Question

How do you find the Maclaurin series for $f(x)={{e}^{3x}}$ ?

Answer 1

As we know that the Maclaurin series is given as $f(x)=\sum\limits_{n=0}^{\infty }{\dfrac{{{f}^{n}}\left( 0 \right)}{n!}}{{x}^{n}}$ . We will expand the series and then find the derivatives of the given function. Then by substituting the values and simplifying the obtained series we will get the desired answer.

Complete step-by-step answer:
We have been given a function $f(x)={{e}^{3x}}$.
We have to find the Maclaurin series for the given function.
We know that the Maclaurin series is given by $f(x)=\sum\limits_{n=0}^{\infty }{\dfrac{{{f}^{n}}\left( 0 \right)}{n!}}{{x}^{n}}$.
The expansion of the above series is given as
$\Rightarrow f\left( x \right)=f\left( 0 \right)+xf'\left( 0 \right)+\dfrac{{{x}^{2}}}{2!}f''\left( 0 \right)+........+\dfrac{{{x}^{n}}}{n!}{{f}^{n}}\left( 0 \right)+....$
Let us assume that $f\left( x \right)={{e}^{3x}}$
Now, let us find the derivatives of the given function. Then we will get
$\begin{aligned} & \Rightarrow f'\left( x \right)=3{{e}^{3x}} \\\ & \Rightarrow f''\left( x \right)=3\times 3{{e}^{3x}}=9{{e}^{3x}} \\\ & \Rightarrow f'''\left( x \right)=3\times 9{{e}^{3x}}=27{{e}^{3x}} \\\ \end{aligned}$
Now, substituting 0 in place of x we will get
$\begin{aligned} & \Rightarrow f\left( 0 \right)={{e}^{3\times 0}}={{e}^{0}}=1 \\\ & \Rightarrow f'\left( 0 \right)=3{{e}^{3\times 0}}=3{{e}^{0}}=3 \\\ & \Rightarrow f''\left( 0 \right)=9{{e}^{3\times 0}}=9{{e}^{0}}=9 \\\ & \Rightarrow f'''\left( 0 \right)=27{{e}^{3\times 0}}=27{{e}^{0}}=27 \\\ \end{aligned}$
Now, substituting the values in the Maclaurin series we will get
$\Rightarrow f\left( x \right)=1+x\times 3+\dfrac{{{x}^{2}}}{2!}\times 9+\dfrac{{{x}^{3}}}{3!}\times 27........$
Now, simplifying the above obtained series we will get
$\begin{aligned} & \Rightarrow {{e}^{3x}}=1+3x+\dfrac{9}{2\times 1}{{x}^{2}}+\dfrac{27}{3\times 2\times 1}{{x}^{3}}........ \\\ & \Rightarrow {{e}^{3x}}=1+3x+\dfrac{9}{2}{{x}^{2}}+\dfrac{27}{6}{{x}^{3}}........ \\\ \end{aligned}$
Hence above is the required Maclaurin series for the given function.

Note: The point to be noted is that the Maclaurin series is derived from the Taylor series and both are infinite series. So to find the pattern of the series we must need to find at least three or four terms in the expansion. Here in this question we can further generalize the obtained series as
$\Rightarrow {{e}^{3x}}=\sum\limits_{n=0}^{\infty }{\dfrac{{{\left( 3x \right)}^{n}}}{n!}}$
Hence above is the Maclaurin series of the given function.