Maclaurin Series of ln(1+x)

In this tutorial we shall derive the series expansion of the trigonometric function $\ln \left( {1 + x} \right)$ by using Maclaurin’s series expansion function.

Consider the function of the form

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

Using $x = 0$, the given equation function becomes

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

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

$$\begin{gathered} f'\left( x \right) = \frac{1}{{1 + x}} = {\left( {1 + x} \right)^{ - 1}},\,\,\,\,\,\,\,\,\,\,f'\left( 0 \right) = {\left( {1 + 0} \right)^{ - 1}} = 1 \\ f''\left( x \right) = - {\left( {1 + x} \right)^{ - 2}},\,\,\,\,\,\,\,\,\,\,f''\left( 0 \right) = - {\left( {1 + x} \right)^{ - 2}} = - 1 \\ f'''\left( x \right) = 2{\left( {1 + x} \right)^{ - 3}},\,\,\,\,\,\,\,\,\,\,f'''\left( 0 \right) = 2{\left( {1 + 0} \right)^{ - 3}} = 2 \\ {f^{\left( {{\text{iv}}} \right)}}\left( x \right) = - 6{\left( {1 + x} \right)^{ - 4}},\,\,\,\,\,\,\,\,\,\,{f^{\left( {{\text{iv}}} \right)}}\left( 0 \right) = - 6{\left( {1 + 0} \right)^{ - 4}} = - 6 \\ \cdots \cdots \cdots \; \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\ \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\ \end{gathered}$$

Now using Maclaurin’s series expansion function, we have

$$f\left( x \right) = f\left( 0 \right) + xf'\left( 0 \right) + \frac{{{x^2}}}{{2!}}f''\left( 0 \right) + \frac{{{x^3}}}{{3!}}f'''\left( 0 \right) + \frac{{{x^4}}}{{4!}}{f^{\left( {{\text{iv}}} \right)}}\left( 0 \right) + \cdots$$

Putting the values in the above series, we have

$$\begin{gathered} \ln \left( {1 + x} \right) = 0 + x\left( 1 \right) + \frac{{{x^2}}}{{2!}}\left( { - 1} \right) + \frac{{{x^3}}}{{3!}}\left( 2 \right) + \frac{{{x^4}}}{{4!}}\left( { - 6} \right) + \cdots \\ \ln \left( {1 + x} \right) = x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6}\left( 2 \right) + \frac{{{x^4}}}{{24}}\left( { - 6} \right) + \cdots \\ \ln \left( {1 + x} \right) = x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3} - \frac{{{x^4}}}{4} + \cdots \\ \end{gathered}$$