Numerical Analysis: Interpolation and Approximation

This text covers the topics of interpolation and approximation in numerical analysis. It includes solved examples and formulas for Lagrange interpolation, Newton's divided difference method, and more.

Ans 1.
(a)
y= y0+ ( y1− y0 ) x−x0
x1−x0
y=−1+ 1∗x+1
1 y=−1+ x +1=x y=x
(b)
f ( x )=f ( x2 ) + ( x −x2 ) ( f ( x3 )−f ( x1 ) )
2∗∆ x + ( x−x2 )2
[ f ( x1 )−2∗f ( x2 ) +f ( x3 ) ]
2∗∆ x2
f ( x ) =0+ ( x −0 ) ( 1+1 )
2 + x2 [ −1−0+1 ]
2 f ( x ) =x
(c)
Let ,f ( x )=a∗x3 +b∗x2+c∗x +d So ,−1=−a+b−c +d0=d1=a+b+ c+d 32=8 a+4 b+2 c+ d
−1=−a+b−c1=a+b+ c Adding we get ,0=−a+ b−c+ a+b+ c=2∗b¿ , b=0So , we get
−1=−a−c 32=8∗a+2∗cSo ,−8=−8 a−8 c Adding the previous two equations we get ,
32−8=8 a+2 c−8 a−8 c24=−6 c c=−4So ,1=a+0−4 +0a=5 So , we get theequation as
f ( x )=5∗x3−4∗x
(d)
y= y0+ ( y1− y0 ) x−x0
x1−x0
¿ , y=1+ ( 32−1 ) ( x−1 )
1 y=1+31 x−31 y=31 x−30
(e)
f ( x )=f ( x2 ) + ( x −x2 ) ( f ( x3 )−f ( x1 ) )
2∗∆ x + ( x−x2 )2
[ f ( x1 )−2∗f ( x2 ) +f ( x3 ) ]
2∗∆ x2
f ( x ) =1+ ( x−1 ) ( 32−1 )
2 + ( x−1 ) 2 [ 0−2∗1+32 ]
2 f ( x ) =1+ 31∗ ( x −1 )
2 + ( x−1 ) 2 30
2
f ( x ) =1+ 31∗ ( x −1 )
2 +15∗( x−1 ) 2
Ans 2.
P1 ( x ) =a0 +a1 ( x−x0 ) P1 ( x ) =−2+0.4 ( x−2 ) P1 ( x ) =−2+0.4 x−0.8 P1 ( x ) =−2.8+0.4 x
So , at x=2.7P1 ( x ) =−2.8+1.08=−1.72

P2 ( x ) =a0 +a1 ( x−x0 ) + a2 ( x −x0 ) ( x−x1 )¿−2+0.4∗( x−2 ) +1.5∗( x−2 ) ( x−3 )
¿−2+0.4 x−0.8+1.5∗x2−7.5∗x+ 9¿ 1.5∗x2 −7.1∗x−6.2 So , at x=2.7 we have ,
P2 ( x ) =1.0935−−19.17−6.2¿−24.2765
P3 ( x ) =a0 + a1 ( x−x0 ) + a2 ( x−x0 ) ( x −x1 ) +a3 ( x −x0 ) ( x− x1 ) ( x−x2 )
P3 ( x ) =−2+0.4∗( x−2 ) +1.5∗( x−2 ) ( x−3 )−0.4∗( x−2 ) ( x−3 ) ( x−5 )
¿−2+0.4 x−0.8+ 1.5∗x2−7.5∗x+9−0.4∗x3 +3.6∗x2−10.4∗x−9.6
¿−0.4∗x3+ 5.1∗x2−17.5∗x−3.4
So , at x=2.7 ; P3 ( x ) =−7.8732+37.179−47.25−3.4=−21.3442
P4 ( x ) =a0 +a1 ( x−x0 ) +a2 ( x−x0 ) ( x−x1 ) +a3 ( x−x0 ) ( x−x1 ) ( x−x2 ) +a4 ( x−x0 ) ( x−x1 ) ( x−x2 ) ( x−x3 )
¿−0.4∗x3+ 5.1∗x2−17.5∗x−3.4+0.0005∗( x3−9∗x2+26∗x−24 ) ( x−5 )
¿−0.4∗x3+5.1∗x2−17.5∗x−3.4+0.0005∗x4−0.007∗x3 +0.0225∗x2+ 0.001∗x−0.005
¿ 0.0005∗x4−0.407∗x3 +5.1225∗x2−17.499∗x−3.405 isthe required equation
Therefore at x=2.7 we get ,
P4 ( x )=0.02657−8.01098+37.34303−47.1123−3.405=−21.15868
Ans3.
a b f(a) f(b) c=(a+b)/2 f(c) Update new b-a
1 2 -4 -1 1.5 -2.75 c=a=1.5 0.5
1.5 2 -2.75 -1 1.75 -1.9375 c=a=1.75 0.25
1.75 2 -1.9375 -1 1.875 -1.48438 c=a=1.875 0.125
1.875 2 -1.48438 -1 1.9375 -1.2461 c=a=1.9375 0.0625
1.9375 2 -1.2461 -1 1.96875 -1.124 c=a=1.96875 0.03125
1.96875 2 -1.124 -1 1.98438 -1.062 c=a=1.98438 0.01562
1.98438 2 -1.06224 -1 1.99219 -1.03118 c=a=1.99219 0.00781