## Thomas' Calculus 13th Edition

$2-\dfrac{1}{2^1 1!}(x+1)+\dfrac{3}{2^3 2!}(x+1)^2+\dfrac{9}{2^5 3!}(x+1)^3...$
Given: $f(x)=\sqrt {3+x^2}$ The taylor polynomial of order $n$ for the function $f(x)$ at the point $a$ can be defined as: $A_n(x)=f(k)+\dfrac{f'(k)}{1!}(x-a)+\dfrac{f''(a)}{2!}(x-a)^2+....+\dfrac{f^{n}(a)}{n!}(x-a)^n$ Now, $f(-1)=2$ and $f'(x)=x(3+x^2)^{-1/2} \implies f'(-1)=\dfrac{-1}{2}\\f''(x)=-x^2(3+x^2)^{-3/2}+(3+x^2)^{-1/2} \implies f''(0)=\dfrac{3}{8}\\f'''(x)=3x^3(3+x^2)^{-5/2}-3x(3+x^2)^{-3/2} \implies f'''(-1)=\dfrac{9}{32}$ Taylor polynomial of order $n$ for the function $f(x)=\sqrt {3+x^2}$ $f(x)=2+(\dfrac{-1}{2})(x+1)+\dfrac{(\dfrac{3}{8})}{2!}(x+1)^2+\dfrac{(\dfrac{9}{32})}{2!}(x+1)^3....=2-\dfrac{1}{2^1 1!}(x+1)+\dfrac{3}{2^3 2!}(x+1)^2+\dfrac{9}{2^5 3!}(x+1)^3...$