Answer :
Sure! Let's go through the steps to find the values of [tex]\( u_2 \)[/tex], [tex]\( u_3 \)[/tex], and the solution to 5 decimal places.
### Part (a)
The sequence is defined by the iterative process:
[tex]\[ u_{n+1} = \frac{3}{u_n + 1} \][/tex]
with the initial value [tex]\( u_1 = 4 \)[/tex].
1. Calculate [tex]\( u_2 \)[/tex]:
[tex]\[ u_2 = \frac{3}{u_1 + 1} \][/tex]
Substitute [tex]\( u_1 = 4 \)[/tex]:
[tex]\[ u_2 = \frac{3}{4 + 1} = \frac{3}{5} = 0.6 \][/tex]
2. Calculate [tex]\( u_3 \)[/tex]:
[tex]\[ u_3 = \frac{3}{u_2 + 1} \][/tex]
Substitute [tex]\( u_2 = 0.6 \)[/tex]:
[tex]\[ u_3 = \frac{3}{0.6 + 1} = \frac{3}{1.6} = 1.875 \][/tex]
So, the values of [tex]\( u_2 \)[/tex] and [tex]\( u_3 \)[/tex] are:
[tex]\[ u_2 = 0.6 \][/tex]
[tex]\[ u_3 = 1.875 \][/tex]
### Part (b)
To find the steady-state solution [tex]\( x \)[/tex], we solve the equation:
[tex]\[ x = \frac{3}{x + 1} \][/tex]
Multiplying both sides by [tex]\( x + 1 \)[/tex] to clear the fraction:
[tex]\[ x(x + 1) = 3 \][/tex]
[tex]\[ x^2 + x = 3 \][/tex]
[tex]\[ x^2 + x - 3 = 0 \][/tex]
This is a quadratic equation. We solve it using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c = -3 \)[/tex]:
[tex]\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-1 \pm \sqrt{1 + 12}}{2} \][/tex]
[tex]\[ x = \frac{-1 \pm \sqrt{13}}{2} \][/tex]
So, we have two solutions:
[tex]\[ x = \frac{-1 + \sqrt{13}}{2} \][/tex]
[tex]\[ x = \frac{-1 - \sqrt{13}}{2} \][/tex]
The second solution is negative, and since [tex]\( x \)[/tex] must be positive (as each term in the sequence is positive), we take:
[tex]\[ x = \frac{-1 + \sqrt{13}}{2} \][/tex]
To five decimal places:
[tex]\[ x \approx 1.3028 \][/tex]
Thus, the steady-state solution to 5 decimal places is:
[tex]\[ x = 1.30280 \][/tex]
### Summary
- [tex]\( u_2 = 0.6 \)[/tex]
- [tex]\( u_3 = 1.875 \)[/tex]
- [tex]\( x = 1.30280 \)[/tex]
Each step breaks down how to iteratively determine the next term in the sequence and find the steady-state solution using algebraic methods.
### Part (a)
The sequence is defined by the iterative process:
[tex]\[ u_{n+1} = \frac{3}{u_n + 1} \][/tex]
with the initial value [tex]\( u_1 = 4 \)[/tex].
1. Calculate [tex]\( u_2 \)[/tex]:
[tex]\[ u_2 = \frac{3}{u_1 + 1} \][/tex]
Substitute [tex]\( u_1 = 4 \)[/tex]:
[tex]\[ u_2 = \frac{3}{4 + 1} = \frac{3}{5} = 0.6 \][/tex]
2. Calculate [tex]\( u_3 \)[/tex]:
[tex]\[ u_3 = \frac{3}{u_2 + 1} \][/tex]
Substitute [tex]\( u_2 = 0.6 \)[/tex]:
[tex]\[ u_3 = \frac{3}{0.6 + 1} = \frac{3}{1.6} = 1.875 \][/tex]
So, the values of [tex]\( u_2 \)[/tex] and [tex]\( u_3 \)[/tex] are:
[tex]\[ u_2 = 0.6 \][/tex]
[tex]\[ u_3 = 1.875 \][/tex]
### Part (b)
To find the steady-state solution [tex]\( x \)[/tex], we solve the equation:
[tex]\[ x = \frac{3}{x + 1} \][/tex]
Multiplying both sides by [tex]\( x + 1 \)[/tex] to clear the fraction:
[tex]\[ x(x + 1) = 3 \][/tex]
[tex]\[ x^2 + x = 3 \][/tex]
[tex]\[ x^2 + x - 3 = 0 \][/tex]
This is a quadratic equation. We solve it using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c = -3 \)[/tex]:
[tex]\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-1 \pm \sqrt{1 + 12}}{2} \][/tex]
[tex]\[ x = \frac{-1 \pm \sqrt{13}}{2} \][/tex]
So, we have two solutions:
[tex]\[ x = \frac{-1 + \sqrt{13}}{2} \][/tex]
[tex]\[ x = \frac{-1 - \sqrt{13}}{2} \][/tex]
The second solution is negative, and since [tex]\( x \)[/tex] must be positive (as each term in the sequence is positive), we take:
[tex]\[ x = \frac{-1 + \sqrt{13}}{2} \][/tex]
To five decimal places:
[tex]\[ x \approx 1.3028 \][/tex]
Thus, the steady-state solution to 5 decimal places is:
[tex]\[ x = 1.30280 \][/tex]
### Summary
- [tex]\( u_2 = 0.6 \)[/tex]
- [tex]\( u_3 = 1.875 \)[/tex]
- [tex]\( x = 1.30280 \)[/tex]
Each step breaks down how to iteratively determine the next term in the sequence and find the steady-state solution using algebraic methods.
