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Solving Nonlinear Equations QA Assignment - Newton-Raphson, Secant, and
Bisection Methods
LIBERTY UNIVERSITY
SCHOOL OF ENGINEERING AND TECHNOLOGY
QA Assignment
DATE: March 15, 2023
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Instructions
Answer the following questions in detail. Provide all necessary steps, calculations, and
explanations. Format any references in APA style if used.
Questions
1. Conceptual Understanding
a. Explain the significance of the Newton-Raphson method in solving nonlinear
equations. What are the advantages and potential drawbacks of using this method?
Answer:
The Newton-Raphson method is an iterative technique used to approximate the roots of a
real-valued function. It is highly efficient and typically converges rapidly when the initial
guess is close to the true root. However, it has drawbacks such as the need for the
derivative of the function and the possibility of divergence if the initial guess is not
sufficiently close to the actual root.
2. Numerical Application
a. Given the equation
f
(
x
)
=x3−3x+1= 0
, use the Newton-Raphson method to find the
root, starting with an initial guess of
x0= 0.5
. Perform three iterations and provide the final
approximation of the root.
Answer:
The Newton-Raphson formula is:
xn+1=xn−f
(
xn
)
f′
(
xn
)
With:
f
(
x
)
=x3−3x+1andf′
(
x
)
= 3x2−3
The iterations are as follows:
• Iteration 1:
x1= 0.5−
(
0.5
)
3−3
(
0.5
)
+1
3
(
0.5
)
2−3≈0.6667
• Iteration 2:
x2= 0.6667−
(
0.6667
)
3−3
(
0.6667
)
+1
3
(
0.6667
)
2−3≈0.6842
• Iteration 3:
x3= 0.6842−
(
0.6842
)
3−3
(
0.6842
)
+1
3
(
0.6842
)
2−3≈0.6850
Final approximation after three iterations:
x≈0.6850
.
3. Secant Method
a. Using the Secant method, find the root of the equation
f
(
x
)
= cos
(
x
)
−x
with initial
guesses
x0= 0.5
and
x1= 0.7
. Perform four iterations.
Answer:
The Secant method formula is:
xn+1=xn−f
(
xn
)
xn−xn−1
f
(
xn
)
−f
(
xn−1
)
Given:
f
(
x
)
= cos
(
x
)
−x
The iterations are:
• Iteration 1:
x2= 0.7−cos
(
0.7
)
−0.7
cos
(
0.7
)
−cos
(
0.5
)
+0.2≈0.7396
• Iteration 2:
x3= 0.7396−cos
(
0.7396
)
−0.7396
cos
(
0.7396
)
−cos
(
0.7
)
+0.0396 ≈0.7391
• Iteration 3:
x4= 0.7391−cos
(
0.7391
)
−0.7391
cos
(
0.7391
)
−cos
(
0.7396
)
+0.0005≈0.7391
Final approximation:
x≈0.7391
.
4. Bisection Method
a. Apply the Bisection method to find the root of the equation
f
(
x
)
=x2−4= 0
in the interval
[0, 3]. Perform five iterations and provide the final approximation of the root.
Answer:
The Bisection method involves:
• Iteration 1:
Midpoint
c1=0+3
2= 1.5
,
f
(
1.5
)
= 1.52−4= −1.75
. Root lies in [1.5, 3].
• Iteration 2:
Midpoint
c2=1.5+3
2= 2.25
,
f
(
2.25
)
= 2.252−4= 1.0625
. Root lies in [1.5, 2.25].
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• Iteration 3:
Midpoint
c3=1.5+2.25
2= 1.875
,
f
(
1.875
)
= 1.8752−4= −0.4844
. Root lies in [1.875,
2.25].
• Iteration 4:
Midpoint
c4=1.875+2.25
2= 2.0625
,
f
(
2.0625
)
= 2.06252−4= 0.2539
. Root lies in
[1.875, 2.0625].
• Iteration 5:
Midpoint
c5=1.875+2.0625
2= 1.96875
,
f
(
1.96875
)
= 1.968752−4= −0.124
. Root lies
in [1.96875, 2.0625].
Final approximation after five iterations:
x≈2.0156
.
Conclusion
In this assignment, we explored various methods for solving nonlinear equations, including
the Newton-Raphson, Secant, and Bisection methods. Each of these methods offers unique
advantages and challenges. The Newton-Raphson method is powerful but requires the
calculation of derivatives and careful selection of initial guesses. The Secant method,
although derivative-free, may converge more slowly. The Bisection method guarantees
convergence but may be less efficient compared to other methods. Through numerical
problems, we have demonstrated how these methods can be applied to obtain approximate
solutions to nonlinear equations, highlighting the importance of selecting appropriate
methods based on the problem at hand.
