Chapter 13.4, Problem 37E

(a)

To determine

To find: The first ten terms of the sequence and the limit of the sequence.

Answer to Problem 37E

The first ten terms of the sequence are $1.732,1.932,1.983,1.996,1.999,2,2,2,2,2$ .

Explanation of Solution

Given:

The function is given by,

  $a_{n+1}=\sqrt{2+a_n}$ and $a_1=0$ .

Calculation:

Substitute $1$ for $n$ and $0$ for $a_1$ , in the formula $a_{n+1}=\sqrt{2+a_n}$ to find the third term of the sequence.

  $\begin{array}{c}{a}_{1+1}=\sqrt{2+a_1}\\ a_2=\sqrt{2+1}\\ =\sqrt{3}\end{array}$

Substitute $2$ for $n$ and $1.732$ for $a_2$ , in the formula $a_{n+1}=\sqrt{2+a_n}$ to find the third term of the sequence.

  $\begin{array}{c}{a}_{2+1}=\sqrt{2+a_2}\\ a_3=\sqrt{2+1.732}\\ =\sqrt{3.732}\end{array}$

Substitute $3$ for $n$ and $1.932$ for $a_3$ , in the formula $a_{n+1}=\sqrt{2+a_n}$ to find the third term of the sequence.

  $\begin{array}{c}{a}_{3+1}=\sqrt{2+a_3}\\ a_4=\sqrt{2+1.932}\\ =\sqrt{3.932}\\ =1.983\end{array}$

Substitute $4$ for $n$ and $1.983$ for $a_4$ , in the formula $a_{n+1}=\sqrt{2+a_n}$ to find the third term of the sequence.

  $\begin{array}{c}{a}_{4+1}=\sqrt{2+a_4}\\ a_4=\sqrt{2+1.983}\\ =\sqrt{3.983}\\ =1.996\end{array}$

Substitute $5$ for $n$ and $1.996$ for $a_5$ , in the formula $a_{n+1}=\sqrt{2+a_n}$ to find the third term of the sequence.

  $\begin{array}{c}{a}_{5+1}=\sqrt{2+a_5}\\ a_6=\sqrt{2+1.996}\\ =\sqrt{3.996}\\ =1.999\end{array}$

Substitute $6$ for $n$ and $1.999$ for $a_6$ , in the formula $a_{n+1}=\sqrt{2+a_n}$ to find the third term of the sequence.

  $\begin{array}{c}{a}_{6+1}=\sqrt{2+a_6}\\ a_7=\sqrt{2+1.999}\\ =\sqrt{3.999}\\ \approx 2\end{array}$

Substitute 7 for $n$ and 2 for $a_6$ , in the formula $a_{n+1}=\sqrt{2+a_n}$ to find the third term of the sequence.

  $\begin{array}{c}{a}_{7+1}=\sqrt{2+a_7}\\ a_7=\sqrt{2+2}\\ =\sqrt{4}\\ =2\end{array}$

So, the next terms will be 2.

Therefore, the first ten terms of the sequence are $1.732,1.932,1.983,1.996,1.999,2,2,2,2,2$ .

(b)

To determine

To find: The value of L.

Answer to Problem 37E

The exact value of L is $1$ .

Explanation of Solution

Given:

The function is given by $L=\sqrt{2+L}$ .

Calculation:

Taking square on both sides.

  $\begin{array}{c}{L}^2={\left(\sqrt{2+L}\right)}^2\\ L^2=2+L\\ L^2-L-2=0\end{array}$

Further, solve the above equation.

  $\begin{array}{c}L=\frac{-\left(-1\right)\pm \sqrt{{\left(-1\right)}^2-4\left(1\right)\left(-2\right)}}{2\left(-1\right)}\\ =-\frac{1\pm \sqrt{1+8}}{2}\\ =-\frac{1\pm 3}{2}\\ =-2,1\end{array}$

As the negative value of L is not true for the equation.

Therefore, the exact value of L is $1$ .