Answer
$p_{0}(x)=2\\ p_{1}(x)=2+\dfrac{(x-4)}{4} \\p_{2}(x)=2+\dfrac{(x-4)}{4}-\dfrac{(x-4)^2}{64}\\p_{3}(x)=2+\dfrac{(x-4)}{4}-\dfrac{(x-4)^2}{64}+\dfrac{1}{512}(x-4)^3$
Work Step by Step
Taylor polynomial of order $n$ for the function $f(x)$ at the point $k$ can be defined as:
$p_n(x)=f(k)+\dfrac{f'(k)}{1!}(x-k)+\dfrac{f''(k)}{2!}(x-k)^2+....+\dfrac{f^{n}(k)}{n!}(x-k)^n$
Here, $f(4)=2 \\ f'(4)=\dfrac{1}{4}\\f''(4)=-\dfrac{1}{32}\\ f'''(4)=\dfrac{3}{256}$
Thus, $p_{0}(x)=2\\ p_{1}(x)=2+\dfrac{(x-4)}{4} \\p_{2}(x)= 2+\dfrac{(x-4)}{4}+(-1/32)\dfrac{(x-4)^2}{2}=2+\dfrac{(x-4)}{4}-\dfrac{(x-4)^2}{64}\\p_{3}(x)=2+\dfrac{(x-4)}{4}-\dfrac{(x-4)^2}{64}+\dfrac{(3/256)}{6}(x-4)^3=2+\dfrac{(x-4)}{4}-\dfrac{(x-4)^2}{64}+\dfrac{1}{512}(x-4)^3$
