Book edition 2nd

Author(s) James Stewart

Pages 830 pages

ISBN 9781133112280

Answers without the blur.

Just sign up for free and you're in.

Short Answer

(a) Use the sum of the first 10 terms and Exercise 33(a) to estimate the sum of the series\(\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} \) . How good is this estimate?

(b) Improve this estimate using Exercise 33(b) with n = 10

(c) Find a value of n that will ensure that the error in the approximation \(S \approx {S_n}\) is less than 0.01

(a) The sum of series \(\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} \) is 1.549768 with error at most 0.1

(b) When n = 10, estimate \(S \approx 1.64522\) with error ≤ 0.005

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

(a) Applying Integral on the Given Function:

Let the associated function \(f(x) = \frac{1}{{{x^2}}}\) is positive and continuous then

\(f(x) = \frac{{ - 2}}{{{x^3}}} < 0{\rm{ for }}x \ge 1\)

Therefore on applying integral it gives an upper bound for the error using the mth partial sum as:

\({R_\infty } = \sum\limits_{n \to 1}^\infty {\frac{1}{{{n^2}}} \le \int\limits_1^\infty {\frac{1}{{{x^2}}}dx} } \)

\({\rm{ }} = \mathop {\lim }\limits_{b \to \infty } \left[ {\frac{{ - 1}}{x}} \right]_m^b = \frac{1}{m} - \mathop {\lim }\limits_{b \to \infty } \frac{1}{b} = \frac{1}{m}\)

The estimate using the sum of the first ten terms is

\(\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}} \approx {S_{10}}} \)

\( = 1 + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + .... + \frac{1}{{{{10}^2}}}\)

\( \approx 1.549768\)

And the error of this estimate is \({R_{10}} \le \frac{1}{{10}}\) , so the error is at most 0.1

(b) Improving Estimate with n = 10

We have \({S_n} + \int\limits_{n + 1}^\infty {f(x)dx \le S \le {S_n} + \int\limits_n^\infty {f(x)dx} } \)

Now with n = 10, we have \({S_{10}} + \int\limits_{11}^\infty {\frac{1}{{{x^2}}}dx

\( = {S_{10}} + \frac{1}{{11}} \le S \le {S_{10}} + \frac{1}{{10}}\)

\( = 1.549768 + 0.090909 \le S \le 1.549768 + 0.1\)

\( = 1.640677 \le S \le 1.6491768\)

So we get S ≈ 1.64522 which is average of 1.640677 and 1.649768, with error ≤ 0.005 (which is rounded up for half 0.1)

(c) Finding Value of n

We have \({S_n} \le \int\limits_n^\infty {\frac{1}{{{x^2}}}dx{\rm{ which is equal to }}\frac{1}{n}} \)

\({\rm{That is, }}{{\rm{S}}_n} \le \int\limits_n^\infty {\frac{1}{{{x^2}}}dx} = \frac{1}{n}\)

\( {\rm{So }}{{\rm{S}}_n} < 0.01{\rm{ if }}\frac{1}{n} < \frac{1}{{1000}}\)

\( {\rm{Hence }}n > 1000\)

Find Study Materials for

Create Study Materials

Select your language

Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

Step by Step Solution

(a) Applying Integral on the Given Function:

(b) Improving Estimate with n = 10

(c) Finding Value of n

Most popular questions for Math Textbooks

Want to see more solutions like these?

Recommended explanations on Math Textbooks

Select your language

Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

Step by Step Solution

(a) Applying Integral on the Given Function:

(b) Improving Estimate with n = 10

(c) Finding Value of n

Most popular questions for Math Textbooks

Want to see more solutions like these?

Recommended explanations on Math Textbooks

Decision Maths

Calculus

Probability and Statistics

Statistics

Mechanics Maths

Geometry

Pure Maths