Ch:4
1. Using a divided differences table for the function ( f(x) = x3 + 2x2 - 3x + 1 )
evaluated at points ( x = 0, 1, 2, 3), compute the approximation of ( f '(2) )
using linear interpolation..
Using the divided differences table for the function ( f(x) = x3 + 2x2 - 3x + 1 )
evaluated at points ( x = 0, 1, 2, 3 ), we have the following divided differences
table:
x
0
1
2
3
F(x)
1
1
7
19
Now, to approximate f'(2) using linear interpolation, we use the formula for first
divided difference:
f [x0, x1] =
f(x1) − f(x0)
𝑥1 – 𝑥0
We'll use ( x0 = 1 ) and \( x1 = 3 ) to approximate ( f'(2) ):
f[1, 3] = f(3) - f(1) / (3 – 1) = (19 – 10) / (3 – 1) = 18 / 2 = 9
So, the approximation of ( f'(2) ) using linear interpolation is ( 9 ).
2. Given the divided differences table for ( f(x) = ex ) at points
( x = 0, 0.2, 0.4, 0.6, 0.8, 1.0 ), estimate ( f ''(0.5) ) using quadratic interpolation
Given the divided differences table for (f (x) = ex) at points (x = 0, 0.2, 0.4, 0.6, 0.8,
1.0 ), we have the following divided differences table:
x
0.0
0.2
0.4
0.6
0.8
1.0
F(x)
1.000
1.221
1.491
1.822
2.225
2.718
To estimate ( f''(0.5) ) using quadratic interpolation, we'll use the central difference
formula for second divided difference:
f[x-1, x0, x1] = f(x1) - 2f(x0) + f(x-1) / (x1 - x0)(x0 - x-1)
We choose ( x-1 = 0.4 ), ( x0 = 0.6 ), and ( x1 = 0.8 ) to approximate ( f''(0.5) ):
F [0.4, 0.6, 0.8] = (1.822 - 2 * 1.491 + 1.221) / (0.8 - 0.6)(0.6 - 0.4)
= (1.822 – 2 * 1.491 + 1.221) / 0.04
= 0.06 * 25
= 1.5
So, the approximation of ( f''(0.5) ) using quadratic interpolation is ( 1.5 ).