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

Mar 4, 2015

If $f \left(x\right) = \frac{1}{\sqrt{x}} = {x}^{- \frac{1}{2}}$
then
$f ' \left(x\right) = - \frac{1}{2} {x}^{- \frac{3}{2}} = - \frac{1}{2} {\left(\frac{1}{\sqrt{x}}\right)}^{3}$
$f ' ' \left(x\right) = \frac{3}{4} {x}^{- \frac{5}{2}} - \frac{1}{2} {\left(\frac{1}{\sqrt{x}}\right)}^{5}$
$f ' ' ' \left(x\right) = - \frac{15}{8} {x}^{- \frac{7}{2}} - \frac{1}{2} {\left(\frac{1}{\sqrt{x}}\right)}^{7}$
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 $f \left(x\right) = \frac{1}{\sqrt{x}}$ 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.