Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do not show that displaystyle{R}{n}{left({x}right)}rightarrow{0}.] displaystyle f{{left({x}right)}}={e}-{5}{x}

Since we are assuming that ff has a power series expansion, we may apply the Mclauren series expansion general formula, which states that the power series expansion of ff is given by:
${\displaystyle f\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}{f}^n\left(0\right)\frac{x^n}{n!}}$
We simply find the general formula for the nth derivative and plug in x=0 to it.
We go by the pattern first:
${\displaystyle f=e^{-5}x={(-1)}^0{5}^0{e}^{-5}x}$
${\displaystyle f^{\prime }=-5e^{-5}x={(-1)}^1{5}^1{e}^{-5}x}$
${\displaystyle f{}^{″}=5^2{e}^{-5}x={(-1)}^2{5}^2{e}^{-5}x}$
...
Generally, we can see the pattern shows that the general formula is:
${\displaystyle f^n\left(x\right)={(-1)}^n{5}^n{e}^{-5}x}$
Therefore:
${\displaystyle f^n\left(0\right)={(-1)}^n{5}^n={(-5)}^n}$
Thus, our Mclaurin series expansion is:
${\displaystyle f\left(x\right)=e^{-5}x=\sum _{n=0}^{\mathrm{\infty }}{(-5)}^n\frac{x^n}{n!}}$