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Given:
[tex]\[ x_{n+1} = \frac{(x_n)^3 - 1}{4} \quad \text{and} \quad x_1 = -1 \][/tex]

a)
(i) Work out the value of [tex]\( x_2 \)[/tex]:
[tex]\[ x_2 = -0.28 \][/tex]

(ii) Work out the value of [tex]\( x_3 \)[/tex]:
[tex]\[ x_3 = \square \][/tex]

b) Work out the solution to 6 decimal places:
[tex]\[ \square \][/tex]



Answer :

GinnyAnswer
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]

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