Question

(a) Using computer or tables (or see Chapter $\mathbf{7}\mathbf{,}$Section $\mathbf{11}$),verify that$$\begin{gathered} \mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\mathbf{(}\mathbf{1}\mathbf{/}{\mathbf{n}}^{\mathbf{2}}\mathbf{)}\mathbf{=}\frac{{\mathbf{\pi }}^{\mathbf{2}}}{\mathbf{6}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{6449}\mathbf{=} \end{gathered}$$,and also verify that the error in approximating the sum of the series by the first five terms is approximately $\mathbf{0}\mathbf{.}\mathbf{1813}$.

(b) By computer or tables verify that $$\begin{gathered} \mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\mathbf{(}\mathbf{1}\mathbf{/}{\mathbf{n}}^{\mathbf{2}}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{/}\mathbf{2}{\mathbf{)}}^{\mathbf{n}}\mathbf{=}\frac{{\mathbf{\pi }}^{\mathbf{2}}}{\mathbf{12}}\mathbf{-}\mathbf{(}\mathbf{1}\mathbf{/}\mathbf{2}\mathbf{\left)}\mathbf{\right(}\mathit{l}\mathit{n}\mathbf{2}{\mathbf{)}}^{\mathbf{2}} \\ \mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5815}\mathbf{+} \end{gathered}$$

the sum of the first five terms is$\mathbf{0}\mathbf{.}\mathbf{5815}\mathbf{+}$

(c) Prove theorem $\mathbf{(}\mathbf{14}\mathbf{.}\mathbf{4}\mathbf{)}$. Hint: The error is $\left|\sum _{N+1}^{\infty }{a}_n{x}^n\right|$.

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 $\mathbf{\left|}{\mathit{a}}_{{}_{\mathbf{n}\mathbf{+}\mathbf{1}}}\mathbf{\right|}\mathbf{\le }\mathbf{\left|}{\mathit{a}}_{\mathbf{n}}\mathbf{\right|}$to replace all ${\mathit{a}}_{\mathbf{n}}$by ${\mathit{a}}_{\mathbf{N}\mathbf{+}\mathbf{1}}$ , and write the appropriate inequality. Sum the geometric series to get the result.

Short answer

The error is approximately $0.1813$

The error is approximately $0.18128$

The error is approximately $0.5818$

Result expression is$\left|S-\sum _{n=0}^N{a}_n{x}^n\right|<\frac{\left|a_{N+1}{x}^{N+1}\right|}{1-\left|x\right|}$

Step-by-step solution

Given Information

The$S=\sum _{n=0}^{\infty }{a}_n{x}^n$converges for$\left|x\right|<1$and$\left|a_{n+1}\right|<\left|a_n\right|$for$n>N$

Definition of Error in alternating series

The difference between any partial sum and the limiting value is the error in an alternating series.

Calculate error in approximating the sum of series

Consider general term.

$\sum _{n=1}^{\infty }\frac{1}{n^2}$

Use general term to calculate error.

$\sum _{n=1}^{\infty }\frac{1}{n^2}=\frac{{\pi }^2}{6}$

Use MATLAB to verify that error

Verify that error is approximately $0.1813$using MATLAB tool,

Calculate error in approximating the sum of series

For first five terms,

$S_5=\sum _{n=1}^5\frac{1}{n^2}$

$$\begin{gathered} =\frac{5269}{3600} \\ =1.46361 \end{gathered}$$

Use $S_5$to calculate error.

Error $=S-S_5$

Error$$\begin{gathered} =\frac{{\mathrm{\pi }}^2}{6}-1.46361 \\ =0.18128 \end{gathered}$$

Use MATLAB to verify that error

Verify that error is approximately $0.18128$using MATLAB tool

Calculate error in approximating the sum of series

Consider general term.

$\sum _{n=1}^{\infty }\left(\frac{1}{n^2}\right)\left(\frac{1}{2^n}\right)$

Use general term to calculate error.

$$\begin{gathered} \sum _{n=1}^{\infty }\left(\frac{1}{n^2}\right)\left(\frac{1}{2^n}\right)=\frac{{\pi }^2}{12}-\frac{1}{2}\left(\ln \right(2){)}^2 \\ =0.5815+ \end{gathered}$$

Use MATLAB to verify that error.

Verify that error is approximately $0.5815$using MATLAB tool.

Prove the expression |S-∑n=0Nanxn|<|aN+1xN+1|1-|x|

Expand the series for first $N$terms,

$$\begin{gathered} S-\sum _{n=0}^N{a}_n{x}^n=\sum _{n=0}^{\infty }{a}_n{x}^n-\sum _{n=0}^N{a}_n{x}^n \\ =\sum _{n=0}^N{a}_n{x}^n+\sum _{N+1}^{\infty }{a}_n{x}^n-\sum _{n=0}^N{a}_n{x}^{n}S-\sum _{n=0}^N{a}_n{x}^n \\ =\sum _{N+1}^{\infty }{a}_n{x}^n \\ \sum _{N+1}^{\infty }{a}_n{x}^n=a_{N+1}{x}^{N+1}+a_{N+2}{x}^{N+2}+a_{N+3}{x}^{N+3}+\dots \end{gathered}$$

Take absolute value and use $\left|a_{n+1}\right|\le |a_n$,

$$\begin{gathered} \left|S-\sum _{n=0}^N{a}_n{x}^n\right|=\left|\sum _{N+1}^{\infty }{a}_n{x}^n\right|<\left|a_{N+1}{x}^{N+1}\right|+\left|a_{N+2}{x}^{N+2}\right|+\left|a_{N+3}{x}^{N+3}\right|+\dots \\ \left|a_{N+1}{x}^{N+1}\right|+\left|a_{N+2}{x}^{N+2}\right|+\left|a_{N+3}{x}^{N+3}\right|+=\left|a_{N+1}{x}^{N+1}\right|+\left|a_{N+1}{x}^{N+2}\right|+\left|a_{N+1}{x}^{N+3}\right|+\dots \\ \left|a_{N+1}{x}^{N+1}\right|+\left|a_{N+2}{x}^{N+2}\right|+\left|a_{N+3}{x}^{N+3}\right|+=\left|a_{N+1}{x}^{N+1}\right|+\left|a_{N+1}{x}^{N+1}\right|\left|x\right|+\left|a_{N+1}{x}^{N+1}\right|\left|x^2\right|+\dots \end{gathered}$$

Consider it as geometric series with $a=\left|a_{N+1}{x}^{N+1}\right|$and $r=\left|x\right|$

$$\begin{gathered} S=\frac{a}{1-r} \\ =\frac{\left|a_{N+1}{x}^{N+1}\right|}{1-\left|x\right|} \\ \left|S-\sum _{n=0}^N{a}_n{x}^n\right|<\frac{\left|a_{N+1}{x}^{N+1}\right|}{1-\left|x\right|} \end{gathered}$$

Hence, the error is approximately $0.1813$

The error is approximately$0.18128$.

The error is approximately $0.5815$ .

Result expression is$\left|S-\sum _{n=0}^N{a}_n{x}^n\right|<\frac{\left|a_{N+1}{x}^{N+1}\right|}{1-\left|x\right|}\mid$