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Assignment # 2
MTH603 (Fall 2012)
Total marks: 10
Lecture # 23  28
Due date: 20012013
DON’T MISS THESE Important instructions:
Lecture # 23 to Lecture # 28.
Question#1 Marks 10
Find an equation of a cubic curve which passes through the points
(4,43), (7, 83), (9,327) and (12, 1053) using Divided Difference Formula.
Question#2 Marks 10
Find interpolation polynomial by Lagrange’s Formula, with the help of following table,
x 
1 
2 
3 
5 
f(x) 
0 
7 
26 
124 
And hence find value of f (6).
Question#3 Marks 10
Find from the following table,
x 
0 
0.2 
0.4 
0.6 
0.8 
1.0 
1.2 
0.710 
1.175 
1.811 
2.666 
3.801 
5.292 
7.232 
Question#4 Marks 10
Find from the following table,
x 
1.3 
1.6 
1.9 
2.2 
2.5 
f(x) 
2.4 
2.9 
3.2 
4.7 
6.4 
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Numerical Analaysis: "Divided Difference Forumula"
Find equation of a "Cubic Curve" that passes through the points (4,43), (7, 83), (9,327) and (12, 1053) using the "Divided Difference Formula".
Points (4,43), (7, 83), (9,327) and (12, 1053)
Using divided difference formula
4 43
42
7 83 16
122 1
9 327 24
242
12 1053
Column 1 and 2 given data
Column 3 obtained by 83(43) / 74 = 83+43 /3 = 126/3 = 42
32783/97 = 244/2 = 122
1053327/129 = 726/3 = 242
Column 4 obtained by 12242/94 = 80/5 = 16
242122/127 = 120/5 = 24
Column 5 obtained by 2416/124 = 8/8 = 1
Third degree approximation given by
f(x) = f(4) + xf(4,7) + x(x4)f(4,7,9)+x(x4)(x7)f(4,7,9,12)
= 43 + x 42 + x(x4) 16 + x(x4)(x7)
= 43 + 42x64x+16x^2 + x(x^211x+28)
= 43 22x +16x^2 + x^311x^2+28x
= 43 +6x + 5x^2 +x^3
Third degree approximation is f(x) = x^3+5x^2+6x43
Cubical curve given above thus ans.
Question No. 02 Complete Solution
Numerical Analaysis: "Lagrange's Formula"
Find interpolation polynomial by "Lagrange’s Formula", with the help of this table:
x 1 2 3 5
f(x) 0 7 26 124
And find the value of f (6).
x 1 2 3 5
f(x) 0 7 26 124
calculating lagranges coefficeitns
L0 = (x2)(x3)(x5)/(12)(13)(15) = (x2)(x^28x+15)/8 = (x^310x^2+31x30)/8
L1 = (x1)(x3)(x5)/(21)(23)(25) = (x1)(x^28x+15)/3 = (x^39x^2+23x15)/3
L2 = (x1)(x2)(x5)/(31)(32)(35) = (x5)(x^23x+2)/4 = (x^38x^2+17x10)/4
L3 = (x2)(x3)(x1)/(51)(52)(53) = (x2)(x^24x+3)/24 = (x^36x^2+11x6)/24
lagranges polynomial is = 0*(x^310x^2+31x30)/8 + 7/3 (x^39x^2+23x15)
26/4 (x^38x^2+17x10) + 124/24 (x^36x^2+11x6)
simplyfying above
poly = x^3 (7/326/4+124/24) + x^2(21+5231 ) +x (23*7/317*26/4+11*124/24)
+ 15*7/3 +10*26/4 31
= x^3 1
f(x) is = x^31
f(6) = 6^31 = 2161 = 215 thus ans.
I have same answer.. Its right
Q.01 and Q.02 are complted.
Q.03 and Q. 04 are also easy easy.... discuss here I will try to solve and post soon
All person having MTH603 must particpate here
yr ya jo solution ap na uper upload keya ha keya ya thek ha? ap keh reha ho k Ist and 2nd are complited and 3rd aur 4th easy
MTH603_Assignment#02_Solution
Q.03 and Q. 04 in lecture no 28.
Q.3 me forward difference use krna hei derivative find krny k ly. or Q.4 me backward
both ex. 28 lec me hen.
Q4 main backward difference k table main aik value negtive main ati ha can any body tel is ka kya krna ha?????????????????
Question#1 Marks 10
Find an equation of a cubic curve which passes through the points
(4,43), (7, 83), (9,327) and (12, 1053) using Divided Difference Formula.
Solution:
X 
Y 
1st divided difference 
2nd divided difference 
3rd divided difference 
4 
43 



7 
83 
42 


9 
327 
122 
16 

12 
1053 
242 
24 
1 
y = f(x) = 43+(x4)42+(x4)(x7)16+(x4)(x7)(x9)(1)
= 43+(x4)42+(x^{2}11x+28)16+(x^{2}11x +28)(x9)
= 43+(x4)42+(x^{2}11x+28)16+(x^{3}–20x^{2}+127x252)
= 43+42x168+16x^{2}176x+448+x^{3}–20x^{2}+127x252
= x^{3}–4x^{2}7x15
Hence y=f(x)= x^{3}–4x^{2}7x–15 is the required cubic curve.
or
Question#2 Marks 10
Find interpolation polynomial by Lagrange’s Formula, with the help of following table,
X 
1 
2 
3 
5 
f(x) 
0 
7 
26 
124 
And hence find value of f (6).
Solution:
Question#3 Marks 10
Find from the following table,
x 
0 
0.2 
0.4 
0.6 
0.8 
1.0 
1.2 
0.710 
1.175 
1.811 
2.666 
3.801 
5.292 
7.232 
Solution:
Question#4 Marks 10
Find from the following table,
X 
1.3 
1.6 
1.9 
2.2 
2.5 
f(x) 
2.4 
2.9 
3.2 
4.7 
6.4 
Solution:
x 
f(x) 

1.3 
2.4 




1.6 
2.9 
0.5 



1.9 
3.2 
0.3 
0.2 


2.2 
4.7 
1.5 
1.2 
1.4 

2.5 
6.4 
1.7 
0.2 
1 
2.4 
= = 2.9
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