COS2633
EXAM
PACK 2023
LATEST
QUESTIONS
AND ANSWERS
For queries or any assignment help
Email:
,COS2633
EXAM PACK
Revision PACK
Questions. Answers
, NUMERICAL METHODS I – COS 233-8
Interval bisection method
Strategy Find two values of x, that is, a and b, which bracket the root by checking whether f ( a ) × f (b) < 0. Then
successively divide the interval in half and replace one endpoint with the midpoint so that the root is
bracketed again.
Requirements The function f(x) must be continuous in the interval.
There should not be multiple roots in the interval.
Advantages The number of iterations to achieve a specified accuracy is known in advance.
It is the method that is recommended for finding a first approximation to the root.
Convergence Slow because the estimate of the root may be better at an earlier iteration than at later ones.
Order of convergence 1
(b − a )
Error formula e=
2n
Note This does not mean that each error is smaller than the previous one.
Secant method
Strategy Choose two values of x, that is, x 0 and x1 , which are close to the root. Draw a straight line through the
points (these points can either be on the same side or on opposites sides of the root). The intersection of
the line with the x-axis should be close to the root. Repeat the process by always using the last two
computed values.
Requirements The function f(x) must be continuous.
The function f(x) must not be far from linear in the vicinity of the root.
Convergence Intermediate because the error is proportional to the product of the previous two errors. It is therefor
faster than a linear method but slower than a quadratic method.
Order of convergence 1.62
f ( xn )
Iteration formula x n +1 = x n − ( x n − x n −1 )
f ( x n ) − f ( x n −1 )
g ′′( ξ1 , ξ2 )
Error formula en +1 = (e n )(en −1 )
2
1+ 5
Note The order of convergence is = 1.62 .
2
, False position (regula falsi) method
Strategy Choose two values of x, that is,x0 and x1 , which bracket the root. Draw a straight line through the
points. The intersection of the line with the x-axis should be close to the root. Repeat the process by
always checking that the root remains bracketed.
Requirement The function f(x) must be continuous in the interval.
Advantage Unlike the interval bisection method, the intersection of the line and the x-axis does not necessarily occur
at the midpoint of the interval.
Convergence Intermediate because, though it is faster than interval bisection, its algorithm is slightly complicated. If
convergence takes place from the end that is farther from the root, it slows down the process.
Order of convergence 1
f ( xn )
Iteration formula x n +1 = x n − ( x n − x n −1 )
f ( x n ) − f ( x n −1 )
Note In most cases, this method converges to the root from one end of the interval.
Newton’s method
Strategy This method is based on a linear approximation of the function but does so by using a tangent to th
curve. Find one starting value that is not too far from the root and move along the tangent to its
intersection with the x-axis so as to obtain the next approximation.
Requirement The function f(x) must be continuous in the interval.
The derivative of the function f(x) must exist.
Convergence This method is rapidly convergent in the neighbourhood of the root. The error of each step approaches a
constant k times the square of the error of the previous step. This means that the number of decimal places
of accuracy nearly doubles at each iteration. However, the method requires two function evaluations per
step.
Order of convergence 2
f ( xn )
Iteration formula x n +1 = x n −
f ′( x n )
Error formula en +1 ∝ e n2
EXAM
PACK 2023
LATEST
QUESTIONS
AND ANSWERS
For queries or any assignment help
Email:
,COS2633
EXAM PACK
Revision PACK
Questions. Answers
, NUMERICAL METHODS I – COS 233-8
Interval bisection method
Strategy Find two values of x, that is, a and b, which bracket the root by checking whether f ( a ) × f (b) < 0. Then
successively divide the interval in half and replace one endpoint with the midpoint so that the root is
bracketed again.
Requirements The function f(x) must be continuous in the interval.
There should not be multiple roots in the interval.
Advantages The number of iterations to achieve a specified accuracy is known in advance.
It is the method that is recommended for finding a first approximation to the root.
Convergence Slow because the estimate of the root may be better at an earlier iteration than at later ones.
Order of convergence 1
(b − a )
Error formula e=
2n
Note This does not mean that each error is smaller than the previous one.
Secant method
Strategy Choose two values of x, that is, x 0 and x1 , which are close to the root. Draw a straight line through the
points (these points can either be on the same side or on opposites sides of the root). The intersection of
the line with the x-axis should be close to the root. Repeat the process by always using the last two
computed values.
Requirements The function f(x) must be continuous.
The function f(x) must not be far from linear in the vicinity of the root.
Convergence Intermediate because the error is proportional to the product of the previous two errors. It is therefor
faster than a linear method but slower than a quadratic method.
Order of convergence 1.62
f ( xn )
Iteration formula x n +1 = x n − ( x n − x n −1 )
f ( x n ) − f ( x n −1 )
g ′′( ξ1 , ξ2 )
Error formula en +1 = (e n )(en −1 )
2
1+ 5
Note The order of convergence is = 1.62 .
2
, False position (regula falsi) method
Strategy Choose two values of x, that is,x0 and x1 , which bracket the root. Draw a straight line through the
points. The intersection of the line with the x-axis should be close to the root. Repeat the process by
always checking that the root remains bracketed.
Requirement The function f(x) must be continuous in the interval.
Advantage Unlike the interval bisection method, the intersection of the line and the x-axis does not necessarily occur
at the midpoint of the interval.
Convergence Intermediate because, though it is faster than interval bisection, its algorithm is slightly complicated. If
convergence takes place from the end that is farther from the root, it slows down the process.
Order of convergence 1
f ( xn )
Iteration formula x n +1 = x n − ( x n − x n −1 )
f ( x n ) − f ( x n −1 )
Note In most cases, this method converges to the root from one end of the interval.
Newton’s method
Strategy This method is based on a linear approximation of the function but does so by using a tangent to th
curve. Find one starting value that is not too far from the root and move along the tangent to its
intersection with the x-axis so as to obtain the next approximation.
Requirement The function f(x) must be continuous in the interval.
The derivative of the function f(x) must exist.
Convergence This method is rapidly convergent in the neighbourhood of the root. The error of each step approaches a
constant k times the square of the error of the previous step. This means that the number of decimal places
of accuracy nearly doubles at each iteration. However, the method requires two function evaluations per
step.
Order of convergence 2
f ( xn )
Iteration formula x n +1 = x n −
f ′( x n )
Error formula en +1 ∝ e n2
