Question

If $f$is a function such that $f\left(0\right)=-3$and localid="1650438953513" role="math" $f\text{'}\left(x\right)=2f\left(x\right)$every value of $x$, find the Maclaurin series for $f$.

Short answer

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

$f\left(x\right)=-3\left(1+2x+\frac{4x^2}{2!}+\frac{8x^3}{3!}+\dots \right)$

Or, it can be written as

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

Step-by-step solution

Given Information

Given equations:

$f\left(0\right)=-3$

$f^{\text{'}}\left(x\right)=2f\left(x\right)$

Finding the Maclaurin series for f  

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, $f\left(0\right)=-3$.

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

So,

$$\begin{gathered} f^{\text{'}\text{'}}\left(0\right)=2f^{\text{'}}\left(0\right) \\ =2·(-6) \\ =-12 \end{gathered}$$

And,

$$\begin{gathered} f^{\text{'}\text{'}}\left(0\right)=2f^{\text{'}\text{'}}\left(0\right) \\ =2·-12 \\ =-24 \end{gathered}$$

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

$f\left(x\right)=-3-6x+\frac{-12x^2}{2!}+\frac{-24x^3}{3!}+\dots$

$=-3\left(1+2x+\frac{4x^2}{2!}+\frac{8x^3}{3!}+\dots \right)$

Or, it can be written as

localid="1650438932580" $f\left(x\right)=\mathbf{-}\mathbf{3}\mathbf{\sum }_{\mathbf{k}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\frac{{\mathbf{2}}^{\mathbf{k}}{\mathbf{x}}^{\mathbf{k}}}{\mathbf{k}\mathbf{!}}$