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Chapter 1: Q6P (page 32)

$x \sqrt{1 + x}$

Short Answer

Expert verified

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

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.

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

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

The general series is given below

$x \sqrt{1 + x} = x {鈭�}_{n = 0}^{鈭�} (\begin{matrix} \frac{1}{2} \\ n \end{matrix}) x^{n}$

$x \sqrt{1 + x} = {鈭�}_{n = 2}^{鈭�} \frac{(- 1)^{n - 1} (2 n - 3)!!}{(2 n)!!} x^{n + 1}$

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

(a) Using computer or tables (or see Chapter $7,$ Section $11$ ),verify that ${鈭�}_{n = 1}^{鈭�} (1 / n^{2}) = \frac{蟺^{2}}{6} = 1.6449 =$ ,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 ${鈭�}_{n = 1}^{鈭�} (1 / n^{2}) (1 / 2)^{n} = \frac{蟺^{2}}{12} - (1 / 2) (l n 2)^{2} = 0.5815 +$

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

(c) Prove theorem $(14.4)$ . Hint: The error is $| {鈭�}_{N + 1}^{鈭�} a_{n} x^{n} |$ .

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 $| a_{_{n + 1}} | 鈮� | a_{n} |$ to replace all $a_{n}$ by $a_{N + 1}$ , and write the appropriate inequality. Sum the geometric series to get the result.

Evaluate the following indeterminate forms by using L鈥橦opital鈥檚 rule and check your results by computer. (Note that Maclaurin series would not be useful here because xdoes not tend to zero, or because a function (In x, for example) is not expandable in a Maclaurin series.

$a) \underset{x 鈫� 蟺}{l i m} \frac{x s i n x}{x - 蟺} b) \underset{x 鈫� 蟺 / 2}{l i m} \frac{l n (2 - s i n x)}{l n (1 + c o s x)} c) \underset{x 鈫� 1}{l i m} \frac{l n (2 - x)}{l n (x - 1)} d) \underset{x 鈫� 鈭�}{l i m} \frac{l n x}{\sqrt{x}} e) \underset{x 鈫� 0}{l i m} x 2 l n x f) \underset{x 鈫� 鈭�}{l i m} x^{n} e^{- x}$

Use Maclaurin series to evaluate each of the following. Although you could do them by computer, you can probably do them in your head faster than you can type them into the computer. So use these to practice quick and skillful use of basic series to make simple calculations.

$\frac{d^{10}}{{dx}^{10}} (x^{8} \tan^{2} x) at x = 0 .$

$x^{2} l n (1 - x)$

Show that the interval of convergence of the series ${鈭�}_{n = 1}^{鈭�} \frac{x^{n}}{n^{2} + n}$ is $| x | 鈮� 1 (For x = 1$ . (For , this is the series of Problem 9.) Using theorem $(14.4)$ , show that for $x = \frac{1}{2}$ , four terms will give two decimal place accuracy.

See all solutions

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