Let f(x) = e-0.5x on [0, 1]. Check the conditions of the fixed point theorem and subsequently conclude whether or not the sequence (Po, P₁, P2,...) defined by Pn = 8(Pn-1), n≥1 with po = 0.7 will converge to a unique fixed point in [0, 1].
Let f(x) = e-0.5x on [0, 1]. Check the conditions of the fixed point theorem and subsequently conclude whether or not the sequence (Po, P₁, P2,...) defined by Pn = 8(Pn-1), n≥1 with po = 0.7 will converge to a unique fixed point in [0, 1].
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Numerical Analysis
Please follow all the steps in the given solution and use the same method. Do NOT skip any steps.
![Question 1 solution: We need to check the conditions:
1. f € C[0, 1]
2. f(x) = [0, 1] for all x = [0, 1]
3. f'(x) exists on (0, 1)
4. There exists a constant k € (0, 1) such that f'(x)| ≤ k for all x € (0, 1).
If all the above are met then, by the Fixed Point Theorem, the sequence will converge
to a unique fixed point of f.
For (a), (c)- clearly f is continuous, and f'(x) = -e* throughout (0, 1).
For (b), we can see that f(x) is decreasing throughout the interval by checking the
derivative: f'(x) < 0 for all x € (0, 1). Therefore f(x) is maximised at x = 0 and
minimised at x = 1, so
0 <
<= 1 € = = f(1) ≤ f(x) ≤ f(0) = 1
hence f(x) = [0, 1] for all x € [0, 1].
For (d),
(²^(x)) = ²1 e-tx;
page 1
if x € [0, 1],
hence we can choose k = 1.
Therefore all the conditions are satisfied and hence the sequence will converge to a
unique fixed point.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4f3fcee3-df4f-43f5-a70c-56c5ba155c30%2F79654504-5791-42cd-8347-11dc705100cf%2F2z2zz36_processed.png&w=3840&q=75)
Transcribed Image Text:Question 1 solution: We need to check the conditions:
1. f € C[0, 1]
2. f(x) = [0, 1] for all x = [0, 1]
3. f'(x) exists on (0, 1)
4. There exists a constant k € (0, 1) such that f'(x)| ≤ k for all x € (0, 1).
If all the above are met then, by the Fixed Point Theorem, the sequence will converge
to a unique fixed point of f.
For (a), (c)- clearly f is continuous, and f'(x) = -e* throughout (0, 1).
For (b), we can see that f(x) is decreasing throughout the interval by checking the
derivative: f'(x) < 0 for all x € (0, 1). Therefore f(x) is maximised at x = 0 and
minimised at x = 1, so
0 <
<= 1 € = = f(1) ≤ f(x) ≤ f(0) = 1
hence f(x) = [0, 1] for all x € [0, 1].
For (d),
(²^(x)) = ²1 e-tx;
page 1
if x € [0, 1],
hence we can choose k = 1.
Therefore all the conditions are satisfied and hence the sequence will converge to a
unique fixed point.
![Let f(x) = e-0.5x on [0, 1].
Check the conditions of the fixed point theorem and subsequently conclude
whether or not the sequence (Po, P1, P2,...) defined by
Pn = 8(Pn-1), n≥1
with po = 0.7 will converge to a unique fixed point in [0, 1].](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4f3fcee3-df4f-43f5-a70c-56c5ba155c30%2F79654504-5791-42cd-8347-11dc705100cf%2F46uswrr_processed.png&w=3840&q=75)
Transcribed Image Text:Let f(x) = e-0.5x on [0, 1].
Check the conditions of the fixed point theorem and subsequently conclude
whether or not the sequence (Po, P1, P2,...) defined by
Pn = 8(Pn-1), n≥1
with po = 0.7 will converge to a unique fixed point in [0, 1].
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Follow-up Question
In condition #3,
The maximum absolute value of g'(x) in the interval [0,1] is approximately 0.39
How did you get 0.39?
Solution
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