To solve for [tex]\( x_3 \)[/tex] using the iterative formula [tex]\( x_{n+1} = 8 + \frac{500}{x_n^2} \)[/tex] and the initial value [tex]\( x_1 = 59 \)[/tex], we proceed as follows:
1. Start with the initial value:
[tex]\[
x_1 = 59
\][/tex]
2. Calculate [tex]\( x_2 \)[/tex] using the iterative formula:
[tex]\[
x_2 = 8 + \frac{500}{x_1^2}
\][/tex]
Substitute [tex]\( x_1 = 59 \)[/tex]:
[tex]\[
x_2 = 8 + \frac{500}{59^2}
\][/tex]
Evaluating [tex]\( 59^2 \)[/tex]:
[tex]\[
59^2 = 3481
\][/tex]
Then, calculate [tex]\(\frac{500}{3481}\)[/tex]:
[tex]\[
\frac{500}{3481} \approx 0.143636885952313
\][/tex]
Now, add this value to 8:
[tex]\[
x_2 = 8 + 0.143636885952313 \approx 8.143636885952313
\][/tex]
3. Calculate [tex]\( x_3 \)[/tex] using the iterative formula:
[tex]\[
x_3 = 8 + \frac{500}{x_2^2}
\][/tex]
Substitute [tex]\( x_2 \approx 8.143636885952313 \)[/tex]:
[tex]\[
x_3 = 8 + \frac{500}{(8.143636885952313)^2}
\][/tex]
Evaluating [tex]\((8.143636885952313)^2 \)[/tex]:
[tex]\[
(8.143636885952313)^2 \approx 66.32271810800071
\][/tex]
Then, calculate [tex]\(\frac{500}{66.32271810800071}\)[/tex]:
[tex]\[
\frac{500}{66.32271810800071} \approx 7.539337807203339
\][/tex]
Now, add this value to 8:
[tex]\[
x_3 = 8 + 7.539337807203339 \approx 15.539337807203339
\][/tex]
4. Finally, round [tex]\( x_3 \)[/tex] to 2 decimal places:
[tex]\[
x_3 \approx 15.54
\][/tex]
Therefore, the value of [tex]\( x_3 \)[/tex] is [tex]\( 15.54 \)[/tex] when rounded to 2 decimal places.
