Sunday, 20 January 2013
Question
No. 01 Complete Solution
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) / 7-4 = 83+43 /3 = 126/3 = 42
327-83/9-7 = 244/2 = 122
1053-327/12-9
= 726/3 = 242
Column
4 obtained by 122-42/9-4 = 80/5 = 16
242-122/12-7 = 120/5 = 24
Column
5 obtained by 24-16/12-4 = 8/8 = 1
Third
degree approximation given by
f(x)
= f(4) + xf(4,7) + x(x-4)f(4,7,9)+x(x-4)(x-7)f(4,7,9,12)
= -43 + x 42 + x(x-4) 16 + x(x-4)(x-7)
= -43 + 42x-64x+16x^2 + x(x^2-11x+28)
= -43 -22x +16x^2 + x^3-11x^2+28x
= -43 +6x + 5x^2 +x^3
Third
degree approximation is f(x) = x^3+5x^2+6x-43
Cubical
curve given above thus ans.
This
is complete solution you can also give ur
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
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 = (x-2)(x-3)(x-5)/(1-2)(1-3)(1-5) = (x-2)(x^2-8x+15)/8 = (x^3-10x^2+31x-30)/8
L1 = (x-1)(x-3)(x-5)/(2-1)(2-3)(2-5) = (x-1)(x^2-8x+15)/3 = (x^3-9x^2+23x-15)/3
L2 = (x-1)(x-2)(x-5)/(3-1)(3-2)(3-5) = (x-5)(x^2-3x+2)/-4 = (x^3-8x^2+17x-10)/-4
L3 = (x-2)(x-3)(x-1)/(5-1)(5-2)(5-3) = (x-2)(x^2-4x+3)/24 = (x^3-6x^2+11x-6)/24
lagranges polynomial is = 0*(x^3-10x^2+31x-30)/8 + 7/3 (x^3-9x^2+23x-15)
-26/4 (x^3-8x^2+17x-10) + 124/24 (x^3-6x^2+11x-6)
simplyfying above
poly = x^3 (7/3-26/4+124/24) + x^2(-21+52-31 ) +x (23*7/3-17*26/4+11*124/24)
+ -15*7/3 +10*26/4 -31
= x^3 -1
f(x) is = x^3-1
f(6) = 6^3-1 = 216-1 = 215 thus ans.
f(x) 0 7 26 124
calculating lagranges coefficeitns
L0 = (x-2)(x-3)(x-5)/(1-2)(1-3)(1-5) = (x-2)(x^2-8x+15)/8 = (x^3-10x^2+31x-30)/8
L1 = (x-1)(x-3)(x-5)/(2-1)(2-3)(2-5) = (x-1)(x^2-8x+15)/3 = (x^3-9x^2+23x-15)/3
L2 = (x-1)(x-2)(x-5)/(3-1)(3-2)(3-5) = (x-5)(x^2-3x+2)/-4 = (x^3-8x^2+17x-10)/-4
L3 = (x-2)(x-3)(x-1)/(5-1)(5-2)(5-3) = (x-2)(x^2-4x+3)/24 = (x^3-6x^2+11x-6)/24
lagranges polynomial is = 0*(x^3-10x^2+31x-30)/8 + 7/3 (x^3-9x^2+23x-15)
-26/4 (x^3-8x^2+17x-10) + 124/24 (x^3-6x^2+11x-6)
simplyfying above
poly = x^3 (7/3-26/4+124/24) + x^2(-21+52-31 ) +x (23*7/3-17*26/4+11*124/24)
+ -15*7/3 +10*26/4 -31
= x^3 -1
f(x) is = x^3-1
f(6) = 6^3-1 = 216-1 = 215 thus ans.
