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COS2633 EXAM PACK 2023
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Cos2633 EXAM PACK2023
QUESTIONS AND ANSWERS
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, 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
, Muller’s method
Strategy This method is based on approximating the function in the neighbourhood of the root by a quadratic
polynomial. A second-degree polynomial is made to fit three points (with x-coordinates x 0 , x1 and x 2 )
near the root. The procedure is developed by writing a quadratic equation, y = ax 2 + bx + c , that fits all
three points. The proper zero of this quadratic is then used as the improved estimate of the root. The
process is then repeated by using the three points that are the closest to the root (that is, if the zero is to
the right of x0 , we take x 0 , x1 and the zero but if it is to the left of x0 , we take x 0 , x2 and the zero.)
Requirement The function f(x) must be continuous.
Convergence The method converges at a rate that is similar to Newton’s method. It needs only one function evaluation
per iteration but the penalty is that one must evaluate the function three times. However, this is overcome
by the time that the required precision is attained.
Order of convergence 1.84
2c γf 1 − f 0 (1 + γ) + f 2 f 1 − f 0 − ah12 x − x2
Iteration formula Root = x 0 − where a = , b = and γ = 0
b ± b 2 − 4ac γh1 (1 + γ)
2
h1 x1 − x 0
If b > 0 (b < 0), choose plus (minus) in the denominator.
Note Though it is very difficult to analyse this method (since it is a function of three variables), Ralston outlined a
technique to show that Muller’s method has a rate of convergence of order 1.84.
Fixed-point iteration
Strategy Rearrange f(x) = 0 into an equivalent form x = g (x ) (this can be done in several ways). Under suitable
conditions, the iterative form x n +1 = g ( x n ) converges to a fixed point which is the intersection of the line
y = x and the curve y = g(x)
Requirement g(x) and g ′(x ) must be continuous on an interval about the root.
The appropriate g(x) must be found so that convergence towards a particular root occurs.
Advantage It is easy to program.
Convergence It is not fast since the method is linearly convergent but it is very cost effective.
Order of convergence 1
Iteration formula x n +1 = g ( x n )
Error term e n +1 = g ( ξn ) × e n where ξn lies between x n and α .
Note The iteration formula x n +1 = g ( x n ) will converge to α if g ′( x) < 1 for all x in the interval provided that x 0 is
chosen in the interval.
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