Post on 23-Feb-2016
description
MATH 175: Numerical Analysis II
Lecturer: Jomar Fajardo RabajanteIMSP, UPLB
2nd Sem AY 2012-2013
Question:
• What if we do not have means of getting an initial bracket?
Let’s start with SECANT METHOD…
Suppose that f is a continuous function . Pick two initial points (not necessarily forming a bracket), then do linear interpolation (not inverse).
4th Method: SECANT METHOD
Initial points:
Interpolating line:
Use x-intercept:
))(,( & ))(,( 2211 xfxxfx
)()()()( 112
121 xx
xxxfxfxfy
)()()(
12
12113 xfxf
xxxfxx
4th Method: SECANT METHOD
Approximate:
- Same as the formula for Regula Falsi- In an iteration, if the points form a bracket, then
the method is similar to Regula Falsi; else the method uses extrapolation. In whatever case, the new points will be
)()()(
12
12113 xfxf
xxxfxx
32
21
::xxxx
Old x2
4th Method: SECANT METHOD
In short: Pick any two distinct points, draw the secant through them, and use the x-intercept (x3) of that secant line as the new estimate of the zero of the function. For the next iteration, discard the oldest point and add (x3, f(x3)) as the new point.
WE DO NOT NEED IZT (IVT) ANYMORE!
4th Method: SECANT METHOD
Secant method can be considerably faster than the previous methods.
However, it may fail to converge. Example: if f(x1)=f(x2), then what would happen?
4th Method: SECANT METHOD
Notice that
can be written as
)()()(
12
12113 xfxf
xxxfxx
)()()(
12
12223 xfxf
xxxfxx
and
can also be written as
You can use any of these formulas for Regula Falsi and Secant Method
4th Method: SECANT METHOD
)()()(
12
12223 xfxf
xxxfxx
)()()()(
12
12213 xfxf
xfxxfxx
4th Method: SECANT METHOD
For Secant Method (not for Regula Falsi), we can generalize the formulas as follows (k=1,2,3,…)
)()()()(
21
2112
kk
kkkkk xfxf
xfxxfxx
)()()(
21
2111
kk
kkkkk xfxf
xxxfxx
4th Method: SECANT METHODExample: Find a zero ofUse 0 & 1 as initial values.
)()()()(
21
2112
kk
kkkkk xfxf
xfxxfxx
1)( 3 xxxf
01 0.5
0.5 0.6363636360.636363636 0.6900523560.690052356 0.68202042
0.68202042 0.6823257810.682325781 0.6823278040.682327804 0.6823278040.682327804 0.6823278040.682327804 0.682327804
=(A1*(A2^3+A2-1)-A2*(A1^3+A1-1))/((A2^3+A2-1)-(A1^3+A1-1))
=B2
4th Method: SECANT METHOD
Assuming that the secant method converges to the root, the order of convergence of the method is
SUPERLINEAR!!! (but not yet quadratic)
62.12
51
p
4th Method: SECANT METHOD
Stopping criterion: You can use
tolxx kk 1,3,3
tol=10^(-m): accurate at least up to m decimal places
5th Method: Newton’s Method/Newton-Raphson Iteration
• What if we make the secant line is a tangent line?
• Hence, we only need one initial point.
• But we add another assumption: f should be differentiable!
5th Method: Newton’s Method/Newton-Raphson Iteration
From Secant Method:
If we use tangent lines: as xk-2 approaches xk-1
)()()(
21
2111
kk
kkkkk xfxf
xxxfxx
)(')()(1
21
21
kkk
kk xfxxxfxf
5th Method: Newton’s Method/Newton-Raphson Iteration
Hence:
)()()(
21
2111
kk
kkkkk xfxf
xxxfxx
)('1)(
111
kkkk xfxfxx
• Newton’s Method: To be continued…
• Assignment: List the advantages and disadvantages of the discussed methods. Research other disadvantages that we did not mention in the class.