text stringlengths 16 3.88k | source stringlengths 60 201 |
|---|---|
• Sentence
• A predicate symbol applied to zero or more terms:
on(a,b), sister(Jane, Joan), sister(Mother-of(John), Jane),
its-raining()
• t1=t2
• For v a variable and Φ a sentence, then
sentences.
v.Φ and
v.Φ are
∃
∀
Lecture 5 • 15
There are two more new constructs. If v is a variable and Phi is a sentence t... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/2e6d5a261d06ac45abdaa2de545b9205_Lecture5FinalPart1Save.pdf |
pretty
informal. In propositional logic, an interpretation is an assignment of truth
values to sentential variables. Now an interpretation's going to have to be
something more complicated. An interpretation is made up of a set and three
mappings.
17
FOL Interpretations
• Interpretation I
• U set of objects; dom... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/2e6d5a261d06ac45abdaa2de545b9205_Lecture5FinalPart1Save.pdf |
OL Interpretations
• Interpretation I
• U set of objects; domain of discourse; universe
• Maps constant symbols to elements of U
• Maps predicate symbols to relations on U (binary
relation is a set of pairs)
• Maps function symbols to functions on U
Lecture 5 • 21
The last mapping is from function symbols to fu... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/2e6d5a261d06ac45abdaa2de545b9205_Lecture5FinalPart1Save.pdf |
don’t worry about it.
Lecture 5 • 24
24
Basic FOL Semantics
Denotation of terms (naming)
• I(Fred)
• I(x)
• I(F(term))
if Fred is constant, then given
undefined
I(F)(I(term))
Lecture 5 • 25
The denotation of a complex term is defined recursively. So, to find the
interpretation of a function symbol applied ... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/2e6d5a261d06ac45abdaa2de545b9205_Lecture5FinalPart1Save.pdf |
to find out which
objects are named by each of the terms. Then, we look up the predicate symbol in
the interpretation, which gives us a mathematical relation on U. Finally, we look to
see if the tuple of objects named by the terms is a member of the relation. If so, the
sentence is true in the given interpretation.... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/2e6d5a261d06ac45abdaa2de545b9205_Lecture5FinalPart1Save.pdf |
find that it names this angry-looking guy.
Lecture 5 • 30
30
Basic FOL Semantics
Denotation of terms (naming)
• I(Fred)
• I(x)
• I(F(term))
if Fred is constant, then given
undefined
I(F)(I(term))
² I P(t1, …, tn) iff <I(t1), …, I(tn)>
brother(John, Joe)??
∈
I(P)
• I(John) =
• I(Joe) =
• I(brother) = {<... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/2e6d5a261d06ac45abdaa2de545b9205_Lecture5FinalPart1Save.pdf |
Semantics of Quantifiers
Extend an interpretation I to bind variable x to
element a
I
U: x/a
∈
Lecture 5 • 34
In order to talk about quantifiers we need the idea of extending an interpretation.
We would like to be able to extend an interpretation to bind variable X to value
A. We'll write that I with X bound to ... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/2e6d5a261d06ac45abdaa2de545b9205_Lecture5FinalPart1Save.pdf |
X such that Phi, it means that Phi has to be true
for some A in U. That is to say, there has to be something in the world such that if
we plug that in for X, then Phi becomes true.
36
Semantics of Quantifiers
Extend an interpretation I to bind variable x to
element a
• ² I ∀
• ² I ∃
U: x/a
I
x.Φ iff ² Ix/a Φ fo... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/2e6d5a261d06ac45abdaa2de545b9205_Lecture5FinalPart1Save.pdf |
• 40
There are four things in our U. Here they are.
40
FOL Example Domain
,
}
• U = {
• Constants: Fred
,
,
The Real
World
Lecture 5 • 41
We have one constant symbol, Fred.
41
FOL Example Domain
,
,
,
}
• U = {
• Constants: Fred
• Preds: above2, circle1, oval1, square1
The Real
World
Lecture 5 ... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/2e6d5a261d06ac45abdaa2de545b9205_Lecture5FinalPart1Save.pdf |
I(Fred) is
the triangle.
44
FOL Example Domain
,
,
,
}
• U = {
• Constants: Fred
• Preds: above2, circle1, oval1, square1
• Function: Hat
• I(Fred) =
• I(above) = {< , >,<
>}
,
The Real
World
Lecture 5 • 45
Now, what kind of a thing is I(above)? Well, above is a predicate symbol, and the
interpretat... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/2e6d5a261d06ac45abdaa2de545b9205_Lecture5FinalPart1Save.pdf |
• U = {
• Constants: Fred
• Preds: above2, circle1, oval1, square1
• Function: Hat
• I(Fred) =
• I(above) = {< , >,<
• I(circle) = {<
• I(oval) = {<
• I(Hat) = {<
>}
>,<
,
>}
,
>,<
>}
>}
,
The Real
World
Lecture 5 • 48
And we’ll say that the hat of the triangle is the square and the hat of the oval... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/2e6d5a261d06ac45abdaa2de545b9205_Lecture5FinalPart1Save.pdf |
. So the sentence is true.
50
FOL Example
• I(Fred) =
• I(above) = {< , >,< , >}
• I(circle) = {<
• I(oval) = {< >,< >}
• I(Hat) = {< , >,< , >}
• I(square) = {< >}
>}
• ² I square(Fred)?
• ² I above(Fred, Hat(Fred))?
What about this one? Does the above relation hold true of Fred and the hat of Fred?
Lectu... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/2e6d5a261d06ac45abdaa2de545b9205_Lecture5FinalPart1Save.pdf |
second. So hat(Fred) is a square.
53
FOL Example
• I(Fred) =
• I(above) = {< , >,< , >}
• I(circle) = {<
• I(oval) = {< >,< >}
• I(Hat) = {< , >,< , >}
• I(square) = {< >}
>}
• ² I square(Fred)?
• ² I above(Fred, Hat(Fred))?
• I(Hat(Fred)) =
• ² I above( ,
) ?
Lecture 5 • 54
Now the question is: does th... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/2e6d5a261d06ac45abdaa2de545b9205_Lecture5FinalPart1Save.pdf |
at(Fred)) =
• ² I above( ,
) ?
• ² I ∃
x. oval(x)?
Okay. What about this sentence: there exists an x such that oval x. Is there a thing
that is an oval? Yes. So how do we show that carefully?
Lecture 5 • 56
56
FOL Example
• I(Fred) =
• I(above) = {< , >,< , >}
• I(circle) = {<
• I(oval) = {< >,< >}
• I(Hat... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/2e6d5a261d06ac45abdaa2de545b9205_Lecture5FinalPart1Save.pdf |
in the same domain and interpretation. Is the
sentence: “For all x there exists a y such that either x is above y or y is above x”
true in I?
58
FOL Example: Continued
• I(Fred) =
• I(above) = {< , >,< , >}
• I(circle) = {<
• I(oval) = {< >,< >}
• I(Hat) = {< , >,< , >}
• I(square) = {< >}
>}
y. above(x,y) ... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/2e6d5a261d06ac45abdaa2de545b9205_Lecture5FinalPart1Save.pdf |
we bind y to the square, then that
makes above(y,x) true, which makes the disjunction true. So, we’ve proved this
existential statement is true. If we can do that for every other binding of x, then
the whole universal sentence is true.
You can verify that it is, in fact, true, by finding the truth value of the sent... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/2e6d5a261d06ac45abdaa2de545b9205_Lecture5FinalPart1Save.pdf |
y. …
• ² I ∀
x.
∀
• ² Ix/
y. above(x,y) v above(y,x)
above(x,y) v above(y,x)
, y/
Lecture 5 • 62
If it’s going to be true, then it has to be true for every possible instantiation of x and
y to elements of U. So, what, in particular, about the case when x is the square
and y is the circle?
62
FOL Example: C... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/2e6d5a261d06ac45abdaa2de545b9205_Lecture5FinalPart1Save.pdf |
² Ix/ ∃
• ² Ix/ , y/ above(x,y) v above(y,x)
y. …
• ² I ∀
x.
∀
• ² Ix/
y. above(x,y) v above(y,x)
above(x,y) v above(y,x)
, y/
And, therefore, neither is the universally quantified statement.
Lecture 5 • 64
64
Recitation Problems: I
For each of the following sentences, determine whether it is true
or fals... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/2e6d5a261d06ac45abdaa2de545b9205_Lecture5FinalPart1Save.pdf |
al(x)
•
∀
→
Lecture 5 • 68
For all x Cat(x) implies Mammal(x). This is saying that every individual in the
cat relation is also in the mammal relation. Or that cats are a subset of
mammals.
68
Writing FOL
• Cats are mammals
x. cat(x)
[
cat1, mammal1]
mammal(x)
•
∀
→
• Jane is a tall surveyor [tall1, surv... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/2e6d5a261d06ac45abdaa2de545b9205_Lecture5FinalPart1Save.pdf |
, mammal1]
mammal(x)
•
→
• Jane is a tall surveyor [tall1, surveyor1, Jane]
• tall(Jane) Æ surveyor(Jane)
• A nephew is a sibling’s son [nephew2, sibling2, son2]
z . [sibling(y,z) Æ son(x,z)]]
xy. [nephew(x,y)
•
↔ ∃
∀
∀
So, the answer is, for all x and y, x is the nephew of y if and only if there
exists a z ... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/2e6d5a261d06ac45abdaa2de545b9205_Lecture5FinalPart1Save.pdf |
, sibling2, son2]
z . [sibling(y,z) Æ son(x,z)]]
xy. [nephew(x,y)
•
• A maternal grandmother … [functions: mgm, mother-of]
z. x=mother-of(z) Æ z=mother-of(y)
xy. x=mgm(y)
•
↔ ∃
↔ ∃
∀
∀
∀
We can say that, for all x and y, x is the maternal grandmother of y if and
only if there exists a z such that x is the moth... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/2e6d5a261d06ac45abdaa2de545b9205_Lecture5FinalPart1Save.pdf |
nephew(x,y)
•
• A maternal grandmother … [functions: mgm, mother-of]
z. x=mother-of(z) Æ z=mother-of(y)
xy. x=mgm(y)
•
↔ ∃
↔ ∃
• Everybody loves somebody [loves2]
•
x.
y. loves(x,y)
∃
∀
∀
∀
∀
Lecture 5 • 76
This one’s fun, because there are really two answers. The usual answer is
for all x, there exists a y... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/2e6d5a261d06ac45abdaa2de545b9205_Lecture5FinalPart1Save.pdf |
Jane. Poor Jane. How can we say that?
Lecture 5 • 78
78
Writing More FOL
• Nobody loves Jane
loves(x,Jane)
x.
•
∀
¬
For all x, not loves(x, Jane). So, for everybody, every single person, that
person doesn't love Jane.
Lecture 5 • 79
79
Writing More FOL
• Nobody loves Jane
loves(x,Jane)
x. loves(x,Jane)... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/2e6d5a261d06ac45abdaa2de545b9205_Lecture5FinalPart1Save.pdf |
)
x. loves(x,Jane)
•
•
x.
¬
∀
¬∃
• Everybody has a father
•
x.
∀
y. father(y,x)
∃
• Everybody has a father and a mother
•
x.
∀
∃
yz. father(y,x) Æ mother(z,x)
Forall x Exists yz. F(y,x) and M(z,x)
Lecture 5 • 84
84
Writing More FOL
• Nobody loves Jane
loves(x,Jane)
x. loves(x,Jane)
•
•
x.
¬
... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/2e6d5a261d06ac45abdaa2de545b9205_Lecture5FinalPart1Save.pdf |
• Nobody loves Jane
loves(x,Jane)
x. loves(x,Jane)
•
•
x.
¬
∀
¬∃
• Everybody has a father
•
x.
∀
y. father(y,x)
∃
• Everybody has a father and a mother
•
x.
∀
∃
yz. father(y,x) Æ mother(z,x)
• Whoever has a father, has a mother
y. father(y,x)]
• [
∃
Now, how do we describe x’s that have a father? ... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/2e6d5a261d06ac45abdaa2de545b9205_Lecture5FinalPart1Save.pdf |
�
• Everybody has a father
•
x.
∀
y. father(y,x)
∃
• Everybody has a father and a mother
•
x.
∀
∃
yz. father(y,x) Æ mother(z,x)
• Whoever has a father, has a mother
x.[[
•
∀
∃
y. father(y,x)]
→
[
∃
y. mother(y,x)]]
Lecture 5 • 89
Note that the two variables named y have separate scopes, and are ent... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/2e6d5a261d06ac45abdaa2de545b9205_Lecture5FinalPart1Save.pdf |
Topic 5 Notes
Jeremy Orloff
5 Introduction to harmonic functions
5.1 Introduction
Harmonic functions appear regularly and play a fundamental role in math, physics and engineering.
In this topic we’ll learn the definition, some key properties and their tight connection to complex
analysis. The key connection to 18.04 i... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/2e739bb156efb0bc7103fc43d0897dda_MIT18_04S18_topic5.pdf |
very strong. In many respects it mirrors
the connection between e and sine and cosine.
Let = + and write () = (, ) + (, ).
Theorem 5.2. If () = (, ) + (, ) is analytic on a region then both and are harmonic
functions on .
Proof. This is a simple consequence of the Cauchy-Riemann equations. Since = we have
Likewise... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/2e739bb156efb0bc7103fc43d0897dda_MIT18_04S18_topic5.pdf |
we know = −, so = . It is clear that = −. Thus
satisfies the Cauchy-Riemann equations, so it is analytic.
3. Let be an antiderivative of :
5
INTRODUCTION TO HARMONIC FUNCTIONS
3
Since is simply connected our statement of Cauchy’s theorem guarantees that () has an
antiderivative in . We’ll need to fus... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/2e739bb156efb0bc7103fc43d0897dda_MIT18_04S18_topic5.pdf |
() = + is analytic then so is () = − + . So, if and are harmonic conjugates
and so are and −.
5.4 A second proof that and are harmonic
This fact is important enough that we will give a second proof using Cauchy’s integral formula.
One benefit of this proof is that it reminds us that Cauchy’s integral formula can tran... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/2e739bb156efb0bc7103fc43d0897dda_MIT18_04S18_topic5.pdf |
corresponding properties of analytic functions. Indeed, we deduce them
from those corresponding properties.
Theorem. (Mean value property) If is a harmonic function then satisfies the mean value property.
That is, suppose is harmonic on and inside a circle of radius centered at
0 =
0 +
0 then
(
,
0
0) =
2
1
2... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/2e739bb156efb0bc7103fc43d0897dda_MIT18_04S18_topic5.pdf |
Then the dot
product of their gradients is 0, i.e.
⋅ = 0.
Proof. The proof is an easy application of the Cauchy-Riemann equations.
⋅ = (, ) ⋅ (, ) = + = − = 0
In the last step we used the Cauchy-Riemann equations to substitute for and − for . □
The lemma holds whether or not the gradients are 0. To guarantee th... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/2e739bb156efb0bc7103fc43d0897dda_MIT18_04S18_topic5.pdf |
ONIC FUNCTIONS
6
5
INTRODUCTION TO HARMONIC FUNCTIONS
7
Example 5.6. Let’s work out the gradients in a few simple examples.
(i) Let
So
() = 2 = (2 − 2) + 2,
= (2, −2)
and = (2, 2).
It’s trivial to check that ⋅ = 0, so they are orthogonal.
5
INTRODUCTION TO HARMONIC FUNCTIONS
8
(ii) Let
So, it’s easy... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/2e739bb156efb0bc7103fc43d0897dda_MIT18_04S18_topic5.pdf |
0 to intersect orthogonally
takes the average and comes into the origin at 45(cid:253).
MIT OpenCourseWare
https://ocw.mit.edu
18.04 Complex Variables with Applications
Spring 2018
For information about citing these materials or our Terms of Use, visit: https://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/2e739bb156efb0bc7103fc43d0897dda_MIT18_04S18_topic5.pdf |
software
studio
asynchronous calls
Daniel Jackson
1 some history
in the 1990s
› most web sites issued a whole page at a time
› clunky for users, excessive bandwidth
idea
› update page incrementally
› do it asynchronously, so browser doesn’t freeze
in 1999, XMLHttpRequest arrives
› Microsoft invents XHR ide... | https://ocw.mit.edu/courses/6-170-software-studio-spring-2013/2e772463eebb035f1b5d38accee189ed_MIT6_170S13_47-asyn-intro.pdf |
8 encoding data for transit
XML
› parsing built into browser (XHR)
› comes back as DOM: not convenient
JSON
› Javascript object literals
› JQuery uses parser, not eval (why?)
<person>
<firstName>John</firstName>
<lastName>Smith</lastName>
<age>25</age>
<address>
<streetAddress>21 2nd
Street</streetAddress>
<c... | https://ocw.mit.edu/courses/6-170-software-studio-spring-2013/2e772463eebb035f1b5d38accee189ed_MIT6_170S13_47-asyn-intro.pdf |
18.409 An Algorithmist’s Toolkit
September 10, 2009
Lecturer: Jonathan Kelner
Scribe: Jesse Geneson (2009)
Lecture 1
1 Overview
The class’s goals, requirements, and policies were introduced, and topics in the class were described. Every
thing in the overview should be in the course syllabus, so please consult th... | https://ocw.mit.edu/courses/18-409-topics-in-theoretical-computer-science-an-algorithmists-toolkit-fall-2009/2e812afd941cb290f887e3a0e53e51df_MIT18_409F09_scribe1.pdf |
=
. . . , vn, and Λ is diagonal, with the
�
n
i=1 λivivT
i .
In Proposition 2, it was important that M was symmetric. No results stated there are necessarily true
in the case that M is not symmetric.
Definition 3 We call the span of the eigenvectors with the same eigenvalue an eigenspace.
3 Matrices for Graphs
D... | https://ocw.mit.edu/courses/18-409-topics-in-theoretical-computer-science-an-algorithmists-toolkit-fall-2009/2e812afd941cb290f887e3a0e53e51df_MIT18_409F09_scribe1.pdf |
definition for the Laplacian
matrix is
LG = DG − AG,
where DG is the diagonal matrix with ith diagonal entry equal to the degree of vi, and AG is the adjacency
matrix.
4
Example Laplacians
Consider the graph H with adjacency matrix
AH =
⎛
⎜
⎜
⎜
⎜
⎝
0
1
0
1
0
1
0
1
0
0
0
1
0
1
1
1
0
1
0
0
⎞
⎟
⎟
⎟
⎟
⎠
0
0
1 ... | https://ocw.mit.edu/courses/18-409-topics-in-theoretical-computer-science-an-algorithmists-toolkit-fall-2009/2e812afd941cb290f887e3a0e53e51df_MIT18_409F09_scribe1.pdf |
⎠
X(1)
X(3)
The action of LG on v is then
1
LGv = −1
⎛
⎝
0
−1
2 −1
1
0 −1
⎞ ⎛
⎠ ⎝
X(1)
X(2)
X(3)
⎞ ⎛
⎠ = ⎝
X(1) − X(2)
2X(2) − X(1) − X(3)
X(3) − X(2)
⎞ ⎛
⎠ = ⎜
⎝ 2
⎞
� X(1) − X(2)
�
X(2) − [ X(1)+X(3) ⎟
] ⎠
2
X(3) − X(2)
For a general Laplacian, we will have
[LGv]i = [di ∗ (X(i) − average o... | https://ocw.mit.edu/courses/18-409-topics-in-theoretical-computer-science-an-algorithmists-toolkit-fall-2009/2e812afd941cb290f887e3a0e53e51df_MIT18_409F09_scribe1.pdf |
envectors.
5
Matlab Demonstration
As remarked before, vectors v ∈ Rn may be construed as maps Xv : V → R. Thus each eigenvector assigns a
real number to each vertex in G. A point in the plane is a pair of real numbers, so we can embed a connected
R2 . The following examples generated in Matlab show that
graph int... | https://ocw.mit.edu/courses/18-409-topics-in-theoretical-computer-science-an-algorithmists-toolkit-fall-2009/2e812afd941cb290f887e3a0e53e51df_MIT18_409F09_scribe1.pdf |
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18.917 Topics in Algebraic Topology: The Sullivan Conjecture
Fall 2007
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
Generating Analytic Functors (Lecture 10)
Let Funan denote the category of analytic functors from Vectf to... | https://ocw.mit.edu/courses/18-917-topics-in-algebraic-topology-the-sullivan-conjecture-fall-2007/2ea8b21ce4498d757a261c7123784662_lecture10.pdf |
we can write F as a filtered colimit of subfunctors
FI0 = im(⊕i∈I0 PVi → F ) ⊆ F.
⊕i∈I PVi → F
where I0 ranges over finite subsets of I. Since the collection of good functors is stable under colimits, it
will suffice to show that each FI0 is a good functor. Proposition 2 is now an immediate consequence of the
followin... | https://ocw.mit.edu/courses/18-917-topics-in-algebraic-topology-the-sullivan-conjecture-fall-2007/2ea8b21ce4498d757a261c7123784662_lecture10.pdf |
⊕α∈A Symnα
where the set A is finite.
For each i ∈ I, let Fi denote the image of F in IVi . Then Fi is a quotient of F , and therefore a polynomial
functor. Moreover, we have an inclusion F �→ ⊕i∈I Fi. It will therefore suffice to prove Proposition 4 after
replacing F by each Fi; in other words, we may suppose that I ... | https://ocw.mit.edu/courses/18-917-topics-in-algebraic-topology-the-sullivan-conjecture-fall-2007/2ea8b21ce4498d757a261c7123784662_lecture10.pdf |
J Fj , and it will suffice to prove the result after applying
F by Fj . In other words, we may reformulate Proposition 5 as follows:
Proposition 6. Let F be a polynomial functor, and suppose there exists an injection
Then there exists an injection
for some m ≥ 0.
F �→ S∞ ⊗ . . . ⊗ S∞.
F � Symm
→
We observe that t... | https://ocw.mit.edu/courses/18-917-topics-in-algebraic-topology-the-sullivan-conjecture-fall-2007/2ea8b21ce4498d757a261c7123784662_lecture10.pdf |
(0) + . . . + dF (m).
We can now Proposition 6 to the following:
Proposition 8. Let F be a functor of the form (Sk)⊗n, where k ≥ 1 and n ≥ 0. Then there exists a
monomorphism F �→ Symm for some m ≥ 0.
Proposition 8 is obvious if n = 0 (in that case, F � Sym0 and we can take m = 0). The main difficulty
is in the case... | https://ocw.mit.edu/courses/18-917-topics-in-algebraic-topology-the-sullivan-conjecture-fall-2007/2ea8b21ce4498d757a261c7123784662_lecture10.pdf |
→
quotient of
(1) If σ ∈ Σd is a permutation, then
v1 ⊗ . . . ⊗ vd = vσ(1) ⊗ . . . ⊗ vσ(d)
in Sk+1(V ).
(2) If d < k, and v ∈ V , then
in Sk+1(V ).
v1 ⊗ . . . ⊗ vd ⊗ v = v1 ⊗ . . . ⊗ vd ⊗ v ⊗ v
We define a map θ : ⊕1≤≤k+1V ⊗d → Sym2k
(V ) by the formula
�
θ(v1 ⊗ . . . ⊗ vd) =
v 2i1 . . . v 2id ,
1
d
2i1 +...+... | https://ocw.mit.edu/courses/18-917-topics-in-algebraic-topology-the-sullivan-conjecture-fall-2007/2ea8b21ce4498d757a261c7123784662_lecture10.pdf |
. , id, j + 1). To complete the proof, it will suffice to show that no other
terms appear on the left hand side. In other words, we must show that if 2i1 + . . . + 2id + 2j = 2k, then
j > 0. If not, we have 2i1 + . . . + 2id = 2k − 1, which has k nonzero digits in its base 2-expansion. Since each
term in the sum is a ... | https://ocw.mit.edu/courses/18-917-topics-in-algebraic-topology-the-sullivan-conjecture-fall-2007/2ea8b21ce4498d757a261c7123784662_lecture10.pdf |
��eld F2. The endomorphism ring S∞ is properly contained in R, and therefore has
dimension 1 over F2. It follows that every nonzero endomorphism of S∞ is an isomorphism.
Suppose F ∩ F � = 0. Then the induced map F → S∞/F � is a monomorphism. Since S∞ is injective, we
can solve the lifting problem depicted in the dia... | https://ocw.mit.edu/courses/18-917-topics-in-algebraic-topology-the-sullivan-conjecture-fall-2007/2ea8b21ce4498d757a261c7123784662_lecture10.pdf |
18.413: ErrorCorrecting Codes Lab
March 4, 2004
Lecturer: Daniel A. Spielman
Lecture 9
9.1 Related Reading
• Fan, Chapter 2, Sections 3 and 4.
• Programming Tips #10: Working with GF (2) matrices.
9.2 LDPC Codes
We will examine the simplest and most natural lowdensity paritycheck (LDPC) codes. These are
spe... | https://ocw.mit.edu/courses/18-413-error-correcting-codes-laboratory-spring-2004/2eb3152fd9bef9307e97d393de620d1b_lect9.pdf |
which each bit node has degree 3 and each check node has degree 6. To choose a
random graph with these parameters, consider the sockets on each side, where a socket is a place
where an edge is attached. So, each bit node has 3 sockets and each check node has 6. You can
form a random graph by matching up the bit node... | https://ocw.mit.edu/courses/18-413-error-correcting-codes-laboratory-spring-2004/2eb3152fd9bef9307e97d393de620d1b_lect9.pdf |
code)
• Each check node is called a parity constraint
• There is a variable on each edge.
At the beginning of the algorithm, each of the interal edges (those with two endpoints) is initialized
to representing equal probabilities of 0 and 1, and the variables on the external edges (those with
degree one attached to... | https://ocw.mit.edu/courses/18-413-error-correcting-codes-laboratory-spring-2004/2eb3152fd9bef9307e97d393de620d1b_lect9.pdf |
them
intrisic instead of prior because they might not be the actual priors.
By the rule above, during each equality phase, a message is sent to the oneendpoint edge attached
to each bit node. This message is almost what the output of the decoder should be. The defect is
that this message is an extrinsic probabilit... | https://ocw.mit.edu/courses/18-413-error-correcting-codes-laboratory-spring-2004/2eb3152fd9bef9307e97d393de620d1b_lect9.pdf |
Eb/N0, and if we are using ±1 signalling, it is
defined to be
Eb
N0
where R is the rate of the code you are using.
=
1
2Rσ2 ,
To complicate things further, SNR is usually recorded in decibels. That is 10 log10(Eb/N0) dB. So,
if I ask you for 1.5dB, that means you should set Eb/N0 so that 10 log10(Eb/N0) = 1.5. ... | https://ocw.mit.edu/courses/18-413-error-correcting-codes-laboratory-spring-2004/2eb3152fd9bef9307e97d393de620d1b_lect9.pdf |
Design of an ESD
Design of an ESD
Core Methodology
Core Methodology
Subject
Subject
February 7, 2007
Dick Larson, Dan Frey, with Roy Welsch
Engineering Systems: At the intersection
of
Engineering, Management & Social Sciences
Management
Social
Sciences
ESD
Engineering
2
For the new Methods
subject
We want to ... | https://ocw.mit.edu/courses/esd-86-models-data-and-inference-for-socio-technical-systems-spring-2007/2ebe106407339f82b451d497f44c9098_lec1_intro.pdf |
This is a
Knowledge
Requirement
> For students whose academic plan is to take MIT
subjects that go much deeper than this subject (in
statistics, probability, quantitative research methods),
the requirement for taking this subject can be
waived.
> This subject represents a 'knowledge requirement' that
will be ass... | https://ocw.mit.edu/courses/esd-86-models-data-and-inference-for-socio-technical-systems-spring-2007/2ebe106407339f82b451d497f44c9098_lec1_intro.pdf |
Deep,
Use all Available
Subjects
> We cannot think that ESD is so unique that no other MIT subjects
can contribute to ESD students' knowledge of 'methods.' In the
course of an ESD doctoral student's studies, she/he will rely most
often on existing subjects at MIT or perhaps Harvard to go deep
in the required metho... | https://ocw.mit.edu/courses/esd-86-models-data-and-inference-for-socio-technical-systems-spring-2007/2ebe106407339f82b451d497f44c9098_lec1_intro.pdf |
2
3
:
n
1n+1
2n+1
3,n+1
4,n+1
n+1
n+1,0
Figure by MIT OCW.
http://www.mathpages.com/home/kmath336/kmath336_files/image001.gif
16
A Real Null Hypothesis: A Sports ‘.500’ Team
> Each game is essentially decided by an independent flip of a
fair coin
> Track the media coverage as certain expected ‘streaks’
during the yea... | https://ocw.mit.edu/courses/esd-86-models-data-and-inference-for-socio-technical-systems-spring-2007/2ebe106407339f82b451d497f44c9098_lec1_intro.pdf |
Lecture 5
8.321 Quantum Theory I, Fall 2017
22
Lecture 5 (Sep. 20, 2017)
5.1 The Position Operator
In the last class, we talked about operators with a continuous spectrum. A prime example is the
position operator. Let’s first consider a particle in d = 1. We define x as the position operator,
with corresponding eigenstat... | https://ocw.mit.edu/courses/8-321-quantum-theory-i-fall-2017/2ecc07c095d8ae56600f7145c6fb4a6a_MIT8_321F17_lec5.pdf |
this means that an initial state
2
ˆ
is sent by the measurement to a state
|ψ(cid:105) =
|ψ(cid:105) → |ψ (cid:105) =
(cid:48)
∞ dx(cid:48) |x(cid:48)(cid:105)(cid:104)x(cid:48)|ψ(cid:105)
−∞
ˆ
x(cid:48)+
∆
2
x(cid:48)
∆
2
−
dx(cid:48)(cid:48) |x(cid:48)(cid:48)(cid:105)(cid:104)x(cid:48)(cid:48)
|ψ(cid:105) .
(5.4)
(5... | https://ocw.mit.edu/courses/8-321-quantum-theory-i-fall-2017/2ecc07c095d8ae56600f7145c6fb4a6a_MIT8_321F17_lec5.pdf |
(5.9)
Lecture 5
8.321 Quantum Theory I, Fall 2017
23
5.1.2 Hilbert Spaces
We now make a brief aside about the actual definition of a Hilbert space. We generalize our notion
of a vector space to the infinite-dimensional case. Consider any sequence of states in the space
that has the property that for any (cid:15) > 0, if... | https://ocw.mit.edu/courses/8-321-quantum-theory-i-fall-2017/2ecc07c095d8ae56600f7145c6fb4a6a_MIT8_321F17_lec5.pdf |
position basis, the momentum operator is given by the familiar expression
pi = −i(cid:126)
∂
∂xi
.
On the space of square-integrable functions ψ(x), this operator acts as
From this, we see that
pψ(x) = −i(cid:126)
dψ
dx
.
[x, p]ψ = −i(cid:126)
x
(cid:18)
dψ
dx
−
d
dx
(cid:19)
(xψ)
= i(cid:126)ψ(x) ,
from which we concl... | https://ocw.mit.edu/courses/8-321-quantum-theory-i-fall-2017/2ecc07c095d8ae56600f7145c6fb4a6a_MIT8_321F17_lec5.pdf |
|ψ(cid:105) =
ˆ
∞ dp(cid:48)
−∞
(cid:12)
(cid:12)p(cid:48)
(cid:12)
(cid:11)(cid:10)p(cid:48) ψ .
(cid:11)
(cid:12)
(5.18)
(5.19)
(5.20)
(5.21)
Analogously to the discussion of measurement of position,
measurements of momentum will
have
Prob(cid:0)momentum is between p(cid:48) and p(cid:48) + dp(cid:48)
(cid:1) = (cid:... | https://ocw.mit.edu/courses/8-321-quantum-theory-i-fall-2017/2ecc07c095d8ae56600f7145c6fb4a6a_MIT8_321F17_lec5.pdf |
11)
(cid:10)
(cid:11)
.
(5.25)
for the momentum operator
The expression on the left-hand side is
in the position basis,since (cid:104)x(cid:48)|p(cid:48)(cid:105) is the position-space wavefunction for the state |p(cid:48)(cid:105), while the
expression on the right-hand side is found by having the momentum operator ac... | https://ocw.mit.edu/courses/8-321-quantum-theory-i-fall-2017/2ecc07c095d8ae56600f7145c6fb4a6a_MIT8_321F17_lec5.pdf |
equal the righ
t-hand side,
(cid:126)
(cid:48)(cid:48)
x )/ = 2π(cid:126)|N |2δ
(cid:0)x(cid:48) − x(cid:48)(cid:48)
(cid:1) ,
(5.29)
2π(cid:126)|N |2δ(cid:0)x(cid:48) − x(cid:48)(cid:48)
(cid:1) = δ(cid:0)x(cid:48) − x(cid:48)(cid:48)
(cid:1) .
(5.30)
This only fixes the modulus of N , but by convention
we have
we choo... | https://ocw.mit.edu/courses/8-321-quantum-theory-i-fall-2017/2ecc07c095d8ae56600f7145c6fb4a6a_MIT8_321F17_lec5.pdf |
:104)p |ψ(cid:105) =
(cid:48)
ˆ
√
dx
(cid:48)
2π(cid:126)
(cid:48)
e ip(cid:48)x /(cid:126)(cid:104)x(cid:48)|ψ(cid:105) .
−
5.3 Normalization of Position and Momentum Eigenstates
What is the wavefunction corresponding to the position eigenstate |x(cid:48)(cid:105)? We have
(cid:10)x(cid:48)(cid:48)
(cid:11) = δ(cid:0)... | https://ocw.mit.edu/courses/8-321-quantum-theory-i-fall-2017/2ecc07c095d8ae56600f7145c6fb4a6a_MIT8_321F17_lec5.pdf |
momentum eigenstates. For finite L, we
require that
ˆ
0
L
dx(cid:48)
(cid:12)
(cid:12)
(cid:10)
x(cid:48)
(cid:12)
(cid:12)p(cid:48)
(cid:11)(cid:12)
2
(cid:12) = 1 ,
which normalizes the momentum eigenstates:
(cid:10)x(cid:48)
(cid:12)
(cid:12)p(cid:48)
(cid:11)
1
= √
L
(cid:48)
eip x(cid:48)/(cid:126) .
(5.38)
(5.39)
... | https://ocw.mit.edu/courses/8-321-quantum-theory-i-fall-2017/2ecc07c095d8ae56600f7145c6fb4a6a_MIT8_321F17_lec5.pdf |
(cid:126)
iap/ =
(cid:88)∞ (−ia/(cid:126))npn
n=0
n!
.
(5.43)
Note that this definition can only make sense when acting on a Hilbert space.
We can see that this operator satisfies the following properties:
1. T −
1(a) = T (−a)
2. T (a(cid:48))T (a(cid:48)(cid:48)) = T (a(cid:48) + a(cid:48)(cid:48))
3. T (a)xT −
1(a) = x... | https://ocw.mit.edu/courses/8-321-quantum-theory-i-fall-2017/2ecc07c095d8ae56600f7145c6fb4a6a_MIT8_321F17_lec5.pdf |
)x(cid:48)
(cid:12)
(cid:11) .
By applying T −
1(a) to each side of this equation, we find
xT −
(cid:0)
1(a)|x(cid:48)(cid:105) = x(cid:48) − a T −
(cid:1)
(cid:12)
1(a) x(cid:48)
(cid:12)
(cid:11)
,
from which we conclude that
T −
(cid:12)
(cid:11)
1(a) x(cid:48) ∝ x(cid:48) − a .
(cid:12)
(cid:12)
(cid:12)
(cid:11)
(5... | https://ocw.mit.edu/courses/8-321-quantum-theory-i-fall-2017/2ecc07c095d8ae56600f7145c6fb4a6a_MIT8_321F17_lec5.pdf |
this story around, by first defining the unitary translation operator and then using it to
define a Hermitian operator p. In the classical limit, the operator p found in this way has the right
properties to be called a momentum. Note that any unitary operator U can be written as eiA for
some Hermitian operator A. This is ... | https://ocw.mit.edu/courses/8-321-quantum-theory-i-fall-2017/2ecc07c095d8ae56600f7145c6fb4a6a_MIT8_321F17_lec5.pdf |
Lecture 7
8.251 Spring 2007
Lecture 7 - Topics
• Area formula for spacial surfaces
Area formula for spatial surfaces
(“spatial” as opposed to “space-time”)
Consider 2D surface in 3D space
3D Space
�x = (x 1 , x 2 , x 3)
Parameter Space: ξ1 , ξ2 (directions along grid lines. Purely arbitrary. No con
nection to... | https://ocw.mit.edu/courses/8-251-string-theory-for-undergraduates-spring-2007/2ed11482f84c747194ee6a8bf3a226ce_lec7.pdf |
�
−
∂�x
∂ξ1
�2
∂�x
∂ξ2
·
2
Lecture 7
8.251 Spring 2007
�
A =
dA
Important that this formula is reparameterization-invariant.
Reparam. Invariance
Choose another coordinate par. (ξ�1 , ξ�2). Can write as functions of our (ξ1, ξ2)
coordinates. Must have:
dξ�1dξ�2
��
∂�x
∂ξ�1
·
∂�x
∂ξ�2
��
∂�x
∂... | https://ocw.mit.edu/courses/8-251-string-theory-for-undergraduates-spring-2007/2ed11482f84c747194ee6a8bf3a226ce_lec7.pdf |
∂�x
∂�x
· ∂ξ1
∂ξ1
∂�x
∂�x
∂ξ1 · ∂ξ2
∂�x
∂�x
· ∂ξ1
∂ξ2
∂�x
∂�x
∂ξ2 · ∂ξ2
3
Lecture 7
8.251 Spring 2007
�
A =
dξ1dξ2√
g
where g = det(gij )
4 | https://ocw.mit.edu/courses/8-251-string-theory-for-undergraduates-spring-2007/2ed11482f84c747194ee6a8bf3a226ce_lec7.pdf |
18.034, Honors Differential Equations
Prof. Jason Starr
Lecture 5
2/13/04
1. Quickly reviewed the proof of existence/uniqueness on a small interval, [t0, t0+C]
2. Explained how to do the same for [t0-c, t0], and then patch the 2 solutions. Checked the
solution is diff. at to.
3. Explained how uniqueness on small... | https://ocw.mit.edu/courses/18-034-honors-differential-equations-spring-2004/2ee7d40005a4ab250cea7594569c523c_lec5.pdf |
Lecture 5
Quantum Mechanical Systems and Measurements
Today’s Program:
1. Wavefunctions in QM
2. Schrodinger’s Equation
3. Representing physical quantities: Observables
4. Hermitian operators
5. Principle of spectral decomposition.
6. Predicting the results of measurements, fourth postulate discrete and continu... | https://ocw.mit.edu/courses/3-024-electronic-optical-and-magnetic-properties-of-materials-spring-2013/2eeeaab73afed1cd3969c79dd22c2a4f_MIT3_024S13_2012lec5.pdf |
Now let’s consider the simplest wavefunction for a particle. In a free space such as vacuum with
no external forces the particle can be approximated as a plane wave: eikxit , where k
p
is the
wavevector. Since there is no external forces then the energy of a particle is its kinetic energy:
E
mv2
2
2
p... | https://ocw.mit.edu/courses/3-024-electronic-optical-and-magnetic-properties-of-materials-spring-2013/2eeeaab73afed1cd3969c79dd22c2a4f_MIT3_024S13_2012lec5.pdf |
e
t
ikxit
e
This exercise results in a curious statement about the propagation of particle ~ plane wave in free
space:
ikxit
e
2 2
2m x 2
ikxit i
e
t
In general in the presence of forces in 3D space energy of a particle would be:
2
E
V x, t , which would yield a foll... | https://ocw.mit.edu/courses/3-024-electronic-optical-and-magnetic-properties-of-materials-spring-2013/2eeeaab73afed1cd3969c79dd22c2a4f_MIT3_024S13_2012lec5.pdf |
inger’s equation. During our derivation of the equation we
have encountered an intermediate landmark equation:
i
e
x
ikxit peikxit
This equation is analogous to an eigenvalue problem: A . Here operator such as a matrix
A acts on it’s eigenfunction or an eigenvector , and is the corresponding ... | https://ocw.mit.edu/courses/3-024-electronic-optical-and-magnetic-properties-of-materials-spring-2013/2eeeaab73afed1cd3969c79dd22c2a4f_MIT3_024S13_2012lec5.pdf |
Properties of Hermitian operators:
1. Aˆ Aˆ (For for matrix operators
A
for function operators Aˆd 3r Aˆ *d 3r
A or
Aˆ *
ˆ
T
ji Aij ,
2. Eigenvalues are real: Aˆi ii, i
3. Eigenvectors belonging to different eigenvalues are orthogonal.
i r
i d 3
r
i ... | https://ocw.mit.edu/courses/3-024-electronic-optical-and-magnetic-properties-of-materials-spring-2013/2eeeaab73afed1cd3969c79dd22c2a4f_MIT3_024S13_2012lec5.pdf |
how to solve mechanical problems using Hamiltonian
approach. Remembering the general form of a Hamiltonian and using our recipe for constructing
of observables we can get a Hamiltonian operator:
H
p2
2m
V
r, t
Hˆ
2
1
2m
2 V r, t
2
2 V r, t
2m
Now if we look ... | https://ocw.mit.edu/courses/3-024-electronic-optical-and-magnetic-properties-of-materials-spring-2013/2eeeaab73afed1cd3969c79dd22c2a4f_MIT3_024S13_2012lec5.pdf |
��
pˆ
r,
2m
Then the Schrodinger’s equation has the following form:
4
2m
2
V
r
r, t
i
r, t
t
This type of differential equation is separable, i.e. we can look for a solution in the following
r, t ... | https://ocw.mit.edu/courses/3-024-electronic-optical-and-magnetic-properties-of-materials-spring-2013/2eeeaab73afed1cd3969c79dd22c2a4f_MIT3_024S13_2012lec5.pdf |
2m
2
V
2m
r r
r
r
r i
t
t
1
t
i
t
t
r and the right side of the
Note that the left side of the equation only depends on position
equation only depends on time t. This can only be true when both sides of t... | https://ocw.mit.edu/courses/3-024-electronic-optical-and-magnetic-properties-of-materials-spring-2013/2eeeaab73afed1cd3969c79dd22c2a4f_MIT3_024S13_2012lec5.pdf |
�r
r
The equation (I) has a solution in a form: t e
i
E
t
.
The equation (II) is an eigenvalue/eigenfunction problem for the Hamiltonian:
Hˆ
r E
r
From the classical Hamiltonian mechanics we remember constructing Hamiltonian based on the
sum of potential and kinetic energy of the s... | https://ocw.mit.edu/courses/3-024-electronic-optical-and-magnetic-properties-of-materials-spring-2013/2eeeaab73afed1cd3969c79dd22c2a4f_MIT3_024S13_2012lec5.pdf |
t
, where CE are the coefficients that can be determined from the initial
E
and boundary conditions.
This is a very important result: If we know the special wavefunctions (Hamiltonian
eigenfunctions) we can easily find time evolution of this conservative system.
If we know E and E r
r
then we know ... | https://ocw.mit.edu/courses/3-024-electronic-optical-and-magnetic-properties-of-materials-spring-2013/2eeeaab73afed1cd3969c79dd22c2a4f_MIT3_024S13_2012lec5.pdf |
d
2
or x
d
2
-d/2
d/2
x
III Obtaining the QM Hamiltonian operator:
d
ˆ p
H x, p Hˆ x, ˆ
pˆ 2
2m
Vˆ xˆ
2
2
2m x 2
V x
For now let us restrict our attention to the inside of the well, where potential is zero (V(x)=0).
Since the Hamiltonian does not depend on time, all we need ... | https://ocw.mit.edu/courses/3-024-electronic-optical-and-magnetic-properties-of-materials-spring-2013/2eeeaab73afed1cd3969c79dd22c2a4f_MIT3_024S13_2012lec5.pdf |
determining E does not determine a and
b). We will narrow down this set by using boundary conditions derived from physical insights
into our problem. Specifically, we will require that our eigenfunctions will be equal to 0 at the
boundary of the well.
The problem of finding the eigenfunctions and eigenvalues of a l... | https://ocw.mit.edu/courses/3-024-electronic-optical-and-magnetic-properties-of-materials-spring-2013/2eeeaab73afed1cd3969c79dd22c2a4f_MIT3_024S13_2012lec5.pdf |
n 1,3,5... or a b,
kd
n
2
2
, n 2,4,6...
Then the solution has the following form:
uE x
n odd
u
n x
n even dn sin knx d sin
c cos k x ccos
n
n
n x
d
n
x
d
, kn
2mEn
2
n
d
Let’s check that un(x) are actually indeed the solutions of to the eigenfunction ... | https://ocw.mit.edu/courses/3-024-electronic-optical-and-magnetic-properties-of-materials-spring-2013/2eeeaab73afed1cd3969c79dd22c2a4f_MIT3_024S13_2012lec5.pdf |
�� 2
n
2md 2
, n 1,2,3,4,5...
V How can we find the coefficients cn and dn: normalization!
Since the probability of finding a particle in any state cannot exceed 1, then it is convenient to
normalize the eigenfunctions to 1.
2 2k 1 x
d
cos 2k 1 z dz 1, z c
2 d
dx 1
x
d
cn
2 c... | https://ocw.mit.edu/courses/3-024-electronic-optical-and-magnetic-properties-of-materials-spring-2013/2eeeaab73afed1cd3969c79dd22c2a4f_MIT3_024S13_2012lec5.pdf |
2
d
sin k n x
2
d
2
d
cos
n x
d
sin
n x
d
Hamiltonian eigenvalues
Ek
2k 2
2m
E n
2k 2
2m
, k
n
d
Important points:
1. Discrete vs. continuous spectrum:
Unlike for the free particle, which can have any energy value, the spectrum of allowed
energy states for the particle in the i... | https://ocw.mit.edu/courses/3-024-electronic-optical-and-magnetic-properties-of-materials-spring-2013/2eeeaab73afed1cd3969c79dd22c2a4f_MIT3_024S13_2012lec5.pdf |
. Used with permission.
How many nodes (zeros) will eigenfunction have? For the first eigenvalue:
E1
E1
E1
2
2
2
2md 2 ,
4 2
2md 2 ,
9 2
2md 2 ,
2
u1 x
cos
x
d
has 0 nodes between the walls of the well.
u2 x
sin
2 x
d
u3 x
cos
3 x
d
has 1 node between the walls of the ... | https://ocw.mit.edu/courses/3-024-electronic-optical-and-magnetic-properties-of-materials-spring-2013/2eeeaab73afed1cd3969c79dd22c2a4f_MIT3_024S13_2012lec5.pdf |
15.081J/6.251J Introduction to Mathematical
Programming
Lecture 6: The Simplex Method II
1 Outline
• Revised Simplex method
• The full tableau implementation
• Anticycling
2 Revised Simplex
Initial data: A, b, c
1. Start with basis B = [AB(1), . . . , AB(m)]
and B−1 .
2. Compute p ′ = c ′
cj = cj − p ′ Aj ... | https://ocw.mit.edu/courses/6-251j-introduction-to-mathematical-programming-fall-2009/2ef2f1dd7045b5f29e5faea299fb0798_MIT6_251JF09_lec06.pdf |
0
0 0
1 0
−1 1
0 0 −1 1
2.2 Practical issues
• Numerical Stability
B−1 needs to be computed from scratch once in a while, as errors accu
mulate
• Sparsity
B−1 is represented in terms of sparse triangular matrices
3 Full tableau implementation
−c ′ B−1b
B
B−1b
c ′ − c ′ B−1A
B
B−1A
or, i... | https://ocw.mit.edu/courses/6-251j-introduction-to-mathematical-programming-fall-2009/2ef2f1dd7045b5f29e5faea299fb0798_MIT6_251JF09_lec06.pdf |
.4
Slide 15
136
4
4
4
x3 =
x1 =
x2 =
0
0
1
0
0
0
0
1
x
3
.
B = (
)
A = (
)
E = (
)
.
.
.
D = (
)
x
1
C = (
.
)
x
2
4 Comparison of implementations
Full tableau
Revised simplex
Memory
Worst-case time
Best-case time
O(mn)
O(mn)
O(mn)
O(m2)
O(mn)
O(m2)
5 Anticycling
5.1 Degeneracy in Pra... | https://ocw.mit.edu/courses/6-251j-introduction-to-mathematical-programming-fall-2009/2ef2f1dd7045b5f29e5faea299fb0798_MIT6_251JF09_lec06.pdf |
0 < ǫ < ǫ1
ǫ
.
..
ǫm
Ax = b +
x ≥ 0
is non-degenerate.
5.2.2 Proof
Let B1, . . . , Br be all the bases.
B−1
r
b +
ǫ
.
..
ǫm
=
r
b1 + Br
r
b
m
+ Br
1m
11ǫ +
· · · + Br ǫm
.
.
.
m1ǫ + · · · + Br
mm
ǫm
where:
•
r
bi
B−1
r =
Br ... | https://ocw.mit.edu/courses/6-251j-introduction-to-mathematical-programming-fall-2009/2ef2f1dd7045b5f29e5faea299fb0798_MIT6_251JF09_lec06.pdf |
i,m ⇒ bi
r
+ Bi
1ǫ + · · · + Br ǫm = 0.
im
•
Let ǫ1 the smallest positive root ⇒ 0 < ǫ < ǫ1 all RHS are
non-degeneracy.
=
0 ⇒
5
Slide 18
Slide 19
Slide 20
Slide 21
�
�
�
5.3 Lexicography
• u is lexicographically larger than v, u > v, if u �
L
=
v and the first
nonzero component of u... | https://ocw.mit.edu/courses/6-251j-introduction-to-mathematical-programming-fall-2009/2ef2f1dd7045b5f29e5faea299fb0798_MIT6_251JF09_lec06.pdf |
0 ⇔
bi + Bi1ǫ + · · · + Bimǫm ≥ 0 ∀ i
⇔ First non-zero component of ui = (bi, Bi1, . . . , Bim) is positive ∀ i.
5.5 Summary
1. We start with: (P ) : Ax = b, x ≥ 0
2. We introduce (Pǫ): Ax = b + (ǫ, . . . , ǫm) ′ , x ≥ 0
3. A basis is feasible + non-degenerate in (Pǫ) ⇔ ui
L
> 0 in (P ).
4. If we maintain ui
... | https://ocw.mit.edu/courses/6-251j-introduction-to-mathematical-programming-fall-2009/2ef2f1dd7045b5f29e5faea299fb0798_MIT6_251JF09_lec06.pdf |
0
4
0
5
3
6 −1
7
9
· · ·
· · ·
· · ·
Slide 27
•
xB(1)/u1 = 1/3 and xB(3)/u3 = 3/9 = 1/3.
•
We divide the first and third rows of the tableau by u1 = 3 and u3 = 9,
respectively, to obtain:
1/3
•
∗
1/3
0
∗
0
5/3
∗
7/9
1
∗
1
· · ·
· · ·
· · ·
•
Since 7/9 < 5/3, the third row is chosen to be th... | https://ocw.mit.edu/courses/6-251j-introduction-to-mathematical-programming-fall-2009/2ef2f1dd7045b5f29e5faea299fb0798_MIT6_251JF09_lec06.pdf |
5.8 Smallest subscript
pivoting rule
1. Find the smallest j for which the reduced cost cj is negative and have the
column Aj enter the basis.
2. Out of all variables xi that are tied in the test for choosing an exiting
variable, select the one with the smallest value of i.
Slide 28
Slide 29
Slide 30
7
... | https://ocw.mit.edu/courses/6-251j-introduction-to-mathematical-programming-fall-2009/2ef2f1dd7045b5f29e5faea299fb0798_MIT6_251JF09_lec06.pdf |
Massachusetts Institute of Technology
6.042J/18.062J, Fall ’05: Mathematics for Computer Science
Prof. Albert R. Meyer and Prof. Ronitt Rubinfeld
September 7
revised August 30, 2005, 956 minutes
InClass Problems Week 1, Wed.
Problem 1. Identify exactly where the bugs are in each of the following bogus proofs.1
... | https://ocw.mit.edu/courses/6-042j-mathematics-for-computer-science-fall-2005/2f40fa3ccb1dfac791414d27201383a6_cp1w.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
6.013/ESD.013J Electromagnetics and Applications, Fall 2005
Please use the following citation format:
Markus Zahn, 6.013/ESD.013J Electromagnetics and Applications, Fall
2005. (Massachusetts Institute of Technology: MIT OpenCourseWare).
http://ocw.mit.edu (accessed MM DD, YYYY... | https://ocw.mit.edu/courses/6-013-electromagnetics-and-applications-fall-2005/2f455332ba1548a8db83ef1d78592088_lec8.pdf |
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