How do you find the power series for #f(x)=int e^(t^3)# from [0,x] and determine its radius of convergence?

Question

Christopher Agresta · Accepted Answer

#int_0^x e^(t^3) =sum_(n=0)^oo x^(3n+1)/((3n+1)n!)# with #R=oo#

Let's start with the MacLaurin expansion of #e^x#:

#e^x = sum_(n=0)^oo x^n/(n!)#

and substitute #x=t^3#

#e^(t^3) = sum_(n=0)^oo t^(3n)/(n!)#

This series converges absolutely in all #RR# so the radius of convergence is #R=oo#.

We can then integrate term by term and obtain a series with the same radius of convergence:

#int_0^x e^(t^3) = int_0^x sum_(n=0)^oo t^(3n)/(n!) = sum_(n=0)^oo int_0^xt^(3n)/(n!)= sum_(n=0)^oo 1/(n!)[t^(3n+1)/(3n+1)]_0^x#

Finally: #int_0^x e^(t^3) =sum_(n=0)^oo x^(3n+1)/((3n+1)n!)#

Charles Arocha · Answer

undefined To find the power series for $ f(x) = \int e^{t^3} $ from $ 0 $ to $ x $, we need to express $ e^{t^3} $ as a power series and then integrate it term by term.

First, let's express $ e^{t^3} $ as a power series:
$$ e^{t^3} = \sum_{n=0}^{\infty} \frac{t^{3n}}{n!} $$

Now, integrate term by term:
$$ f(x) = \int e^{t^3} dt = \int \sum_{n=0}^{\infty} \frac{t^{3n}}{n!} dt $$
$$ f(x) = \sum_{n=0}^{\infty} \frac{1}{n!} \int t^{3n} dt $$

The integral $ \int t^{3n} dt $ can be evaluated as:
$$ \int t^{3n} dt = \frac{t^{3n+1}}{3n+1} + C $$

Therefore, the power series for $ f(x) $ is:
$$ f(x) = \sum_{n=0}^{\infty} \frac{x^{3n+1}}{(3n+1)n!} $$

To determine the radius of convergence, we use the ratio test:
$$ R = \lim_{n \to \infty} \left| \frac{a_{n}}{a_{n+1}} \right| $$
Where $ a_n $ is the coefficient of $ x^n $ in the series.

In our series, $ a_n = \frac{1}{(3n+1)n!}x^{3n+1} $.
$$ \lim_{n \to \infty} \left| \frac{a_{n}}{a_{n+1}} \right| = \lim_{n \to \infty} \left| \frac{(3n+2)x^3}{3n+1} \right| = |x^3| $$

For the series to converge, this limit must be less than 1. Hence, the radius of convergence $ R $ is 1.