Question

Find the Maclaurin series for the functions in Exercises 51–60

by substituting into a known Maclaurin series. Also, give the

interval of convergence for the series

$e^{x^3}$

Short answer

The required Maclaurin series is $e^{x^3}=\sum _{k=0}^{\infty }\frac{x^{3k}}{k!}$with an interval of convergence $\mathrm{ℝ}$

Step-by-step solution

Step 1. Given information

Given function$e^{x^3}$

we have to find the Maclaurin series for the given functions and the interval of convergence for the series

Step 2. Explanation

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

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

Therefore,

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

Implies that

$e^{x^3}=\sum _{k=0}^{\infty }\frac{x^{3k}}{k!}$