This method, Newton raphson helps to approximate the root of a non linear equation. The presentation also tells about the Advantages and disadvantages of the method.

1. RRoooottss ooff aa NNoonnlliinneeaarr
EEqquuaattiioonn
NNeewwttoonn--RRaapphhssoonn
MMeetthhoodd
1

2. NNeewwttoonn--RRaapphhssoonn MMeetthhoodd
x = x - f(x )
i
f (x )
2
i
i+1 i ¢
f(x)
f(xi)
f(xi-1)
q
[ ( ) ] i i x f x ,
xi+2 xi+1 xi X

3. DDeerriivvaattiioonn
tan(a ) = AB
AC
f x f x
i i x x
+ -
1
'( ) ( )
( )
i
=
i i
i i f x
3
f(x)
f(xi)
xi+1 xi
X
B
1
C a A '( )
i
x = x - f x +

4. AAllggoorriitthhmm ffoorr NNeewwttoonn--
RRaapphhssoonn MMeetthhoodd
4

5. SStteepp 11
Evaluate f¢(x) symbolically
5

6. SStteepp 22
6
Calculate the next estimate of the root
x = x - f(x )
i
f'(x )
i
i+1 i
Find the absolute relative approximate error
x 100
= x - x
i i
1
1
x
i
a
+
Î +

7. SStteepp 33
Find if the absolute relative approximate error is greater
than the pre-specified relative error tolerance.
If so, go back to step 2, else stop the algorithm.
Also check if the number of iterations has exceeded the
maximum number of iterations.
7

8. EExxaammppllee
Solve the given equation using Newton’s method using 3 iterations:
8
f ( x) = x3-0.165x2+3.993x10-4

9. GGrraapphh ooff ffuunnccttiioonn ff((xx))
9
f ( x) = x3-0.165x2+3.993x10-4

10. IItteerraattiioonn ##11
x x f x
0
=
= -
0.02 3.413x10
10
( )
( )
0.08320
75.96%
5.4x10
'
0.02
-
4
3
0
1
0
1 0
=
Î =
-
= -
-
a
x
f x
x

11. IItteerraattiioonn ##22
0.08320
x x f x
1
=
= - -
0.08320 1.670x10
11
( )
( )
0.05824
42.86%
-
6.689x10
'
4
3
1
2
1
2 1
=
Î =
-
= -
-
a
x
f x
x

12. IItteerraattiioonn ##33
0.05824
2
=
x = x -
f x
0.05284 3.717x10
12
( )
( )
0.06235
6.592%
-
9.043x10
'
2
a
3
5
2
3 2
=
Î =
-
= -
-
f x
x

13. AAddvvaannttaaggeess
CCoonnvveerrggeess ffaasstt,, iiff iitt ccoonnvveerrggeess
RReeqquuiirreess oonnllyy oonnee gguueessss
13

14. DDrraawwbbaacckkss
14
10
5
0
f(x)
-2 -1 0 1 2
-5
-10
-15
-20
1
2 3
x
f ( x) = ( x -1)3 = 0
IInnfflleeccttiioonn PPooiinntt

15. DDrraawwbbaacckkss
-0.03 -0.02 -0.01 0 0.01 0.02 0.03 0.04
f ( x) = x3 - 0.03x2 + 2.4x10-6 = 0
15
1.00E-05
7.50E-06
5.00E-06
2.50E-06
0.00E+00
-2.50E-06
-5.00E-06
-7.50E-06
-1.00E-05
x
f(x)
0.02
Division by zero

16. DDrraawwbbaacckkss
16
Root Jumping
f ( x) = Sin x = 0

17. 17