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Quiz 3: The Newton-Raphson method

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Question 1

 
 
Solve the equation ex - 4x = 0 using Newton-Raphson iteration.
a) x = 0.61906 and x = 1.51213.   b) x = 0.35 and x = 2.1.
c) x = 0.35740 and x = 2.15329.   d) Newton-Raphson iteration cannot be used
since the answer oscillates between 2 and -2.

 

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Your answer is correct.
Sketch the curve of f(x) = ex - 4x and see that there is a solution between x = 0 and x = 1 and between x = 2 and x = 3. Now f′(x) = ex - 4 so the iterative function is
F(x) = x - ex --4x-= ex(x--1), ex - 4 ex - 4
giving x0 = 0.5, x1 = 0.35060, x2 = 0.35739, x3 = 0.35740, x4 = 0.35740, and x0 = 2.5, x1 = 2.23327, x2 = 2.15874, x3 = 2.15332, x4 = 2.15329, x5 = 2.15329.
Not correct. Choice (d) is false.
 

Question 2

 
 
Use the Newton-Raphson method to solve 2x3 - 6x2 + 6x - 1 = 0 to 4 decimal places.
a) There is no solution since the curve is always increasing.   b) x = 0.2063.
c) x = 0.7351.   d) Newton-Raphson cannot be used because the tangents to the
curve do not cut the axes on the interval 0 ≤ x ≤ 1.

 

Not correct. Choice (a) is false.
Your answer is correct.
Sketching the curve f(x) = 2x3 - 6x2 + 6x - 1 we see there is a root between 0 and 0.5. f(0) = -1, f(0.5) = 0.75 so the function changes sign on the interval 0 ≤ x ≤ 0.5. f′(x) = 6x2 - 12x + 6 = 6(x - 1)2 so there is a critical value at x = 1. f′′(x) = 12x - 12 so the second derivative changes sign at x = 1. The tangent at x = 0 is y = 6x - 1 which cuts the x-axis at x = 1 6 . The tangent at x = 0.5 is y = 3 2x which cuts the x-axis at x = 0. Therefore we can use Newton-Raphson with the iterative function
2x3-- 6x2-+-6x--1 4x3 --6x2-+-1 F(x) = x- 6x2 - 12x+ 6 = 6x2 - 12x + 6,
so x0 = 0.5, x1 = 0, x2 = 1 6 , x3 = 0.20444, x4 = 0.206295, x5 = 0.20630, x6 = 0.2063.
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.
 

Question 3

 
 
Newton-Raphson method will always converge to a solution for f(x) = 0 on the interval a ≤ x ≤ b if certain conditions are met. Which of the following is not one of these conditions ?
a) f is continuous on the interval a ≤ x ≤ b.   b) f(a) and f(b) have opposite signs.
c) f′′(x) does not change sign on the interval a ≤ x ≤ b.   d) f′(x) = 0 on the interval a ≤ x ≤ b.

 

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
Your answer is correct.
The condition should be that f′(x)⁄=0 on the interval a ≤ x ≤ b.
 

Question 4

 
 
The function f(x) = 2x3 - 2x2 - 3x + 2 has a root between 0 and 1. Which of the following conditions fail ?
a) f(0) and f(1) have opposite signs.   b) f′(x)⁄=0 on 0 ≤ x ≤ 1.
c) f′′(x) does not change sign on the interval 0 ≤ x ≤ 1.   d) The tangents at 0 and 1 cut the axes in the interval 0 ≤ x ≤ 1.

 

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Your answer is correct.
f′′(x) does change sign on the interval since f′′(0) = -4 and f′′(1) = 8.
Not correct. Choice (d) is false.
 

Question 5

 
 
As we saw in question 4, we cannot use the Newton-Raphson method to find the root of the function f(x) = 2x3 - 2x2 - 3x + 2 on the interval 0 ≤ x ≤ 1. Find a suitable function to use the Gregory-Dary iteration method and find the solution.
a) x = φ(x) = 1 - 3(2x3 - 2x2 + 2) is a suitable function and the solution is x = 0.57318.   b) x = φ(x) = 1 3(2x3 - 2x2 + 2) is a suitable function and the solution is x = 0.57138.
c) x = φ(x) = 1 + 3 --- 2x - 1 --2 x is a suitable function and the solution is x = 0.57318.   d) Gregory-Dary cannot be used either.

 

Your answer is correct.
x = φ(x) = 13(2x3 - 2x2 + 2) is a suitable function since ∣φ′(x)∣ < 1 on 0 ≤ x ≤ 1 and starting with x0 = 0.5 it takes 6 iterations to get the solution x5 = x6 = 0.57318.
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.
 

Question 6

 
 
Consider the function f(x) = x4 + 3x2 - x - 5. It has one root at x = -1. Use the Newton-Raphson method to find the other real root to 9 decimal places.
a) The other root is x = 1.185037375.   b) The other root is x = -1.000003000.
c) The other root is x = 1.185038651.   d) The iterations do not converge.

 

Your answer is correct.
xn+1 = xn -f(′xn)- f (xn)
= 3xn4-+-3xn2 +-5 4xn3 + 6xn - 1.
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.
 

Question 7

 
 
Consider the function f(x) = 3sin2x - 2e3x + x. Use the Newton-Raphson method starting at x0 = -2 and x0 = -3 to find the two roots to 9 decimal places (NB: For this question you should use Mathematica or a programmable calculator to ensure accurate answers).
a) The two roots are x = -1.919158972 and x = -2.617643267.   b) The root starting at x0 = -2 converges to x = -1.919255184 but the other does not converge.
c) The two roots are x = -1.919255184 and x = -2.612781943.   d) The iterations do not converge for either starting values.

 

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Your answer is correct.
xn+1 = xn --f′(xn)- f (xn)
= 6xncos2xn-- (6xn---2)e3xn --3sin2xn 6 cos2xn - 6e3xn + 1.
Not correct. Choice (d) is false.
 

Question 8

 
 
How many times do we need to iterate using the Newton-Raphson method to find the root of the function f(x) = 4xe2x - 2 to 4 decimal places starting with x0 = 0.5 ?
a) The method cannot be used since the derivative is zero when x = -0.5.   b) 15 iterations.
c) 6 iterations.   d) 4 iterations.

 

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
Your answer is correct.
xn+1 = xn - f(xn) f′(xn)
= 22xn -8x2xn-e---+2-- 4e n(1+ 2xn),
so x0 = 0.5, x1 = 0.3420, x2 = 0.2887, x3 = 0.2836
x4 = 0.2836. The answer is obtained in 4 iterations when starting with x0 = 0.5. Compare this with the Gregory-Dary iteration method for the same function where it took 15 iterations to get the answer.
 

Question 9

 
 
Use the Newton-Raphson method to find the second iterate to 5 dec. places of the solution of
2 -x f(x) = (37x + 1)e - 40 cos 2x = 0
starting at x0 = 0.5 .
a) x2 = 8.57612   b) x2 = 0.67150
c) x2 = 5.56270   d) x2 = 4.84173

 

Not correct. Choice (a) is false.
Your answer is correct.
The iterative function is
-x 3 2 F(x) = e-(--37x-+x-37x--2-x-- 1)-+80x-sin2x-+-40cos2x. e (- 37x + 74x - 1)+ 80sin2x
Note that our starting point will determine which of the many possible solutions we find.
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.
 

Question 10

 
 
The amount of insulin in microunits per mL in a diabetic patient is given by the function I(t) = 4.5 + 40.5te-0.26t where t is the number of hours since the last injection. If the next insulin injection must be given after the insulin has peaked and then fallen to 45 microunits per mL, at what time, to the nearest minute, must the next insulin injection be given ?
a) Using Newton-Raphson we find the injection must be given after 1 hour and 23 minutes.   b) Using Newton-Raphson we find the injection must be given after 8 hours exactly.
c) Using Newton-Raphson we find the injection must be given after 1 hour and 43 minutes.   d) Using the Gregory-Dary method, with t = φ(t) = lnt we find the injection must given after 7 hours and 16 minutes.

 

Not correct. Choice (a) is false.
Your answer is correct.
Consider the graph of I(t) shown below.
PIC
 
We wish to solve f(t) = 40.5te-0.26t - 40.5 = 0. The first root is between 0 and 1 and the second appears to be between 7 and 8.5. Try Newton-Raphson with x0 = 7.5 and we get x = 7.994 or 8 hours to the nearest minute.
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.