Question

b8-taylor and maclaurin series: p
(2 points)
find the maclaurin series for ( f(x)=x^{7} e^{x^{5}} ).
( x^{7} e^{x^{5}}=sum_{n = 0}^{infty} )
on what interval is the expansion valid? give your answer using interval notation
is the only point in the interval of convergence, you would answer with 0.
the expansion is valid on
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Explanation:

Step1: Recall the Maclaurin series of \(e^t\)

The Maclaurin series of \(e^t=\sum_{n = 0}^{\infty}\frac{t^{n}}{n!}\), which converges for all \(t\in(-\infty,\infty)\)

Step2: Substitute \(t = x^{5}\) into the series of \(e^t\)

When \(t=x^{5}\), we have \(e^{x^{5}}=\sum_{n = 0}^{\infty}\frac{(x^{5})^{n}}{n!}=\sum_{n = 0}^{\infty}\frac{x^{5n}}{n!}\)

Step3: Multiply by \(x^{7}\)

$$
LATEXBLOCK0
$$

Step4: Determine the interval of convergence

Since the Maclaurin series of \(e^t\) converges for all \(t\in(-\infty,\infty)\), and we substitute \(t = x^{5}\) (a polynomial in \(x\)). The interval of convergence of the series \(\sum_{n = 0}^{\infty}\frac{x^{5n + 7}}{n!}\) is the set of all real numbers \(x\) such that \(x^{5}\in(-\infty,\infty)\). Since \(x^{5}\) is defined for all real \(x\), the interval of convergence is \((-\infty,\infty)\)

Answer:

1. $\sum_{n = 0}^{\infty}\frac{x^{5n + 7}}{n!}$
2. $(-\infty,\infty)$