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By
Ahmed Haider Ahmed
Pre-M.Sc. Physics – ASU
To My father ,
May Allah
enter you his
paradise
1 – Bisection Method
 If F(wi+1) is negative we put wi+1 instead of ai or bi
2
1
ii
i
ba
w
2 – Falsie Method
)()(
)()(
1
ii
iiii
i
bfaf
afbbfa
x
3 – Newton Method
)(
)(
1
i
i
ii
xf
xf
xx
4 – Secant Method
)()(
))((
1
1
1
ii
iii
ii
xfxf
xxxf
xx
Examples
Find the positive root of x – cos x = 0 using
bisection method.
 We will take points that give positive value
with the ne...
)(249.0)75.0(
75.0
2
15.0
150
)(5.0)5.0(
)(5.0
2
10
10
2
1
vef
x
and.enroot betweHence the
vef
vex
andenroot betweHence the
062.0)9375.0(
9375.0
2
1875.0
18750
)(12488.0)875.0(
875.0
2
75.01
1750
4
3
f
x
and.enroot betweHence the
vef
x
and.enroot...
Solve equation x3 – x – 1 = 0 using
falsie method.
5128)2(
1111)1(
1)( 3
f
f
xxxf
2,1
)()(
)()(
00
00
0000
1
ba
bfaf
afbbf...
)(279351.0)2547.1(
2547.1
5684.05
5684.02517.1
217.1from
)(5684.0)17.1(
17.1
6
25
51
1251
2
2
1
vef
x
tox
vef
x
)(0578.0)311.1(
311.1
127.05
127.025294.1
2294.1from
)(127.0)294.1(
294.1
279351.05
279351.0252547.1
22547.1from
4
4
3
3
v...
3238.1
0109.05
0109.0253223.1
23223.1from
0109.00)3223.1(
3223.1
024255.05
024255.025319.1
2319.1from
)(024255.0)319.1(
31...
Solve equation x3 – x – 1 = 0 using
Newton method at x0 = 1
213)(
1111)(
13)(
1)(
)(
)(
0
0
2
3
1
xf
xf
xxf
xxxf
xf
xf
xx
...
4497.4)3478.1(
1006.0)3478.1(
3478.1
75.5
875.0
5.1
)(
)(
75.5)5.1(
875.0)5.1(
5.1
2
1
1
)(
)(
1
1
12
0
0
01
f
f
xf
xf
xx
...
000077.0)3247.1(
3247.1
269.4
00206.0
3252.1
)(
)(
269.4)3252.1(
00206.0)3252.1(
3252.1
4497.4
1006.0
3478.1
)(
)(
3
3
34
...
REFERANCES
 Lecture notes on numerical methods ,
Dr. Shemi 2011, Minia university written
by Aya Hassan.
 Lecture notes ...
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Numerical methods

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Bisection method, Newton method, Falsi method and Secant method with examples.

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Numerical methods

  1. 1. By Ahmed Haider Ahmed Pre-M.Sc. Physics – ASU
  2. 2. To My father , May Allah enter you his paradise
  3. 3. 1 – Bisection Method  If F(wi+1) is negative we put wi+1 instead of ai or bi 2 1 ii i ba w
  4. 4. 2 – Falsie Method )()( )()( 1 ii iiii i bfaf afbbfa x
  5. 5. 3 – Newton Method )( )( 1 i i ii xf xf xx
  6. 6. 4 – Secant Method )()( ))(( 1 1 1 ii iii ii xfxf xxxf xx
  7. 7. Examples
  8. 8. Find the positive root of x – cos x = 0 using bisection method.  We will take points that give positive value with the negative one i.e 0 and 1 )(1)2( )(00015.0)1( )(1)0( cos)( vef vef vef xxxf
  9. 9. )(249.0)75.0( 75.0 2 15.0 150 )(5.0)5.0( )(5.0 2 10 10 2 1 vef x and.enroot betweHence the vef vex andenroot betweHence the
  10. 10. 062.0)9375.0( 9375.0 2 1875.0 18750 )(12488.0)875.0( 875.0 2 75.01 1750 4 3 f x and.enroot betweHence the vef x and.enroot betweHence the
  11. 11. Solve equation x3 – x – 1 = 0 using falsie method. 5128)2( 1111)1( 1)( 3 f f xxxf 2,1 )()( )()( 00 00 0000 1 ba bfaf afbbfa x
  12. 12. )(279351.0)2547.1( 2547.1 5684.05 5684.02517.1 217.1from )(5684.0)17.1( 17.1 6 25 51 1251 2 2 1 vef x tox vef x
  13. 13. )(0578.0)311.1( 311.1 127.05 127.025294.1 2294.1from )(127.0)294.1( 294.1 279351.05 279351.0252547.1 22547.1from 4 4 3 3 vef x tox vef x tox
  14. 14. 3238.1 0109.05 0109.0253223.1 23223.1from 0109.00)3223.1( 3223.1 024255.05 024255.025319.1 2319.1from )(024255.0)319.1( 319.1 0578.05 0578.025311.1 2311.1from 7 7 6 6 5 5 x tox f x tox vef x tox
  15. 15. Solve equation x3 – x – 1 = 0 using Newton method at x0 = 1 213)( 1111)( 13)( 1)( )( )( 0 0 2 3 1 xf xf xxf xxxf xf xf xx i i ii
  16. 16. 4497.4)3478.1( 1006.0)3478.1( 3478.1 75.5 875.0 5.1 )( )( 75.5)5.1( 875.0)5.1( 5.1 2 1 1 )( )( 1 1 12 0 0 01 f f xf xf xx f f xf xf xx
  17. 17. 000077.0)3247.1( 3247.1 269.4 00206.0 3252.1 )( )( 269.4)3252.1( 00206.0)3252.1( 3252.1 4497.4 1006.0 3478.1 )( )( 3 3 34 2 2 23 f xf xf xx f f xf xf xx
  18. 18. REFERANCES  Lecture notes on numerical methods , Dr. Shemi 2011, Minia university written by Aya Hassan.  Lecture notes on computational physics for Pre-M.Sc. Students , Prof. Dr. S.Hendawi 2013, Ain Shams University.

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