q3 part 1 sec01 sol
TRANSCRIPT
-
8/8/2019 Q3 Part 1 Sec01 Sol
1/4
-
8/8/2019 Q3 Part 1 Sec01 Sol
2/4
650:231 M.E. Computational Analysis and Design, Spring 2010Quiz 3_Sec 01
1. [15 points] In (a)-(b) below, you will obtain the quadratic Lagrange interpolation
polynomial for 32)( 2 ! xxf in the interval [0.00, 2.00] using the data:
j 1 2 3
Xj 0.0 1.0 2.0
fj f(0.0)= f(1.0)= 5 f(2.0)= 11
(a) [10 pts.] Write the basic Lagrange interpolation polynomials )(),(),( 321 xLxLxL , where the
subscript j in )(xLj indicates the interpolation point j in the table. (no MATLAB codes, just
mathematical formulas)
)23(2
1
)20)(10(
)2)(1()( 21 !
! xx
xxx
)2()21)(01(
)2)(0()( 22 xx
xxx
!
!
)(2
1
)12)(02(
)1)(0()( 23 xx
xxx
!
!
(b) [5 pts.] Give the interpolation function using the Lagrange interpolation polynomial.
Expand the products to give the interpolation in the final form of . No needto evaluate the square root function.
3)2
115433(
2
)11523(
3)2
11152
2
33()
2
1115
2
13(
)(2
111)2(5)23(
2
13
)()2()()1()()0()(
2
2
222
3212
!
!
!
!
xx
xx
xxxxxx
xLfxLfxLfxP
-
8/8/2019 Q3 Part 1 Sec01 Sol
3/4
650:231 M.E. Computational Analysis and Design, Spring 2010Quiz 3_Sec 01
2. [20 points] Consider the numerical integration of the function 32)( 2 ! xxf over an
interval 0
-
8/8/2019 Q3 Part 1 Sec01 Sol
4/4
650:231 M.E. Computational Analysis and Design, Spring 2010Quiz 3_Sec 01
3. [15 points] Consider the function 32)( 2 ! xxf again.
(a) [7.5 points] Write down the Newton representation of the quadratic interpolationpolynomial with interpolation points, X1 = 0, X2 = 2, X3 = 1, in terms of the divided
differences. It is not necessary to expand the polynomials. Note that the interpolationpoints are not arranged in an increasing order.
(b) [7.5 points] Obtain the numerical values of the divided differences used in (a) andsubstitute these values to determine the Newton representation explicitly. It is not
necessary to expand the polynomials. No need to evaluate the square root function.
(a)
f(x)= f[x1]+ f[x1,x2](x-x1)+f[x1,x2,x3](x-x1)(x-x2)= f[x1]+ f[x1,x2]x+f[x1,x2,x3]x(x-2)
(b)
Construct divided differences table.
f[x1]= 3 , f[x2]= 11 , f[x3]= 5 ,
f[x1,x2]=( 11 - 3 )/(2-0)= ( 11 - 3 )/2,
f[x2,x3]=( 5 - 11 )/(1-2)= 11 - 5 ,
F[x1,x2,x3]=( 11 - 5 -( 11 - 3 )/2)/(1-0)=( 11 -2 5 + 3 )/2
= 3 + x( 11 - 3 )/2+x(x-2) ( 11 -2 5 + 3 )/2