Chapter 1: Q8P (page 36)

Estimate the Error if $\sum_{n = 1}^{\infty} \frac{x^{n}}{n^{3}}$ is approximated by the sum of its first three terms for $| x | < \frac{1}{2}$ .

Short Answer

Expert verified

The Error is 0.001953.

Step by step solution

Given Information

The function is $\sum_{n = 1}^{\infty} \frac{x^{n}}{n^{3}}$ .

Definition of the Error of the Series

The calculated Series is Underestimate if the Error is less than zero. The calculated Series is Overestimate if the Error is more than zero.

Verify the statement

$The function is :$

$\sum_{n = 1}^{\infty} \frac{x^{n}}{n^{3}}$

The formula states that:

$\sum_{n = 1}^{\infty} \frac{x^{n}}{n^{3}} = x + \frac{x^{2}}{2^{3}} + \frac{x^{3}}{2^{3}} + \frac{x^{4}}{2^{3}} + . . . .$

The theorem states that:

$If S = \sum_{n = 0}^{\infty} a_{n} x^{n} coverages for |x| < 1, and if |a_{n + 1}| < |a_{n}| for n > N, then |S - \sum_{n = 0}^{N} a_{n} x^{n}| < \frac{|a_{N + 1} x^{N + 1}|}{(1 - |x|)} |S - \sum_{n = 0}^{N} a_{n} x^{n}| < \frac{|a_{N + 1} x^{N + 1}|}{(1 - |x|)} |\sum_{n = 1}^{\infty} \frac{x^{n}}{n^{3}} - (x + \frac{x^{2}}{2^{3}} + \frac{x^{3}}{3^{3}})| < \frac{|a_{N + 1} x^{N + 1}|}{(1 - |x|)} |\sum_{n = 1}^{\infty} \frac{x^{n}}{n^{3}} - (x + \frac{x^{2}}{2^{3}} + \frac{x^{3}}{3^{3}})| < \frac{|\frac{1}{4^{3}} . \frac{1}{2^{4}}|}{1 - |\frac{1}{2}|} |\sum_{n = 1}^{\infty} \frac{x^{n}}{n^{3}} - (x + \frac{x^{2}}{2^{3}} + \frac{x^{3}}{3^{3}})| < \frac{0.000976}{0.5} Solve further, and we get, |\sum_{n = 1}^{\infty} \frac{x^{n}}{n^{3}} - (x + \frac{x^{2}}{2^{3}} + \frac{x^{3}}{3^{3}})| < 0.001953 Error \approx 0.001953 Hence, the Error is 0.001953 .$

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Estimate the Error if $\sum_{n = 1}^{\infty} \frac{x^{n}}{n^{3}}$ is approximated by the sum of its first three terms for $| x | < \frac{1}{2}$ .

Short Answer

Step by step solution

Given Information

Definition of the Error of the Series

Verify the statement

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Definition of the Error of the Series

Verify the statement

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Most popular questions from this chapter

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Estimate the Error if $\sum_{n = 1}^{\infty} \frac{x^{n}}{n^{3}}$ is approximated by the sum of its first three terms for $| x | < \frac{1}{2}$ .