1) Outline Pros and Cons and Winners and Losers of Trade Using information from
this Economist Article.
After Nafta trade with Mexico increased as did trade with China after they joined the WTO in 2001.
Prolots of new work and investment in Mexico and Chi
1) Outline Pros and Cons and Winners and Losers of Trade Using information from this
Economist Article. W
2) Below the first article is another one from the NYTimes (most of text is below but some
graphs are missing, you can visit the website if you want
MAP 4341
Exam 2, Take-Home
You are allowed any help from a non-human source. DO NOT discuss this exam with any
classmate, friend, or professor (other than me). Exam is due at 2:30 on December 10. You
may drop it o at my oce, or you may send it to me elect
MAP 4341
Exam 1, Take-Home
You are allowed any help from a non-human source. DO NOT discuss this exam with any
classmate, friend, or professor (other than me). Exam is due at the beginning of class on
Ocotber 21.
Number
1
2
3
4
5
6
7
8
9
10
Total
Points S
Review: Exam 2
MAP 4341
In-Class Portion: The in-class portion of the exam will be during our regularly scheduled
exam time: Tuesday, December 7, 1:00-2:50. The exam will be closed book, closed note,
closed calculator. The in-class portion will consist of
Review Sheet for Exam 1
MAP 4341
In-Class:
For the in-class portion of the exam, you may bring in a calculator (any type) and a
writing instrument. While I am allowing calculators of any type, I need to see all work
on the integrals only giving me the ou
MAP 6385
Homework 4
September 18, 2009
1) By hand (yes, by hand), calculate z , where z = [1, 0, 3, 2].
2) I have said that z z is a linear transformation. This means we can represent it as a
matrix multiplication. Find the matrix for when M = 4.
3) a) As
MAP 6385
Homework 3
September 9, 2009
WARNING: There are some big Mathematica problems on here. One of them is an openended question that requires experimentation. You probably shouldnt wait until the last
minute to do these.
0) In class, we found the gen
MAP 6385 Homework 2
September 2, 2009
1) Graph the rst ve Legendre polynomials (P0 (x) through P4 (x) on the same graph. Then plot the rst ve Chebyshev polynomials (T0 (x) through T4 (x) on the same graph. Compare the two results. What similarities do you
MAP 6385
Homework 1
August 27, 2009
1) The following table gives a list of data points. Find the least squares polynomials of
degrees 1, 2, and 3 for this data set. For each degree, calculate the error E , and plot the
polynomial versus the data set.
xi
y
MAP 6385
Homework Rules
1) You are allowed to work together. HOWEVER, if you do work with someone else,
then when you submit your homework, I want to know with whom you worked. If
someone else solved a problem and then helped you, tell me. Give credit whe
Huffman Compression
Steps for the Huffman Compression 1) Write down all of the letters appearing in the text in a list, most frequent to least frequent. For each letter, put the frequency as a subscript (it will look like a scrabble tile). For our example
MAP 6385 Exam 2
November 5, 2009
Same rules apply as from Exam 1. The exam is due on November 12 at 4:30 PM. Note that Problems 7 and 8 do not have explicit answers: they are more experimentation problems. For credit, you need to describe what you tried t
MAP 6385 DWT, Part 3
Denition 18. Let M be an even number. Dene the down operator D : CM CM/2 by D(z )(k ) = z (2k 1) Furthermore, if M is any number, then dene the up operator U : CM C2M by k+1 z ( 2 ) if k is odd U (z )(k ) = 0 if k is even Theorem 19.
MAP 6385 DWT, Part 2
Example 14. Let M be a number divisible by 4. The First Stage Shannon Wavelet Basis is the wavelet basis generated by the father and mother wavelets u and v satisfying: 2 if 1 n M or 3M + 1 n M 4 4 u(n) = 0 if M + 1 n 3M 4 4 0 if 1 n
MAP 6385 DWT
Definition 1. A vector z CM is time localized if its entires with large magnitudes are clustered around a particular index. A vector z is frequency localized if the entries of z with ^ large magnitudes are clustered around a particular index.
MAP 6385 DFT, Part 2
Denition: Let z and w be vectors in CM . The convolution of z and w, written as z w, is the vector dened by
M
z w(m) =
n=1
z (m n + 1)w(n)
Theorem: Let z , w, and y be vectors in CM , and let c C. Then: i) z w = w z ii) (z w) y = z (w
MAP 6385 DFT
Definition: Let CM be the vector m-dimensional complex vector space. We represent a typical element by z, written out as: z = [z(1), z(2), . . . , z(M )] The inner product on CM is defined as
M
z, w =
k=1
z(k)w(k)
We make our vector space cyc
MAP 6385 Convolution and Image Processing
1
Convlution
k = 1/2
Denition 1. Let z, w l2 (R), the set of sequences indexed on Z such that z (m)2
k=
<
The convolution z w of z and w is the sequence dened by the equation
k =
(z w)(m) =
k=
z (k )w(m k )
Deniti
96
Chap. 15 Limits to Computation
15.8 Represent a real number in a bin as an innite column of binary digits, similar to the representation of functions in Figure 15.4. Now we can use a simple
diagonalization proof. Assume that some assignment of real num
95
(b) It is possible to compute xn in log n time, and the rest of the formula
requires a constant number of multiplications. Thus, the number of
multiplications required is polynomial on the input size.
15.5 First, we should note that, for both problems,
15
Limits to Computation
15.1 This reduction provides an upper bound of O(n log n) for the problem of
maximum nding (that is the time for the entire process), and a lower bound
of constant time for SORTING (since that is the time spent by Maximum
Finding
93
(b) Fill an array with 2n + 1 elements, which forces a nal growth to an
array of size 2n+1 . Now, do an arbitrarily long series of alternating
inserts and deletes. This will cause the array to repeatedly shrink and
grow, for bad (n2 ) performance.
(c)
92
Chap. 14 Analysis Techniques
B elements are repeatedly reinserted one fewer times, the next 2B elements
2 fewer times, etc. Thus, once we insert 2i B elements, we have done a total
number of insertions costing 2i B + 2i1 B + 2i2 2B + + 2i1 B . This
wor
91
n log n, so we will guess that
T(n) = ( n log n.
To complete the proof we must show that T(n) is in ( n log n), but this
turns out to be impossible.
On the other hand, the recurrence is clearly n. So, lets guess that T(n)
is in O( n) for c = 4, n = 2.
89
14.3 From Equation 2.2 we know that
n
i2 =
i=1
2n3 + 3n2 + n
.
6
Thus, when summing the range a i b, we get
b
2b3 + 3b2 + b 2a3 + 3a2 + a
6
6
i2 =
i=a
2(b3 a3 ) + 3(b2 a2 ) + (b a)
.
6
=
14.4 We need to do some rearranging of the summation to get somet
14
Analysis Techniques
14.1 Guess that the solution is of the form
an3 + bn2 + cn + d.
Since when n = 0 the summation is 0, we know that d = 0. We have the
following simultaneous equations available:
a+b+c = 1
8a + 4b + 2c = 5
27a + 9b + 3c = 14
Solving t