Maclaurin Series of Sqrt(1+x)

In this tutorial we shall derive the series expansion of \sqrt {1 + x} by using Maclaurin’s series expansion function.

Consider the function of the form

f\left( x \right) = \sqrt {1 + x}

Using x = 0, the given equation function becomes

f\left( 0 \right) = \sqrt {1 + 0} = \sqrt 1 = 1

Now taking the derivatives of the given function and using x = 0, we have

\begin{gathered} f'\left( x \right) = \frac{1}{{2\sqrt {1 + x} }} = \frac{1}{2}{\left( {1 + x} \right)^{ - \frac{1}{2}}},\,\,\,\,\,\,\,\,\,\,f'\left( 0 \right) = \frac{1}{{2\sqrt {1 + 0} }} = \frac{1}{{2\sqrt 1 }} = \frac{1}{2} \\ f''\left( x \right) = - \frac{1}{4}{\left( {1 + x} \right)^{ - \frac{3}{2}}},\,\,\,\,\,\,\,\,\,\,f''\left( 0 \right) = - \frac{1}{4}{\left( {1 + 0} \right)^{ - \frac{3}{2}}} = - \frac{1}{4} \\ f'''\left( x \right) = \frac{3}{8}{\left( {1 + x} \right)^{ - \frac{5}{2}}},\,\,\,\,\,\,\,\,\,\,f'''\left( 0 \right) = \frac{3}{8}{\left( {1 + 0} \right)^{ - \frac{5}{2}}} = \frac{3}{8} \\ \cdots \cdots \cdots \; \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\ \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\ \end{gathered}

Now using Maclaurin’s series expansion function, we have

f\left( x \right) = f\left( 0 \right) + xf'\left( 0 \right) + \frac{{{x^2}}}{{2!}}f''\left( 0 \right) + \frac{{{x^3}}}{{3!}}f'''\left( 0 \right) + \cdots

Putting the values in the above series, we have

\begin{gathered} \sqrt {1 + x} = 1 + x\left( {\frac{1}{2}} \right) + \frac{{{x^2}}}{{2!}}\left( { - \frac{1}{4}} \right) + \frac{{{x^3}}}{{3!}}\left( { - \frac{3}{8}} \right) + \cdots \\ \sqrt {1 + x} = 1 + \frac{x}{2} - \frac{{{x^2}}}{8} + \frac{{{x^3}}}{{16}} - \cdots \\ \end{gathered}