Question

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

Short answer

So, The Maclaurin series for the function is:

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

Or, it can be written as

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

Step-by-step solution

given information

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

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

calculation

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$f\left(0\right)=1,$ therefore $f^{\text{'}}\left(0\right),f^{\text{'}\text{'}}\left(0\right)and{f}^{\text{'}\text{'}}\left(0\right)$are 1 .