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

Skilled2021-02-22Added 169 answers

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:

$f={e}^{-5}x={(-1)}^{0}{5}^{0}{e}^{-5}x$

${f}^{\prime}=-5{e}^{-5}x={(-1)}^{1}{5}^{1}{e}^{-5}x$

$f{}^{\u2033}={5}^{2}{e}^{-5}x={(-1)}^{2}{5}^{2}{e}^{-5}x$

...

Generally, we can see the pattern shows that the general formula is:

${f}^{n}\left(x\right)={(-1)}^{n}{5}^{n}{e}^{-5}x$

Therefore:

$f}^{n}\left(0\right)={(-1)}^{n}{5}^{n}={(-5)}^{n$

Thus, our Mclaurin series expansion is:

$f\left(x\right)={e}^{-5}x=\sum _{n=0}^{\mathrm{\infty}}{(-5)}^{n}\frac{{x}^{n}}{n!}$

We simply find the general formula for the nth derivative and plug in x=0 to it.

We go by the pattern first:

...

Generally, we can see the pattern shows that the general formula is:

Therefore:

Thus, our Mclaurin series expansion is:

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