Discover the answers you need at Westonci.ca, where experts provide clear and concise information on various topics. Explore thousands of questions and answers from knowledgeable experts in various fields on our Q&A platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.
Sagot :
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.
Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.
