At Westonci.ca, we provide reliable answers to your questions from a community of experts. Start exploring today! Discover a wealth of knowledge from professionals across various disciplines on our user-friendly Q&A platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
Let's work through each part of the question step-by-step:
### Part a) i)
#### Calculating [tex]\( x_2 \)[/tex]:
Given the recursive formula:
[tex]\[ x_{n+1} = \frac{(x_n)^3 - 1}{4} \][/tex]
Starting with:
[tex]\[ x_1 = -1 \][/tex]
We substitute [tex]\( x_1 \)[/tex] into the formula to find [tex]\( x_2 \)[/tex]:
[tex]\[ x_2 = \frac{(x_1)^3 - 1}{4} \][/tex]
[tex]\[ x_2 = \frac{(-1)^3 - 1}{4} \][/tex]
[tex]\[ x_2 = \frac{-1 - 1}{4} \][/tex]
[tex]\[ x_2 = \frac{-2}{4} \][/tex]
[tex]\[ x_2 = -0.5 \][/tex]
So:
[tex]\[ x_2 = -0.5 \][/tex]
___
### Part a) ii)
#### Calculating [tex]\( x_3 \)[/tex]:
Now we use [tex]\( x_2 \)[/tex] to find [tex]\( x_3 \)[/tex]:
[tex]\[ x_3 = \frac{(x_2)^3 - 1}{4} \][/tex]
[tex]\[ x_3 = \frac{(-0.5)^3 - 1}{4} \][/tex]
[tex]\[ x_3 = \frac{-0.125 - 1}{4} \][/tex]
[tex]\[ x_3 = \frac{-1.125}{4} \][/tex]
[tex]\[ x_3 = -0.28125 \][/tex]
So:
[tex]\[ x_3 = -0.28125 \][/tex]
___
### Part b)
#### Calculating the solution to 6 decimal places:
We start with the initial value [tex]\( x_1 = -1 \)[/tex] and compute successive values [tex]\( x_n \)[/tex] using the recursive formula until the values converge within a tolerance of [tex]\( 10^{-6} \)[/tex].
Upon iterating the process, the value converges to:
[tex]\[ x \approx -0.254102 \][/tex]
So the final solution to 6 decimal places is:
[tex]\[ \text{Solution} = -0.254102 \][/tex]
___
Here is the completed result for the problem:
### Final Answers:
a)
(i) [tex]\( x_2 = -0.5 \)[/tex]
(ii) [tex]\( x_3 = -0.28125 \)[/tex]
b) [tex]\( \text{Solution to 6 decimal places} = -0.254102 \)[/tex]
### Part a) i)
#### Calculating [tex]\( x_2 \)[/tex]:
Given the recursive formula:
[tex]\[ x_{n+1} = \frac{(x_n)^3 - 1}{4} \][/tex]
Starting with:
[tex]\[ x_1 = -1 \][/tex]
We substitute [tex]\( x_1 \)[/tex] into the formula to find [tex]\( x_2 \)[/tex]:
[tex]\[ x_2 = \frac{(x_1)^3 - 1}{4} \][/tex]
[tex]\[ x_2 = \frac{(-1)^3 - 1}{4} \][/tex]
[tex]\[ x_2 = \frac{-1 - 1}{4} \][/tex]
[tex]\[ x_2 = \frac{-2}{4} \][/tex]
[tex]\[ x_2 = -0.5 \][/tex]
So:
[tex]\[ x_2 = -0.5 \][/tex]
___
### Part a) ii)
#### Calculating [tex]\( x_3 \)[/tex]:
Now we use [tex]\( x_2 \)[/tex] to find [tex]\( x_3 \)[/tex]:
[tex]\[ x_3 = \frac{(x_2)^3 - 1}{4} \][/tex]
[tex]\[ x_3 = \frac{(-0.5)^3 - 1}{4} \][/tex]
[tex]\[ x_3 = \frac{-0.125 - 1}{4} \][/tex]
[tex]\[ x_3 = \frac{-1.125}{4} \][/tex]
[tex]\[ x_3 = -0.28125 \][/tex]
So:
[tex]\[ x_3 = -0.28125 \][/tex]
___
### Part b)
#### Calculating the solution to 6 decimal places:
We start with the initial value [tex]\( x_1 = -1 \)[/tex] and compute successive values [tex]\( x_n \)[/tex] using the recursive formula until the values converge within a tolerance of [tex]\( 10^{-6} \)[/tex].
Upon iterating the process, the value converges to:
[tex]\[ x \approx -0.254102 \][/tex]
So the final solution to 6 decimal places is:
[tex]\[ \text{Solution} = -0.254102 \][/tex]
___
Here is the completed result for the problem:
### Final Answers:
a)
(i) [tex]\( x_2 = -0.5 \)[/tex]
(ii) [tex]\( x_3 = -0.28125 \)[/tex]
b) [tex]\( \text{Solution to 6 decimal places} = -0.254102 \)[/tex]
Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.