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Chapter 1: Q6P (page 32)

$x \sqrt{1 + x}$

Short Answer

Expert verified

The series is $x \sqrt{1 + x} = \sum_{n = 2}^{\infty} \frac{(- 1)^{n - 1} (2 n - 3)!!}{(2 n)!!} x^{n + 1}$

The sum is given below.

$S_{1} = x$

$S_{2} = x + \frac{1}{2} x^{2}$

$S_{3} = x + \frac{1}{2} x^{2} - \frac{1}{8} x^{3}$

$S_{4} = x + \frac{1}{2} x^{2} - \frac{1}{8} x^{3} + \frac{1}{16} x^{4}$

The graphs are shown below.

Step by step solution

01

Given Information

The Maclaurin series.

02

Definition of the MaclaurinSeries.

A Maclaurin series is a function with an expansion series that gives the sum of the function's derivatives.

03

Find the series and sum.

The Maclaurin series is given below.

$(1 + x)^{1 / 2} = 1 + \frac{1}{2} x - \frac{1}{8} x^{2} + \frac{1}{16} x^{3} - \frac{5}{128} x^{4} + \dots$

$x (1 + x)^{1 / 2} = x + \frac{1}{2} x^{2} - \frac{1}{8} x^{3} + \frac{1}{16} x^{4} - \frac{5}{128} x^{5} + \dots$

The general series is given below

$x \sqrt{1 + x} = x \sum_{n = 0}^{\infty} (\begin{matrix} \frac{1}{2} \\ n \end{matrix}) x^{n}$

$x \sqrt{1 + x} = \sum_{n = 2}^{\infty} \frac{(- 1)^{n - 1} (2 n - 3)!!}{(2 n)!!} x^{n + 1}$

The sum is given below.

localid="1657352249901" $S_{1} = x$

$S_{2} = x + \frac{1}{2} x^{2}$

$S_{3} = x + \frac{1}{2} x^{2} - \frac{1}{8} x^{3}$

$S_{4} = x + \frac{1}{2} x^{2} - \frac{1}{8} x^{3} + \frac{1}{16} x^{4}$

04

Plot the graphs.

The graph of the function with $S_{1}, S_{2}, S_{3}, S_{4}$ is given below

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Most popular questions from this chapter

Solve for all possible values of the real numbers xand y in the following equations. $x + i y = {(1 - i)}^{2}$

Show that the binomial coefficients $(\begin{array}{l} - 1 \\ n \end{array}) = (- 1)^{n}$

(a) Using computer or tables (or see Chapter $7,$ Section $11$ ),verify that $\sum_{n = 1}^{\infty} (1 / n^{2}) = \frac{π^{2}}{6} = 1.6449 =$ ,and also verify that the error in approximating the sum of the series by the first five terms is approximately $0.1813$ .

(b) By computer or tables verify that $\sum_{n = 1}^{\infty} (1 / n^{2}) (1 / 2)^{n} = \frac{π^{2}}{12} - (1 / 2) (l n 2)^{2} = 0.5815 +$

the sum of the first five terms is $0.5815 +$

(c) Prove theorem $(14.4)$ . Hint: The error is $| \sum_{N + 1}^{\infty} a_{n} x^{n} |$ .

Use the fact that the absolute value of a sum is less than or equal to the sum of the absolute values. Then use the fact that $| a_{_{n + 1}} | \leq | a_{n} |$ to replace all $a_{n}$ by $a_{N + 1}$ , and write the appropriate inequality. Sum the geometric series to get the result.

The series $\sum_{n - 1}^{\infty} 1 / n^{s}$ , $s > 1,$ is called the Riemann Zeta function, $ζ (s)$ . (In Problem $14.2 (a)$ you found $ζ (2) = \frac{π^{2}}{6}$ . When $n$ is an even integer, these series can be summed exactly in terms of $π$ .) By computer or tables, find

$(a) ζ (4) = \sum_{n = 1}^{\infty} \frac{1}{n^{4}} (b) ζ (3) = \sum_{n = 1}^{\infty} \frac{1}{n^{3}} (c) ζ (\frac{3}{2}) = \sum_{n = 1}^{\infty} \frac{1}{n^{3 / 2}}$

$\sqrt{1 + \ln (1 + x)}$

See all solutions

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