algebric solutions by newton raphson method and secant method

Solution of algebraic & transdental equation
with Newton Raphson method & Secant
method
1S.N.P.I.T. & R.C.

Presented by:
 130490106065 – Modi Nagma
 130490106085 – Patel Pinal
 140493106015 – Padhiyar Sagar
 140493106025 – Taylor Kishan
2S.N.P.I.T. & R.C.

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.

Newton Raphson’s Method
 The equation for
Newton’s Method can
be determined
graphically!
4S.N.P.I.T. & R.C.

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.

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.

 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.

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.

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.

Equation 2 becomes:
Continue…
10S.N.P.I.T. & R.C.
3
2
2 5
1
3 2
n n
n n
n
x x
x x
x
 
  


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
 
 

 
 



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
 
 

 
 



 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.

 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


 

 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.

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

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 

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







Secant Method
19S.N.P.I.T. & R.C.

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.

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.

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.

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.

Continue....
 Iteratively this is
written as:
   
   nn
nn
nnn
xfxf
xx
xfxx






1
1
1
24S.N.P.I.T. & R.C.

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….

x
f(x)
1
2
new est.
x
f(x)
1
new est.
2
FALSE POSITION
SECANT METHOD
The new estimate
is selected from the
intersection with the
x-axis
26S.N.P.I.T. & R.C.

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.

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.

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

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

31
Iteration 1
   
   
 
 
0660.0
10662.210993.3
10662.2010993.311.0
44
44









UL
ULLU
m
xfxf
xfxxfx
x
         
5
423
101944.3
10993.30660.0165.00660.00660.0



 fxf m
           00660.00  ffxfxf mL
0660.0,0  UL xx
S.N.P.I.T. & R.C.

32
Iteration
1 0.0000 0.1100 0.0660 N/A -3.1944x10-5
2 0.0000 0.0660 0.0611 8.00 1.1320x10-5
3 0.0611 0.0660 0.0624 2.05 -1.1313x10-7
4 0.0611 0.0624 0.0632377619 0.02 -3.3471x10-10
Lx Ux mx %a  mxf
  010993.3165.0 423
 
xxxfTable 1: Root of
for secant Method.
S.N.P.I.T. & R.C.

algebric solutions by newton raphson method and secant method

More Related Content

What's hot

Viewers also liked

Similar to algebric solutions by newton raphson method and secant method

More from Nagma Modi

Recently uploaded

In this document

algebric solutions by newton raphson method and secant method