Zoe Oneal

2021-02-21

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 $Rn\left(x\right)\to 0$.]
$f\left(x\right)=e-5x$

Jaylen Fountain

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

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