Question

Find the Maclaurin series for the functions in Exercises 51–60 by substituting into a known Maclaurin.

Also, give the interval of convergence for the series.

$e^{-3x^2}$

Short answer

The required Maclaurin series is $e^{-3x^2}=\sum _{k=0}^{\infty }\frac{(-3{)}^k{x}^{2k}}{k!}$

Step-by-step solution

Step 1. Given Information

Consider the function$f\left(x\right)=e^{-3x^2}$

Step 2

We know that the Maclaurin series for the function $g\left(x\right)=e^x$is $e^x=\sum _{k=0}^{\infty }\frac{x^k}{k!}$

So, to find the Maclaurin series of the function $f\left(x\right)=e^{-3x^2}$, we replace $x$by $-3x^2$in the Maclaurin series of the function $e^x$

Therefore, $e^{-3x^2}=\sum _{k=0}^{\infty }\frac{{\left(-3x^2\right)}^k}{k!}$implies that $e^{-3x^2}=\sum _{k=0}^{\infty }\frac{(-3{)}^k{x}^{2k}}{k!}$