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An iterative formula is shown below.

[tex]\[ x_{n+1} = \frac{7 - x_n^3}{8} \][/tex]

Starting with [tex]\( x_1 = 2.9 \)[/tex], work out the values of [tex]\( x_3 \)[/tex] and [tex]\( x_4 \)[/tex]. Give each answer to 1 d.p.


Sagot :

To solve this problem, we will follow the iteration process step by step as outlined by the given formula [tex]\( x_{n+1} = \frac{7 - x_n^3}{8} \)[/tex].

Given:
[tex]\( x_1 = 2.9 \)[/tex]

We need to find [tex]\( x_3 \)[/tex] and [tex]\( x_4 \)[/tex] with each answer rounded to 1 decimal place.

### Step-by-Step Solution:

1. Calculate [tex]\( x_2 \)[/tex] using [tex]\( x_1 \)[/tex]:
[tex]\[ x_2 = \frac{7 - x_1^3}{8} \][/tex]
Substituting [tex]\( x_1 = 2.9 \)[/tex]:
[tex]\[ x_2 = \frac{7 - (2.9)^3}{8} \][/tex]

2. Calculate [tex]\( x_3 \)[/tex] using [tex]\( x_2 \)[/tex]:
[tex]\[ x_3 = \frac{7 - x_2^3}{8} \][/tex]
Substituting the calculated value of [tex]\( x_2 \)[/tex]:
[tex]\[ x_3 = \frac{7 - (-2.173625)^3}{8} \][/tex]

3. Calculate [tex]\( x_4 \)[/tex] using [tex]\( x_3 \)[/tex]:
[tex]\[ x_4 = \frac{7 - x_3^3}{8} \][/tex]
Substituting the calculated value of [tex]\( x_3 \)[/tex]:
[tex]\[ x_4 = \frac{7 - (2.1587009850754395)^3}{8} \][/tex]

4. Rounding to 1 decimal place:

- The value of [tex]\( x_3 \)[/tex] calculated is approximately 2.1587009850754395, which rounds to 2.2.
- The value of [tex]\( x_4 \)[/tex] calculated is approximately -0.3824406100402049, which rounds to -0.4.

### Conclusion:
- [tex]\( x_3 \)[/tex] is approximately 2.2 when rounded to 1 decimal place.
- [tex]\( x_4 \)[/tex] is approximately -0.4 when rounded to 1 decimal place.
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