^-a +(x-a)(e^a +e^-a)+ (x-a)^2 (e^a-e^-a)/(2!)`+-----

=`e^a {1 + (x-a) +(x-a)^2 /(2!) + ........} - e^-a {1- (x-a)+ (x-a)^2 /(2!) .....}`