MATH 280: Homework 10
1. Find the unique cubic polynomial with p(1) = 2, p(0) = 1, p(1) =
3, p(4) = 0. Use the Newton interpolation form.
2. Recall given a function f and distinct nodes x0 , x1 , . . . , xn , the unique
degree n polynomial p(x) with p(xi
Math 280: Solutions of Homework IX
1. Solution. (a)
16 0 0
A A = 0 0 0
0 0 49
So 1 = 4, 2 = 0 and 3 = 7. u1 = (1, 0, 0)T , u2 = (0, 1, 0)T and u1 = (0, 0, 1)T .
Rank(A A) = 2.
(b)
42
21
A A =
The characteristic polynomial is 2 5. So 1 = 5 and 2 = 0. u1 =
Math 280: Solutions of Homework VIII
1. Solution. (a) Suppose x = c1 u1 + c2 u2 , if x is not generic, then c1 = 0, i.e.,
x = c2 u2 . Then x is the eigenvector of eigenvalue 2 . If x0 is nongeneric, by the
above discussion, x0 = cu2 for some c Cn . We hav
Math 280: Solutions of Homework VII
1. Solution. (a) By denition, we have
n
2
2
2
(|x|) = (maxcfw_|xi | : 1 i n) = maxcfw_|xi | : 1 i n
i=1
|xi |2 = (|x|2)2 .
And
n
n
2
(|x|2 ) =
i=1
|xi |
n
2
2
i=1
|xi | +
i,j =1
n
2|xi |xj | = (
i=1
|xi |)2 = (|x|1 )2
Math 280: Solutions of Homework VI
1. Solution. (a) A has Doolittles decomposition
10
21
A=
12
01
which is also its Choleskys factorization.
(b) A has Doolittles decomposition
100
411
2
1
A = 1 1 0 0 1 8
8
1
1
001
1
4
8
4
Therefore
L1
And
Finally,
1
00
1
Math 280: Solutions of Homework V
1. Solution. (a) f (x) =
x [0, 10].
2x
.
(1+x2 )2
We want to nd the maximal value of |f (x)| for
|f (x)| =
2 6x2
.
(1 + x2 )3
1
So |f (x)| = 0 has only one root 3 in [0, 10]. |f (0)| = 0, |f (10)| 0.0002 and
3
1
|f ( 3 )|
Math 280: Solutions of Homework IV
1. Proof. (a) Since |cN r |
1
(b a0 ) < . So
2N +1 0
1
(b
2N +1 0
a0 ), to guarantee that |cN r | < we need
b0 a0
Apply logarithm function on both sides, we have
2N +1 >
(N + 1) log(2) > log(b0 a0 ) log()
which implies
Math 280: Solutions of Homework II
1. Solution. (a) Denote x = (x1 , x2 , x3 , ). If E (x) = x, namely, (x2 , x3 , x4 , ) =
(x1 , x2 , x3 , ), then we have xn = xn1 . So xn = n1 x1 = n1 C .
(b) For any C, dene = (1, , 2, ). We have = 0 and E ( ) = .
x
x
x
Math 280: Solutions of Homework I
1. Proof. (a) It is trivial when x = 0. For the case x = 0, we have
| sin(1/x)| 1. Hence |f (x)| = |x| sin(1/x)| |x|. Moreover, for xed
> 0, if |x 0| < , then |f (x) f (0)| = |f (x)| |x| = |x 0| < .
Hence, f (x) is conti