## Calculus 8th Edition

$\frac{f^{8}(5)}{40,320}$
Since, the Taylor's series of $f$ centered at $a$ is $f(x)=\Sigma_{n=0}^{\infty}\frac{f^{n}(a)(x-a)^{n}}{n!}$ Thus, $b_{n}=\frac{f^{n}(a)}{n!}$ Now, $b_{8}=\frac{f^{8}(5)}{8!}=\frac{f^{8}(5)}{40,320}$
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