Q. 9 - Calculus

Question

If $f$ is a function such that $f\left(0\right)=2$ and $f^{\text{'}}\left(x\right)=-3f\left(x\right)$ for every value of $x$, find the Maclaurin series for $f$.

Step-by-Step Solution

Answer

The Maclaurin series for the function $f\left(x\right)$ is.

$f\left(x\right)=2\left(1-3x+\frac{9x^2}{2!}-\frac{27x^3}{3!}+\dots \right)$

Or, it can be written as

$f\left(x\right)=2\sum _{k=0}^{\infty }\frac{(-1{)}^k{3}^k{x}^k}{k!}$

step 1: given information

Let us consider the function such that that $f\left(0\right)=1$ and $f^{\text{'}}\left(x\right)=-3f\left(x\right)$

step 2: calculation

The general formula to calculate the Maclaurin series for the function is:

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

As, $f\left(0\right)=2$

$$\begin{gathered} f^{\text{'}}\left(0\right)=-3f\left(0\right) \\ =-3·2 \\ =-6 \end{gathered}$$

So,

$$\begin{gathered} f^{\text{'}\text{'}}\left(0\right)=-3f^{\text{'}}\left(0\right) \\ =-3·(-6) \\ =18 \end{gathered}$$

And

$$\begin{gathered} f^m\left(0\right)=-3f^{\text{'}\text{'}}\left(0\right) \\ =-3·18 \\ =-54 \end{gathered}$$