Answer
$A_{0}(x)=1\\A_{1}(x)=1-\dfrac{1}{2}x \\A_{2}(x)=1-\dfrac{1}{2}x-\dfrac{1}{4}x^2\\A_{3}(x)=1-\dfrac{1}{2}x-\dfrac{1}{4}x^2-\dfrac{1}{16}x^3$
Work Step by Step
The Taylor polynomial is defined for the function $f(x)$ at the point $a$ of order $n$ as:
$A_n(x)=f(a)+\dfrac{f'(a)}{1!}(x-a)+\dfrac{f''(a)}{2!}(x-a)^2+....+\dfrac{f^{n}(a)}{n!}(x-a)^n$
Now, we get $f(0)=1 ;f'(0)=\dfrac{-1}{2};f''(0)=-\dfrac{1}{4};f'''(0)=\dfrac{3}{8}$
Thus, $A_{0}(x)=1\\A_{1}(x)=1-\dfrac{1}{2}x \\A_{2}(x)= 1-\dfrac{1}{2}x+(-\dfrac{1}{4})\dfrac{(x-0)^2}{2}=1-\dfrac{1}{2}x-\dfrac{1}{4}x^2\\A_{3}(x)=1-\dfrac{1}{2}x-\dfrac{1}{8}x^2+(\dfrac{-3}{8})(\dfrac{1}{6})x^3=1-\dfrac{1}{2}x-\dfrac{1}{4}x^2-\dfrac{1}{16}x^3$
Hence, $A_{0}(x)=1\\A_{1}(x)=1-\dfrac{1}{2}x \\A_{2}(x)=1-\dfrac{1}{2}x-\dfrac{1}{4}x^2\\A_{3}(x)=1-\dfrac{1}{2}x-\dfrac{1}{4}x^2-\dfrac{1}{16}x^3$
