Answer
The Taylor series centered at $c=4$ is
$T\left( x \right) = \mathop \sum \limits_{n = 0}^\infty \left( {\begin{array}{*{20}{c}}
{\dfrac{1}{2}}\\
n
\end{array}} \right)\dfrac{2}{{{4^n}}}{\left( {x - 4} \right)^n}$
It is valid for $\left| {x - 4} \right| \lt 4$.
Work Step by Step
Write $f\left( x \right) = \sqrt x = \sqrt {4 + \left( {x - 4} \right)} = 2\sqrt {1 + \left( {\dfrac{{x - 4}}{4}} \right)} $.
We have from Table 2:
${\left( {1 + x} \right)^a} = \mathop \sum \limits_{n = 0}^\infty \left( {\begin{array}{*{20}{c}}
a\\
n
\end{array}} \right){x^n} = 1 + ax + \dfrac{{a\left( {a - 1} \right)}}{{2!}}{x^2} + \dfrac{{a\left( {a - 1} \right)\left( {a - 2} \right)}}{{3!}}{x^3} + \cdot\cdot\cdot$
converges for $\left| x \right| \lt 1$.
Substituting $\left( {\dfrac{{x - 4}}{4}} \right)$ for $x$ in the series above, we get
$\sqrt {1 + \left( {\dfrac{{x - 4}}{4}} \right)} = \mathop \sum \limits_{n = 0}^\infty \left( {\begin{array}{*{20}{c}}
{\dfrac{1}{2}}\\
n
\end{array}} \right){\left( {\dfrac{{x - 4}}{4}} \right)^n}$, ${\ \ \ }$ converges for $\left| {\dfrac{{x - 4}}{4}} \right| \lt 1$
Thus, the Taylor series centered at $c=4$ is $T\left( x \right) = \mathop \sum \limits_{n = 0}^\infty \left( {\begin{array}{*{20}{c}}
{\dfrac{1}{2}}\\
n
\end{array}} \right)\dfrac{2}{{{4^n}}}{\left( {x - 4} \right)^n}$. It is valid for $\left| {x - 4} \right| \lt 4$.
