Respuestas
Given ( , ), 0 0 x y ( , ), 1 1 x y and ( , ), 2 2 x y fit a quadratic interpolant through the data. Note that
y = f (x), ( ), 0 0 y = f x ( ), 1 1 y = f x and ( ), 2 2 y = f x assume the quadratic interpolant ( ) 2 f x
given by
( ) ( ) ( )( ) 2 0 1 0 2 0 1 f x = b + b x − x + b x − x x − x
At 0 x = x
( ) ( ) ( ) ( )( ) 0 2 0 0 1 0 0 2 0 0 0 1 f x = f x = b + b x − x + b x − x x − x
= b0
( ) 0 0 b = f x
At 1 x = x
( ) ( ) ( ) ( )( ) 1 2 1 0 1 1 0 2 1 0 1 1 f x = f x = b + b x − x + b x − x x − x
( ) ( ) ( ) 1 0 1 1 0 f x = f x + b x − x
then
1 0
1 0
1
( ) ( )
x x
f x f x b −
− =
At 2 x = x
( ) ( ) ( ) ( )( ) 2 2 2 0 1 2 0 2 2 0 2 1 f x = f x = b + b x − x + b x − x x − x
( ) ( )( ) ( ) ( ) ( ) ( ) 2 0 2 2 0 2 1
1 0
1 0
2 0 x x b x x x x
x x
f x f x f x f x − + − − −
− = +
then
2 0
1 0
1 0
2 1
2 1
2
( ) ( ) ( ) ( )
x x
x x
f x f x
x x
f x f x
b −
−
− − −
−
=
Hence the quadratic interpolant is given by
( ) ( ) ( )( ) 2 0 1 0 2 0 1 f x = b + b x − x + b x − x x − x
( )( )