2 Finding square root of any number

Given any number, to find its sqrt, we find a number whose sqrt is known and closest to the number we want to find its sqrt. For an example, to find \(\sqrt {17}\), we know \(\sqrt {16}=4\). Hence we start by writing

\[ f\left ( x\right ) =\sqrt {x}\]

And now Taylor series is used to approximate \(\sqrt {17}\). The expansion point is always the point where \(\sqrt {x}\) is known, which is \(x=16\) in this example.

Therefore we have

\[ f\left ( 17\right ) =f\left ( 16\right ) +\left . f^{\prime }\left ( x\right ) \right \vert _{x=16}h+\left . f^{\prime \prime }\left ( x\right ) \right \vert _{x=16}\frac {h^{2}}{2!}+\left . f^{\prime \prime \prime }\left ( x\right ) \right \vert _{x=16}\frac {h^{3}}{3!}+\cdots \]

So all what we have to do is find derivatives of \(\sqrt {x}\) and evaluate that at \(x=16\). In the above \(h\) is the distance between \(x=17\) and \(x=16\) which is one. The above gives

\begin{align*} f\left ( 17\right ) & =f\left ( 16\right ) +\left ( \frac {1}{2\sqrt {x}}\right ) _{x=16}-\left ( \frac {1}{4x^{\frac {3}{2}}}\right ) _{x=16}\frac {1}{2}+\left ( \frac {3}{8x^{\frac {5}{2}}}\right ) \frac {1}{6}+\cdots \\ & =4+\left ( \frac {1}{2\left ( 4\right ) }\right ) -\left ( \frac {1}{4\left ( 4\right ) ^{3}}\right ) \frac {1}{2}+\left ( \frac {3}{8\left ( 4\right ) ^{5}}\right ) \frac {1}{6}+\cdots \\ & =4+\frac {1}{8}-\left ( \frac {1}{4\left ( 64\right ) }\right ) \frac {1}{2}+\left ( \frac {3}{8\left ( 1024\right ) }\right ) \frac {1}{6}+\cdots \\ & =4+\frac {1}{8}-\frac {1}{512}+\frac {1}{16\,384}+\cdots \\ & =4+0.125-0.0019531+0.000061035-\cdots \\ & =4.1231\cdots \end{align*}

We see that to get good approximation, only 3 terms should be enough. The closer the expansion point is to the actual \(x\) we are trying to find sqrt at, the better the approximation is and less terms will be needed.