Maclaurin Series of a^x

In this tutorial we shall drive the series expansion of trigonometric function {a^x} by using Maclaurin’s series expansion function.

Consider the function of the form

f\left( x \right) = {a^x}

Put x = 0, the given equation function becomes

f\left( 0 \right) = {a^x} = {a^0} = 1

Now take derivatives of the given function and put x = 0, we have

\begin{gathered} f'\left( x \right) = {a^x}\ln a,\,\,\,\,\,\,\,\,\,\,f'\left( 0 \right) = {a^0}\ln a = \ln a \\ f''\left( x \right) = {a^x}{\left( {\ln a} \right)^2},\,\,\,\,\,\,\,\,\,\,f''\left( 0 \right) = {a^0}{\left( {\ln a} \right)^2} = {\left( {\ln a} \right)^2} \\ f'''\left( x \right) = {a^x}{\left( {\ln a} \right)^3},\,\,\,\,\,\,\,\,\,\,f'''\left( 0 \right) = {a^0}{\left( {\ln a} \right)^3} = {\left( {\ln a} \right)^3} \\ {f^{\left( {{\text{iv}}} \right)}}\left( x \right) = {a^x}{\left( {\ln a} \right)^4},\,\,\,\,\,\,\,\,\,\,{f^{\left( {{\text{iv}}} \right)}}\left( 0 \right) = {a^0}{\left( {\ln a} \right)^4} = {\left( {\ln a} \right)^4} \\ \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 above series, we have

\begin{gathered} {a^x} = 1 + x\left( {\ln a} \right) + \frac{{{x^2}}}{{2!}}{\left( {\ln a} \right)^2} + \frac{{{x^3}}}{{3!}}{\left( {\ln a} \right)^3} + \frac{{{x^4}}}{{4!}}{\left( {\ln a} \right)^4} + \cdots \\ {a^x} = 1 + x\left( {\ln a} \right) + \frac{{{{\left( {x\ln a} \right)}^2}}}{{2!}} + \frac{{{{\left( {x\ln a} \right)}^3}}}{{3!}} + \frac{{{{\left( {x\ln a} \right)}^4}}}{{4!}} + \cdots \\ \end{gathered}