Question

Calculate the fourth degree Taylor polynomial about x0 = 0.5 for the function f(x) = sin^(-1) √x

Answer 1

Find the fourth  Taylor polynomial approximation of $\displaystyle{sin^{-1}\left(\sqrt{{x}}\right)}$ at $x=\frac{1}{2}$

$f(\frac{1}{2})=\frac{\pi}{ 4}$

$f'(x)=\frac{1}{ 2\cdot \sqrt{{x}-{{x}}^{2}}} \implies f'(\frac{1}{2})=$ 1

$f''(x)=\frac{1}{ 4} (2x-1)(x-x^2)^{-\frac{3}{2}} \implies f''(\frac{1}{2})=0$

$f^{(3)}(x)=\frac{1}{ 2} (x-x^2)^{-\frac{3}{2}} + \frac{3}{8} (2x-1)^2(x-x^2)^{-\frac{5}{2}} \implies f^{(3)}(\frac{1}{2})=4$

$f^{(4)}(x)=\frac{3}{ 4} (x-x^2)^{-\frac{5}{2}}(2x-1) + \\ + \frac{3}{8} (4(2x-1)(x-x^2)^{-\frac{5}{2}} +\frac{5}{2}(2x-1)^3 (x-x^2)^{-\frac{7}{2}}) \implies f^{(4)}(\frac{1}{2})=0$

The fourth degree Taylor polynomia:

$\displaystyle{sin^{-1}\left(\sqrt{{x}}\right)} \approx \frac{\pi}{ 4}-(x-\frac{1}{2})+\frac{2}{3}(x-\frac{1}{2})^3$

close to $x=\frac{1}{2}$