Question

How do you find the first terms of the taylor series `f(x)=1/sqrt(x)` centered at `x=4`?

Answer 1

If `f(x) = 1/(sqrt(x)) = x^(-1/2)`
then
`f'(x) = - 1/2x^(-3/2) = - 1/2 (1/sqrt(x))^3`
`f''(x) = 3/4x^(-5/2) - 1/2 (1/sqrt(x))^5`
`f'''(x) = -(15)/8x^(-7/2) - 1/2 (1/sqrt(x))^7`
and so on

The Taylor series is given by
`sum_(n=0)^oo = (f^('n)(a))/(n!) * (x - a)^n`

so the first `4` terms of the Taylor series of `f(x) = 1/(sqrt(x))` centered at `a=4` would be
(after noting that `sqrt(4) = 2`)

`1//2 + (- 1/2 (1/2)^3)(x-4) + ((3/4 (1/2)^5)(x-4)^2)/2 + ((- 7/2 (1/2)^7)(x-4)^3)/(3!)`

From there on, it's just arithmetic.