4. Rotation of the sphere by 180 about the X -axis has the eect : (X, Y, Z ) (X, Y, Z ). Recalling that the projection from the sphere to the plane is given by : (X, Y, Z ) Thus Y X +i . 1Z 1Z X Y +i
5.3
From Partial to Total Orders
Given a nite partial order, we can extend it to a total order. For example, suppose we have a set of tasks T to perform. We wish to decide in what order to perform the
5.2
Analysing Partial Orders
The shape of the partial orders (N , ) and (Z , ) are different from each other. The number 0 is the smallest element with respect to the natural numbers, but not with res
5.1
Hasse Diagrams
Since partial orders are special binary relations on a set A, we can represent them by directed graphs. However, these graphs get rather cluttered if every arrow is drawn. We theref
In the denition of partial order, notice that anti-symmetric is not the opposite of symmetric, since a relation can be both symmetric and anti-symmetric: for example, the identity relation or the empt
Comment The rational numbers are also countable. In contrast, Cantor showed that there are uncountable sets: that is, innite sets that are too large to be countable. An important example is the set of
are bigger than the set of natural numbers. The set of natural numbers is one of the simplest innite sets. We can build it up by stages: 0 0, 1 0, 1, 2 . . . in such a way that any number n will appea
1. Program modules can import other modules. [You have seen this already in the Haskell course.] They can also depend indirectly on modules via some chain of importation, so that for instance M depend
Manchester Knock London Dubin Paris
Edinburgh
Rome Madrid
Dene the relation R+ by a R+ b if and only if there is a trip from a to b. Then clearly a R+ b if and only if there is some path from a to b i
The equivalence classes of a set A can be represented by a Venn diagram: for example,
A2
A1 A
A3 A5 A4
In this case, there are ve equivalence classes, illustrated by the ve disjoint subsets. In fact,
Examples 1. Given n N , the binary relation R on Z dened by a R b if and only if n divides into (b a) is an equivalence relation. 2. The binary relation S on the set Z N dened by (z 1 , n1 ) S (z2 , n
Relations with these properties occur naturally: the equality relation on sets is reexive, symmetric and transitive; the relations on numbers and on sets are reexive and transitive, but not symmetric;
This program is a well-known example in computer science, since the function computed by this program grows extremely rapidly. We wish to prove that this program always terminates, and therefore defin
well-founded. Even though a member such as (4, 3) has infinitely many other elements below it (for example, (3, y) for every y), any decreasing chain must be finite. Hence, the Ack program always give
CONNECTED COMPONENTS
Recall the definition of connectedness (2.45 in [1]). Definition 1. Let (X, d) be a metric space. A subset E X is called disconnected (or separated) if there exists nonempty A, B
roots, so it must be the constant zero polynomial, i.e. p(z ) = 0 for all z . Thus (z 0 ) (z n1 ) = z n 1, for all z . (b) This follows immediately be equating the degree n 1 coecients in the equation
we have
n
cos(j ) =
j =0
n j =0
= = =
1 ei/2 ei/2 ei(n+ 2 ) ei/2 ei/2 ei/2
1 ei(n+1) 1 ei
eij
=
ei/2 ei(n+ 2 ) ei/2 ei/2
1
i ei/2 ei(n+ 2 )
1
i ei/2 ei/2
Now we know
ei ei = sin(), 2i
so i ei
2
I.2
1. Note that the problem asks for all solutions. (c) If z = 4 1, then z 4 = 1, thus z 8 = 1, so we only need consider 8th roots of unity. Since z 4 = 1, this rules out 1, 1, i, i. A simple calcu
Thus the equation |z |2 2 (az ) + |a|2 ) + |a|2 = 2 , becomes (x u)2 + (y v )2 = 2 , which is the equation for a circle of radius centered at (u, v ), which in complex notation is the point a = u + iv
Homework 1 From Gamelin : Complex Analysis
January 21, 2007
Throughout this assignment we make use of the notation z = x + yi where x, y R, i.e. z = x, z = y .
1
1.
I.1
a. |z 1 i| = 1, so |(x 1) + i(y
5
Let us conclude these notes with a discussion of the most disconnected set possible the Cantor set. Example 20. Let C denote the Cantor set (cf. .244 in [1]). Recall that C may be described as the s
4
hand, E = A B, and so A B B; in particular, A B. But A B = , and this is only possible if A = , contra our assumptions. Whence, A1 = , and indeed, neither A1 nor B1 are empty. Therefore, E = A1 B1 i
3
Example 12. In Example 8, we saw that C1 = [0, 1] in the metric space [0, 1] (2, 3). Following Lemma 9, this means that for any point x [0, 1], Cx = [0, 1] (indeed, this is easy to show directly). O
2
Hence, what we have shown is: if G = A B is a separation of G, then each set U G lies entirely inside either A or B. But, by assumption, there is a non-empty common intersection for all sets U G say
CONNECTED COMPONENTS
Recall the definition of connectedness (2.45 in [1]). Definition 1. Let (X, d) be a metric space. A subset E X is called disconnected (or separated) if there exists nonempty A, B
name . Brown, B Smith, J . We have introduced three database operationsjoin, projection, selection and have seen how each operation has a counterpart in our formalism. More details about relational da
We might wish to modify Student Exam by hiding the number information to get a new relation Results. This can be done by the operation projection, to yield the following result: name . Brown, B Smith,
For example, let A = cfw_1, 3, 5, 7, 9 and B = cfw_3, 5, 6, 10, 11. Then A B = cfw_3, 5 A B = cfw_1, 3, 5, 6, 7, 9, 10, 11
A B = cfw_1, 7, 9 A
B = cfw_1, 7, 9, 6, 10, 11
It is often helpful to illustr
Approach (2) can be generalised to a general axiom of comprehension, which constructs sets by taking all elements of a set which satises some property P (x). In fact, you have seen a similar construct