Sure, let's explore the sequence defined by the iterative process [tex]\( x_{n+1} = (x_n)^2 - x_n \)[/tex] for the given initial values in parts (a), (b), and (c).
### Part a) Describe the sequence of numbers when [tex]\( x_1 = 1 \)[/tex]
1. Start with [tex]\( x_1 = 1 \)[/tex].
2. Calculate the next term [tex]\( x_2 \)[/tex]:
[tex]\[
x_2 = (x_1)^2 - x_1 = (1)^2 - 1 = 1 - 1 = 0
\][/tex]
3. Now calculate [tex]\( x_3 \)[/tex]:
[tex]\[
x_3 = (x_2)^2 - x_2 = (0)^2 - 0 = 0
\][/tex]
4. The sequence from this point onwards stabilizes at 0, as subsequent terms continue to be 0:
[tex]\[
x_4 = (x_3)^2 - x_3 = 0
\][/tex]
and so on.
Therefore, the sequence when [tex]\( x_1 = 1 \)[/tex] is: [tex]\( \{1, 0, 0, 0, \ldots\} \)[/tex].
### Part b) Describe the sequence of numbers when [tex]\( x_1 = -1 \)[/tex]
1. Start with [tex]\( x_1 = -1 \)[/tex].
2. Calculate the next term [tex]\( x_2 \)[/tex]:
[tex]\[
x_2 = (x_1)^2 - x_1 = (-1)^2 - (-1) = 1 + 1 = 2
\][/tex]
3. Now calculate [tex]\( x_3 \)[/tex]:
[tex]\[
x_3 = (x_2)^2 - x_2 = (2)^2 - 2 = 4 - 2 = 2
\][/tex]
4. The sequence stabilizes at 2 because each subsequent term remains 2:
[tex]\[
x_4 = (x_3)^2 - x_3 = 2
\][/tex]
and so on.
Therefore, the sequence when [tex]\( x_1 = -1 \)[/tex] is: [tex]\( \{-1, 2, 2, 2, \ldots\} \)[/tex].
### Part c) Work out the value of [tex]\( x_2 \)[/tex] when [tex]\( x_1 = 1 - \sqrt{2} \)[/tex]
1. Start with [tex]\( x_1 = 1 - \sqrt{2} \)[/tex].
2. Calculate the next term [tex]\( x_2 \)[/tex]:
[tex]\[
x_2 = (x_1)^2 - x_1 = (1 - \sqrt{2})^2 - (1 - \sqrt{2})
\][/tex]
3. First, expand [tex]\( (1 - \sqrt{2})^2 \)[/tex]:
[tex]\[
(1 - \sqrt{2})^2 = 1 - 2\sqrt{2} + 2 = 3 - 2\sqrt{2}
\][/tex]
4. Now substitute into the expression for [tex]\( x_2 \)[/tex]:
[tex]\[
x_2 = (3 - 2\sqrt{2}) - (1 - \sqrt{2}) = 3 - 2\sqrt{2} - 1 + \sqrt{2} = 2 - \sqrt{2}
\][/tex]
5. This simplifies to:
[tex]\[
x_2 \approx 0.585786
\][/tex]
Therefore, the value of [tex]\( x_2 \)[/tex] when [tex]\( x_1 = 1 - \sqrt{2} \)[/tex] is approximately [tex]\( 0.5857864376269051 \)[/tex].
