1. 4 Numerical Methods Root Finding Secant Method Modified Secant
False Position Method
Newton Method

2. Secant method Math 685/CSI 700 Spring 08
George Mason University, Department of Mathematical Sciences

3. Secant method Math 685/CSI 700 Spring 08
George Mason University, Department of Mathematical Sciences

4. Requires two initial estimates
Secant method
Requires two initial estimates
f(x) is not required to change signs, therefore this is not a bracketing method

5. { Secant method Select two estimates. Note: f(xi) and f(xi+1) f(x)
are not opposite
signs.
f(x) { x
initial estimates

6. Secant method
f(x)
slope
between
two
estimates { x
initial estimates

7. { Secant method f(x) slope between two estimates x initial estimates
new estimate

8. To get xn+1 from xn and xn-1 we write out the equation of the secant line and using the points xn and xn-1. We then plug in the point (xn+1,0) and solve for xn+1.
equation of secant
substitute (xn+1,0)
solve for xn+1
xn+1=h(xn-1,xn)
The equation above gives the recursively defined sequence for xn. This is what is used for the Secant Method. The halting condition is usually given by the Standard Cauchy Error.

9. Secant Illustration F(x) = x2 - 10 1 (a=1, fa=-9) (b=10, fb=90)
 int = 1.8, fint = -6.7
2 (a=10, fa=90) (b=1.8, fb= -6.7)
int = 0.88, fint =
3 (a=1.8, fa=-6.7) (b=0.88, fb=-9.22)
int = 4.25, fint = 8
4 (a=0.88, fa=-9.22) (b=4.25, fb=8)
Int =2.68, fint = -2.8
Etc… 1 2 3 4

10. Example
Determine the root of f(x) = e-x - x using the secant method. Use the starting points x0 = 0 and x1 = 1.0.

11. Choose two starting points
Solution
Choose two starting points
x0 = 0 f(x0 ) =1
x1 = f(x1) =
Calculate x2
x2 = 1 - (-0.632)(0 - 1)/( ) =

12. NOTE: f(x) are the same sign. OK here.
Solution
Second iteration
x1 = f(x1) =
x2 = f(x2) =
NOTE: f(x) are the same sign. OK here.
x3 = ( )( )/( )
x3 = f(x3) =
ea = abs[( )/(0.564)] x 100 = 8.23%

13. ea = 0.59% et = 0.0048% Third iteration x2 = 0.613 f(x2) = -0.0708
Solution
Third iteration
x2 = f(x2) =
x3 = f(x3) =
x4 = f(x4) =
ea = 0.59% et = %
Know the difference between these error terms

14. The change from ordinary
double secant(double c, int iterations, double tol)…
for ( int i = 0; i < iterations; i++) {
double x = ( fa*b - fb*a ) / ( fa - fb );
double fx = func( x, c );
if ( fabs( fx ) < tol )
return x;
a = b;
fa = fb;
b = x;
fb = fx; }
return -1;
SECANT METHOD
The change from ordinary
regula is that the sign check is dropped and points are just “shifted over”

15. Math 685/CSI 700 Spring 08
Example
George Mason University, Department of Mathematical Sciences

16. Use secant method to estimate the root of f(x) = ln x.
Ex.
Use secant method to estimate the root of f(x) = ln x.
Start the computation with value of
xl = xi-1 = 0.5
xu = xi = 5.0
Solution
The secant method

17. Example 1
You are making a bookshelf to carry books that range from 8 ½ ” to 11” in height and would take 29”of space along length. The material is wood having Young’s Modulus Msi, thickness 3/8 ” and width 12”. You want to find the maximum vertical deflection of the bookshelf. The vertical deflection of the shelf is given by
where x is the position where the deflection is maximum. Hence to find
the maximum deflection we need to find where and conduct the second derivative test.

18. Example 1 Cont.
The equation that gives the position x where the deflection is maximum is given by
Figure 2 A loaded bookshelf.
Use the secant method of finding roots of equations to find the position where the deflection is maximum. Conduct three iterations to estimate the root of the above equation.
Find the absolute relative approximate error at the end of each iteration and the number of significant digits at least correct at the end of each iteration.

19. Example 1 Cont.
Figure 3 Graph of the function f(x).

20. Example 1 Cont.
Solution
Let us take the initial guesses of the root of as and
Iteration 1
The estimate of the root is

21. Example 1 Cont.

22. Example 1 Cont.
The absolute relative approximate error at the end of Iteration 1 is
The number of significant digits at least correct is 1, because the absolute relative approximate error is less than 5%.

23. Example 1 Cont.
Iteration 2
The estimate of the root is

24. Example 1 Cont.
The absolute relative approximate error at the end of Iteration 2 is
The number of significant digits at least correct is 2, because the absolute relative approximate error is less than 0.5%.

25. Example 1 Cont.
Iteration 3
The estimate of the root is

26. Example 1 Cont.
The absolute relative approximate error at the end of Iteration 3 is
The number of significant digits at least correct is 6, because the absolute relative approximate error is less than %.

27. Problems With the Secant Method
The number of iterations required can not be determined before the algorithm begins.
The algorithm will halt (program termination by division by zero if not checked for) if a horizontal secant line is encountered.
The secant method will sometimes find an extraneous root.

28. Modified Secant Method
Root Finding
Modified Secant Method

29. Modified Secant Method

30. Modified Secant Method
Original Secant Method
Modified Secant Method

31. x0 = 1 f(x0) = -0.63212 x1+ x0 = 1.01 f(x1+ x0) = -0.64578 Ex.
Use the secant method to estimate the root of f(x) = e-x – x. Use a value of 0.01 for  and start with x0 = 1.0.
Solution (true root = …)
First iteration
x0 = f(x0) =
x1+ x0 = f(x1+ x0) =
Calculate t = 5.3%

32. Applied Problem You buy a $20 K piece of equipment for nothing down
and $5K per year for 5 years. What interest rate are you
paying? The formula relating present worth (P), annual
payments (A), number of years (n) and the interest rate
(i) is:

33. Root Finding
False Position Method

34. “Brute Force” of bisection method is inefficient
False Position Method
“Brute Force” of bisection method is inefficient
Join points by a straight line
Improves the estimate
Replacing the curve by a straight line gives the “false position”

35. Based on
similar
triangles
next estimate, xr
f(xu) xl xu
f(xl)

36. The False-Position Method (Regula-Falsi)
We can approximate the solution by doing a linear interpolation between f(xu) and f(xl)
Find xr such that l(xr)=0, where l(x) is the linear approximation of f(x) between xl and xu
Derive xr using similar triangles

37. Example
Determine the root of the following equation using the false position method starting with an initial estimate of xl=4.55 and xu=4.65
f(x) = x3 - 98

38. Pitfalls of False Position Method

39. Comparison of False Position and Secant Method2
f(x)
f(x) 2 1 1 x x
new est.
new est.

40. FALSE POSITION
f(x)
SECANT METHOD
f(x) 1 1 x x
- select first
estimate

41. FALSE POSITION
f(x) 2
SECANT METHOD 2
f(x) 1 1 x
- select second estimate x

42. FALSE POSITION
f(x) 2
SECANT METHOD 2
f(x) 1 1 x
Note the sign
of f(x) in each method.
False position must
bracket the root. x

43. FALSE POSITION
f(x) 2
SECANT METHOD 2
f(x) 1 1 x
- Connect the
two points
with a line x

44. FALSE POSITION x
f(x) 1
new est. 2
SECANT METHOD 2
f(x) 1
The new estimate
is selected from the
intersection with the
x-axis x
new est.

45. Root Finding
Newton Method

46. Newton’s Method Open solution, that requires only one current guess.
Root does not need to be bracketed.
Consider some point x0.
If we approximate f(x) as a line about x0, then we can again solve for the root of the line.

47. Newton Raphson
f(xi) xi
tangent
xi+1

48. Solving, leads to the following iteration:
Newton’s Method
Solving, leads to the following iteration:

49. Secant method Substitute finite difference approximation for the
first derivative into this equation for Newton Raphson
Secant method

50. Math 685/CSI 700 Spring 08
Newton’s method
George Mason University, Department of Mathematical Sciences

51. Newton Raphson Pitfalls
solution diverges
f(x)
(x)

52. Finding a square-root
Example: 2 = Let x0 be one and apply Newton’s method.

53. Finding a square-root
Example: 2 = Note the rapid convergence Note, this was done with the standard Microsoft calculator to maximum precision.

54. Math 685/CSI 700 Spring 08
Newton’s method
George Mason University, Department of Mathematical Sciences

55. Use the Newton Raphson method to determine the root of
Example
Use the Newton Raphson method to determine the root of
f(x) = x using an initial guess of xi = 3

56. f(x) = x2 - 11 f '(x) = 2x initial guess xi = 3 f(3) = -2 f '(3) = 6
Solution
f(x) = x2 - 11
f '(x) = 2x
initial guess xi = 3
f(3) = -2
f '(3) = 6

57. Solution In this method, we begin to use a numbering system: x0 = 3
Continue to determine x2, x3 etc.

58. Solution

59. Newton’s Algorithm
Requires the derivative function to be evaluated, hence more function evaluations per iteration.
A robust solution would check to see if the iteration is stepping too far and limit the step.
Most uses of Newton’s method assume the approximation is pretty close and apply one to three iterations blindly.

60. Division by Multiplication
Newton’s method has many uses in computing basic numbers. For example, consider the equation: Newton’s method gives the iteration:

61. CONVERGENCE
If the numbers xn become closer and closer to r as n becomes large, then we say that the sequence converges to r and we write:

62. Applied Problem The concentration of pollutant bacteria C in a lake
decreases according to:
Determine the time required for the bacteria to be reduced to 10 ppm.

63. Summary Method Pros Cons Bisection Newton Secant
4/26/2017
Summary
Method
Pros
Cons
Bisection
- Easy, Reliable, Convergent
- One function evaluation per iteration
- No knowledge of derivative is needed
- Slow
- Needs an interval [a,b] containing the root, i.e., f(a)f(b)<0
Newton
- Fast (if near the root)
- Two function evaluations per iteration
- May diverge
- Needs derivative and an initial guess x0 such that f’(x0) is nonzero
Secant
- Fast (slower than Newton)
- One function evaluation per iteration
- Needs two initial points guess x0, x1 such that
f(x0)- f(x1) is nonzero