Partial Solutions to Homework X
Y. Zhou
Section 6.1
2.
Proof. The right-hand side is the Lagrange polynomial interpolating f (x) at n + 1 nodes {xi }.
Denote this polynomial by q(x), also denote Lf (x) = p(x). Both polynomials are of degree at most
n. Note p(x) − q(x) = 0 at n + 1 points {xi } since they interpolate the same function f (x) at the
same set of nodes. Then then p(x) − q(x) ≡ 0 since they have n + 1 zeros while they are polynomials
of degree at most n. This proves that Lf = q as required.
Proof.
L(af + bg) =
n
X
af (xi ) + bg(xi ) li
i=0
=
n
X
af (xi )li +
i=0
n
X
= a
n
X
bg(xi )li
i=0
n
X
f (xi )li + b
i=0
g(xi )li
i=0
= aLf + bLg
4.
Proof. Let p = Lq. Then p is a polynomial of degree at most n. Since q is a polynomial of degree
at most n, so is p − q. But then p − q ≡= 0 as p(xi ) − q(xi ) = 0 at n + 1 nodes {xi }.
11.
Proof. The Lagrange interpolation of p(x) at n + 1 nodes {xi } is given by
n
n
X
Y
x − xj
Lp(x) =
.
p(xi )
xi − xj
i=0
j=0,j6=i
This is a polynomial of degree at most n. Since p(x) is a polnomial of degree at most n − 1, the
coefficient of xn in this Lagrange polynomial
n
n
X
Y
1
p(xi )
x
−
xj
i
i=0
j=0,j6=i
must be zero. This prove the claim in the argument.
Section 6.2
3.
Proof. Since x0 is in (a, b) and x1 , ·xn converge to x0 , there is a moment at x0 , ·xn are all located
in the interval (a, b) so we can apply Theorem 4 in Section 6.2.
From the proof of Theorem 2 in Section 6.1 we know that ξx is located in the interval
(min{x, x0 , · · · , xn }, max{x, x0 , · · · , xn }).
We then know that ξ in Theorem 4 of Section 6.2 is located in the interval
(min{x0 , · · · , xn }, max{x0 , · · · , xn }),
in other words,
min{x0 , · · · , xn } ≤ ξ ≤ max{x0 , · · · , xn }.
Thus ξ → x0 as x1 , · · · , xn all converge to x0 . The conclusion then follows this Theorem 4.
1
2
5.
Proof. Let
q(x) =
n
X
i=0
p[x0 , · · · , xi ]
i−1
Y
(x − xj ),
j=0
i.e, q(x) is the Newton interpolation of p(x) at n + 1 nodes. But then q(x) is a polynomial of degree
at most n. Since p(x) is also an such polynomial, so is q(x) − p(x). But then q(x) − p(x) ≡ 0 since
it has n + 1 zeros at {xi }. This proves that q(x) = p(x) as required.
8.(hints) Note in the left-hand side is the Lagrange interpolation and in the right-hand size is
the Newton interpolation, these two interpolation polynomials have to the same (as they are all
polynomials of degree at most n, while have n + 1 common zeros).
This can be proved by following the mathematical induction.
9.(hints) This says the coefficients of xn in the Lagrange interpolation polynomial and the Newton
interpolation polynomial are the same, and thus is a corollary of Problem 8.
This can also be proved by directly using the mathematical induction.