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MATH1013 Quizzes

Quiz 2: The Gregory-Dary method
Question 1 Questions
The function f(x) = x3 + 3x2 + 4x + 1 has a root between x = 0 and x = 1. Which of the functions below are in the form x = ϕ(x) and will lead to finding a solution to f(x) = 0 using the Gregory-Dary method ?
a)
ϕ(x) = 1 4(x3 + 3x2 + 1)
 b)
ϕ(x) = 3 + 4 x + 1 x2
c)
ϕ(x) = 1 3 x2 + 4 + 1 x
 d)
ϕ(x) = 1 4 x3 3x2 1

Choice (a) is correct!
ϕ(x) = 3 4x2 3 2x so |ϕ(x)| < 1 when 1 < x < 0.
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
The function f(x) = x3 + 3x2 + 4x + 1 has a root at x = 0.3177, correct to 4 decimal places. Using the Gregory-Dary method, how many steps are required to find this root, starting at x = 0.5 ?
a)
3
 b)
5
c)
7
 d)
9

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
x1 = 1 4(x03 + 3x 02 + 1) where x0 = 0.5 x1 = 0.4063, x2 = 0.3570, x3 = 0.3342, x4 = 0.3244, x5 = 0.3204 x6 = 0.3188, x7 = 0.3181, x8 = 0.3178, x9 = 0.3177,x10 = 0.3177.
Rearrange the equation f(x) = 3x 2 3sinx into the form x = ϕ(x) and determine if the solution to f(x) = 0 can be found using the Gregory-Dary iterative method.
a)
x = ϕ(x) = sinx + 2 and the Gregory-Dary iterative method can be used since |ϕ(x)| 1 for all x.
 b)
x = ϕ(x) = sinx + 2 3 and the Gregory-Dary iterative method can be used since |ϕ(x)| 1 for all x.
c)
x = ϕ(x) = sinx + 2 and the Gregory-Dary iterative method can be used since |ϕ(x)| 1 for all x.
 d)
x = ϕ(x) = sinx + 2 3 and the Gregory-Dary iterative method can be used since |ϕ(x)| 1 for all x.

Choice (a) is incorrect
Choice (b) is correct!
If x = ϕ(x) = sinx + 2 3 then |ϕ(x)| = |cosx| 1 for all x.
Choice (c) is incorrect
Choice (d) is incorrect
Find the solution to f(x) = 3x 2 3sinx = 0 correct to 4 decimal places using the Gregory-Dary method.
a)
x = 1.6647
 b)
x = 1.6621
c)
x = 1.6625
 d)
The solution cannot be found since |ϕ(x)| 1 for all x.

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
Using x = ϕ(x) = sinx + 2 3 and starting at x0 = 1, we find x1 = 1.5081, x2 = 1.6647, x3 = 1.6623, x4 = 1.6625, x5 = 1.6625.
Choice (d) is incorrect
Rearrange the equation f(x) = x2 4x 2cos2x = 0 into the form x = ϕ(x) and determine if the solution of f(x) = 0 can be found using the Gregory-Dary iterative method for the root between x = 0.3 and x = 0.6.
a)
x = ϕ(x) = 1 4x2 + 1 2cos2x
 b)
x = ϕ(x) = 4x 2 cos 2x
c)
x = ϕ(x) = 1 2arccos(2x 1 2x2)
 d)
No re-arrangements of the equation are possible to find a solution.

Choice (a) is incorrect
Choice (b) is correct!
Note that |ϕ(x)| = |1 2x + cos2x| < 1 for x between x = 0.3 and x = 0.6.
Choice (c) is incorrect
Choice (d) is incorrect
Rearrange the equation f(x) = x2 4x 2cos2x = 0 into the form x = ϕ(x) and determine if the solution to f(x) = 0 can be found using the Gregory-Dary iterative method for the root between x = 4 and x = 5.
a)
x = ϕ(x) = 1 4x2 + 1 2cos2x
 b)
x = ϕ(x) = 4x 2 cos 2x
c)
x = ϕ(x) = 1 2arccos(2x 1 2x2)
 d)
No re-arrangements of the equation are possible to find a solution.

Choice (a) is incorrect
Choice (b) is correct!
Note that ϕ(x) = 2 2sin2x 4x 2 cos 2x < 1 when 4 < x < 5.
Choice (c) is incorrect
Choice (d) is incorrect
The function f(x) = x2 4x + 2 has two roots, x = α and x = β. Suppose α < β. Determine the functions ϕ1(x) and ϕ2(x) which would be suitable to find both α and β using the Gregory-Dary method.
a)
ϕ1(x) = x = 1 4(x2 + 2) is suitable to find both α and β.
 b)
ϕ1(x) = x = 4 2 x is suitable to find both α and β.
c)
ϕ1(x) = x = 1 4(x2 + 2) is suitable to find α and
ϕ2(x) = x = 4 2 x is suitable to find β.
 d)
ϕ2(x) = x = 4 2 x is suitable to find α and
ϕ1(x) = x = 1 4(x2 + 2) is suitable to find β.

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
x2 4x + 2 = 0 has two roots x = 2 ±2 so α 0.5858 and β 3.4142.
Consider ϕ1(x) = 1 4(x2 + 2), then ϕ1(x) = 1 2x and |ϕ1(x)| < 1 when 0 < x < 2 so ϕ1(x) is suitable to find α. Now consider ϕ2(x) = 4 2 x, then ϕ2(x) = 2 x2 and |ϕ2(x)| < 1 when x > 2 so ϕ2(x) is suitable to find β.
Choice (d) is incorrect
The function f(x) = 4xe2x 2 has a root between x = 0 and x = 1. Which of the functions below are in the form x = ϕ(x) and will lead to finding a solution of f(x) = 0 using the Gregory-Dary method ?
a)
x = 1 2e2x
 b)
2x = e2x
c)
x = 1 2ln2x
 d)
The function |ϕ(x)| > 1 on 0 < x < 1 for all choices of ϕ(x).

Choice (a) is correct!
ϕ(x) = 1 2e2xhence ϕ(x) = e2xand|ϕ(x)| < 1on0 < x < 1.
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Use the Gregory-Dary method to find the root of the function f(x) = 4xe2x 2 to 4 decimal places.
a)
x = 0.2835
 b)
x = 0.2836
c)
x = 0.2837
 d)
x = 0.2805

Choice (a) is incorrect
Choice (b) is correct!
Iteration needs to be performed 15 times to obtain the solution correct to 4 decimal places and 19 for 5 decimal places.
Choice (c) is incorrect
Choice (d) is incorrect
The equation x4 + 3x2 x 5 = 0 has two solutions, x = 1 and a solution between 1 and 2. Which of the following statements is correct ?
a)
The solution x = 1 can be found using the Gregory-Dary method with x = x4 + 3x2 5 and starting at x = 0.5 but the other solution cannot be found.
 b)
The solution x = 1 cannot be found using the Gregory-Dary method with x = x4 + 3x2 5 but the other solution can be found starting at x = 1.5.
c)
Neither solution can be found using the Gregory-Dary method with x = x4 + 3x2 5 since |ϕ(x)| < 1 on the intervals containing the solutions.
 d)
Neither solution can be found using the Gregory-Dary method with x = x4 + 3x2 5 since |ϕ(x)| > 1 on the intervals containing the solutions.

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
ϕ(x) = 4x3 + 6x > 1for1 < x < 2and |ϕ(x)| = |4x3 + 6x| > 1for 1.5 < x < 0.5. Note that it may be possible to calculate the roots using other methods, in particular the Newton-Raphson method.