In this problem use paper and pencil except for evaluating the function you can use calculator. Consider the fixed point iteration method to f (x) = x² – x – 8 = x with initial value . With tolerance 0.0001 and maximum number of iterations set to N = 100 the first three steps of the methoc %3D result in X1 X2 X3 =
In this problem use paper and pencil except for evaluating the function you can use calculator. Consider the fixed point iteration method to f (x) = x² – x – 8 = x with initial value . With tolerance 0.0001 and maximum number of iterations set to N = 100 the first three steps of the methoc %3D result in X1 X2 X3 =
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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Fixed point iterations
Transcribed Image Text:Fixed Point Iteration
In this problem use paper and pencil except for evaluating the function you can use calculator.
Consider the fixed point iteration method to f (x) = x² – x – 8
With tolerance 0.0001 and maximum number of iterations set to N
= x with initial value .
100 the first three steps of the method
result in
X1 =
X2 =
X3
![Fixed point iterations
On this problem you should do the calculation by hand, and submit your answer in form of a formula instead of
decimal numbers.
Consider the function g(x) = x° -
- 6x2 + 15x – 20.
|
a. Calculate: g(3)
and g(3.5) =
%3|
Note that the function is continuous and changes sign, hence it has a root on the interval 3, 3.5|.
But calculating the root of a cubic polynomial is nontrivial.
b. From this root finding problem x° – 6x + 15x – 20 = 0 create a fixed point finding problem by
adding x to both sides.
You get the equation
c. Using notation f(x) =
for the left hand side of the previous equation calculate
f'(x)=|
and f'(3)=
and f"(3)=
d. Calculate f"(x)=
Since f"(3) > 0 and f"(x) is an increasing function on the interval 3, 3.5], we have that f"(3) > 0
on the interval 3, 3.5], and hence f'(x) > 1 on the interval 3, 3.5 containing the fixed point of f.
Hence the fixed point iteration ?
O to the fixed point on the interval 3, 3.5|.
e. From the root finding problem x° – 6x + 15x – 20 = 0 create another fixed point finding problem by
|
first rearraning the equations to get
= 2³.
Then take the cube root (1/3 power ) to get
= x.
f. Using notation h(x) =
for the left hand side of the previous equation calculate
h'(x)=
and h' (3.5)=
g. With more calculations it would be also possible to show that f'(x) < 1 on the interval 3, 3.5|
containing the fixed point of f. Hence the fixed point iteration ?
O to the fixed point
on the interval 3, 3.5].](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F60115176-7765-4199-b768-9f01752190d9%2Fecf524ae-da14-4762-82d2-129ad073c43c%2F8yvzyco_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Fixed point iterations
On this problem you should do the calculation by hand, and submit your answer in form of a formula instead of
decimal numbers.
Consider the function g(x) = x° -
- 6x2 + 15x – 20.
|
a. Calculate: g(3)
and g(3.5) =
%3|
Note that the function is continuous and changes sign, hence it has a root on the interval 3, 3.5|.
But calculating the root of a cubic polynomial is nontrivial.
b. From this root finding problem x° – 6x + 15x – 20 = 0 create a fixed point finding problem by
adding x to both sides.
You get the equation
c. Using notation f(x) =
for the left hand side of the previous equation calculate
f'(x)=|
and f'(3)=
and f"(3)=
d. Calculate f"(x)=
Since f"(3) > 0 and f"(x) is an increasing function on the interval 3, 3.5], we have that f"(3) > 0
on the interval 3, 3.5], and hence f'(x) > 1 on the interval 3, 3.5 containing the fixed point of f.
Hence the fixed point iteration ?
O to the fixed point on the interval 3, 3.5|.
e. From the root finding problem x° – 6x + 15x – 20 = 0 create another fixed point finding problem by
|
first rearraning the equations to get
= 2³.
Then take the cube root (1/3 power ) to get
= x.
f. Using notation h(x) =
for the left hand side of the previous equation calculate
h'(x)=
and h' (3.5)=
g. With more calculations it would be also possible to show that f'(x) < 1 on the interval 3, 3.5|
containing the fixed point of f. Hence the fixed point iteration ?
O to the fixed point
on the interval 3, 3.5].
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