Question

approximate the value of e^3x when x=1 using the first six terms of the Maclaurin expansion

Answer 1

Given: A function

$f(x)=e^{3x}$

Required: To approximate the value of the function when x=1 using the first six terms of the Maclaurin expansion.

Explanation: Maclaurin's expansion is a special case of Taylor's theorem. The terms of the expansion are as follows:

$f(x)+\frac{f^{\prime}(x)}{1!}+\frac{f^{\prime\prime}(x)}{2!}+...$

So we need to find the following terms at x=1

$\begin{gathered} f(1)=e^3 \\ f^{\prime}(1)=3e^3 \\ f^{\prime}^{\prime}(1)=9e^3 \\ f^{\prime}^{\prime}^{\prime}(1)=27e^3 \end{gathered}$

Further

$e^x=1+x+\frac{x^2}{2!}+...$

We can write the given function as

$e^{3x}=e^{-3}e^{3(x+1)}$

The series would be

$e^{3(x+1)}=1+3(x+1)+\frac{3^2(x+1)^2}{2}+\frac{3^3(x+1)^3}{6}+\frac{3^4(x+1)^4}{24}+\frac{3^5(x+1)^5}{120}$

Solving further at x=1 gives

$e^{3(x+1)}=1+3(x+1)+\frac{9}{2}(x+1)^2+\frac{27(x+1)^3}{6}+\frac{81(x+1)^4}{24}+\frac{243(x+1)^5}{120}$

Hence the required function is-

$e^{-3}e^{3(x+1)}=e^{-3}(1+3(x+1)+\frac{9}{2}(x+1)^2+\frac{27(x+1)^3}{6}+\frac{81(x+1)^4}{24}+\frac{243(x+1)^5}{120})$

Now for x=1 we have

$=e^{-3}(1+6+18+36+54+64.8)$

Which gives

$\begin{gathered} =0.0497\times179.8 \\ =8.93606 \end{gathered}$

Final Answer: 8.93606