SOLUTIONS TO NON-LINEAR EQUATION
1
Continuation…
d. NEWTON RAPHSON ST 1 METHOD
f ( xi ) f ' ( xi ) = xi − xi +1 xi − xi +1
f ( xi ) = f ' ( xi )
f ( xi ) xi +1 = xi − f ' ( xi )
Example: 1.
Solve for the roots of f(x) = sin x + cos(1 + with an interval of (1,2). Solution: 2 sin x + cos( 1 + x ) −1 f(x) = x∈(1,2) 2 f’(x) =cos x − 2 x sin(1 + x )
x2 ) −1
f ( xi ) xi +1 = xi − f ' ( xi ) i
Xi
f(Xi)
f’(Xi)
Xi+1
0
2
0.1930
3.4196
1.9436
1
1.9436
-0.0035
3.5147
1.9446
2
1.9446
0
3.5148
1.9446
Example: 1.
Solve for the roots of f(x) = sin x + cos(1 + with an interval of (1,2). Solution: 2 sin x + cos( 1 + x ) −1 f(x) = x∈(1,2) 2 f’(x) =cos x − 2 x sin(1 + x )
x2 ) −1
f ( xi ) xi +1 = xi − f ' ( xi ) i
Xi
f(Xi)
f’(Xi)
Xi+1
0
2
0.1930
3.4196
1.9436
1
1.9436
-0.0036
3.5147
1.9446
2
1.9446
0.0000
3.5148
1.9446
NEXT TOPIC:
e. NEWTON RAPHSON ND 2 METHOD
•Uses the 2nd derivative of the function
f ' ' ( xi ) f ' ( xi ) xi +1 = xi + − 2 f ' ( xi ) f ( xi )
−1
Example: 1.
Solve for the roots of f(x) = sin x + cos(1 + x with an interval of (1,2). Solution: 2 sin x + cos( 1 + x ) −1 f(x) = x∈(1,2) 2 cos x − 2 x sin( 1 + x ) f’(x) = 2 2 2 − sin x − 4 x cos( 1 + x ) − 2 sin( 1 + x ) f” (x) = −1
f ' ' ( xi ) f ' ( xi ) xi +1 = xi + − 2 f ' ( x ) f ( x ) i i
2
) −1
C D E = − 2C B
−1
+A
A
B
C
D
E
i
Xi
f(Xi)
f’(Xi)
f”(Xi)
Xi+1
0
2
0.1930
3.4196
-3.5300
1.9452
1
1.9452
0.0021
3.5148
-0.0158
1.9446
2
1.9446
0
3.5148
0.0202
1.9446
Example: 1.
Solve for the roots of f(x) = sin x + cos(1 + x with an interval of (1,2). Solution: 2 sin x + cos( 1 + x ) −1 f(x) = x∈(1,2) 2 cos x − 2 x sin( 1 + x ) f’(x) = 2 2 2 − sin x − 4 x cos( 1 + x ) − 2 sin( 1 + x ) f” (x) =
2
) −1
A
B
C
D
E
i
Xi
f(Xi)
f’(Xi)
f”(Xi)
Xi+1
0
2
0.1930
3.4196
-3.5300
1.9452
1
1.9452
0.0020
3.5148
-0.0139
1.9446
2
1.9446
0.0000
3.5148
0.0197
1.9446
f ' ' ( xi ) f ' ( xi ) xi +1 = xi + − 2 f ' ( x ) f ( x ) i i
−1
C D E = − 2C B
−1
+A
NEXT TOPIC:
f. SECANT METHOD
− f ( x i −1 ) f ( xi ) = xi +1 −x i −1 xi − xi +1
− xi f ( xi −1 ) + xi + 1 f ( xi −1 ) = xi + 1 f ( x i ) − xi −1 f ( xi ) xi −1 f ( xi ) − xi f ( xi −1 ) = xi +1 [ f ( xi ) − f ( xi −1 )] xi −1 f ( xi ) − xi f ( xi −1 ) xi +1 = f ( xi ) − f ( xi −1 )
Example: 1.
Solve for the roots of f(x) = sin x + cos(1 + with an interval of (1,2). Solution: 2 sin x + cos( 1 + x ) −1 f(x) = x∈(1,2)
xi −1 f ( xi ) − xi f ( xi −1 ) xi +1 = f ( xi ) − f ( xi −1 )
C
x ) −1 2
E=
AD-CB D-B
D
E
A
B
i
Xi-1
f(Xi-1)
(Xi)
f(Xi)
Xi+1
0
1
-0.5747
2
0.1930
1.7486
1
2
0.1930
1.7486
-0.6248
1.9407
2
1.7486
-0.6248
1.9407
-0.0137
1.9450
3
1.9407
-0.0137
1.9450
0.0014
1.9446
4
1.9450
0.0014
1.9446
0
1.9446
Example: 1.
Solve for the roots of f(x) = sin x + cos(1 + with an interval of (1,2). Solution: 2 sin x + cos( 1 + x ) −1 f(x) = x∈(1,2) A
B
i
Xi-1
f(Xi-1)
(Xi)
f(Xi)
Xi+1
0
1
-0.5747
2
0.1930
1.7486
1
2
0.1930
1.7486
-0.6247
1.9407
2
1.7486
-0.6247
1.9407
-0.0138
1.9450
3
1.9407
-0.0138
1.9450
0.0014
1.9446
4
1.9450
0.0014
1.9446
0.0000
1.9446
xi −1 f ( xi ) − xi f ( xi −1 ) xi +1 = f ( xi ) − f ( xi −1 )
C
x ) −1
D
2
E=
AD-CB D-B
E
NEXT TOPIC:
g. MODIFIED SECANT METHOD
•Uses a fractional perturbation, δ
mathematical methods that give approximate solutions to problems that cannot be solved exactly
δ xi f ( xi ) xi +1 = xi − f ( xi + δ xi ) − f ( xi )
Example: 1.
Solve for the roots of f(x) = sin x + cos(1 + x ) − 1 with an interval of (1,2) and δ=0.02 Solution: 2 sin x + cos( 1 + x ) −1 f(x) = x∈(1,2) δ=0.02 2
δ xi f ( xi ) xi +1 = xi − f ( xi + δ xi ) − f ( xi )
CB E = A − D − B
A
B
C
i
Xi
f(Xi)
δXi
D f(Xi+δXi )
E
0
2
0.1930
0.0400
0.3262
1.9421
1
1.9421
-0.0088
0.0388
0.1272
1.9446
2
1.9446
0
Xi+1
1.9446
Example: 1.
Solve for the roots of f(x) = sin x + cos(1 + x ) − 1 with an interval of (1,2) and δ=0.02 Solution: 2 sin x + cos( 1 + x ) −1 f(x) = x∈(1,2) δ=0.02 2
A
B
i
Xi
f(Xi)
δXi
D f(Xi+δXi )
0
2
0.1930
0.0400
0.3262
1.9421
1
1.9421
-0.0089
0.0388
0.1271
1.9446
2
1.9446
0.0000
0.0389
0.1361
1.9446
δ xi f ( xi ) xi +1 = xi − f ( xi + δ xi ) − f ( xi )
C
E
Xi+1
CB E = A − D − B
End of Solutions to Non-Linear Equation
GOD BLESS TO YOUR EXAM…
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