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Chapter 1: Q5P (page 32)

x2ln(1-x)

Short Answer

Expert verified

The series isx2ln(1-x)=-∑n=1∞xn+2n

The sum is given below.

S1=-x3

S2=-x3--x42

S3=-x3--x42--x53

The graphs are shown below.

Step by step solution

01

Given Information

The Maclaurin series.

02

Definition of the Maclaurin Series.

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.

ln(1+x)=x-x22+x33-x44+…

ln(1-x)=-x-x22-x33-x44+…

x2ln(1-x)=-x3-x42-x53-…

The general series is given below x2ln(1-x)=-x3-x42-x53-…-xn+2n

x2ln(1-x)=-∑n=1∞xn+2n

The sum is given below.

S1=-x3

S2=-x3--x42

S3=-x3--x42--x53

04

Plot the graphs.

The graph of the function with S1,S2,S3is given below.

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Most popular questions from this chapter

Use power series to evaluate the function at the given point. Compare with computer results, using the computer to find the series, and also to do the problem without series. Resolve any disagreement in results (see Example 1).earcsinx+In(1-xe)atx=0.0003.

∫0xsintdtt

The velocityof electrons from a high energy accelerator is very near the velocityof light. Given the voltage Vof the accelerator, we often want to calculate the ratio v / c. The relativistic formula for this calculation is (approximately, forV≫1)

vc=1-(0.511V)2, V=Number of million volts

Use two terms of the binomial series (13.5) to find1 - v/cin terms ofV. Use your result to find 1 - v/cfor the following values of V. Caution: V= the number of millionvolts.

(a) V =100 million volts

(b)V =500 million volts

(c)V =25,000 million volts

(d)V =100 gigavolts (100109 volts105 million volts)

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.

d4dx4ln(1+x3)atx=0 .

(a) Using computer or tables (or see Chapter 7,Section 11),verify that∑n=1∞(1/n2)=π26=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 ∑n=1∞(1/n2)(1/2)n=π212-(1/2)(ln2)2=0.5815+

the sum of the first five terms is0.5815+

(c) Prove theorem (14.4). Hint: The error is |∑N+1∞anxn|.

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 |an+1|≤|an|to replace all anby aN+1 , and write the appropriate inequality. Sum the geometric series to get the result.
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