Question

How do you find the taylor series of `1 + sin(x)`?

Answer 1

The general form of the Taylor series is
`f(x) = Sigma_(n=0)^oo (f^(|n|) (a))/(n!)*(x-a)^n`
where `f^(|n|)` denotes the `n^(th)` derivative of `f`.

Set `f(x) = sin(x)`
and use `a=0`

Remember
`( dsin(x))/(dx) = cos(x)`
`(d cos(x))/(dx) = -sin(x)`
`sin(0) = 0`
`cos(0)=1`

Then the Taylor expansion of `1 + sin(x)`
becomes
`1 + x -(x^3)/(3!) + (x^5)/(5!) - (x^7)/(7!) + (x^9)/(9!) ...`