Question

Use the first three terms of the series f(x)=((x^n)(n^n))/n! from n=1 to infinity to approximate f(-1/3).

Answer 1

@SithsAndGiggles

Answer 2

\[f(x)=\sum_{n=1}^\infty \frac{x^nn^n}{n!}\approx \frac{x^11^1}{1!}+\frac{x^22^2}{2!}+\frac{x^33^3}{3!}\]
which means
\[f\left(-\frac{1}{3}\right)\approx \frac{\left(-\frac{1}{3}\right)^11^1}{1!}+\frac{\left(-\frac{1}{3}\right)^22^2}{2!}+\frac{\left(-\frac{1}{3}\right)^33^3}{3!}\]

Answer 3

Thank you so much.

Answer 4

yw