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

2. Root Finding
False Position Method

3. 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

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

5. k xk (Bisection) fk xk (False Position) 1 0.5 0.471 2 0.25 0.372 3
Example k
xk (Bisection) fk
xk (False Position) 1
0.5
0.471 2
0.25
0.372 3
0.375
0.362 4
0.3125
0.360 5
-0.042
2.93×10-5

6. 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

7. Pitfalls of False Position Method

8. Bisection Method (Converge quicker)
Iteration xl xu xr
εa (%)
εt (%) 1
1.3
0.65 35 2
0.975
33.3 25 3
1.1375
14.3
13.8 4
7.7
5.6 5
4.0
1.6
False-position Method
Iteration xl xu xr
εa (%)
εt (%) 1
1.3
90.6 2
48.1
81.8 3
30.9
73.7 4
22.3
66.2 5
17.1
59.2

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

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

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

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

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

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

15. 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.

16. Secant method table n xn-1 f(xn-1) xn f(xn) xn+1 1 4.0000 -3.4800
6.0000
9.8800
4.5210 2
6.0000
9.8800
4.5210
4.7303 3
4.5210
4.7303
4.8513 4
4.7303
4.8513
4.8368 5
4.8513
4.8373
-.0032
4.8373
We will assume initial guesses for xn-1 = 4.00 and xn = 6.00 (the same as we used for the bisection method). The value of the function at 4.00 is and at 6.00 it is Using these values the root approximation equation yields: xn+1 = 6.00 – 9.88*( )/(9.88-(-3.48)) = We then replace xn-1 with xn and xn with xn+1 and repeat the procedure. After 6 iterations the root converges to and the absolute value of the function is less than
The secant method is almost as fast as Newton’s method, and has the advantage of not requiring an analytical derivative. However, it has the same problem as Newton’s that when the root approaches a point near a maximum or minimum of the function, the new root calculation will approach infinity.

17. 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

18. Example
Determine the root of f(x) = e-x - x using the secant method. Use the starting points x0 = 0 and x1 = 1.0.error = 0.01 or the root =

19. 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)/( ) =

20. 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%

21. 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

22. 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”

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

24. 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

25. 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.

26. Modified Secant Method

27. Modified Secant Method
Original Secant Method
Modified Secant Method

28. 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%

29. 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:

30. Root Finding
Newton Method

31. 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.

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

33. Newton’s Method (cont)
Derived using Taylor’s expansion

34. The error rate proof
Ek = x* - xk Ek+1 = x* - xk+1 = x* - [ xk – f(xk) / f’ (xk)] = (x* - xk) + f / f’

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

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

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

38. 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

39. 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

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

41. Solution

42. Example: Newton’s Method
f(x)= x3–3x2 – x+9=0 k xk 1 9 2
6.41 46
Worse than bisection !?

43. 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.

44. 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.

45. Roots of Equations Chemical Engineering (C&C 8.1, p. 187):
van der Waals equation; v = V/n (= volume/# moles)
Find the molal volume v such that
p = pressure,
T = temperature,
R = universal gas constant,
a & b = empirical constants

46. Roots of Equations Civil Engineering (C&C Prob. 8.17, p. 205):
Find the horizontal component of tension, H, in a cable that passes through (0,y0) and (x,y)
w = weight per unit length of cable

47. Roots of Equations Electrical Engineering (C&C 8.3, p. 194):
Find the resistance, R, of a circuit such that the charge reaches q at specified time t
L = inductance,
C = capacitance,
q0 = initial charge

48. Roots of Equations Mechanical Engineering (C&C 8.4, p. 196):
Find the value of stiffness k of a vibrating mechanical system such that the displacement x(t) becomes zero at t= 0.5 s. The initial displacement is x0 and the initial velocity is zero. The mass m and damping c are known, and λ = c/(2m).
in which

49. Comparison
Newton’s method
False Position
(Newton) xk
xk+1 ‧
f ’(xk)

50. Summary Method Pros Cons Bisection Newton Secant
5/10/2018
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