text stringlengths 30 4k | source stringlengths 60 201 |
|---|---|
lawyer
questioned
the witness
about
the
revolver
• Step 1: represent a tree as a sequence of decisions d1 . . . dm
T = ∈d1, d2, . . . dm→
• First n decisions are tagging decisions
∈d1 . . . dn→ = ∈ DT, NN, Vt, DT, NN, IN, DT, NN →
Layer 2: Chunks
NP
Vt
NP
IN
NP
DT NN
questioned
DT
NN
about
DT
NN
... | https://ocw.mit.edu/courses/6-864-advanced-natural-language-processing-fall-2005/0abc5d7ab6458ec55a14c9f7c300438b_lec10.pdf |
constituent with label X
(always acts on leftmost constituent with no start or join label above it)
• Check=NO does nothing
• Check=YES takes previous Join or Start action, and converts
it into a completed constituent
NP
Vt
NP
IN
NP
DT NN
questioned
DT
NN
about
DT
NN
the
lawyer
the witness
the
revolv... | https://ocw.mit.edu/courses/6-864-advanced-natural-language-processing-fall-2005/0abc5d7ab6458ec55a14c9f7c300438b_lec10.pdf |
DT
NN
about
the
revolver
the
lawyer
the witness
Start(S)
Start(VP)
Join(VP)
Start(PP)
NP
NP
Vt
NP
IN
DT
NN
DT NN
questioned
DT
NN
about
the
revolver
the
lawyer
the witness
Check=NO
Start(S)
Start(VP)
Join(VP)
Start(PP)
Join(PP)
NP
Vt
NP
IN
NP
DT NN
questioned
DT
NN
about
DT
NN
t... | https://ocw.mit.edu/courses/6-864-advanced-natural-language-processing-fall-2005/0abc5d7ab6458ec55a14c9f7c300438b_lec10.pdf |
, Vt, DT, NN, IN, DT, NN,
Start(NP), Join(NP), Other, Start(NP), Join(NP),
Other, Start(NP), Join(NP),
Start(S), Check=NO, Start(VP), Check=NO,
Join(VP), Check=NO, Start(PP), Check=NO,
Join(PP), Check=YES, Join(VP), Check=YES,
Join(S), Check=YES →
A General Approach: (Conditional) History-Based Models
• Step 1:... | https://ocw.mit.edu/courses/6-864-advanced-natural-language-processing-fall-2005/0abc5d7ab6458ec55a14c9f7c300438b_lec10.pdf |
d1 . . . di−1, S) =
�(∗d1...di−1 ,S∈,di )·W
e
d�A e�(∗d1 ...di−1,S∈,d)·W
�
• The big question: how do we define �?
• Ratnaparkhi’s method defines � differently depending on
whether next decision is:
– A tagging decision
(same features as before for POS tagging!)
– A chunking decision
– A start/join decision after c... | https://ocw.mit.edu/courses/6-864-advanced-natural-language-processing-fall-2005/0abc5d7ab6458ec55a14c9f7c300438b_lec10.pdf |
th tree
relative to the decision, where n = 0, 1, 2
• Looks at bigram features of the above for (-1,0) and (0,1)
• Looks at trigram features of the above for (-2,-1,0), (-1,0,1)
and (0, 1, 2)
• The above features with all combinations of head words
excluded
• Various punctuation features
Layer 3: Check=NO or C... | https://ocw.mit.edu/courses/6-864-advanced-natural-language-processing-fall-2005/0abc5d7ab6458ec55a14c9f7c300438b_lec10.pdf |
3 Representations of finite groups: basic results
Recall that a representation of a group G over a field k is a k-vector space V together with a
group homomorphism δ : G
GL(V ). As we have explained above, a representation of a group G
over k is the same thing as a representation of its group algebra k[G].
⊃
In thi... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0abca81c8a50d3d529fa0fc3d3e5319a_MIT18_712F10_ch3.pdf |
nite-dimensional representation of G and W
as representations.
subrepresentation W �
V such that V = W
→
W �
→
Choose any complement ˆW of W in
ˆ
W as vector spaces, but not necessarily
V . (Thus V = W
as representations.) Let P be the projection along W onto W , i.e., the operator on V defined by
|W = Id and ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0abca81c8a50d3d529fa0fc3d3e5319a_MIT18_712F10_ch3.pdf |
, then the characteristic of k does not divide
G
.
|
|
Proof. Write k[G] =
trivial one-dimensional representation. Then
�
r
i=1 End Vi, where the Vi are irreducible representations and V1 = k is the
k[G] = k
End Vi = k
r
�
�
i=2
r
�
�
i=2
diVi,
where di = dim Vi. By Schur’s Le... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0abca81c8a50d3d529fa0fc3d3e5319a_MIT18_712F10_ch3.pdf |
��eld k of characteristic p is trivial.
Let G be a group of order pn. Show that every irreducible representation of G over
3.2 Characters
If V is a finite-dimensional representation of a finite group G, then its character νV : G
k
is defined by the formula νV (g) = tr
|V (δ(g)). Obviously, νV (g) is simply the restr... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0abca81c8a50d3d529fa0fc3d3e5319a_MIT18_712F10_ch3.pdf |
gh
hg
|
�
f (gh) = f (hg)
−
ker �
g, h
g, h
⊕
,
G
G
}
�
|
⊕
�
}
Corollary 3.6. The number of isomorphism classes of irreducible representations of G equals the
number of conjugacy classes of G (if
= 0 in k).
G
|
|
⇒
Exercise. Show that if
G
= 0 in k then the number of isomorphism c... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0abca81c8a50d3d529fa0fc3d3e5319a_MIT18_712F10_ch3.pdf |
group
representations then so are δ1(g)δ2(g) and δ1(g)−
of G.
|
is an abelian group: if δ1, δ2 : G
1. G∗
G∗|
=
⊃
G
|
|
For given n
⊂
so Z∗n =∪ Zn. In general,
1, define δ : Zn
⊃
C×
by δ(m) = e2νim/n. Then Z∗n =
δk : k = 0, . . . , n
−
1
,
}
{
(G1 G2
×
× · · · ×
Gn) = G1∗ ×
∗
G∗2 × · · · ×
G∗n ,
=∪ G... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0abca81c8a50d3d529fa0fc3d3e5319a_MIT18_712F10_ch3.pdf |
ations over C. If their dimensions are d1, d2, d
3, then d1+d2 +d3 = 6, so S3 must have
two 1-dimensional and one 2-dimensional representations. The 1-dimensional representations
are the trivial representation C+ given by δ(ε) = 1 and the sign representation C given by
δ(ε)
−
(
2
2
2
= 1)ε.
−
The 2-dimensional repr... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0abca81c8a50d3d529fa0fc3d3e5319a_MIT18_712F10_ch3.pdf |
defining relations
k
}
k = ij =
ji,
−
−
1 = i2 = j2 = k2.
−
The 5 conjugacy classes are
, so there are 5 different irreducible
,
{±
representations, the sum of the squares of whose dimensions is 8, so their dimensions must
be 1, 1, 1, 1, and 2.
,
{±
,
{±
,
{−
k
}
{
}
}... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0abca81c8a50d3d529fa0fc3d3e5319a_MIT18_712F10_ch3.pdf |
the space of functions f : Q8
multiplication, g
C such that f (gi) = ∀
f (x) = f (xg)).
⊃
−
∞
−
{
×
×
Z2 is
Z2, where Z2
e, (12)(34), (13)(24), (14)(23)
4. The symmetric group S4. The order of S4 is 24, and there are 5 conjugacy classes:
e, (12), (123), (1234), (12)(34). Thus the sum of the squares of the dim... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0abca81c8a50d3d529fa0fc3d3e5319a_MIT18_712F10_ch3.pdf |
−
Note that another realization of C3
is by action of S4 by symmetries (not necessarily rotations)
−
of the regular tetrahedron. Yet another realization of this representation is the space of
functions on the set of 4 elements (on which S4 acts by permutations) with zero sum of
values.
+ and C3 are different,
for... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0abca81c8a50d3d529fa0fc3d3e5319a_MIT18_712F10_ch3.pdf |
also a representation, via
�
δV W (g) = δV (g)
�
δ
W (g).
�
Therefore, νV
�
W (g) = νV (g)νW (g).
An interesting problem discussed below is to decompose V
direct sum of irreducible representations.
W (for irreducible V, W ) into the
�
3.5 Orthogonality of characters
We define a positive definite Hermitian inne... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0abca81c8a50d3d529fa0fc3d3e5319a_MIT18_712F10_ch3.pdf |
.
P
|X =
Id, if X = C,
� 0, X =
C.
Therefore, for any representation X the operator P
X G of G-invariants in X. Thus,
|X is the G-invariant projector onto the subspace
Tr
|V W � (P ) = dim HomG(C, V
= dim(V
W ⊕
�
�
W ⊕)
)G = dim HomG(W, V ).
�
⇒
Theorem 3.8 gives a powerful method... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0abca81c8a50d3d529fa0fc3d3e5319a_MIT18_712F10_ch3.pdf |
onality formula” for characters, in which summation is taken over irre
ducible representations rather than group elements.
Theorem 3.9. Let g, h
�
G, and let Zg denote the centralizer of g in G. Then
νV (g)νV (h) =
if g is conjugate to h
Zg|
|
0, otherwise
�
�
V
where the summation is taken over all irreducib... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0abca81c8a50d3d529fa0fc3d3e5319a_MIT18_712F10_ch3.pdf |
jugate to G. Thus, by Theorem 3.8, the rows of the matrix U are orthonormal. This means
that U is unitary and hence its columns are also orthonormal, which implies the statement.
. Note that the conjugacy class of g is G/Zg, thus
Zg|
Zg|
�
G
/
|
|
|
|
3.6 Unitary representations. Another proof of Maschke’s theorem f... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0abca81c8a50d3d529fa0fc3d3e5319a_MIT18_712F10_ch3.pdf |
irreducible representation and B1, B2 are two positive definite Hermitian forms on V, then
B1(v, w) = B2(Av, w) for some homomorphism A : V
V (since any positive definite Hermitian
form is nondegenerate). By Schur’s lemma, A = ∂Id, and clearly ∂ > 0.
⊃
Theorem 3.11 implies that if V is a finite dimensional representa... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0abca81c8a50d3d529fa0fc3d3e5319a_MIT18_712F10_ch3.pdf |
V = W
W �.
�
Theorems 3.11 and 3.12 imply Maschke’s theorem for complex representations (Theorem 3.1).
Thus, we have obtained a new proof of this theorem over the field of complex numbers.
Remark 3.13. Theorem 3.12 shows that for infinite groups G, a finite dimensional representation
may fail to admit a unitary stru... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0abca81c8a50d3d529fa0fc3d3e5319a_MIT18_712F10_ch3.pdf |
invariant Hermitian
wi}
W ⊕ be the linear function on W defined by taking the inner product with
wi⊕ �
vi}
{
wi: wi⊕
1
G
x
|
|
(u) = (u, wi). Then for x
G x, we have
�
G we have (xwi⊕, wj⊕) = (xwi, wj ). Therefore, putting P =
�
⎨
V W
(tij , ti⊗j⊗ )
�
1
=
G
|
|
−
�
G
x
�
(xvi, v
j )(xwi⊗ , wj⊗ ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0abca81c8a50d3d529fa0fc3d3e5319a_MIT18_712F10_ch3.pdf |
⊕
vj⊗ ) =
ζ ζ
ii⊗ jj⊗
dim V
�
which finishes the proof of (i) and (ii). The last statement follows immediately from the sum of
squares formula.
�
3.8 Character tables, examples
The characters of all the irreducible representations of a finite group can be arranged into a char
acter table, with conjugacy classes of ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0abca81c8a50d3d529fa0fc3d3e5319a_MIT18_712F10_ch3.pdf |
of 4 items. There are three
one-dimensional representations (as A4 has a normal subgroup Z2
Z2 = Z3).
Since there are four conjugacy classes in total, there is one more irreducible representation of
dimension 3. Finally, the character table is
Z2, and A4/Z2
�
�
A4
#
C
C
C
ρ2
C3
ρ
Id
1
... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0abca81c8a50d3d529fa0fc3d3e5319a_MIT18_712F10_ch3.pdf |
-1
1
1
1
1
1
-2
i
2
1
1
-1
-1
0
j
2
1
-1
1
-1
0
k
2
1
-1
-1
1
0
S4
#
C+
C
−
C2
C3
+
C3
−
A5
#
C
C3
+
C3
−
C4
C5
Id
1
1
1
2
3
3
Id
1
1
3
3
4
5
(12)
6
1
-1
0
-1
1
(123)
20
1
0
0
1
-1
(12)(34)
3
1
1
2
-1
-1
(123)
8
1
1
-1
0
0
(1234)
6
1
-1
0
... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0abca81c8a50d3d529fa0fc3d3e5319a_MIT18_712F10_ch3.pdf |
S3 via the surjective homomorphism S4
character from the character table of S3.
⊃
The character of the
+ is computed from its geometric realization
by rotations of the cube. Namely, by rotating the cube, S4 permutes the main diagonals. Thus
rotation by 1800 around an axis that is perpendicular to two opposite edge... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0abca81c8a50d3d529fa0fc3d3e5319a_MIT18_712F10_ch3.pdf |
1440, around axes going through two
opposite vertices. The character of this representation is computed from this description in a
straightforward way.
, in which (12)(34) is the rotation by 1800
Another representation of A5, which is also 3-dimensional, is C3
+ twisted by the automorphism
of A5 given by conjugati... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0abca81c8a50d3d529fa0fc3d3e5319a_MIT18_712F10_ch3.pdf |
representation is computed similarly to the character of C4, or from the orthogonality formula.
3.9 Computing tensor product multiplicities using character tables
Character tables allow us to compute the tensor product multiplicities Nij using
k
Example 3.16. The following tables represent computed tensor product m... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0abca81c8a50d3d529fa0fc3d3e5319a_MIT18_712F10_ch3.pdf |
�
C3
+
C4
C4
C4
C4
C
C3
− �
C3
+
�
C3
� − � �
C5
�
C5
�
C4
C3
+
C5
�
C5
C5
C3
C3
C4
+ �
− � �
C3
C3
C4
+
� −
�
�
C3
C3
2C5
+ �
�
− �
C3
C3
2C4
C5
C5
C4
+ �
− �
�
C
�
2C5
Problem 3.17. Let G be the group of symmetries of a regular N-gon (it has 2N elements).
(a) Describe all irred... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0abca81c8a50d3d529fa0fc3d3e5319a_MIT18_712F10_ch3.pdf |
1
⎝
(δ
⎧
�
f )(x) = f (x
−
1),
(note that zx makes
sense since zp = 1).
1 0 0
0 1 1
�
0 0 1
⎝
(δ
⎧
�
f )(x) = z
xf (x).
(a) Show that such a representation exists and is unique, and compute δ(g) for all g
�
(b) Denote this representation by Rz. Show that Rz is irreducible if and only if z = 1.
G.
(c) Classif... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0abca81c8a50d3d529fa0fc3d3e5319a_MIT18_712F10_ch3.pdf |
it is spanned by a subset S of a basis of eigenvectors of δ(H). Use the
invariance of W under the operators δ(1 + Eij ) (where Eij is defined by Eij ek = ζjkei) for all i = j
to show that if the subset S is nonempty, it is necessarily the entire basis.
{
Problem 3.20. Recall that the adjacency matrix of a graph � (wit... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0abca81c8a50d3d529fa0fc3d3e5319a_MIT18_712F10_ch3.pdf |
space of functions on the set of faces and
the set of edges of the icosahedron.
Problem 3.22. Let Fq be a finite field with q elements, and G be the group of nonconstant inho
Fq). Find all irreducible
mogeneous linear transformations, x
complex representations of G, and compute their characters. Compute the tensor ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0abca81c8a50d3d529fa0fc3d3e5319a_MIT18_712F10_ch3.pdf |
i2 = j2 = k2 =
ji = k, jk =
j. Thus we have that Q8 is a subgroup of the group H× of invertible elements
kj = i, ki =
1, ij =
ik =
−
−
−
of H under multiplication.
−
The algebra H is called the quaternion algebra.
(d) For q = a+bi+cj +dk, a, b, c, d
=
Show that q1q2 = q¯2q¯1, and
q1q2||
||
R, let q¯ = a
�
... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0abca81c8a50d3d529fa0fc3d3e5319a_MIT18_712F10_ch3.pdf |
It is known that the classification of finite subgroups of SO(3) is as follows:
1) the cyclic group Z/nZ, n
⊂
1, generated by a rotation by 2β/n around an axis;
2) the dihedral group Dn of order 2n, n
a plane containing a regular n-gon5;
⊂
2 (the group of rotational symmetries in 3-space of
3) the group of rotati... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0abca81c8a50d3d529fa0fc3d3e5319a_MIT18_712F10_ch3.pdf |
Now let P1, ..., Pk be the poles representing all the orbits of G on the set
of poles, and m1, ..., mk be their orders. By counting nontrivial elements of G, show that
G
|
|
2(1
−
1
n
) =
(1
−
1
mi
).
�
i
Then find all possible mi and n that can satisfy this equation and classify the corresponding groups.
(b) Us... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0abca81c8a50d3d529fa0fc3d3e5319a_MIT18_712F10_ch3.pdf |
of representation theory to physics (elasticity
theory). We first describe the physical motivation and then state the mathematical problem.
Imagine a material which occupies a certain region U in the physical space V = R3 (a space
with a positive definite inner product). Suppose the material is deformed. This means, w... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0abca81c8a50d3d529fa0fc3d3e5319a_MIT18_712F10_ch3.pdf |
v centered at P of area
the part of the material on the v-side of ε acts on the part on the opposite side. It is easy to deduce
from Newton’s laws that Fv is linear in v, so there exists a linear operator SP : V
V such that
Fv = SP v. It is called the stress tensor.
⊃
||
||
v
An elasticity law is an equation SP =... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0abca81c8a50d3d529fa0fc3d3e5319a_MIT18_712F10_ch3.pdf |
W one has f (x + y) = Kx + µy for some
R, y
lemma that SP is always symmetric, and for x
real numbers K, µ.
�
�
In fact, it is clear from physics that K, µ are positive. Physically, the compression modulus K
characterises resistance of the material to compression or dilation, while the shearing modulus µ
charact... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0abca81c8a50d3d529fa0fc3d3e5319a_MIT18_712F10_ch3.pdf |
18.212: Algebraic Combinatorics
Andrew Lin
Spring 2019
This class is being taught by Professor Postnikov.
February 27, 2019
This is a reminder that the problem set is due on Monday, so we should start it soon. A few bonus problems were also
added that are a bit more challenging.
Next Wednesday, we will discuss t... | https://ocw.mit.edu/courses/18-212-algebraic-combinatorics-spring-2019/0b039163b47d51f947e6fdbea5b99844_MIT18_212S19_lec10.pdf |
Definition 2
The generating function
is called the Eulerian polynomial.
X
w ∈Sn
des(w )
x
So we have a number of inversions, cycles, and descents. Here’s a meta-mathematical claim: interesting permutation
statistics are likely equidistributed with one of these classes!
Fact 3
Things related to inversions are c... | https://ocw.mit.edu/courses/18-212-algebraic-combinatorics-spring-2019/0b039163b47d51f947e6fdbea5b99844_MIT18_212S19_lec10.pdf |
records of w .
For example, w = (2, 5, 7, 3, 1, 6, 8, 4) has records 2, 5, 7, 8, so rec(w ) = 4.
Theorem 7
rec(w ) and cyc(w ) are equidistributed.
This can be proved by induction: for example, show the generating functions satisfy the same recurrence relation. But
we prefer combinatorial proofs: maybe we can find ... | https://ocw.mit.edu/courses/18-212-algebraic-combinatorics-spring-2019/0b039163b47d51f947e6fdbea5b99844_MIT18_212S19_lec10.pdf |
the number of cycles of w is the number of records of w˜, as desired.
This is bijective (to go backwards, find the records and put parentheses back), so we’re done!
Definition 8
An exceedance in w is an index 1 ≤ i ≤ n such that wi > i . Let exc(w ) be the number of exceedances of w .
For example,
in (2, 5, 7, 3, 1,... | https://ocw.mit.edu/courses/18-212-algebraic-combinatorics-spring-2019/0b039163b47d51f947e6fdbea5b99844_MIT18_212S19_lec10.pdf |
anti-exceedances in w
is the number of descents in w˜.
This is because i being a descent in w˜ means that because the i + 1th entry is not larger than w˜, i and i + 1 are in the
same cycle. Then that means that i goes to something smaller than itself, making it an anti-exceedance!
Let’s look a bit at Stirling numbe... | https://ocw.mit.edu/courses/18-212-algebraic-combinatorics-spring-2019/0b039163b47d51f947e6fdbea5b99844_MIT18_212S19_lec10.pdf |
first expression is called “raising power of x,” while the second is called the “falling power of x.” The latter is
k=0
sometimes denoted (x)n.
Definition 13
Define the Stirling number of the second kind
S(n, k) = number of set-partitions of [n] into k non-empty blocks.
3
We also use the convention S(0, 0) = 1 and... | https://ocw.mit.edu/courses/18-212-algebraic-combinatorics-spring-2019/0b039163b47d51f947e6fdbea5b99844_MIT18_212S19_lec10.pdf |
LP Example
Stanley B. Gershwin�
Massachusetts Institute of Technology
Consider the factory in Figure 1 that consists of three parallel machines. It makes a single
product which can be produced using any one of the machines. The possible material flows are
indicated.
Assume that the cost ($/part) of using machine Mi... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/0b2ff52cac042a60403e72ddce64c3cf_MIT2_854F16_LpExample.pdf |
1
LP Example
M1
M2
M3
Figure 1: Factory
x2
2µ
D <
1µ
1µ
< D <
1µ
+
2µ
1µ
x1
3µ
x3
Figure 2: Constraint space
2
LP Example
LP in Standard Form: Define x4 and x5 as the slack variables associated with the upper bounds
on x1 and x2. Then
min c1x1 + c2x2 + c3x3
such that
x1 + x2 + x3 = D
x1 + x4 = µ1
x2 ... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/0b2ff52cac042a60403e72ddce64c3cf_MIT2_854F16_LpExample.pdf |
B =
⎛
c1 0 0
⎜
c
T
N =
⎛
c2 c3
⎜
It is easy to show that
A−1
B
AN =
⎞
⎟
⎠
1
1
−1 −1
0
1
�
�
�
This is demonstrated at the end of this note.
Therefore,
T
c
R
= c
T
N
−c
T
B
A−1
B
AN =
c2 c3 − c1 0 0
⎜ ⎛
⎜
⎛
⎞
⎟
⎠
1
1
−1 −1
0
1
=
⎛
�
�
�
3
c2 c3 − c1 c1 =
⎜ ⎛
⎜
c2 − c1 c3 − c1
⎛
⎜
... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/0b2ff52cac042a60403e72ddce64c3cf_MIT2_854F16_LpExample.pdf |
T
N =
⎛
c3 0
⎜
A−1
B AN =
0
−1
1
⎞
⎟
⎠
1
−1
−1
�
�
�
c
T
R
= c
T
N
− c
T
B
BA−1
AN = −c2 + c3 c1 − c2 + c3
⎛
⎜
Then
and
which is also componentwise positive.
3. µ1 + µ2 < D:
We have guessed that x1 = µ1, x2 = µ2, x3 = D − µ1 − µ2. Then x4 = x5 = 0 and x1, x2, x3 are
the basic variables. Therefore,
AB = ... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/0b2ff52cac042a60403e72ddce64c3cf_MIT2_854F16_LpExample.pdf |
µ1+ µ2
Figure 3: Cost
5
LP Example
which is also componentwise positive.
Note that the cost as a function of D is shown in Figure 3.
To show that, in the first case,
A−1
B AN
=
⎞
⎟
⎠
1
1
−1 −1
0
1
�
�
�
A−1AN is a matrix of two columns. The first column satisfies
B
1
0
1
⎞ � ⎞
=
⎟ � ⎟
⎠
⎠
�
1 0 0
1 ... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/0b2ff52cac042a60403e72ddce64c3cf_MIT2_854F16_LpExample.pdf |
⎟
⎠
1
1
−1 −1
0
1
�
�
�
6
MIT OpenCourseWare
https://ocw.mit.edu
2.854 / 2.853 Introduction To Manufacturing Systems
Fall 2016
For information about citing these materials or our Terms of Use, visit: https://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/0b2ff52cac042a60403e72ddce64c3cf_MIT2_854F16_LpExample.pdf |
Chapter 1
Root Concepts of the Standard
Model
The standard model of particle physics accurately describes a vast range of phenomena
using a small number of parameters. Much of the power of the standard model
arises from the fact that it embodies a few deep physical and mathematical concepts
which are difficult, but ... | https://ocw.mit.edu/courses/8-325-relativistic-quantum-field-theory-iii-spring-2003/0b475fe7464cbd47a63424d0261905b8_chap1.pdf |
effects is the spontaneous breaking of local gauge
symmetry.
It is a form of super-conductivity, but operating in (what we perceive
as) empty space. In conventional super-conductivity, roughly speaking, the electrons
in a metal, which normally behave as a gas of independent particles, condense into
a liquid of over... | https://ocw.mit.edu/courses/8-325-relativistic-quantum-field-theory-iii-spring-2003/0b475fe7464cbd47a63424d0261905b8_chap1.pdf |
second of these dynamical effects is confinement of quarks. Confinement
should be considered together with the closely related property of asymptotic free
dom, which is in a sense its inverse. Confinement and asymptotic freedom occur in
the sector of the standard model dealing with the strong interaction, quantum chro
... | https://ocw.mit.edu/courses/8-325-relativistic-quantum-field-theory-iii-spring-2003/0b475fe7464cbd47a63424d0261905b8_chap1.pdf |
consider processes that involve large changes in energy and mo
mentum, the forces are feeble an d the radiation is rare. This is asymptotic freedom.
These unusual behaviors – fundamental forces that grow with distance, radiators that
become quiescent as they are shaken violently – were once thought to be paradoxi
c... | https://ocw.mit.edu/courses/8-325-relativistic-quantum-field-theory-iii-spring-2003/0b475fe7464cbd47a63424d0261905b8_chap1.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
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.
Free Modules (Lecture 7)
We first recall a bit of notation: If I = (i1, . . . , ik) is a sequence of intege... | https://ocw.mit.edu/courses/18-917-topics-in-algebraic-topology-the-sullivan-conjecture-fall-2007/0b65ed5433a1570711db5113fe4576a0_lecture7.pdf |
this, we observe that the freeness of F (n) implies that there is a (unique) map φ : F (n) M with
φ(νn) = x. Consequently, any linear relation among the expressions {SqI νn} would entail a linear relation
among the expressions {SqI x}.
→
It will therefore suffice to choose M to be some sufficiently nontrivial unstable A-... | https://ocw.mit.edu/courses/18-917-topics-in-algebraic-topology-the-sullivan-conjecture-fall-2007/0b65ed5433a1570711db5113fe4576a0_lecture7.pdf |
formula:
Sqk(t1
a1 . . . tn
an ) =
�
k=k1+...+kn
� � � �
an t1
. . .
kn
a1
k1
a1+k1 . . . tn
an+kn .
We now make a crucial observation about the formula above. Suppose that each exponent ai is a power
� �
of 2. The binomial coefficient a
is equal to 1 if ki = 0 or ki = ai, and vanishes otherwise (since we ar... | https://ocw.mit.edu/courses/18-917-topics-in-algebraic-topology-the-sullivan-conjecture-fall-2007/0b65ed5433a1570711db5113fe4576a0_lecture7.pdf |
the subspace of F2[t1, . . . , tn] consisting of those polynomials which satisfy (a) and (b)
above. We observe that M is invariant under the action of the Steenrod algebra A, and is therefore an
unstable A-module in its own right. Moreover, M contains the element x = t1 . . . tn of degree n. To
complete the proof of... | https://ocw.mit.edu/courses/18-917-topics-in-algebraic-topology-the-sullivan-conjecture-fall-2007/0b65ed5433a1570711db5113fe4576a0_lecture7.pdf |
� = (�0, . . . , �k) with �0 + . . . + �k = n, there is a unique sequence 0 ≤ b1 ≤ . . . ≤ bn
bn ). Thus M has a basis
b1 . . . t2
such that �i is the cardinality of the set {j : bj = i}. We then set f� = σ(t2
n
1
consisting of the polynomials {f�}, where � ranges over sequences of nonnegative integers (�0, . . . ,... | https://ocw.mit.edu/courses/18-917-topics-in-algebraic-topology-the-sullivan-conjecture-fall-2007/0b65ed5433a1570711db5113fe4576a0_lecture7.pdf |
+ �1, . . . , 2�k + �k−1, �k) of excess
≤ n. We will denote this admissible sequence by I(�).
We now wish to compare the expressions {SqI(�)(x)} with the basis {f�} for M . They do not coincide,
but we get the next best thing: the translation between these two bases is upper triangular. To be more
precise, we need ... | https://ocw.mit.edu/courses/18-917-topics-in-algebraic-topology-the-sullivan-conjecture-fall-2007/0b65ed5433a1570711db5113fe4576a0_lecture7.pdf |
1 . . . tn) + lower order
t�k +1 . . . tn)
Sq�k (x) = σ(t1
2 . . . t4
1t4
�k
x = σ(t1 . . . tn)
2t2
2
2 . . . t�
k
t2
�k +1 . . . t2
. . .
SqI(�)(x) = f� + lower order
�k +�k−1
We now wish to reformulate some of the above ideas, using Kuhn’s theory of “generic representations”.
In what follows, we let V de... | https://ocw.mit.edu/courses/18-917-topics-in-algebraic-topology-the-sullivan-conjecture-fall-2007/0b65ed5433a1570711db5113fe4576a0_lecture7.pdf |
Remark 4. Kuhn refers to objects of Fun as generic representations. If F : Vectf
Vect is a functor,
then for every finite dimensional vector space V ∈ Vectf , we obtain a new vector space F (V ) which is
equipped with an action of Aut(V ) � GLn(F2). In other words, we can think of F as providing a family of
represen... | https://ocw.mit.edu/courses/18-917-topics-in-algebraic-topology-the-sullivan-conjecture-fall-2007/0b65ed5433a1570711db5113fe4576a0_lecture7.pdf |
(Symn(V ), Sym∗(V )).
Let V be the free vector space generated by a basis {t1, . . . , tn}, and let x = t1 . . . tn; then it will suffice
to show that the elements {SqI (x)} are linearly independent in Sym∗(V ). This follows immediately from
Proposition 2.
We now wish to prove that HomFun(Symn , Sym∗) is spanned by t... | https://ocw.mit.edu/courses/18-917-topics-in-algebraic-topology-the-sullivan-conjecture-fall-2007/0b65ed5433a1570711db5113fe4576a0_lecture7.pdf |
linear map determines a
commutative diagram
Symn(V )
φ
� Sym∗(V )
� �
Symn(W )
αW
� �
� Sym∗(W ),
so that αW (w1 . . . wn) = f (w1, . . . , wn) = 0 ∈ Sym∗(W ).
We now wish to describe the image of the map φ. Fix α : Symn
Sym∗, and let f = φ(α). Since
x = t1 . . . tn ∈ Symn(V ) is invariant under the permutatio... | https://ocw.mit.edu/courses/18-917-topics-in-algebraic-topology-the-sullivan-conjecture-fall-2007/0b65ed5433a1570711db5113fe4576a0_lecture7.pdf |
f can be written as a sum of monomials of the
form t2b1 . . . t2bn . Since f is symmetric, we conclude that f ∈ M ⊆ F2[t1, . . . , tn].
1
n
We therefore have a factorization
φ : HomFun(Symn , Sym∗) �→ M ⊆ F2[t1, . . . , tn].
The map φ carries SqI to SqI (x). Proposition 2 implies that M is generated by these expre... | https://ocw.mit.edu/courses/18-917-topics-in-algebraic-topology-the-sullivan-conjecture-fall-2007/0b65ed5433a1570711db5113fe4576a0_lecture7.pdf |
LECTURE 6
• Readings: Sections 2.4-2.6
Lecture outline
• Review PMF, expectation, variance
• Conditional PMF
• Geometric PMF
• Total expectation theorem
• Joint PMF of two random variables
• Independence
Review
• Random variable X: function from
sample space to the real numbers
• PMF (for discrete random variables):... | https://ocw.mit.edu/courses/6-041-probabilistic-systems-analysis-and-applied-probability-spring-2006/0ba32c276891edcfb233b7dd35e2d594_lec06.pdf |
MEASURE AND INTEGRATION: LECTURE 12
Appoximation of measurable functions by continuous func
tions. Recall Lusin’s theorem.Let f : X → C be measurable, A ⊂ X,
µ(A) < ∞, and f (x) = 0 if x �∈ A. Given � > 0, there exists g ∈ Cc(X)
such that µ({x f (x) = g(x)}) < � and
|
|
|
sup g(x) ≤ sup f (x) .
x∈X
x∈X
|
|
A... | https://ocw.mit.edu/courses/18-125-measure-and-integration-fall-2003/0bb49f0e3bf706941074216dcee6e3ab_18125_lec12.pdf |
µ(Ek ) = 0. Let E =
Ek . Then µ(E) = 0. Redefine fk = 0 on E; this does not change
g a.e., and we can apply the regular LDCT.
�
Theorem 0.2. Let f1, f2, . . . : X → C with each fk ∈ L1(µ) and as
sume that ∞
� �
�
X |fk dµ < ∞. Then ∞ fk exists a.e. and
|
� �
∞ �
�
∞ �
�
| ≤ | |
k=1
k=1
fk dµ =
fk dµ.
X
k... | https://ocw.mit.edu/courses/18-125-measure-and-integration-fall-2003/0bb49f0e3bf706941074216dcee6e3ab_18125_lec12.pdf |
dµ = lim
j→∞
j→∞
�
X
Fj dµ
fk dµ = lim
j→∞
j �
�
k=1
X
�
j
�
X
k=1
fk dµ.
= lim
j→∞
∞ �
�
=
k=1
X
fk dµ
�
Countable additivity of the integral. Let E1, E2, . . . be a countable
C be
sequence of measurable sets. Let E = ∪∞ Ek and f : X
measurable. Assume either f ≥ 0 or f ∈ L1(E) (i.e., � →
dµ ... | https://ocw.mit.edu/courses/18-125-measure-and-integration-fall-2003/0bb49f0e3bf706941074216dcee6e3ab_18125_lec12.pdf |
Ek dµ
|f | dµ
k=1 Ek
�
f dµ < ∞,
|
|
=
=
because of the case when f ≥ 0.
E
� | https://ocw.mit.edu/courses/18-125-measure-and-integration-fall-2003/0bb49f0e3bf706941074216dcee6e3ab_18125_lec12.pdf |
20.110/5.60 Fall 2005
Lecture #7
page
1
Entropy, Reversible and Irreversible Processes and
Disorder
Examples of spontaneous processes
T1
T2
Connect two metal blocks thermally in
an isolated system
(∆U = 0)
Initially
2T T≠
1
dS dS dS
+
=
2
1
=
1
đ
đ
q q
2
−
T T
2
1
=
đ
q
1
)
(
−
T T
1
2
TT
1 2
(
đ
q
1
= −
đ 2
... | https://ocw.mit.edu/courses/20-110j-thermodynamics-of-biomolecular-systems-fall-2005/0bea388d6ba3e32ddda64a22f6574964_l07.pdf |
mol gas (V,T)
∆
S
backwards
=
∫
rev
đ
q
T
= −
∫
đ
w
T
=
V
∫
V
2
RdV
V
=
R
ln
1
2
∴ ∆ =
S R
>
ln2 0
spontaneous
IMPORTANT!! To calculate ∆S for the irreversible process, we
needed to find a reversible path so we could determine
đ
q
rev
and
rev
đ
q
T
∫
.
• Mixing of ideal gases at constant T and p
nA A (g, VA, T... | https://ocw.mit.edu/courses/20-110j-thermodynamics-of-biomolecular-systems-fall-2005/0bea388d6ba3e32ddda64a22f6574964_l07.pdf |
gas
∴ ∆
S
demix
=
∫
rev
dq
T
=
V
A
∫
V
pdV
A A
T
+
V
B
∫
V
pdV
B B
T
=
R
n
A
ln
V
A
V
+
nR
B
ln
V
B
V
Put in terms of mole fractions
X
A
=
n
A
n
X
B
=
n
B
n
Ideal gas ⇒
X
A
=
V
A
V
X
B
=
V
B
V
∴ ∆
demixS
=
nR X X X X
]
+
A B
ln
ln
[
A
B
⇒ ∆
S
mix
= −
nR X X X X
]
+
A B
ln
ln
[
A
B
Since
A BX X
,
<
1
⇒ ∆
Smix
>
0
... | https://ocw.mit.edu/courses/20-110j-thermodynamics-of-biomolecular-systems-fall-2005/0bea388d6ba3e32ddda64a22f6574964_l07.pdf |
a
maximally disordered state.
Microscopic understanding: Boltzmann Equation of Entropy:
ΩlnS k
=
Where k is Boltzmann’s constant (k=R/NA).
And Ω is the number of equally probable microscopic
arrangements for the system.
This can also be used to calculate ∆S
In the case of the Joule expansion of an ideal gas in v... | https://ocw.mit.edu/courses/20-110j-thermodynamics-of-biomolecular-systems-fall-2005/0bea388d6ba3e32ddda64a22f6574964_l07.pdf |
T
∫
(a) Mixing of ideal gases at constant T and p
nA A (g, VA, T) + nB A (g, VB, T) = n (A + B) (g, V = VA + VB, T)
∆
S
mix
=
−
nR X X X X ]
B
+
A B
ln
ln
[
A
(b) Heating (or cooling) at constant V
A (T1, V) = A (T2, V)
∆ =
S
rev
đ
q
T
∫
=
T
∫ 2
T
1
CdT
V
T
T
if
VC
is
=
-independent
C
V
ln
T
2
T
1
[Note ∆ >
S
0 ... | https://ocw.mit.edu/courses/20-110j-thermodynamics-of-biomolecular-systems-fall-2005/0bea388d6ba3e32ddda64a22f6574964_l07.pdf |
2O (s, -10∞C, 1 bar)
This is spontaneous and irreversible.
\ We need to find a reversible path between the two states
to calculate ∆S.
H2O (l, -10∞C, 1 bar) =
irreversible
H2O (s, -10∞C, 1 bar)
đ
q C dT
( )
(cid:65)
=
rev
p
đ
( )
q C dT
sp
=rev
H2O (l, 0∞C, 1 bar) =
reversible
H2O (s, 0∞C, 1 bar)
rev
q
p
= −... | https://ocw.mit.edu/courses/20-110j-thermodynamics-of-biomolecular-systems-fall-2005/0bea388d6ba3e32ddda64a22f6574964_l07.pdf |
Lecture 4
Contents
1 de Broglie wavelength and Galilean transformations
2 Phase and Group Velocities
3 Choosing the wavefunction for a free particle
1 de Broglie wavelength and Galilean transformations
B. Zwiebach
February 18, 2016
1
4
6
We have seen that to any free particle with momentum p, we can associate a plane w... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/0c07cbdc9c352c39eb9539b31ded90d7_MIT8_04S16_LecNotes4.pdf |
along the +x direction of S with constant velocity v. At time equal zero, the origins of the
two reference frames coincide.
(cid:48)
(cid:48)
(cid:48)
The time and spatial coordinates of the two frames are related by a Galilean transformation, which
states that
(cid:48)
x0 = x (cid:0) vt,
−
(cid:48)
t0 = t .
(1.3)
Inde... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/0c07cbdc9c352c39eb9539b31ded90d7_MIT8_04S16_LecNotes4.pdf |
S will obtain rather different de Broglie wavelengths λ0 and λ! Indeed,
(cid:48)
(cid:48)
(cid:48)
(cid:48)
(cid:48)
λ0 =
h
p0 =
(cid:48)
h
p −
mv
(cid:54)=
6= λ,
(1.7)
This is very strange! As we review now, for ordinary waves that propagate in the rest frame of
a medium (like sound waves or water waves) Galilean obser... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/0c07cbdc9c352c39eb9539b31ded90d7_MIT8_04S16_LecNotes4.pdf |
48)(x(cid:48)0, t(cid:48)0) = φ(x, t)
(1.9)
2
(cid:54)
where the coordinates and times are related by a Galilean transformation. Therefore
(cid:48)
(cid:48)
(cid:48)
φ0(x0, t0) =
2π
λ
(x (cid:0) V t) =
−
2π
λ
(cid:48)
(x0 + vt0 (cid:0) V t0) =
(cid:48) − (cid:48)
2π
λ
(cid:48) −
x0 (cid:0)
−
2π(V (cid:0) v)
λ
(cid:48)... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/0c07cbdc9c352c39eb9539b31ded90d7_MIT8_04S16_LecNotes4.pdf |
:48)
Ψ(x, t) 6= Ψ0(x0, t0) ,
(cid:48)
(cid:48)
(1.13)
where (x, t) and (x0, t0) are related by Galilean transformations and thus represent the same point and
time. You will figure out in Homework the correct relation between Ψ(x, t) and Ψ0(x0, t0).
(cid:48)
(cid:48)
(cid:48)
(cid:48)
(cid:48)
What is the frequency ω of ... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/0c07cbdc9c352c39eb9539b31ded90d7_MIT8_04S16_LecNotes4.pdf |
.18)
(cid:54)
(cid:54)
Similarly, for waves whose phases are relativistic invariant we have another four-vector
(cid:17)
, k
(cid:16) ω
c
(1.19)
Setting two four-vectors equal to each other is a con
frames. As you can see, both de Broglie relations follow from
sistent choice: it would be valid in all Lorentz
(cid:19)
... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/0c07cbdc9c352c39eb9539b31ded90d7_MIT8_04S16_LecNotes4.pdf |
which the packet ψ(x, t) takes large values.
We use the stationary phase principle: since only for k (cid:24) k0 the integral over k has a chance to give a
non-zero contribution, the phase factor must be stationary at k = k0. The idea is simple: if a function
is multiplied by a rapidly varying phase, the integral washe... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/0c07cbdc9c352c39eb9539b31ded90d7_MIT8_04S16_LecNotes4.pdf |
−
(cid:0) k0)
(cid:1)
2
.
Then we find, neglecting the O((k (cid:0) k )2
−
0
) terms
(cid:90)
ψ(x, t) =
dk Φ(k) eikx e(cid:0)iω(k0)te
−
− −
i(k k ) dω
(cid:0) (cid:0) 0 dk jk0 .
|
t
It is convenient to take out of the integral all the factors that do not depend on k:
ψ(x, t) = e
−
(cid:0)
−
(cid:0)
= e
iω(k )t+ik dω
0
|... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/0c07cbdc9c352c39eb9539b31ded90d7_MIT8_04S16_LecNotes4.pdf |
|
(cid:16)
(cid:12)
(cid:12)
(cid:12)ψ
−
x (cid:0)
dω
dk
(cid:12)
(cid:12)
(cid:12)
(cid:12)
k0
t,
0
(cid:17)(cid:12)
(cid:12).
(cid:12)
If ψ(x, 0) peaks at some value x it is clear from the abov
e equation
0
|
that jψ(x, t)j peaks for
|
−
x (cid:0)
(cid:12)
dω
(cid:12)
(cid:12)
dk k
(cid:12)
0
t = x0 ! x = x0 +
→
dω
d... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/0c07cbdc9c352c39eb9539b31ded90d7_MIT8_04S16_LecNotes4.pdf |
∝
−
−
−
In the third and fourth options we have indicated that the time dependence could come with either
sign. We will use superposition to decide which is the right one! We are looking for a wave-function
which is non-zero for all values of x.
Let’s take them one by one:
1. Starting from (1), we build a superposition... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/0c07cbdc9c352c39eb9539b31ded90d7_MIT8_04S16_LecNotes4.pdf |
−
(3.5)
(3.6)
(3.7)
This wavefunction meets our criteria! It is never zero for all values of x because e(cid:0)iωt is never
zero.
−
6
4. A superposition of exponentials from (4) also meets our criteria
−
−
Ψ(x, t) = e(cid:0)i(kx(cid:0)ωt) + e(cid:0)i((cid:0)kx(cid:0)ωt)
= (eikx + e(cid:0)ikx) eiωt
= 2 cos kx eiωt .
− ... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/0c07cbdc9c352c39eb9539b31ded90d7_MIT8_04S16_LecNotes4.pdf |
ei(k(cid:1)x(cid:0)ωt) ,
· −
representing a particle with
p = (cid:126) k ,
and E = (cid:126)ω .
(3.12)
(3.13)
(3.14)
(3.15)
Andrew Turner and Sarah Geller transcribed Zwiebach’s handwritten notes to create the first LaTeX
version of this document.
7
MIT OpenCourseWare
https://ocw.mit.edu
8.04 Quantum Physics I
Spring ... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/0c07cbdc9c352c39eb9539b31ded90d7_MIT8_04S16_LecNotes4.pdf |
Compression in Bayes nets
• A Bayes net compresses the joint
probability distribution over a set of
variables in two ways:
– Dependency structure
– Parameterization
• Both kinds of compression derive from
causal structure:
– Causal locality
– Independent causal mechanisms
Dependency structure
,
MJAEBP
,
)
)
,
AMPAJ... | https://ocw.mit.edu/courses/9-66j-computational-cognitive-science-fall-2004/0c18b39d66e8cc6e2125099193dc722e_oct_12_2004_fin.pdf |
)
0 0 0
0 1 wB = 0.29
1 0 wE = 0.94
1 1 wB +(1-wB )wE
JohnCalls
A P(J|A)
0 0.05
1 0.90
MaryCalls
FIX FOR NEXT
YEAR Wb and We
A P(M|A)
0 0.01
1 0.70
Outline
• The semantics of Bayes nets
– role of causality in structural compression
• Explaining away revisited
– role of causality in probabilistic inference
• Sampling a... | https://ocw.mit.edu/courses/9-66j-computational-cognitive-science-fall-2004/0c18b39d66e8cc6e2125099193dc722e_oct_12_2004_fin.pdf |
Global factorization is equivalent to a set of constraints
on pairwise relationships between variables.
“Markov property”: Each node is conditionally
independent of its non-descendants given its parents.
U1
Um
Z1j
X
Znj
Y1
Yn
Figure by MIT OCW.
Local semantics
Global factorization is equivalent to a set of constra... | https://ocw.mit.edu/courses/9-66j-computational-cognitive-science-fall-2004/0c18b39d66e8cc6e2125099193dc722e_oct_12_2004_fin.pdf |
Suppose we get the direction of causality wrong...
Burglary
Earthquake
Alarm
JohnCalls
MaryCalls
• Inserting a new arrow captures this correlation.
• This model is too complex: does not believe that
P(JohnCalls, MaryCalls|Alarm) =
P(JohnCalls|Alarm) P(MaryCalls|Alarm)
An alternative basis
Suppose we get the directio... | https://ocw.mit.edu/courses/9-66j-computational-cognitive-science-fall-2004/0c18b39d66e8cc6e2125099193dc722e_oct_12_2004_fin.pdf |
|B,E)
0 0 0
0 1 1
1 0 1
1 1 1
Explaining away
• Logical OR: Independent deterministic causes
Burglary
Earthquake
Alarm
A priori, no correlation between B and E:
ebP
),(
=
ePbP
)(
)(
B E P(A|B,E)
0 0 0
0 1 1
1 0 1
1 1 1
Explaining away
• Logical OR: Independent deterministic causes
Burglary
Earthquake
Alarm
B E P(A|B,... | https://ocw.mit.edu/courses/9-66j-computational-cognitive-science-fall-2004/0c18b39d66e8cc6e2125099193dc722e_oct_12_2004_fin.pdf |
OR: Independent deterministic causes
Burglary
Earthquake
Alarm
After observing A = a, E= e, …
B E P(A|B,E)
0 0 0
0 1 1
1 0 1
1 1 1
eabP
|
),
(
=
|
(
ebPebaP
),
(
)|
eaP
(
)|
Both terms = 1
Explaining away
• Logical OR: Independent deterministic causes
Burglary
Earthquake
Alarm
After observing A = a, E= e, …
B E P(A|... | https://ocw.mit.edu/courses/9-66j-computational-cognitive-science-fall-2004/0c18b39d66e8cc6e2125099193dc722e_oct_12_2004_fin.pdf |
)
1
])
Annoyance = C1* w1 +C2* w2
Contrast w/ conditional reasoning
Rain
Sprinkler
Grass Wet
• Formulate IF-THEN rules:
– IF Rain THEN Wet
– IF Wet THEN Rain
IF Wet AND NOT Sprinkler
THEN Rain
• Rules do not distinguish directions of inference
• Requires combinatorial explosion of rules
Spreading activation or recurr... | https://ocw.mit.edu/courses/9-66j-computational-cognitive-science-fall-2004/0c18b39d66e8cc6e2125099193dc722e_oct_12_2004_fin.pdf |
More on the relation between causality and
probability:
Causal structure
Statistical dependencies
• Learning causal graph structures.
• Learning causal abstractions (“diseases
cause symptoms”)
• What’s missing from graphical models
Outline
• The semantics of Bayes nets
– role of causality in structural compression
... | https://ocw.mit.edu/courses/9-66j-computational-cognitive-science-fall-2004/0c18b39d66e8cc6e2125099193dc722e_oct_12_2004_fin.pdf |
(
)
“marginalization”
A more complex system
Battery
Radio
Ignition
Gas
Starts
On time to work
• Joint distribution sufficient for any inference:
SOPGISPGPBIPBRPBPOSGIRBP
)
=
)
(
)
)
(
(
)
)
(
(
(
(
,
,
,
,
,
,
|
|
|
|
)
GOP
|
(
)
=
)
GOP
,
(
GP
(
)
=
∑
SIRB
,
,
(
(
)
SOPGISPGPBIPBRPBP
,
|(
)
)
)
)
(
(
(
,
|
|
|
)
GP
(... | https://ocw.mit.edu/courses/9-66j-computational-cognitive-science-fall-2004/0c18b39d66e8cc6e2125099193dc722e_oct_12_2004_fin.pdf |
sprinkler
,
,
,
,
¬
rain
¬
rain
wet
>
wet
>
,
,
¬
rain
,
¬
wet
>
rain
,
wet
>
,
¬
sprinkler
,
¬
rain
,
¬
wet
>
¬<
cloudy
. . .
As the sample gets larger,
the frequency of each
possible world approaches
its true prior probability
under the model.
How do we use these
samples for inference?
Summary
• Exact inference me... | https://ocw.mit.edu/courses/9-66j-computational-cognitive-science-fall-2004/0c18b39d66e8cc6e2125099193dc722e_oct_12_2004_fin.pdf |
8.06 Spring 2016 Lecture Notes
2. Time-dependent approximation methods
Aram Harrow
Last updated: March 12, 2016
Contents
1 Time-dependent perturbation theory
1.1 Rotating frame . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Perturbation expansion . . . . . . . . . . . . . . . . . . . ... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
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