Chapter 14.8, Problem 1E

a.

To determine

The approximations for $\pi$ for given pair of (a,b) for each value of N.

Explanation of Solution

Given information:

Here ,

  $\begin{array}{l}(a,b)=\left(\frac{1}{4},\frac{3}{5}\right)\\ N=5,10\&15\end{array}$

Calculations:

Here first we check the value of $\frac{a+b}{1-ab}$

If the value is 1, then we proceed further,

Therefore,

  $\frac{a+b}{1-ab}=\frac{\frac{1}{4}+\frac{3}{5}}{1-\frac{1}{4}\times \frac{3}{5}}=\frac{\frac{5+12}{20}}{1-\frac{3}{20}}=\frac{\frac{17}{20}}{\frac{17}{20}}=1$

Therefore we can write,

  $\frac{\pi }{4}={\tan }^{-1}1=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\mathrm{....}+{(-1)}^{n-1}\frac{x^{2n-1}}{2n-1}+\mathrm{...}$

Now taking,

  $\begin{array}{l}N=5\\ \frac{\pi }{4}={\tan }^{-1}1=1-\frac{1^3}{3}+\frac{1^5}{5}-\frac{1^7}{7}+\frac{1^9}{9}=0.8349206\\ Therefore,\\ \pi =4\times 0.8349206=3.339682\end{array}$

Now for $N=10$ .

  $\begin{array}{l}N=10\\ \frac{\pi }{4}={\tan }^{-1}1=1-\frac{1^3}{3}+\frac{1^5}{5}-\frac{1^7}{7}+\frac{1^9}{9}-\frac{1^{11}}{11}+\frac{1^{13}}{13}-\frac{1^{15}}{15}+\frac{1^{17}}{17}-\frac{1^{19}}{19}=0.7604599\\ Therefore,\\ \pi =4\times 0.7604599=3.04183\end{array}$

Now for $N=15$ .

  $\begin{array}{l}N=15\\ \frac{\pi }{4}={\tan }^{-1}1=1-\frac{1^3}{3}+\frac{1^5}{5}-\frac{1^7}{7}+\frac{1^9}{9}-\frac{1^{11}}{11}+\frac{1^{13}}{13}-\frac{1^{15}}{15}+\frac{1^{17}}{17}-\frac{1^{19}}{19}+\frac{1^{21}}{21}-\frac{1^{23}}{23}+\frac{1^{25}}{25}-\frac{1^{27}}{27}+\frac{1^{29}}{29}=0.8020464\\ Therefore,\\ \pi =4\times 0.8020464=3.20818\end{array}$

b.

To determine

The approximations for $\pi$ for given pair of (a,b) for each value of N.

Explanation of Solution

Given information:

Here ,

  $\begin{array}{l}(a,b)=\left(\frac{6}{11},\frac{5}{17}\right)\\ N=5,10\&15\end{array}$

Calculations:

Here first we check the value of $\frac{a+b}{1-ab}$

If the value is 1, then we proceed further,

Therefore,

  $\frac{a+b}{1-ab}=\frac{\frac{6}{11}+\frac{5}{17}}{1-\frac{6}{11}\times \frac{5}{17}}=1$

Therefore we can write,

  $\frac{\pi }{4}={\tan }^{-1}1=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\mathrm{....}+{(-1)}^{n-1}\frac{x^{2n-1}}{2n-1}+\mathrm{...}$

Now taking,

  $\begin{array}{l}N=5\\ \frac{\pi }{4}={\tan }^{-1}1=1-\frac{1^3}{3}+\frac{1^5}{5}-\frac{1^7}{7}+\frac{1^9}{9}=0.8349206\\ Therefore,\\ \pi =4\times 0.8349206=3.339682\end{array}$

Now for $N=10$ .

  $\begin{array}{l}N=10\\ \frac{\pi }{4}={\tan }^{-1}1=1-\frac{1^3}{3}+\frac{1^5}{5}-\frac{1^7}{7}+\frac{1^9}{9}-\frac{1^{11}}{11}+\frac{1^{13}}{13}-\frac{1^{15}}{15}+\frac{1^{17}}{17}-\frac{1^{19}}{19}=0.7604599\\ Therefore,\\ \pi =4\times 0.7604599=3.04183\end{array}$

Now for $N=15$ .

  $\begin{array}{l}N=15\\ \frac{\pi }{4}={\tan }^{-1}1=1-\frac{1^3}{3}+\frac{1^5}{5}-\frac{1^7}{7}+\frac{1^9}{9}-\frac{1^{11}}{11}+\frac{1^{13}}{13}-\frac{1^{15}}{15}+\frac{1^{17}}{17}-\frac{1^{19}}{19}+\frac{1^{21}}{21}-\frac{1^{23}}{23}+\frac{1^{25}}{25}-\frac{1^{27}}{27}+\frac{1^{29}}{29}=0.8020464\\ Therefore,\\ \pi =4\times 0.8020464=3.20818\end{array}$

c.

To determine

The approximations for $\pi$ for given pair of (a,b) for each value of N.

Explanation of Solution

Given information:

Here ,

  $\begin{array}{l}(a,b)=\left(\frac{5}{6},\frac{1}{11}\right)\\ N=5,10\&15\end{array}$

Calculations:

Here first we check the value of $\frac{a+b}{1-ab}$

If the value is 1, then we proceed further,

Therefore,

  $\frac{a+b}{1-ab}=\frac{\frac{5}{6}+\frac{1}{11}}{1-\frac{5}{6}\times \frac{1}{11}}=1$

Therefore we can write,

  $\frac{\pi }{4}={\tan }^{-1}1=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\mathrm{....}+{(-1)}^{n-1}\frac{x^{2n-1}}{2n-1}+\mathrm{...}$

Now taking,

  $\begin{array}{l}N=5\\ \frac{\pi }{4}={\tan }^{-1}1=1-\frac{1^3}{3}+\frac{1^5}{5}-\frac{1^7}{7}+\frac{1^9}{9}=0.8349206\\ Therefore,\\ \pi =4\times 0.8349206=3.339682\end{array}$

Now for $N=10$ .

  $\begin{array}{l}N=10\\ \frac{\pi }{4}={\tan }^{-1}1=1-\frac{1^3}{3}+\frac{1^5}{5}-\frac{1^7}{7}+\frac{1^9}{9}-\frac{1^{11}}{11}+\frac{1^{13}}{13}-\frac{1^{15}}{15}+\frac{1^{17}}{17}-\frac{1^{19}}{19}=0.7604599\\ Therefore,\\ \pi =4\times 0.7604599=3.04183\end{array}$

Now for $N=15$ .

  $\begin{array}{l}N=15\\ \frac{\pi }{4}={\tan }^{-1}1=1-\frac{1^3}{3}+\frac{1^5}{5}-\frac{1^7}{7}+\frac{1^9}{9}-\frac{1^{11}}{11}+\frac{1^{13}}{13}-\frac{1^{15}}{15}+\frac{1^{17}}{17}-\frac{1^{19}}{19}+\frac{1^{21}}{21}-\frac{1^{23}}{23}+\frac{1^{25}}{25}-\frac{1^{27}}{27}+\frac{1^{29}}{29}=0.8020464\\ Therefore,\\ \pi =4\times 0.8020464=3.20818\end{array}$

b.

To determine

The approximations for $\pi$ for given pair of (a,b) for each value of N.

Explanation of Solution

Given information:

Here ,

  $\begin{array}{l}(a,b)=\left(\frac{2}{5},\frac{3}{7}\right)\\ N=5,10\&15\end{array}$

Calculations:

Here first we check the value of $\frac{a+b}{1-ab}$

If the value is 1, then we proceed further,

Therefore,

  $\frac{a+b}{1-ab}=\frac{\frac{2}{5}+\frac{3}{7}}{1-\frac{2}{5}\times \frac{3}{7}}=1$

Therefore we can write,

  $\frac{\pi }{4}={\tan }^{-1}1=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\mathrm{....}+{(-1)}^{n-1}\frac{x^{2n-1}}{2n-1}+\mathrm{...}$

Now taking,

  $\begin{array}{l}N=5\\ \frac{\pi }{4}={\tan }^{-1}1=1-\frac{1^3}{3}+\frac{1^5}{5}-\frac{1^7}{7}+\frac{1^9}{9}=0.8349206\\ Therefore,\\ \pi =4\times 0.8349206=3.339682\end{array}$

Now for $N=10$ .

  $\begin{array}{l}N=10\\ \frac{\pi }{4}={\tan }^{-1}1=1-\frac{1^3}{3}+\frac{1^5}{5}-\frac{1^7}{7}+\frac{1^9}{9}-\frac{1^{11}}{11}+\frac{1^{13}}{13}-\frac{1^{15}}{15}+\frac{1^{17}}{17}-\frac{1^{19}}{19}=0.7604599\\ Therefore,\\ \pi =4\times 0.7604599=3.04183\end{array}$

Now for $N=15$ .

  $\begin{array}{l}N=15\\ \frac{\pi }{4}={\tan }^{-1}1=1-\frac{1^3}{3}+\frac{1^5}{5}-\frac{1^7}{7}+\frac{1^9}{9}-\frac{1^{11}}{11}+\frac{1^{13}}{13}-\frac{1^{15}}{15}+\frac{1^{17}}{17}-\frac{1^{19}}{19}+\frac{1^{21}}{21}-\frac{1^{23}}{23}+\frac{1^{25}}{25}-\frac{1^{27}}{27}+\frac{1^{29}}{29}=0.8020464\\ Therefore,\\ \pi =4\times 0.8020464=3.20818\end{array}$

e.

To determine

The best approximation from given values of N.

Explanation of Solution

For N=15, the approximation is very close and gives more accurate result. So for N=15 is the best approximation.