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.