Homework 10 solutions
MATH 2283 Spring 2014
2.14
(6 pts)
1
(f) Since for all n we have sin n sin 1, then
sin
n=1
1
n
n
n
(sin 1) .
n=1
Since the series on the left is geometric with ratio sin 1 < 1 it must converge,
so the series on the right converges as
Homework 3 solutions
MATH 2283 Spring 2014
4.6 P (n) : 52n 1 is divisible by8.
Basis case: 52(1) 1 = 24 is divisible by 8 .
Induction hypothesis: assume P (n) is true.
Induction step: show P (n + 1) : 52(n+1) 1 is divisible by 8.
52(n+1) 1 = 52n+2 1 = 52
Homework 8 solutions
MATH 2283 Spring 2014
6.9
(a) Here are the rst few terms:
n
xn
1
2.646
2
3.106
3
3.179
4
3.190
5
.
3.192
(b) We show that the sequence is monotonic and bounded then conclude that
it is convergent.
Step 1. We use induction to show that
Homework 9 solutions
MATH 2283 Spring 2014
6.14
(3 pts)
(a) We prove xn > r for all n by induction:
Basis case: x1 > r by hypothesis.
Assume xn > r.
By the Mean Value Theorem, c [r, xn ] such that
f (c) =
f (xn ) f (r)
f (xn )
=
,
xn r
xn r
where we used
Homework 4 solutions
MATH 2283 Spring 2014
1.12 ( = )
We show that x = sup S implies statements a) and b). By denition of the
supremum, x must be an upper bound of S, hence a) holds true. Suppose now
that b) is false. Thus
> 0 such that s S[x s].
This im
Homework 2 solutions
MATH 2283 Spring 2014
2.6
(a) False, since for k = 4, 2k + 1 = 9 is not prime
(b) False, since for k = 3, 2k + 1 = 9 is not prime
(c) False, since for k = 3, 22k + 1 = 65 is not prime
2.12 Let P (x, y) be the statement x < y. Thus it
Homework 6 solutions
MATH 2283 Spring 2014
2.12(b)
We want to nd out for which values of n the following holds true:
|sn 1| < .
On the left hand side, we have
|sn 1| =
2n + n
2n + n (2n + 1)
n1
= n
1 =
.
n+1
2
2n + 1
2 +1
Since 2n > n2 for all n > 4,
n
n
Homework 5 solutions
MATH 2283 Spring 2014
1.7 By the Archimedean property, > 0, n0 N s.t.
But for any n, 2n > n. Thus n > n0 ,
|sn L| =
1.18
Let
|sn L| =
1
n0
< .
1
1
1
< <
< .
n
2
n
n0
> 0. Then n,
17
17
1
2n + 3 2
5(2n + 3) 2(5n 1)
=
=
<
<
5n 1 5
5(5n
Homework 1 solutions
MATH 2283 Spring 2014
1.3
(a) P &Q&R
(b) P Q R
(c) (P &R) Q
(d) P &(Q R)
(e) (P &Q) R
1.4
(a) Q = P
(b) P = Q
(c) P Q
(d) P if not Q, or in other words Q = P
1.6 Consider for example P = n is a prime greater than 3 and Q = n is
an odd
Homework 7 solutions
MATH 2283 Spring 2014
4.4
(3 pts)
(b) Note that
sn =
1+
n
1
n
.
We recognize the sequence that converges to ex with x = 1, thus
lim sn = e1 .
n
(c) Note that
1
=
sn
Thus limn
4.5
1
sn
n
2n + 1
2n
=
1+
1/2
n
n
.
= e1/2 , so limn sn = e
Homework 11 solutions
MATH 2283 Spring 2014
3.7
(4pts)
(b) By (), we have
Tn = 1 +
1
1
+ . + ln n 1
2
n
ln n
Tn = 1 +
1
1
1
1
+ ln n > 0
+ . +
2
n1 n
n
ln n
Hence Tn is bounded below by 0 and above by 1.
(c) By the mean value theorem, using f (x) = ln x,
function [lu, pvt] = LUfactor ( A )
%LUFACTOR
compute the LU decomposition for the matrix A
%
%
calling sequence:
%
[lu, pvt] = LUfactor ( A )
%
%
input:
%
A
coefficient matrix for linear system
%
(matrix must be square)
%
%
outputs:
%
lu
matrix containin