MATH 609-600, Homework #3, Solutions
(1) (20 pts) Problem 14, page 325 from the book.
Solution: Let P (x) be the interpolation polynomial and x0 = 0, x1 , . . . , xn−1 be the nodes
of interpolation. Then we have
f (x) − P (x) =
f (n) (ξ)
(x − 0)(x − x1 ) · · · (x − xn−1 ),
n!
where ξ ∈ [−1, 1]. We need the following two facts:
(i) |f (n) (ξ)| ≤ 0.5(e + e−1 ) for any ξ ∈ [−1, 1];
(ii) if f (x) = sinh(x) then |x| ≤ sinh(x) for any x ∈ [−1, 1].
Both of these are easy to show using that sinh(x) = 0.5(ex − e−x ) and observing that
g(x) = 0.5(ex − e−x ) − x is increasing for x ≥ 0. Using the above estimates we get
|f (x)−P (x)| ≤
0.5(e + e−1 )
2
2n
|x||(x−x1 ) · · · (x−xn−1 )| ≤ |f (x)||(x−x1 ) · · · (x−xn−1 )| ≤
|f (x)|
n!
n!
n!
because each term |x − xi | is not greater than 2 if x, xi ∈ [−1, 1].
(2) (20 pts) Problem 25, page 326.
Solution: Using the Lagrange formula we have
P (x) = −2
(x − 1)(x − 2)(x − 6) (x − 1)(x − 3)(x − 6)
(x − 1)(x − 3)(x − 2)
(x − 3)(x − 2)(x − 6)
−22
−
−37
.
(1 − 3)(1 − 2)(1 − 6)
(3 − 1)(3 − 2)(3 − 6) (2 − 1)(2 − 3)(2 − 6)
(6 − 1)(6 − 2)(6 − 3)
(3) (20 pts) Problem 9, page 336.
Solution: Using the previos problem (solved in class), we have the two forms of the same
Lagrange polynomials
P (x) =
n
X
i=0
f (xi )
Y (x − xj )
= f (x0 )+f [x0 , x1 ](x−x0 )+· · · f [x0 , x1 , . . . , xn ](x−x0 ) · · · (x−xn−1 ).
(xi − xj )
j6=i
Then the coefficient in from of xn must be the same
n
X
i=0
f (xi )
Y
j6=i
1
= f [x0 , x1 , . . . , xn ](x − x0 ).
(xi − xj )
(4) (20 pts) Problem 2, page 348.
Solution: Using the divided difference formula (easy to compute by hand with the table
format, see pages 342 and 343 in the book), we obtain
P (x) = 2 − 9x + 3x2 + 7x2 (x − 1) + 5x2 (x − 1)2 −
1
153 2
x (x − 1)2 (x − 2).
18
(5) (20 pts) Problem 4, page 348.
Solution: It is clear that x0 6= x1 . Otherwise, the interpolation problem will be not solvable
for c00 6= c10 . The space of all cubic polynomials has the following basis if x0 6= x1 :
{x − x0 , x − x1 , (x − x0 )(x − x1 ), x(x − x0 )(x − x1 )}.
The interpolation conditions p(xi ) = ci0 give:
(x − x0 )
(x − x1 )
+ c10
+ (ax + b)(x − x0 )(x − x1 ).
(x0 − x1 )
(x1 − x0 )
The coefficients a and b should be determined by the other two interpolation conditions
p00 (xi ) = ci2 . The corresponding system for a and b can always be solved and the solution
is
c02 − c12
a=
, b = 0.5(c02 − (4x0 − 2x1 )a).
6(x0 − x1 )
Therefore, if x0 6= x1 the inerpoation problem is always solvable by a cubic polynomial.
Moreover, the polynomiac is exactly degree three if c02 6= c12 .
p(x) = c00