6.4 Working with taylor series  (Page 6/11)

${\left(1+x^2\right)}^{\mathrm{-1}\text{/}3}={\displaystyle \sum _{n=0}^{\infty }\left(\begin{array}{c}-\frac{1}{3}\hfill \\ \hfill n\hfill \end{array}\right)x^{2n}}$

${\left(1-2x\right)}^{2\text{/}3}={\displaystyle \sum _{n=0}^{\infty }{\left(\mathrm{-1}\right)}^n{2}^n}\left(\begin{array}{c}\frac{2}{3}\hfill \\ n\hfill \end{array}\right)x^n$

$\sqrt{2+x^2}={\displaystyle \sum _{n=0}^{\infty }{2}^{\left(1\text{/}2\right)-n}}\left(\begin{array}{c}\frac{1}{2}\hfill \\ n\hfill \end{array}\right)x^{2n};\left(\left|x^2\right|<2\right)$

$\sqrt{2x-x^2}=\sqrt{1-{\left(x-1\right)}^2}$ so $\sqrt{2x-x^2}={\displaystyle \sum _{n=0}^{\infty }{\left(\mathrm{-1}\right)}^n}\left(\begin{array}{c}\frac{1}{2}\hfill \\ n\hfill \end{array}\right){\left(x-1\right)}^{2n}$

$\sqrt{x}=2\sqrt{1+\frac{x-4}{4}}$ so $\sqrt{x}={\displaystyle \sum _{n=0}^{\infty }{2}^{1-2n}\left(\begin{array}{c}\frac{1}{2}\hfill \\ n\hfill \end{array}\right){\left(x-4\right)}^n}$

$\sqrt{x}={\displaystyle \sum _{n=0}^{\infty }{3}^{1-3n}\left(\begin{array}{c}\frac{1}{2}\hfill \\ n\hfill \end{array}\right){\left(x-9\right)}^n}$

$10{\left(1+\frac{x}{1000}\right)}^{1\text{/}3}={\displaystyle \sum _{n=0}^{\infty }{10}^{1-3n}}\left(\begin{array}{c}\frac{1}{3}\hfill \\ n\hfill \end{array}\right)x^n.$ Using, for example, a fourth-degree estimate at $x=1$ gives $\begin{array}{cc}\hfill {\left(1001\right)}^{1\text{/}3}& \approx 10\left(1+\left(\begin{array}{c}\frac{1}{3}\hfill \\ 1\hfill \end{array}\right){10}^{\mathrm{-3}}+\left(\begin{array}{c}\frac{1}{3}\hfill \\ 2\hfill \end{array}\right){10}^{\mathrm{-6}}+\left(\begin{array}{c}\frac{1}{3}\hfill \\ 3\hfill \end{array}\right){10}^{\mathrm{-9}}+\left(\begin{array}{c}\frac{1}{3}\hfill \\ 4\hfill \end{array}\right){10}^{\mathrm{-12}}\right)\hfill \\ & =10\left(1+\frac{1}{{3.10}^3}-\frac{1}{{9.10}^6}+\frac{5}{{81.10}^9}-\frac{10}{{243.10}^{12}}\right)=\mathrm{10.00333222...}\hfill \end{array}$ whereas ${\left(1001\right)}^{1\text{/}3}=10.00332222839093....$ Two terms would suffice for three-digit accuracy.

The approximation is $2.3152;$ the CAS value is $2.23\text{…}.$

The approximation is $2.583\text{…};$ the CAS value is $2.449\text{…}.$

$\sqrt{1-x^2}=1-\frac{x^2}{2}-\frac{x^4}{8}-\frac{x^6}{16}-\frac{5x^8}{128}+\text{⋯}.$ Thus

${\displaystyle {\int }_{\mathrm{-1}}^1\sqrt{1-x^2}}dx=x-\frac{x^3}{6}-\frac{x^5}{40}-\frac{x^7}{7·16}-\frac{5x^9}{9·128}+\text{⋯}{|}_{\mathrm{-1}}^1\approx 2-\frac{1}{3}-\frac{1}{20}-\frac{1}{56}-\frac{10}{9·128}+\text{error}=1.590...$ whereas $\frac{\pi }{2}=1.570...$

${\left(1+x\right)}^{4\text{/}3}=\left(1+x\right)\left(1+\frac{1}{3}x-\frac{1}{9}{x}^2+\frac{5}{81}{x}^3-\frac{10}{243}{x}^4+\text{⋯}\right)=1+\frac{4x}{3}+\frac{2x^2}{9}-\frac{4x^3}{81}+\frac{5x^4}{243}+\text{⋯}$

${\left(1+{\left(x+3\right)}^2\right)}^{1\text{/}3}=1+\frac{1}{3}{\left(x+3\right)}^2-\frac{1}{9}{\left(x+3\right)}^4+\frac{5}{81}{\left(x+3\right)}^6-\frac{10}{243}{\left(x+3\right)}^8+\text{⋯}$