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Chapter 8: Q. 51 (page 692)

$e^{x^{3}}$

Short Answer

Expert verified

The answer is $e^{x^{3}} = \sum_{k = 0}^{\infty} \frac{x^{3 k}}{k!}$

Step by step solution

01

Step 1. Given Information

Consider the function, $f (x) = e^{x^{3}}$

02

Step 2

We know that the Maclaurin series for the function $g (x) = e^{x}$ is

$e^{x} = \sum_{k = 0}^{\infty} \frac{x^{k}}{k!}$

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

Therefore, $e^{x^{3}} = \sum_{k = 0}^{\infty} \frac{{(x^{3})}^{k}}{k!}$ implies thatrole="math" localid="1649860232493" $e^{x^{3}} = \sum_{k = 0}^{\infty} \frac{x^{3 k}}{k!}$

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Most popular questions from this chapter

Find the interval of convergence for each power series in Exercises 21–48. If the interval of convergence is finite, be sure to analyze the convergence at the endpoints.

$\sum_{k = 0}^{\infty} {(- 1)}^{k} \frac{k^{2}}{3^{k}} {(x - \frac{1}{2})}^{k}$

In Exercises 49–56 find the Taylor series for the specified function and the given value of $x_{0}$ . Note: These are the same functions and values as in Exercises 41–48.

$52 . \sqrt{x}, 1$

What is the relationship between a Maclaurin series and a power series in x?

Find the interval of convergence for power series: $\sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{2 k + 1} x^{2 k + 1}$

Find the interval of convergence for each power series in Exercises 21–48. If the interval of convergence is finite, be sure to analyze the convergence at the endpoints.

$\sum_{k = 0}^{\infty} \frac{{(k!)}^{2}}{(2 k!)} {(x + 2)}^{k}$

See all solutions

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