Secant Method

SECANT METHOD Although the Bisection Method always converges, the speed of convergence is usually too slow for general use. The secant method does not have the rootbracketing property of the Bisection Method. As a consequence, the method does not always converge, but when it does converge, it generally does so much faster than the bisection method. We use two stopping conditions in the Secant Method. First, we assume that pi is sufficiently accurate when / pi – pi-1 / is within a given tolerance. Also, a safeguard exit based upon a maximum number of iterations is given in case the method fails to converge as expected. The approximation pn+1 for n>1, to a root of f(x)=0 is computed from the approximations pn and pn-1 using the equation: f(pn) (pn - pn-1) Pn+1 = pn - --------------------f(Pn) - f(pn-1)

Examples: 1. Find the root of f(x) = X6 – x – 1 accurate to within error tolerance of E = 0.001. Set p0=2, p1=1. 2. Find the root of f(x) = x3 – x2 –x -1 accurate to within error tolerance of E=0.0001. Set p0=2, p1=1.5 3. Find the root of f(x) = x2 – 6 accurate to within error tolerance E=0.001. Set p0=2, p1=3

1. Find the root of f(x) = X6 – x – 1 accurate to within error tolerance of E = 0.001. Set p0=2, p1=1.

n

pn

0

2

1 2

f(pn) (a)

(b) 1.01

-1 (c)

(d)

4

1.118

5

(g) 1.13

(e) (f)

-0.07292

-0.0224

(h) (i)

(j)

0.01613

0.6575

5

7

-

-1

6

3

6

pn - pn-1

0.00229

0.0000

(k)

At P=_______________, tolerance of 0.001 is met. 2. Find the root of f(x) = x3 – x2 –x -1 accurate to within error tolerance of E=0.0001. Set p0=2, p1=1.5 n

pn

f(pn)

0

2

1

1

1.5

2 3

5 6

(a) (b)

(c) 1.85 7

4

pn - pn-1

-0.2614 (e)

(f) 1.83 9

(d) 0.06795

-0.0042 (h)

(i)

-0.5

(g) 0.00075

0.0000

At P=_______________, tolerance of 0.0001 is met.

(j)

3. Find the root of f(x) = x2 – 6 accurate to within error tolerance E=0.001. Set p0=2, p1=3 n

pn

f(pn)

pn - pn-1

0 1 2 3 4 5

At P=_______________, tolerance of 0.001 is met.

Answers:

Answers:

1) n

pn

f(pn)

pn - pn-1

0

2

61

-

1

1

-1

-1

-0.9154

0.01613

0.6575

0.17445

-0.1685

-0.07292

-0.0224

0.01488

0.0010

0.00229

5

0.0000

-0.00009

n

pn

f(pn)

pn - pn-1

0

2

1

-

-1.3750

-0.5

-0.2614

0.28947

0.1007

0.06795

-0.0042

-0.01890

-0.0001

0.00075

0.0000

0.00001

1.01 2

6 1.19

3

1

4 5

1.118 1.13 3 1.13

6

5 1.13

7 2)

1 2 3 4 5 6

1.5 1.78 9 1.85 7 1.83 9 1.83 9 1.83 9

3) n

pn

f(pn)

pn - pn-1

0

2

-2

-

1 2 3 4 5

3 2.4000 0 2.4444 4 2.4495 4 2.4494 9

3

1

-0.24000

-0.60000

-0.02469

0.04444

0.00025

0.00510

0.00000

-0.00005