Question

Obtain the Taylor series for f(x) = √x at a=4

Answer 1

Given

$f(x)= \sqrt{x},\ \ \ \ \ \ \ \ a= 4$

Since

$\begin{aligned} f(x) &=\sqrt{x} \ \ \ \ \ \ \ \ \ \ f(4)=2 \\ f'(x) &=\frac{1}{2}x^{-1/2} \ \ \ \ \ \ \ \ \ \ f'(4)=\frac{1}{4}\\ f''(x) &=\frac{-1}{4}x^{-3/2} \ \ \ \ \ \ \ \ \ \ f''(4)=\frac{-1}{32}\\ f'''(x) &=\frac{3}{8}x^{-5/2} \ \ \ \ \ \ \ \ \ \ f'''(4)=\frac{3}{256}\\ f^{(4)}(x) &=\frac{-15}{16}x^{-7/2} \ \ \ \ \ \ \ \ \ \ f^{(4)}(4)=\frac{-15}{2048}\\ f^{(5)}(x) &=\frac{105}{32}x^{-9/2} \ \ \ \ \ \ \ \ \ \ f^{(5)}(4)=\frac{105}{16384}\\ \end{aligned}$

Then  Taylor series for f(x) 

$f(x) \approx P(x)\\ =\sum_{k=0}^{n} \frac{f^{(k)}(a)}{k !}(x-a)^{k}\\ = 2+\frac{1}{4}(x-4)-\frac{1}{64}(x-4)^{2}+\frac{1}{512}(x-4)^{3}-\\-\frac{5}{16384}(x-4)^{4} +\frac{7}{131072}(x-4)^{5}+\cdots$