How do you write the taylor series for the function $f\left( x \right)=\sqrt{x}$ at $a=16$?

Hint: We start solving the problem by recalling the fact that the taylor series of the function $f\left( x \right)$ at $x=a$ is defined as $f\left( x \right)=f\left( a \right)+{{f}^{'}}\left( a \right)\times \dfrac{\left( x-a \right)}{1!}+{{f}^{''}}\left( a \right)\times \dfrac{{{\left( x-a \right)}^{2}}}{2!}+.......{{f}^{n}}\left( a \right)\times \dfrac{{{\left( x-a \right)}^{n}}}{n!}+.....$. We then use this result for the given function at $a=16$. We then find the general form of all the derivatives of the function $f\left( x \right)$ and substitute in the formula. We then make the necessary calculations to get the required answer.

Complete step-by-step solution:
According to the problem, we are asked to find the taylor series for the function $f\left( x \right)=\sqrt{x}$ at $a=16$ and also the radius of convergence.
We know that the taylor series of the function $f\left( x \right)$ at $x=a$ is defined as $f\left( x \right)=f\left( a \right)+{{f}^{'}}\left( a \right)\times \dfrac{\left( x-a \right)}{1!}+{{f}^{''}}\left( a \right)\times \dfrac{{{\left( x-a \right)}^{2}}}{2!}+.......{{f}^{n}}\left( a \right)\times \dfrac{{{\left( x-a \right)}^{n}}}{n!}+.....$.
Now, let us use this result to find the Taylor series for the given function at $a=16$.
$f\left( x \right)=f\left( 16 \right)+{{f}^{'}}\left( 16 \right)\times \dfrac{\left( x-16 \right)}{1!}+{{f}^{''}}\left( 16 \right)\times \dfrac{{{\left( x-16 \right)}^{2}}}{2!}+.......{{f}^{n}}\left( 16 \right)\times \dfrac{{{\left( x-16 \right)}^{n}}}{n!}+.....$ ---(1).
Now, let us find the derivatives of the functions $f\left( x \right)$.
We have $f\left( x \right)=\sqrt{x}$.
$\Rightarrow {{f}^{'}}\left( x \right)=\dfrac{d\left( {{x}^{\dfrac{1}{2}}} \right)}{dx}$ ---(2).
We know that $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$. Let us use this result in equation (2).
$\Rightarrow {{f}^{'}}\left( x \right)=\dfrac{1}{2}{{x}^{\dfrac{-1}{2}}}$ ---(3).
Now, ${{f}^{''}}\left( x \right)=\dfrac{d\left( \dfrac{1}{2}{{x}^{\dfrac{-1}{2}}} \right)}{dx}$.
$\Rightarrow {{f}^{''}}\left( x \right)=\dfrac{-1}{4}{{x}^{\dfrac{-3}{2}}}$ ---(4).
Now, ${{f}^{3}}\left( x \right)=\dfrac{d\left( \dfrac{-1}{4}{{x}^{\dfrac{-3}{2}}} \right)}{dx}$.
$\Rightarrow {{f}^{3}}\left( x \right)=\dfrac{3}{8}{{x}^{\dfrac{-5}{2}}}$ ---(5).
Now, ${{f}^{4}}\left( x \right)=\dfrac{d\left( \dfrac{3}{8}{{x}^{\dfrac{-5}{2}}} \right)}{dx}$.
$\Rightarrow {{f}^{4}}\left( x \right)=\dfrac{-15}{16}{{x}^{\dfrac{-7}{2}}}$ ---(6).
Similarly, we get ${{f}^{n}}\left( x \right)=\dfrac{{{\left( -1 \right)}^{n}}\left( {{2}^{n}}-1 \right)}{\left( {{2}^{n}} \right)}{{x}^{\dfrac{-2n+3}{2}}}$ ---(7).
Now, let us substitute equations (2), (3), (4), (5), (6), and (7) in equation (1).
$f\left( x \right)=\sqrt{16}+\left( \dfrac{1}{2}{{\left( 16 \right)}^{\dfrac{-1}{2}}} \right)\times \dfrac{\left( x-16 \right)}{1!}+\left( \dfrac{-1}{4}{{\left( 16 \right)}^{\dfrac{-3}{2}}} \right)\times \dfrac{{{\left( x-16 \right)}^{2}}}{2!}+.......\left( \dfrac{{{\left( -1 \right)}^{n}}\left( {{2}^{n}}-1 \right)}{\left( {{2}^{n}} \right)}{{\left( 16 \right)}^{\dfrac{-2n+3}{2}}} \right)\times \dfrac{{{\left( x-16 \right)}^{n}}}{n!}+.....$.
$\Rightarrow f\left( x \right)=4+\dfrac{\left( x-16 \right)}{8}-\dfrac{{{\left( x-16 \right)}^{2}}}{512}+.......\left( \dfrac{{{\left( -1 \right)}^{n}}\left( {{2}^{n}}-1 \right)}{\left( {{2}^{n}} \right)}{{\left( 16 \right)}^{\dfrac{-2n+3}{2}}} \right)\times \dfrac{{{\left( x-16 \right)}^{n}}}{n!}+.....$.
So, we have found the taylor series of the function $f\left( x \right)$ at $x=a$ as
$\Rightarrow f\left( x \right)=4+\dfrac{\left( x-16 \right)}{8}-\dfrac{{{\left( x-16 \right)}^{2}}}{512}+.......\left( \dfrac{{{\left( -1 \right)}^{n}}\left( {{2}^{n}}-1 \right)}{\left( {{2}^{n}} \right)}{{\left( 16 \right)}^{\dfrac{-2n+3}{2}}} \right)\times \dfrac{{{\left( x-16 \right)}^{n}}}{n!}+.....$.

Note: We should perform each step carefully to avoid confusion and calculation mistakes. We should confuse between the variables a and x while solving this type of problem. We can also find the Maclaurin series for the given function $f\left( x \right)$ by taking $a=1$. Similarly, we can expect problems to find the Taylor series of the function $g\left( x \right)=\sin 3x$.