3. Two Fundamental Approaches
1. Bracketing or Closed Methods
- Bisection Method
- False-position Method (Regula falsi).
2. Open Methods
- Newton-Raphson Method
- Secant Method
- Fixed point Methods
Roots of Equations
3S.N.P.I.T. & R.C.
4. Newton Raphson’s Method
The equation for
Newton’s Method can
be determined
graphically!
4S.N.P.I.T. & R.C.
5. Continue.....
The equation for Newton’s
Method can be determined
graphically!
From the diagram tan Ө =
ƒ'(x0) = ƒ(x0)/(x0 – x1)
5S.N.P.I.T. & R.C.
6. Continue....
The equation for Newton’s
Method can be determined
graphically!
From the diagram tan Ө =
ƒ'(x0) = ƒ(x0)/(x0 – x1)
Thus, x1=x0 -ƒ(x0)/ƒ'(x0).
6S.N.P.I.T. & R.C.
7. In general, if the nth approximation is xn and f’(xn) ≠
0, then the next approximation is given by:
3
1
( )
12
'( )
n
n n
n
f x
x x
f x
Equation/Formula 2
Continue…
7S.N.P.I.T. & R.C.
8. Starting with x1 = 2, find the third
approximation x3 to the root of
the equation
x3 – 2x – 5 = 0
Example Of Newton Raphson’s
method
Example 1
8S.N.P.I.T. & R.C.
9. We apply Newton’s method with
f(x) = x3 – 2x – 5 and f’(x) = 3x2 – 2
Newton himself used this equation to illustrate
his method.
He chose x1 = 2 after some experimentation
because f(1) = -6, f(2) = -1 and f(3) = 16
Continue…
Example 1
9S.N.P.I.T. & R.C.
11. With n = 1, we have:
Continue…..
11S.N.P.I.T. & R.C.
3
1 1
2 1 2
1
3
2
2 5
3 2
2 2(2) 5
2
3(2) 2
2.1
x x
x x
x
12. With n = 2, we obtain:
It turns out that this third approximation x3 ≈ 2.0946
is accurate to four decimal places.
Continue….
12S.N.P.I.T. & R.C.
3
2 2
3 2 2
2
3
2
2 5
3 2
2.1 2(2.1) 5
2.1
3(2.1) 2
2.0946
x x
x x
x
13. Use Newton’s method to find correct to eight
decimal places.
First, we observe that finding is equivalent to
finding the positive root of the equation x6 – 2 = 0
So, we take f(x) = x6 – 2
Then, f’(x) = 6x5
Example 2
Newton Raphson’s method
6
2
6
2
13S.N.P.I.T. & R.C.
14. So, Formula 2 (Newton’s method)
becomes:
Continue…
14S.N.P.I.T. & R.C.
6
1 5
2
6
n
n n
n
x
x x
x
15. Choosing x1 = 1 as the initial approximation, we obtain:
As x5 and x6 agree to eight decimal places, we
conclude that to eight decimal places.
Continue…
2
3
4
5
6
1.16666667
1.12644368
1.12249707
1.12246205
1.12246205
x
x
x
x
x
6
2 1.12246205
15S.N.P.I.T. & R.C.
16. Newton Raphson’s method
Use Newton’s method to find correct to four
decimal places.
Where, N=12
q=3
= 1
S.N.P.I.T. & R.C. 16
1 1
1
1n n q
n
N
x x q
q x
0x
3
12
17. continue….
So, Formula 2 (Newton’s method) for find out qth
root
becomes:
S.N.P.I.T. & R.C. 17
1 3 1
1 12
3 1 1
3
n n
n
x x
x
1 4.6667x
18. Continue….
Choosing x1 = 1 as the initial approximation, we obtain:
As x5 and x6 agree to four decimal places, we
conclude that to four decimal places.
S.N.P.I.T. & R.C. 18
3
12 2.2894
2
3
4
5
6
7
3.2948
2.5650
2.3180
2.2898
2.2894
2.2894
x
x
x
x
x
x
20. Continue....
First we find two
points(x0,x1), which are
hopefully near the root
(we may use the
bisection method).
A line is then drawn
through the two points
and we find where the
line intercepts the x-axis,
x2.
20S.N.P.I.T. & R.C.
21. Continue....
If f(x) were truly linear,
the straight line would
intercept the x-axis at
the root.
However since it is not
linear, the intercept is
not at the root but it
should be close to it.
21S.N.P.I.T. & R.C.
22. Continue....
From similar triangles
we can write that,
10
10
1
21
xfxf
xx
xf
xx
1x 0x2x
1xf
0xf
22S.N.P.I.T. & R.C.
23. Continue....
From similar triangles
we can write that,
Solving for x2 we get:
10
10
1
21
xfxf
xx
xf
xx
10
10
112
xfxf
xx
xfxx
23S.N.P.I.T. & R.C.
25. S.N.P.I.T. & R.C. 25
Approx. f '(x) with backward FDD:
Substitute this into the N-R equation:
to obtain the iterative expression:
i 1 i
i 1 i
f (x ) f (x )
f '(x)
x x
i
i 1 i
i
f (x )
x x
f '(x )
i i 1 i
i 1 i
i 1 i
f(x )(x x )
x x
f(x ) f(x )
Continue….
27. Example of Secant Method
The floating ball has a specific gravity of 0.6 and has a
radius of 5.5cm.
You are asked to find the depth to which the ball is
submerged when floating in water.
The equation that gives the depth x to which the ball is
submerged under water is given by
010993.3165.0 423
xx
27S.N.P.I.T. & R.C.
28. Use the secant method of finding roots of
equations to find the depth to which the ball is
submerged under water. 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 converged iteration.
28S.N.P.I.T. & R.C.
29. S.N.P.I.T. & R.C. 29
Secant Method
From the physics of the problem
11.00
)055.0(20
20
x
x
Rx
x
water
Figure 2 :
Floating ball
problem
30. S.N.P.I.T. & R.C. 30
Let us assume
11.0,0 UL xx
4423
4423
10662.210993.311.0165.011.011.0
10993.310993.30165.000
fxf
fxf
U
L
Hence,
010662.210993.311.00 44
ffxfxf UL
