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The False-Position Method
Introduction
The poor convergence of the bisection
method as well as its poor adaptability
to higher dimensions motivate the use
of better techniques. One such
method is the Method of False
Position.
Methodology
we start with an initial interval [x1,x2], and we
assume that the function changes sign only once
in this interval. Now we find an x3 in this
interval, which is given by the intersection of the
x axis and the straight line passing through
(x1,f(x1)) and (x2,f(x2)). It is easy to verify that
x3 is given by
Methodology
Now, we choose the new
interval from the two choices
[x1,x3] or [x3,x2] depending on in
which interval the function
changes sign.
Note
The False-Position and Bisection
algorithms are
quite similar. The only difference is
the formula used to
calculate the new estimate of the
root x3
Graphical look
Numerical Example
Find a root of 3x + sin(x) - exp(x) =0. The
graph of this equation is given in the
figure.
From this it's clear that there is a
root between 0
and 0.5 and also
another root between 1.5 and
2.0. Now let us consider the function f
(x) in the
interval [0, 0.5] where f (0) * f (0.5) is
less than
zero and use the regula-falsi scheme to
obtain the
zero of f (x) = 0.
So one of the roots of 3x +
sin(x) - exp(x) = 0 is
approximately 0.36. Note :
Although the length of the
interval is getting smaller in
each iteration, it is possible
that it may not go to zero. If
the graph y = f(x) is concave
near the root 's', one of the
endpoints becomes fixed and
the other end marches
towards the root.
Matlab implementation
Thank You…!
Presented by
Tayyaba Abbas
Roll no 5

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The False-Position Method

  • 2. Introduction The poor convergence of the bisection method as well as its poor adaptability to higher dimensions motivate the use of better techniques. One such method is the Method of False Position.
  • 3. Methodology we start with an initial interval [x1,x2], and we assume that the function changes sign only once in this interval. Now we find an x3 in this interval, which is given by the intersection of the x axis and the straight line passing through (x1,f(x1)) and (x2,f(x2)). It is easy to verify that x3 is given by
  • 4. Methodology Now, we choose the new interval from the two choices [x1,x3] or [x3,x2] depending on in which interval the function changes sign.
  • 5. Note The False-Position and Bisection algorithms are quite similar. The only difference is the formula used to calculate the new estimate of the root x3
  • 7. Numerical Example Find a root of 3x + sin(x) - exp(x) =0. The graph of this equation is given in the figure. From this it's clear that there is a root between 0 and 0.5 and also another root between 1.5 and 2.0. Now let us consider the function f (x) in the interval [0, 0.5] where f (0) * f (0.5) is less than zero and use the regula-falsi scheme to obtain the zero of f (x) = 0.
  • 8. So one of the roots of 3x + sin(x) - exp(x) = 0 is approximately 0.36. Note : Although the length of the interval is getting smaller in each iteration, it is possible that it may not go to zero. If the graph y = f(x) is concave near the root 's', one of the endpoints becomes fixed and the other end marches towards the root.