# How do you find the taylor series series for  (4-x)^(1/2)?

Jun 3, 2015

There's a few different ways this can be done. Probably the most straight-forward approach is to let $f \left(x\right) = {\left(4 - x\right)}^{\frac{1}{2}}$ and write the Taylor series about $x = 0$ as f(0)+f'(0)x+(f''(0))/(2!)x^2+(f'''(0))/(3!)x^3+\cdots

Since $f \left(x\right) = {\left(4 - x\right)}^{\frac{1}{2}}$, $f \left(0\right) = {4}^{\frac{1}{2}} = 2$, $f ' \left(x\right) = \setminus \frac{1}{2} {\left(4 - x\right)}^{- \frac{1}{2}} \cdot \left(- 1\right)$, $f ' \left(0\right) = - \setminus \frac{1}{2} \cdot {4}^{- \frac{1}{2}} = - \setminus \frac{1}{4}$,

$f ' ' \left(x\right) = \setminus \frac{1}{4} {\left(4 - x\right)}^{- \frac{3}{2}} \cdot \left(- 1\right)$, $f ' ' \left(0\right) = - \setminus \frac{1}{4} \cdot {4}^{- \frac{3}{2}} = - \setminus \frac{1}{32}$, etc...

Eventually this leads to the Taylor series $2 - \setminus \frac{1}{4} x - \setminus \frac{1}{64} {x}^{2} - \setminus \frac{1}{512} {x}^{3} - \setminus \cdots$, which converges to $f \left(x\right) = {\left(4 - x\right)}^{\frac{1}{2}}$ for $| x | < 4$.

Another approach is to use a bit of slick algebra along with the generalized binomial theorem (discovered by Newton) that (1+x)^{p}=1+px+\frac{p(p-1}}{2!}x^2+\frac{p(p-1)(p-2)}{3!}x^{3}+\cdots for $| x | < 1$.

We have: ${\left(4 - x\right)}^{\frac{1}{2}} = {4}^{\frac{1}{2}} {\left(1 - \frac{x}{4}\right)}^{\frac{1}{2}} = 2 {\left(1 + \left(- \frac{x}{4}\right)\right)}^{\frac{1}{2}}$. Now use the generalized binomial theorem with $- \frac{x}{4}$ in place of $x$ and $p = \frac{1}{2}$ to get, for $| - \frac{x}{4} | < 1 \setminus \Leftrightarrow | x | < 4$

${\left(4 - x\right)}^{\frac{1}{2}}$

$= 2 {\left(1 + \left(- \frac{x}{4}\right)\right)}^{\frac{1}{2}}$

=2(1+(1/2)(-x/4)+\frac{1/2*-1/2}{2!}(-x/4)^2+\frac{1/2*-1/2*-3/2}{3!}(-x/4)^3+\cdots)

$= 2 - \frac{x}{4} - \setminus \frac{1}{64} {x}^{2} - \setminus \frac{1}{512} {x}^{3} - \setminus \cdots$