Question

If $f$is a function such that $f\left(0\right)=1$and $f^{\text{'}}\left(x\right)=f\left(x\right)$for every value of $x$, find the Maclaurin series for $f$.

Short answer

The Maclaurin series for the function is:

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

Or, it can be written as

${\mathit{e}}^{\mathbf{x}}\mathbf{=}\mathbf{\sum }_{\mathbf{k}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\frac{\mathbf{1}}{\mathbf{k}\mathbf{!}}{\mathit{x}}^{\mathbf{k}}$

Step-by-step solution

Given Information

Given equations :

$$\begin{gathered} f\left(0\right)=1 \\ f\text{'}\left(x\right)=f\left(x\right) \end{gathered}$$

Finding the Maclaurin series for f

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

So, the function must be $f\left(x\right)=e^x$

Since, 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 role="math" localid="1650439558986" $f\left(0\right)=1$, therefore role="math" localid="1650439565403" $f^{\text{'}}\left(0\right),f^{\text{'}\text{'}}\left(0\right)\mathrm{and}f\text{'}\text{'}\left(0\right)\mathrm{are}\boxed{1}$

So, The Maclaurin series for the function is:

role="math" localid="1650439570178" $$\begin{gathered} f\left(x\right)=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots \\ \Rightarrow {\mathit{e}}^{\mathbf{x}}\mathbf{=}\mathbf{\sum }_{\mathbf{k}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\frac{\mathbf{1}}{\mathbf{k}\mathbf{!}}{\mathit{x}}^{\mathbf{k}} \end{gathered}$$