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Assignment # 2
MTH603 (Spring 2017)
Total marks: 20
Lectures # 23 - 28
Due date: 19-07-2017
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Lecture # 23 to Lecture # 28.
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Question#1 Marks 10
Find an equation of a cubic curve which passes through the points
(0, -1), (0, -4), (1, 5) and (2, -6) 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
0 7 26 124
And hence find value of .
Question#3 Marks 10
Find from the following table,
x 1.3 1.6 1.9 2.2 2.5
2.4 2.9 3.2 4.7 6.4
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+ http://bit.ly/vucodes (Vu Study Groups By Subject Codes Wise)i don't understand the simplification part
how you made
3+(x+1)(-7) = 3 - 7x -14 ?
it should be 3-7x-7 ?
3 - 7x -14 + 8x2 + 8x + (x2+x)(x-1)(-6)
this step is wrong..
it is 3-7x-7
Solution:
The value of table for X and Y
X | 1 x0=1 | 2 x1=2 | 3 x2=3 | 5 x3=5 |
---|---|---|---|---|
Y | 0 y0=0 | 7 y1=7 | 26 y2=26 | 124 y3=124 |
Langrange's Interpolating Polynomial
The value of x at you want to find Pn(x):x=4
Y(4)=(4-2)(4-3)(4-5)(1-2)(1-3)(1-5)×0+(4-1)(4-3)(4-5)(2-1)(2-3)(2-5)×7+(4-1)(4-2)(4-5)(3-1)(3-2)(3-5)×26+(4-1)(4-2)(4-3)(5-1)(5-2)(5-3)×124
Y(4)=(2)(1)(-1)(-1)(-2)(-4)×0+(3)(1)(-1)(1)(-1)(-3)×7+(3)(2)(-1)(2)(1)(-2)×26+(3)(2)(1)(4)(3)(2)×124
Y(4)=-2-8×0+-33×7+-6-4×26+624×124
Y(4)=63
i think you made mistake
it should be
Y(4)=(4-2)(4-3)(4-5)/(1-2)(1-3)(1-5)×0+(4-1)(4-3)(4-5)/(2-1)(2-3)(3-5)×7
Please recheck
can you please explain question 2 simplification ?
this qs is wrong.first find the polynomial i.e equation nd then put value of x=4 answer will be 63
i solve the question but my answer is different i wanna compare and understand.
Question No. 2
my answer is 63
Question No. 3
my answer is
y'(2.5)=2.8889 and
y''(2.5)= -25.370
for question no. 2 my answer is 59.5 i have issue in last two lines of simplification. can you share simplification step by step in last two lines
bhai full solution share kr do plzzzz
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