# Taylor expansion of $\sqrt{1+x}$

The Taylor series for  $f(x)=\sqrt{1+x}$  using the

 $T(x)=\sum_{k=0}^{\infty}\frac{f^{(k)}(a)}{k!}\,(x-a)^{k}$

is given in the table below for the first few .

The general coefficient of the expansion, for $n\geq 2$ is:

 $\displaystyle\frac{f^{(n)}(a)}{n!}$ $\displaystyle=\binom{\frac{1}{2}}{n}\,(1+a)^{\frac{1}{2}-n}$ $\displaystyle=\frac{\frac{1}{2}\left(-\frac{1}{2}\right)\left(-\frac{3}{2}% \right)\cdots\left(\frac{1}{2}-(n-1)\right)}{n(n-1)(n-2)\cdots 1}\quad(1+a)^{-% \frac{2n-1}{2}}$ $\displaystyle=\frac{\frac{1}{2}\,(-\frac{1}{2})^{n-1}\quad 1\cdot 3\cdots(2n-3% )}{n(n-1)(n-2)\cdots 1}\quad(1+a)^{-\frac{2n-1}{2}}$ $\displaystyle=\frac{(-1)^{n-1}}{2^{n}\>n!}\frac{(2n-3)!}{(2n-4)(2n-6)\cdots 2}% \quad(1+a)^{-\frac{2n-1}{2}}$ $\displaystyle=(-1)^{n-1}\>\frac{(2n-3)!}{2^{2n-2}\>n!\,(n-2)!}\quad(1+a)^{-% \frac{2n-1}{2}}\,.$
Title Taylor expansion of $\sqrt{1+x}$ TaylorExpansionOfsqrt1x 2013-03-22 15:45:52 2013-03-22 15:45:52 stevecheng (10074) stevecheng (10074) 11 stevecheng (10074) Example msc 26A09 ExamplesOnHowToFindTaylorSeriesFromOtherKnownSeries