## University Calculus: Early Transcendentals (3rd Edition)

$2-\dfrac{(x+1)}{2^1 1!}+\dfrac{3(x+1)^2}{2^3 2!}+\dfrac{9(x+1)^3}{2^5 3!}...$
We are given that $f(x)=\sqrt {3+x^2}$ The Taylor polynomial of order $n$ for the function $f(x)$ at the point $k$ can be defined as: $p_n(x)=f(k)+\dfrac{f'(k)}{1!}(x-k)+\dfrac{f''(k)}{2!}(x-k)^2+....+\dfrac{f^{n}(k)}{n!}(x-k)^n$ Now, $f'(x)=x(3+x^2)^{-1/2}\\f''(x)=-x^2(3+x^2)^{-3/2}+(3+x^2)^{-1/2}\\f'''(x)=3x^3(3+x^2)^{-5/2}-3x(3+x^2)^{-3/2}$ Here, $f(-1)=2 \\ f'(-1)=\dfrac{-1}{2} \\f''(0)=\dfrac{3}{8}\\ f'''(-1)=\dfrac{9}{32}$ Thus, the given series follows the pattern of $f(x)=\sqrt {3+x^2}=2+(\dfrac{-1}{2})(x+1)+\dfrac{(\dfrac{3}{8})}{2!}(x+1)^2+\dfrac{(\dfrac{9}{32})}{2!}(x+1)^3....=2-\dfrac{(x+1)}{2^1 1!}+\dfrac{3(x+1)^2}{2^3 2!}+\dfrac{9(x+1)^3}{2^5 3!}...$