text stringlengths 16 3.88k | source stringlengths 60 201 |
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suppose both ∈ F. Let
G = G(Y ) ⊂ F be the field generated by Y . We define E[X|Y ] to be
E[X|G].
1.3 Proof of existence
We now give a proof sketch of Theorem 1.
Proof. Given two probability measures P1, P2 defined on the same (Ω, F), P2
is defined to be absolutely continuous with respect to P1 if for every set A ∈ F... | https://ocw.mit.edu/courses/15-070j-advanced-stochastic-processes-fall-2013/69dc2223b9957bbcc754de94dfc64cfd_MIT15_070JF13_Lec9.pdf |
of E to emphasize that the expectation operator is
with respect to the original measure P. Check that this is indeed a probability
measure on (Ω, G). Now P also induced a probability measure on (Ω, G). We
claim that P2 is absolutely continuous with respect to P. Indeed if P(A) = 0
then the numerator is zero. By the... | https://ocw.mit.edu/courses/15-070j-advanced-stochastic-processes-fall-2013/69dc2223b9957bbcc754de94dfc64cfd_MIT15_070JF13_Lec9.pdf |
b ≤ φ(x), ∀ x}.
Then φ(x) = sup{ax + b : (a, b) ∈ A}.
Now we prove the Jensen’s inequality. For any pair of rationals a, b ∈
Q satisfying the bound above, we have, by monotonicity that E[φ(X)|G] ≥
aE[X|G] + b, a.s., implying E[φ(X)|G] ≥ sup{aE[X|G] + b : (a, b) ∈ A} =
φ(E[X|G]) a.s.
Tower property. Suppose G1 ⊂ G... | https://ocw.mit.edu/courses/15-070j-advanced-stochastic-processes-fall-2013/69dc2223b9957bbcc754de94dfc64cfd_MIT15_070JF13_Lec9.pdf |
E[E[X|G]] = E[X].
5
3 Filtration and martingales
3.1 Definition
A family of σ-fields {Ft} is defined to be a filtration if Ft1 ⊂ Ft2 whenever
t1 ≤ t2. We will consider only two cases when t ∈ Z+ or t ∈ R+. A stochastic
process {Xt} is said to be adapted to filtration {Ft} if Xt ∈ Ft for every t.
Definition 2. A stoc... | https://ocw.mit.edu/courses/15-070j-advanced-stochastic-processes-fall-2013/69dc2223b9957bbcc754de94dfc64cfd_MIT15_070JF13_Lec9.pdf |
Sn − µn =
Xk − µn is a martingale. Indeed Sn is adapted to
Fn, and
0≤k≤n
i
E[Sn+1 − (n + 1)µ|Fn] = E[Xn+1 − µ + Sn − nµ|Fn]
= E[Xn+1 − µ|Fn] + E[Sn − nµ|Fn]
a = E[Xn+1 − µ] + Sn − nµ
= Sn − nµ.
Here in (a) we used the fact that Xn+1 is independent from Fn and Sn ∈
Fn.
2. Random walk squared. Under the same s... | https://ocw.mit.edu/courses/15-070j-advanced-stochastic-processes-fall-2013/69dc2223b9957bbcc754de94dfc64cfd_MIT15_070JF13_Lec9.pdf |
Kinetics of Rigid Bodies
1
2.003J/1.053J Dynamics and Control I, Spring 2007
Professor Thomas Peacock
3/7/2007
Lecture 9
2D Motion of Rigid Bodies: Kinetics, Poolball
Example
Kinetics of Rigid Bodies
Angular Momentum Principle for a Rigid Body
Figure 1: Rigid Body rotating with angular velocity ω. Figure by MIT ... | https://ocw.mit.edu/courses/2-003j-dynamics-and-control-i-spring-2007/6a37882fb3147897a4ebbcee39bd791a_lec09.pdf |
with mass M . Figure by MIT OCW.
If one takes angular momentum about the center of mass:
H B
�
= r
c
× P + Icω
H c = Icω
(Angular Momentum about B) = (Angular Momentum about C) + (Moment
of Linear Momentum about B)
Therefore:
H B = H c
�
c × P
+ r
Special Case of Fixed Axis of Rotation about B
�
i.e. v c = v... | https://ocw.mit.edu/courses/2-003j-dynamics-and-control-i-spring-2007/6a37882fb3147897a4ebbcee39bd791a_lec09.pdf |
center of mass.
ext = H C and H C = IC ω
τB
d
dt
Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J Dynamics and
Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology.
Downloaded on [DD Month YYYY].
Cue hitting a pool ball... | https://ocw.mit.edu/courses/2-003j-dynamics-and-control-i-spring-2007/6a37882fb3147897a4ebbcee39bd791a_lec09.pdf |
ed on [DD Month YYYY].
Cue hitting a pool ball
5
Kinetics: Free Body Diagrams
Figure 5: Free Body Diagram of Cue Ball. Figure by MIT OCW.
Impulse force that provides impulse J
0+
J =
�
0−
F dt =
0+
�
0−
Jδ(t)dt i.e. F = Jδ(t)
(i) Linear Momentum Principle
F ext = P
d
dt
y-direction: C always at same ... | https://ocw.mit.edu/courses/2-003j-dynamics-and-control-i-spring-2007/6a37882fb3147897a4ebbcee39bd791a_lec09.pdf |
0+
F hdt =
0+
�
0−
IC
dω
dt
dt
Jδ(t)dt = IC ω(0+) − IC ω(0
−
)
0−
�
0+
�
0−
IC ω(0−) = 0 because ω(0−) = 0.
Jh = IC ω(0+)
Impulsive torque about center of mass = Change of angular momentum caused
by the torque
Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J Dy... | https://ocw.mit.edu/courses/2-003j-dynamics-and-control-i-spring-2007/6a37882fb3147897a4ebbcee39bd791a_lec09.pdf |
6 . 2 7 0 : A U T O N O M O U S R O B O T
D E S I G N C O M P E T I T I O N
• Assignment 1: General
•
Comments
Sensor store and $30
Electronics Rule
Electronics review
•
• Handy Board hardware and
interface
Sensors and motors
Interactive C development
environment
Assignment 2 handed out
•
•
•
LECTURE 2... | https://ocw.mit.edu/courses/6-270-autonomous-robot-design-competition-january-iap-2005/6a3fdcd640dba5716eb4d7d07e77ec3f_lecture2_slides.pdf |
price you’re claiming
– If you’re getting free samples, they don’t count as free
• Make sure you know what you are doing
• Check out our stuff before you start looking elsewhere
• Tools do not count towards the allotment
• Replacements for servos do not count towards your $30 allotment—
only based on what’s on you... | https://ocw.mit.edu/courses/6-270-autonomous-robot-design-competition-january-iap-2005/6a3fdcd640dba5716eb4d7d07e77ec3f_lecture2_slides.pdf |
ADC
GND
HandyBoard Sensor Inputs
• HandyBoard has digital and analog hardware
ports
– You can read both types of ports as digital or analog
values
– BUT, digital ports read as analog values always give 0
or 255
– Analog ports read as digital values use a cutoff to
decide if it is 0 or 1
– Analog ports: 0-6, 1... | https://ocw.mit.edu/courses/6-270-autonomous-robot-design-competition-january-iap-2005/6a3fdcd640dba5716eb4d7d07e77ec3f_lecture2_slides.pdf |
third pin
• Current can flow
either direction,
motor runs in either
direction
Motors
Interactive C
• Originally developed for 6.270 by Randy
Sargent
• Supports Handy Board, RugWarrior and
RugWarrior Pro
Where to Develop
• Work at Lab
– IC 3.1
– Laptops in lab
• Work at Home
– IC 3.2 for Windows
– http:/... | https://ocw.mit.edu/courses/6-270-autonomous-robot-design-competition-january-iap-2005/6a3fdcd640dba5716eb4d7d07e77ec3f_lecture2_slides.pdf |
-C206): 7 loaded
Returned <int> 60
• Can also evaluate series of expressions by creating a block with
curly braces
– Example:
C> { int i = 10; printf( “%d\n”, i ); }
Downloading 22 bytes (addresses C200-C215): 22 loaded
(LCD displays “10”)
Program Development
• Develop in favorite editor (vi, emacs,
Notepad)
• In I... | https://ocw.mit.edu/courses/6-270-autonomous-robot-design-competition-january-iap-2005/6a3fdcd640dba5716eb4d7d07e77ec3f_lecture2_slides.pdf |
code, ensure that file
containing them is listed first
• load <listfile> loads files in the
order prescribed
Printing to LCD
• Only 31 characters ( 16 x 2 – ♥ )
• Characters printed beyond final character
position are cut off
• printf() treats the LCD screen as one
long line instead of two
• Cannot print long... | https://ocw.mit.edu/courses/6-270-autonomous-robot-design-competition-january-iap-2005/6a3fdcd640dba5716eb4d7d07e77ec3f_lecture2_slides.pdf |
0 milliseconds
• long mseconds()
• float seconds()
– 1 millisecond resolution
– int vs. float (the period)
• void sleep(float sec)
– At sec or a little longer than sec seconds
• void msleep(long msec)
– At msec or longer than msec milliseconds
Tones
• void beep()
• And for the bored:
• void tone(float freq, f... | https://ocw.mit.edu/courses/6-270-autonomous-robot-design-competition-january-iap-2005/6a3fdcd640dba5716eb4d7d07e77ec3f_lecture2_slides.pdf |
18.212: Algebraic Combinatorics
Andrew Lin
Spring 2019
This class is being taught by Professor Postnikov.
March 13, 2019
Remember from last lecture: if we have a poset P , then J(P ) is the poset of order ideals in P , ordered by inclusion. (Order
ideals are closed downward.)
Lemma 1
J(P ) is a distributive lat... | https://ocw.mit.edu/courses/18-212-algebraic-combinatorics-spring-2019/6a5b3667bc3139abefe8adabc0ada96d_MIT18_212S19_lec16.pdf |
if it is not the minimal element of L [which exists] and we cannot express it as
x = y ∨ z for y , z < x (≤ and not equal).
For example, if we take the lattice above, everything except the bottom and top element is join-irreducible.
But it turns out we can just construct P to be the subposet of L of all join-irreduc... | https://ocw.mit.edu/courses/18-212-algebraic-combinatorics-spring-2019/6a5b3667bc3139abefe8adabc0ada96d_MIT18_212S19_lec16.pdf |
P , then the rank is just ρ(I) = |I|. So our
goal is to show that I l J, they only di˙er in one element: this is true because I < J means I is strictly contained in J.
Definition 6
Let P be a finite ranked poset. Define ri to be the number of elements in P with rank i : call this a rank number
of P . These rank number... | https://ocw.mit.edu/courses/18-212-algebraic-combinatorics-spring-2019/6a5b3667bc3139abefe8adabc0ada96d_MIT18_212S19_lec16.pdf |
just an increasing chain of n elements. Clearly this is rank-symmetric
and unimodular.
Example 9
What does [m] × [n] look like? We have a grid, but we have to rotate it by 45 degrees. Then (m, n) has rank m + n.
2
The rank numbers look like (1, 2, · · · , k − 1, k, · · · , k, k − 1, · · · , 2, 1), where k is the ... | https://ocw.mit.edu/courses/18-212-algebraic-combinatorics-spring-2019/6a5b3667bc3139abefe8adabc0ada96d_MIT18_212S19_lec16.pdf |
):
Theorem 10
L(m, n) is rank-symmetric and unimodular.
Let’s go back to looking at some examples! What is J([2] × [n])? We want the Young diagrams that fit inside a 2 × n
rectangle. This Hasse diagram looks like a triangle:
But let’s look at the order ideals of this triangular Hasse diagram: what is
J(J([2] × [n)... | https://ocw.mit.edu/courses/18-212-algebraic-combinatorics-spring-2019/6a5b3667bc3139abefe8adabc0ada96d_MIT18_212S19_lec16.pdf |
opposite: no two elements are
compatible.
Definition 12
Let P be a finite ranked poset with rank numbers r0, r1, · · · , rN . P is Sperner if M, the maximal size of an antichain
in P , is max(r0, · · · , rN ).
It’s clear that M should be at least the maximum of r0, · · · , rN : just take all elements with some fixed ... | https://ocw.mit.edu/courses/18-212-algebraic-combinatorics-spring-2019/6a5b3667bc3139abefe8adabc0ada96d_MIT18_212S19_lec16.pdf |
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Physics Department
Physics 8.07: Electromagnetism II
Prof. Alan Guth
October 11, 2012
LECTURE NOTES 8
THE TRACELESS SYMMETRIC TENSOR EXPANSION
AND STANDARD SPHERICAL HARMONICS
These notes are an addendum to Lecture 14, Wednesday October 10, 2012. The
notes will describe a topic tha... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/6a67040c6a90f70c8164cdea63473a30_MIT8_07F12_ln8.pdf |
)mY(cid:1)m(θ, φ) .
We have shown that
(cid:2)
((cid:1))
θ Ci1i2...i n
(cid:1) ˆi1
nˆi2
. . . nˆ
(cid:3)
(cid:1) = −(cid:12)((cid:12) + 1)
i
(
(cid:1)
)Ci1i2...i n
(cid:1) ˆi1
nˆi2
. . . nˆi(cid:1)
,
∇2
where
∇2
θ =
1
∂
sin θ ∂θ
(cid:5)
(cid:4)
sin θ
∂
∂θ
+
1
sin2
∂2
θ ∂φ2
.
In the standard approach one would show that... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/6a67040c6a90f70c8164cdea63473a30_MIT8_07F12_ln8.pdf |
(cid:1) = Y(cid:1)m(θ, φ
) .
(8.7)
((cid:1),m)
Ci1...i(cid:1) explicitly. We have already shown that the number of
Our goal is to construct
linearly independent traceless symmetric tensors of rank (cid:12) (i.e., with (cid:12) indices) is given
by 2(cid:12) + 1, which is not surprisingly equal to the number of Y(cid:1)... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/6a67040c6a90f70c8164cdea63473a30_MIT8_07F12_ln8.pdf |
vector, which is the same as a tensor of rank 1. It is obvious
that the only vector that is invariant under rotations about the z-axis is a vector that
points along the z axis. I will let zˆ be a unit vector in the z-direction (which I have also
called eˆz and eˆ3), and then for azimuthal symmetry we have
(1)Ci
(1)Ci =... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/6a67040c6a90f70c8164cdea63473a30_MIT8_07F12_ln8.pdf |
zˆ } = ˆz zˆ zˆ − 1
5 ˆiδjk + ˆzj δik + ˆzkδij
zˆizˆjδkm + ˆzizˆkδmj
(cid:10)
i ˆjzˆkzˆm } = ˆzizˆjzˆkzˆm −
i j k
i j k
1
7
z
(cid:9)
(cid:10)
{ zˆ z
+ ˆzjzˆmδik + ˆz
kzˆmδij +
+ ˆzizˆmδjk + ˆzj zˆkδim
(cid:9)
δij δkm + δikδjm
1
35
+ δimδjk
(8.14)
(cid:10)
,
where the coefficients are all determined b
We will
argue later... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/6a67040c6a90f70c8164cdea63473a30_MIT8_07F12_ln8.pdf |
nˆi1
. . . nˆi(cid:1)
,
(8.16)
where the constant is yet to be determined.
Both sides of Eq. (8.16) are polynomials in cos θ, where the highest power is cos(cid:1) θ.
If we can find the coefficients of this highest power on each side of the equation, we
can determine the constant. On the right-hand side, the highest power... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/6a67040c6a90f70c8164cdea63473a30_MIT8_07F12_ln8.pdf |
)
(cid:4)
d
x
d
(cid:5)(cid:1)−2 (cid:7)
(2(cid:12))(2(cid:12)
− 1)x2(cid:1)−2 + (lower powers)
(cid:8)
(cid:7)
(2(cid:12))(2(cid:12)
−
1) . . . ((cid:12) + 1)x(cid:1) + (lower powers)
(cid:8)
x(cid:1) + (lower powers) .
(8.17)
Matching these coefficients, we see that
P(cid:1)(cos θ) =
(2(cid:12))!
2(cid:1)((cid:12)!)
2
... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/6a67040c6a90f70c8164cdea63473a30_MIT8_07F12_ln8.pdf |
) = vx cos ψ − vy sin ψ
vx
(cid:3) = vx sin ψ + vy cos ψ .
vy
(8.20)
8.07 LECTURE NOTES 8, FALL 2012
We thus seek an eigenvector of the matrix
(cid:4)
R =
cos ψ − sin ψ
cos ψ
sin ψ
(cid:5)
.
The eigenvalues λ of the matrix are determined by the characteristic equation
det(R − λI) = 0 ,
p. 5
(8.21)
(8.22)
where I is th... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/6a67040c6a90f70c8164cdea63473a30_MIT8_07F12_ln8.pdf |
�(j) = δij .
We can complete a basis for three-dimensional vectors by adding
uˆ(3) ≡ zˆ = ˆez .
(8.27)
(8.28)
(8.29)
You might ask how one should visualize a vector with imaginary components. What
direction does it point? It certainly points in a definite direction in complex three-
dimensional space, which is equivalen... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/6a67040c6a90f70c8164cdea63473a30_MIT8_07F12_ln8.pdf |
the right-hand side is expanded and all the indices are summed to give dot
products, the only terms that survive are those for which every uˆ+ is dotted into zˆ.
Thus, the right-hand side of Eq. (8.32) is proportional to eimφ. From Eq. (8.4), we know
that the right-hand side of Eq. (8.32) is an eigenfunction of ∇2
θ wi... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/6a67040c6a90f70c8164cdea63473a30_MIT8_07F12_ln8.pdf |
�)meimφ
(cid:7)
(cos θ)(cid:1)−m + (lower powers of cos θ)
(cid:8)
.
(8.35)
8.07 LECTURE NOTES 8, FALL 2012
p. 7
To compare Eq. (8.35) with the leading term in the expansion for the standard
It is given in Jackson as Eq. (3.53),
function Y(cid:1)m, we need a formula for Y(cid:1)m(θ, φ).
p. 108, as
(cid:12)
Y(cid:1)m(θ... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/6a67040c6a90f70c8164cdea63473a30_MIT8_07F12_ln8.pdf |
(cid:1)
. . . n
ˆi(cid:1)
,
where
with
Ci i
((cid:1),m)
1 2...i(cid:1) =
d(cid:1)m u+ . . . u+ zi
ˆim ˆ m+1
{ ˆi1
(cid:12)
. . . zi
ˆ (cid:1)
}
,
d(cid:1)m =
(−1)m(2(cid:12))!
2(cid:1)(cid:12)!
2m (2(cid:12) + 1)
4π ((cid:12) + m)! ((cid:12) − m)!
.
(8.39)
(8.40)
(8.41)
For negative values of m, the calculation is iden... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/6a67040c6a90f70c8164cdea63473a30_MIT8_07F12_ln8.pdf |
{ uˆ+
i
1
= ˆu+
i
1
i zˆi
. . . uˆ+
ˆ+
i z
. . . u
m ˆim+1
mm
. . .
zˆi(cid:1)
} {nˆ i1
. . . nˆi(cid:1)
} (8.44a)
+1
. . . zˆi(cid:1)
{ nˆi1
. . . nˆi(cid:1)
} ,
(8.44b)
8.07 LECTURE NOTES 8, FALL 2012
p. 8
. . . nˆi(cid:1)
} differs from nˆi1
where the top line is justified because { nˆi1
. . . nˆi(cid:1) only by term... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/6a67040c6a90f70c8164cdea63473a30_MIT8_07F12_ln8.pdf |
to eimφ, and that both are
eigenfunctions of ∇2
θ with eigenvalue −(cid:12)((cid:12) + 1). We claimed that, up to a multiplicative
constant, there is only one function that has these properties. Assuming that the power
series representation of Eq. (8.1) always exists, the uniqueness that we need is easy to
see. We show... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/6a67040c6a90f70c8164cdea63473a30_MIT8_07F12_ln8.pdf |
6.088 Intro to C/C++
Day 6: Miscellaneous Topics
Eunsuk Kang & Jean Yang
In the last lecture...
Inheritance
Polymorphism
Abstract base classes
Today’s topics
Polymorphism (again)
Namespaces
Standard Template Library
Copying objects
Integer overflow
Polymorphism revisited
Polymorphism revisited
Recall: Abil... | https://ocw.mit.edu/courses/6-088-introduction-to-c-memory-management-and-c-object-oriented-programming-january-iap-2010/6a77c25467108764576f82f7f0f4e12a_MIT6_088IAP10_lec06.pdf |
Change the type hierarchy!
Think about behaviors, not just characteristics of an
object!
Solutions
Ugly: Modify setWidth and setLength in Rectangle to
return a boolean.They always return “true” in
Rectangle, and “false” in Square. Define a separate
method “setDimension” in Square.
Better: Maybe Square shouldn’t... | https://ocw.mit.edu/courses/6-088-introduction-to-c-memory-management-and-c-object-oriented-programming-january-iap-2010/6a77c25467108764576f82f7f0f4e12a_MIT6_088IAP10_lec06.pdf |
protected:
int id;
std::string name;
std::string address;
public:
MITPerson(int id, std::string name, std::string address);
void displayProfile();
void changeAddress(std::string newAddress);
};
Using namespace std
#include <string>
class MITPerson {
protected:
int id;
std::string name;
std::string addre... | https://ocw.mit.edu/courses/6-088-introduction-to-c-memory-management-and-c-object-oriented-programming-january-iap-2010/6a77c25467108764576f82f7f0f4e12a_MIT6_088IAP10_lec06.pdf |
primitives initialized to
some default value
objects created using
default constructor
Student class from the last lecture
#include <iostream>
#include <vector>
#include "MITPerson.h"
#include "Class.h"
class Student : public MITPerson {
int course;
int year;
std::vector<Class*> classesTaken;
// 1 = freshman,... | https://ocw.mit.edu/courses/6-088-introduction-to-c-memory-management-and-c-object-oriented-programming-january-iap-2010/6a77c25467108764576f82f7f0f4e12a_MIT6_088IAP10_lec06.pdf |
vector
classesTaken.push_back(c1);
classesTaken.push_back(c2);
// accessing an element
Class* c3 = classesTaken[0];
Class* c4 = classesTaken.at(1);
std::cout << c3.getName() << “\n”; // prints “6.01”
std::cout << c4.getName() << “\n”; // prints “6.005”
// removing elements from the back of the vector
classesTaken.pop... | https://ocw.mit.edu/courses/6-088-introduction-to-c-memory-management-and-c-object-oriented-programming-january-iap-2010/6a77c25467108764576f82f7f0f4e12a_MIT6_088IAP10_lec06.pdf |
Traversing a vector using an iterator
// display a list of classes taken by the student
// create an iterator
std::vector<Class*>::iterator it;
std::cout << "Classes taken:\n";
// step through every element in the vector
for (it = classesTaken.begin(); it != classesTaken.end(); it++){
Class* c = *it;
std::cout <... | https://ocw.mit.edu/courses/6-088-introduction-to-c-memory-management-and-c-object-oriented-programming-january-iap-2010/6a77c25467108764576f82f7f0f4e12a_MIT6_088IAP10_lec06.pdf |
Person p){
p.displayProfile();
}
int main() {
2. pass by value, so make a copy
MITPerson p1(921172, “James Lee”, “32 Vassar St.”);
MITPerson p2 = p1;
print(p2);
}
(3. could also return an object as a return value)
37
Copying objects using constructors
void print(MITPerson p){
p.displayProfile();
}
int mai... | https://ocw.mit.edu/courses/6-088-introduction-to-c-memory-management-and-c-object-oriented-programming-january-iap-2010/6a77c25467108764576f82f7f0f4e12a_MIT6_088IAP10_lec06.pdf |
Copy assignment operator in MITPerson
#include <string>
class MITPerson {
protected:
int id;
std::string name;
std::string address;
public:
MITPerson(int id, std::string name, std::string address);
MITPerson(const MITPerson& other);
MITPerson& operator=(const MITPerson& other);
void displayProfile();
void chang... | https://ocw.mit.edu/courses/6-088-introduction-to-c-memory-management-and-c-object-oriented-programming-january-iap-2010/6a77c25467108764576f82f7f0f4e12a_MIT6_088IAP10_lec06.pdf |
pb = new B(); }
~A() { delete pb; } // destructor
void printB() { pb->print(); }
};
void foo(A a) {
a.printB();
}
int main() {
A a1(5);
a1.printB();
foo(a1);
return 0;
}
Double free!
How do we fix this?
47
Default copy constructor - caution!
class B {
public:
void print() { std::cout << "Hello World!\n”... | https://ocw.mit.edu/courses/6-088-introduction-to-c-memory-management-and-c-object-oriented-programming-january-iap-2010/6a77c25467108764576f82f7f0f4e12a_MIT6_088IAP10_lec06.pdf |
= length - 1;
while (low <= high) {
int mid = (low + high) / 2;
int midVal = a[mid];
if (midVal < key)
low = mid + 1
else if (midVal > key)
high = mid - 1;
else
return mid;
// key found
}
return -(low + 1); // key not found
}
Can you find the bug?
Courtesy of Joshua Bloch. Used with permission.
52
Binary s... | https://ocw.mit.edu/courses/6-088-introduction-to-c-memory-management-and-c-object-oriented-programming-january-iap-2010/6a77c25467108764576f82f7f0f4e12a_MIT6_088IAP10_lec06.pdf |
Object-Oriented Analysis and Design with
Applications (G. Booch, et. al)
57
Congratulations!
Now you know enough about C/C++ to
embark on your own journey!
58
MIT OpenCourseWare
http://ocw.mit.edu
6.088 Introduction to C Memory Management and C++ Object-Oriented Programming
January IAP 2010
For information about ... | https://ocw.mit.edu/courses/6-088-introduction-to-c-memory-management-and-c-object-oriented-programming-january-iap-2010/6a77c25467108764576f82f7f0f4e12a_MIT6_088IAP10_lec06.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
2.161 Signal Processing: Continuous and Discrete
Fall 2008
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
Massachusetts Institute of Technology
Department of Mechanical Engineering
2.161 Signal Processing - Continuous and Di... | https://ocw.mit.edu/courses/2-161-signal-processing-continuous-and-discrete-fall-2008/6a9dd6d7fe4f6102720e6dcded0c76e9_lecture_02.pdf |
14) (cid:5)
����
�
�
�
�
�
�
�
�
�
�
�
�
����
�
�����
�
�
� � �� �� ����
�
����
�
�
�
�
�
�
�
�
�
�
�
�
����
�����
�
�
�����
�
� � �� �� ����
�
����
u˜T (t) = u(nT )
for nT ≤ t < (n + 1)T.
• The approximation ˜uT (t) is written as a superposition of non-overlapping pulses
1copyright (cid:3) D.Rowell... | https://ocw.mit.edu/courses/2-161-signal-processing-continuous-and-discrete-fall-2008/6a9dd6d7fe4f6102720e6dcded0c76e9_lecture_02.pdf |
nated hT (t) as shown below. If the system is linear and time-invariant, the response
to a delayed unit pulse, occurring at time nT is simply a delayed version of the pulse
response:
yn(t) = hT (t − nT )
(cid:2)
(cid:3) (cid:4)
(cid:3)
(cid:3) (cid:7)
(cid:6) (cid:10) (cid:3)
(cid:2)
(cid:3) (cid:4)
(cid:3)
(cid... | https://ocw.mit.edu/courses/2-161-signal-processing-continuous-and-discrete-fall-2008/6a9dd6d7fe4f6102720e6dcded0c76e9_lecture_02.pdf |
not contribute to the sum, so that the upper limit of the
summation may be rewritten:
(cid:2)
N
y˜T (t) =
u(nT )hT (t − nT )T
for N T ≤ t < (N + 1)T.
n=−∞
2–2
(cid:2)
(cid:8)
(cid:9)
(cid:2)
(cid:4)
(cid:8)
(cid:8)
(cid:9)
(cid:2)
(cid:2)
(cid:4)
(cid:13)
(cid:4)
(cid:8)
(cid:7)
(cid:2)
(cid:4)
(cid:2)
(cid:5)... | https://ocw.mit.edu/courses/2-161-signal-processing-continuous-and-discrete-fall-2008/6a9dd6d7fe4f6102720e6dcded0c76e9_lecture_02.pdf |
u(t) ⊗ h(t) =
u(τ )h(t − τ )dτ.
(2)
Equation (??) is in the form of a linear operator, in that it transforms, or maps, an input
function to an output function through a linear operation.
−∞
(cid:2) (cid:12) (cid:3) (cid:13)
(cid:2) (cid:12) (cid:3) (cid:13)
(cid:19) (cid:18) (cid:17) (cid:9)
(cid:3)
(cid:16) ... | https://ocw.mit.edu/courses/2-161-signal-processing-continuous-and-discrete-fall-2008/6a9dd6d7fe4f6102720e6dcded0c76e9_lecture_02.pdf |
by t to form h(t − τ ). The product u(t)h(t − τ ) is then evaluated and integrated to find
the response. This graphical representation is useful for defining the limits necessary in the
integration. For example, since for a physical system the impulse response h(t) is zero for all
t < 0, the reflected and shifted impul... | https://ocw.mit.edu/courses/2-161-signal-processing-continuous-and-discrete-fall-2008/6a9dd6d7fe4f6102720e6dcded0c76e9_lecture_02.pdf |
the
form u(t) = A sin(Ωt + φ), where A is the amplitude, Ω is the angular frequency (rad/s),
and φ is a phase angle (rad). (We note that we can also write u(t) = A sin(2πF t + φ), where
F is the frequency in Hz.)
We begin by noting that a sinusoid may be expressed in terms of complex exponentials
through the Euler... | https://ocw.mit.edu/courses/2-161-signal-processing-continuous-and-discrete-fall-2008/6a9dd6d7fe4f6102720e6dcded0c76e9_lecture_02.pdf |
BejΩt
(cid:9)
= bm(jΩ)m + bn−1(jΩ)m−1 + · · · + b1(jΩ) + b0 ejΩt
(cid:10)
(cid:10)
so that
where
an(jΩ)n + an−1(jΩ)n−1 + · · · + a1(jΩ) + a0
B = bm(jΩ)m + bn−1(jΩ)m−1 + · · · + b1(jΩ) + b0
y(t) = H(jΩ)ejΩt
H(jΩ) =
N (jΩ)
D(jΩ)
=
an(jΩ)n + an−1(jΩ)n−1 + · · · + a1(jΩ) + a0
bm(jΩ)m + bn−1(jΩ)m−1 + · · · + b1... | https://ocw.mit.edu/courses/2-161-signal-processing-continuous-and-discrete-fall-2008/6a9dd6d7fe4f6102720e6dcded0c76e9_lecture_02.pdf |
complex conjugate.
The response to the real sinusoid
u(t) = A sin (Ωt + φ) =
(cid:10)
ej(Ωt+φ) − e −j(Ωt+φ)
A (cid:9)
2j
may be found from the principle of superposition by summing the response to each compo
nent:
y(t) =
=
A
2j
A
2j
(cid:9)
(cid:10)
H(jΩ)ej(Ωt+φ) − H(−jΩ)e −j(Ωt+φ)
(cid:11)
(cid:12)
H(... | https://ocw.mit.edu/courses/2-161-signal-processing-continuous-and-discrete-fall-2008/6a9dd6d7fe4f6102720e6dcded0c76e9_lecture_02.pdf |
:8)
(cid:9)
(cid:2)
2–6
(cid:3)
(cid:4) (cid:5) (cid:2) (cid:3) (cid:6) (cid:4)
(cid:7)
is described by the differential equation
RC
dvo
dt
+ vo = Vin(t)
Find the frequency response function.
By inspection
and
H(jΩ) = jRCΩ + 1
1
|H(jΩ)| =
(cid:4) H(jΩ) =
|1|
|1 + jRCΩ|
(cid:4) (1) − (cid:4) (1 + jRCΩ) ... | https://ocw.mit.edu/courses/2-161-signal-processing-continuous-and-discrete-fall-2008/6a9dd6d7fe4f6102720e6dcded0c76e9_lecture_02.pdf |
dt
Find the frequency response function.
By inspection
and
jRCΩ
H(jΩ) = jRCΩ + 1
|H(jΩ)| =
|jRCΩ|
|1 + jRCΩ|
= (cid:13)
RCΩ
(RCΩ)2 + 1
(cid:4) H(jΩ) =
(cid:4) (jRCΩ) − (cid:4) (1 + jRCΩ) =
2–7
π − tan−1 (RCΩ)
2
Clearly, as Ω → 0, |H(jΩ)| → 0, and � H(jΩ) → π/2 rad (90◦). As Ω → ∞,
|H(jΩ)| → 1, and � H(... | https://ocw.mit.edu/courses/2-161-signal-processing-continuous-and-discrete-fall-2008/6a9dd6d7fe4f6102720e6dcded0c76e9_lecture_02.pdf |
MEASURE AND INTEGRATION: LECTURE 14
Convex functions. Let ϕ : (a, b) → R, where −∞ ≤ a < b ≤ ∞.
Then ϕ is convex if ϕ((1 − t)x + ty) ≤ (1 − t)ϕ(x) + tϕ(y) for all
x, y ∈ (a, b) and t ∈ [0, 1]. Looking at the graph of ϕ, this means that
(t, ϕ(t)) lies below the line segment connecting (x, ϕ(x)) and (y, ϕ(y))
for x ... | https://ocw.mit.edu/courses/18-125-measure-and-integration-fall-2003/6aa2562c8e9e459dd82775b0a4f01aea_18125_lec14.pdf |
t = Ω f dµ. Since a < f < b,
�
Ω
Ω
�
a = a · µ(Ω) <
f dµ < b · µ(Ω) = b,
so a < t < b. Conversely,
Ω
Fix t, and let
ϕ(t) − ϕ(s)
t − s
≤
ϕ(u) − ϕ(t)
.
u − t
B = sup
a<s<t
ϕ(t) − ϕ(s)
t − s
.
Then ϕ(t) − ϕ(s) ≤ B(t − s) for s < t. We have
Date: October 21, 2003.
B ≤
ϕ(u) − ϕ(t)
u − t
1
2
MEASURE... | https://ocw.mit.edu/courses/18-125-measure-and-integration-fall-2003/6aa2562c8e9e459dd82775b0a4f01aea_18125_lec14.pdf |
a finite set of points and define µ({pi}) =
1/n. Then µ(Ω) = 1. Let f : Ω R with f (pi) = xi. Then
→
�
Ω
f dµ =
n
�
f (pi)µ({pi})
i=1
1
n
(x1 + · · ·
+ xn).
=
Thus
exp
(x1 + · · ·
+ xn) ≤
e f dµ
� �
�
1
n
Ω
1
n
(e x1
≤
+ · · ·
+ e xn ).
xi
Let yi = e . Then
(y1 + · · ·
+ yn)1/n
≤
1
n
(y1... | https://ocw.mit.edu/courses/18-125-measure-and-integration-fall-2003/6aa2562c8e9e459dd82775b0a4f01aea_18125_lec14.pdf |
p dµ
gq dµ
(H¨older’s)
X
X
X
(f + g)pdµ
�1/p ��
≤
�1/p ��
+
f p dµ
�1/p
gp dµ
(Minkowski’s).
X
X
�
� older’s. Without loss of generality we may assume that
Proof. H¨
1 and X gq = 1. Indeed, if f p = 0 and
gq = 0, then let
�
�
X
f p =
and
��
X
f =
��
X
f
f p
,
�1/p
g =
g
�1/p
f p
.
��
X
... | https://ocw.mit.edu/courses/18-125-measure-and-integration-fall-2003/6aa2562c8e9e459dd82775b0a4f01aea_18125_lec14.pdf |
Thus,
�
(f g)dµ ≤
X
�
1
p X
ap dµ +
�
1
q X
1
bq dµ = + = 1.
p
1
q
Minkowski’s. Observe that
(f + g)p = f (f + g)p−1 + g(f + g)p−1 .
�
�
4
MEASURE AND INTEGRATION: LECTURE 14
Since p and q are conjugate exponents, q = p/(p − 1). Thus,
�
f (f + g)p−1 ≤
��
�1/p ��
f p
(f + g)(p−1)p/(p−1)
�(p−1)/p
... | https://ocw.mit.edu/courses/18-125-measure-and-integration-fall-2003/6aa2562c8e9e459dd82775b0a4f01aea_18125_lec14.pdf |
Large Sample Theory of Maximum Likelihood Estimates
Maximum Likelihood
Large Sample Theory
MIT 18.443
Dr. Kempthorne
Spring 2015
MIT 18.443
Maximum LikelihoodLarge Sample Theory
1 Large Sample Theory of Maximum Likelihood Estimates
Asymptotic Distribution of MLEs
Confidence Intervals Based on MLEs
Outline
1... | https://ocw.mit.edu/courses/18-443-statistics-for-applications-spring-2015/6aa8f060093ad081f930b31592274962_MIT18_443S15_LEC5.pdf |
θn)
θn
Limiting Distribution:
√
L
n(θˆn − θ) −−→ N(0, κ(θ)).
MIT 18.443
Maximum LikelihoodLarge Sample Theory
3
Large Sample Theory of Maximum Likelihood Estimates
Asymptotic Distribution of MLEs
Confidence Intervals Based on MLEs
Theorems for Asymptotic Results
Setting:
n
x1, . . . , xn a realization of... | https://ocw.mit.edu/courses/18-443-statistics-for-applications-spring-2015/6aa8f060093ad081f930b31592274962_MIT18_443S15_LEC5.pdf |
n is close to θ∗ maximizing E [log f (x | θ) | θ0]
Under smoothness conditions on f (x | θ),
θ∗ maximizes E [log f (x | θ) | θ0]
if θ∗ solves
d (E [log f (x | θ) | θ0]) = 0
dθ
Claim: θ∗ = θ0:
d
dθ
(E [log f (x | θ) | θ0]) =
=
(cid:82)
d
dθo
(cid:82)
d
(dθ
(cid:82)
log[f (x | θ)]f (x | θ0)dx
{
log[f (x | θ)] ... | https://ocw.mit.edu/courses/18-443-statistics-for-applications-spring-2015/6aa8f060093ad081f930b31592274962_MIT18_443S15_LEC5.pdf |
2
Proof: Using the Taylor approximation to £'(θ), centered at θ0
consider the following development:
0 = £'(θˆ) ≈ £'(θ0) + (ˆ
(θˆ − θ) ≈
θ − θ)£''(θ0)
=⇒
£'(θ0)
−£''(θ0)
√
√
=⇒ n(θˆ − θ) ≈
n[ 1 £'(θ0)]
n
1 [−£''(θ0)]
n
L
n[ 1 £'(θ0)] −−→ N(0, I(θ0))
L
By the CLT
n
By the WLLN 1 [−£''(θ0)] −−→ I(θ0)
√
√ ... | https://ocw.mit.edu/courses/18-443-statistics-for-applications-spring-2015/6aa8f060093ad081f930b31592274962_MIT18_443S15_LEC5.pdf |
U(X ; θ) | θ] = I(θ)
o
= E [U(X ; θ)]2 | θ]
{
Proof:
(cid:82)
f (x | θ)dx = 1 with respect to θ two times.
Differentiate
Interchange the order of differentiation and integration.
(a) follows from the first derivative.
(b) follows from the second derivative.
MIT 18.443
Maximum LikelihoodLarge Sample Theory
7
La... | https://ocw.mit.edu/courses/18-443-statistics-for-applications-spring-2015/6aa8f060093ad081f930b31592274962_MIT18_443S15_LEC5.pdf |
Parameter Estimate: X (sample mean)
2)
2), unknown mean µ
X =
n
11
n
E [X ] = µ
i=1
Xi
Var [X ] = σ2 = σ0
2/n
X
A 95% confidence interval for µ is a random interval,
calculated from the data, that contains µ with probability
0.95, no matter what the value of the true µ.
MIT 18.443
Maximum LikelihoodLarg... | https://ocw.mit.edu/courses/18-443-statistics-for-applications-spring-2015/6aa8f060093ad081f930b31592274962_MIT18_443S15_LEC5.pdf |
Theory of Maximum Likelihood Estimates
Asymptotic Distribution of MLEs
Confidence Intervals Based on MLEs
Confidence Interval for a Normal Mean
Important Properties/Qualifications
The confidence interval is random.
The parameter µ is not random.
100(1 − α)%: the confidence-level of the confidence interval is
the prob... | https://ocw.mit.edu/courses/18-443-statistics-for-applications-spring-2015/6aa8f060093ad081f930b31592274962_MIT18_443S15_LEC5.pdf |
(α/2)) = α/2
By symmetry of Student’s t distribution
(Xi − X )2 .
(cid:80)
n
i=1
1
n−1
[
P [−tn−1(α/2) < T < +tn−1(α/2)] = 1 − α
X − µ
S n
P −tn−1(α/2) < √ < +tn−1(α/2) = 1 − α
i.e.,
MIT 18.443
Maximum LikelihoodLarge Sample Theory
13
Large Sample Theory of Maximum Likelihood Estimates
Asymptotic Distri... | https://ocw.mit.edu/courses/18-443-statistics-for-applications-spring-2015/6aa8f060093ad081f930b31592274962_MIT18_443S15_LEC5.pdf |
variance σ2 .
X1, . . . , Xn i.i.d. N(µ, σ2), with MLEs:
µˆ = X
1
n
σˆ2 =
n
1
(Xi − X )2
i=1
Confidence interval for σ2 based on the sampling distribution
of the MLE ˆσ2 .
nσˆ2
σ2 ∼ χ2
n−1.
Ω =
n−1 is Chi-squared distribution with (n − 1) d.f.
−1(α∗)) = α∗
n−1(α∗) : P(χ2
2
n−1 > χn
where χ2
Define ... | https://ocw.mit.edu/courses/18-443-statistics-for-applications-spring-2015/6aa8f060093ad081f930b31592274962_MIT18_443S15_LEC5.pdf |
− α/2)
(α/2)
P[
<
<
<
<
χ2
n
−1
χ2
n
−1
−1(α/2)] = 1 − α
(α/2)] = 1 − α
] = 1 − α
The 100(1 − α)% confidence interval for σ2 is given by
nσˆ2
[
χ2
n
−1
(α/2)
< σ2 <
nσˆ2
(1 − α/2)
]
χ2
n
−1
Properties of confidence interval for σ2
Asymmetrical about the MLE ˆσ2 .
Width proportional to ˆσ2 (random!)
100... | https://ocw.mit.edu/courses/18-443-statistics-for-applications-spring-2015/6aa8f060093ad081f930b31592274962_MIT18_443S15_LEC5.pdf |
of Maximum Likelihood Estimates
Asymptotic Distribution of MLEs
Confidence Intervals Based on MLEs
Confidence Intervals Based On Large Sample Theory
Asymptotic Framework (Re-cap)
Data Model : Xn = (X1, X2, . . . , Xn) i.i.d. sample with
pdf/pmf
f (x1, . . . , xn | θ) =
Likelihood of θ given Xn = xn = (x1, . . . ,... | https://ocw.mit.edu/courses/18-443-statistics-for-applications-spring-2015/6aa8f060093ad081f930b31592274962_MIT18_443S15_LEC5.pdf |
dence Intervals Based on Large Sample Theory
Large-Sample Confidence Interval
Exploit the limiting pivotal quantity
Zn = nI(θ)(θˆn − θ) −−→ N(0, 1).
L
(cid:112)
I.e.
P(−z(α/2) <
⇐⇒ P(−z(α/2) <
(cid:112)
(cid:113)
Zn
< +z(α/2)) ≈ 1 − α
nI(θ)(θˆn − θ) < +z(α/2)) ≈ 1 − α
nI(θˆn)(θˆn − θ) < +z(α/2)) ≈ 1 − α
⇐⇒ P(... | https://ocw.mit.edu/courses/18-443-statistics-for-applications-spring-2015/6aa8f060093ad081f930b31592274962_MIT18_443S15_LEC5.pdf |
f (x | λ) = λx
ex!
(cid:80)
n
£(λ) =
i=1
[xi ln(λ) − λ − ln(x!)]
−λ
MLE λˆ
λˆ solves:
d£(λ)
dλ
d 2
dλ2
n
1
=
i=1
xi
[ − 1] = 0; λˆ = X .
λ
log f (x | λ)] = E [
ˆλ − λ L
(cid:113)
ˆλ/n
Xi
λ2
] =
1
λ
−−→ N(0, 1)
I(λ) = E [−
(cid:113)
Zn = nI(λˆ)(λˆ − λ) =
Large-Sample Confidence Interval for λ
Approx... | https://ocw.mit.edu/courses/18-443-statistics-for-applications-spring-2015/6aa8f060093ad081f930b31592274962_MIT18_443S15_LEC5.pdf |
�
For an 95% confidence interval, z(α/2) = 1.96 giving
0.61 ± (1.96)(.0552) = [.5018, .7182]
= λMLE /n = .0552
λMLE
MIT 18.443
Maximum LikelihoodLarge Sample Theory
21
Large Sample Theory of Maximum Likelihood Estimates
Asymptotic Distribution of MLEs
Confidence Intervals Based on MLEs
Confidence Interval for Mu... | https://ocw.mit.edu/courses/18-443-statistics-for-applications-spring-2015/6aa8f060093ad081f930b31592274962_MIT18_443S15_LEC5.pdf |
22
Large Sample Theory of Maximum Likelihood Estimates
Asymptotic Distribution of MLEs
Confidence Intervals Based on MLEs
MLEs of Multinomial Parameter
Maximum Likelihood Estimation for Multinomial
Likelihood function of counts
lik(p1, . . . , pm) =
=
log [f (x1, . . . , xm | p1, . . . , pm)]
m
j=1 xj ... | https://ocw.mit.edu/courses/18-443-statistics-for-applications-spring-2015/6aa8f060093ad081f930b31592274962_MIT18_443S15_LEC5.pdf |
inomial Parameter
Example 8.5.1.A Hardy-Weinberg Equilibrium
Equilibrium frequency of genotypes: AA, Aa, and aa
P(a) = θ and P(A) = 1 − θ
Equilibrium probabilities of genotypes: (1 − θ)2, 2(θ)(1 − θ),
and θ2 .
Multinomial Data: (X1, X2, X3) corresponding to counts of
AA, Aa, and aa in a sample of size n.
Sample... | https://ocw.mit.edu/courses/18-443-statistics-for-applications-spring-2015/6aa8f060093ad081f930b31592274962_MIT18_443S15_LEC5.pdf |
2x1 + x2)
1 − θ
(2x3 + x2)
θ
=⇒ θ =
ˆ
2x3 + x2
2x1 + 2x2 + 2x3
=
2x3 + x2
2n
= 0.4247
MIT 18.443
Maximum LikelihoodLarge Sample Theory
25 Large Sample Theory of Maximum Likelihood Estimates
Asymptotic Distribution of MLEs
Confidence Intervals Based on MLEs
Asymptotic variance of MLE θˆ:
Var (θˆ) −→
1
E [... | https://ocw.mit.edu/courses/18-443-statistics-for-applications-spring-2015/6aa8f060093ad081f930b31592274962_MIT18_443S15_LEC5.pdf |
y-Weinberg Model
Approximate 95% Confidence Interval for θ
Interval : θ ± z(α/2) × σˆ ˆ, with
θ
ˆ
ˆθ = 0.4247
σˆθˆ = 0.0109
z(α/2) = 1.96 (with α = 1 − 0.95)
Interval : 0.4247 ± 1.96 × (0.0109) = [0.4033, .4461]
Note (!!) : Bootstrap simulation in R of θˆ the RMSE (θˆ) = 0.0109
is virtually equal to ˆσθˆ
MIT ... | https://ocw.mit.edu/courses/18-443-statistics-for-applications-spring-2015/6aa8f060093ad081f930b31592274962_MIT18_443S15_LEC5.pdf |
Writing Kimmo Lexicons
1.0 First, two general tips on rule writing and lexicon writing.
Besides tracing a rule, there are two other PCKIMMO commands
that you can use.
1. The first is SET RULE <rule number> {ON|OFF} this lets you turn
individual rules on or off. This is very helpful if you find that
your recognize... | https://ocw.mit.edu/courses/6-863j-natural-language-and-the-computer-representation-of-knowledge-spring-2003/6ad96b72dd085923176112e82960ba8f_kimmolex.pdf |
:x). followed by +:0. And the right context is indeed simply s
# (on the surface -- we don't have to mention the hash mark # boundary
symbol in the lexical or underlying string, really - it is assumed to
be the same as the surface string.) So we are really lining up the
following pair of characters, where I have wr... | https://ocw.mit.edu/courses/6-863j-natural-language-and-the-computer-representation-of-knowledge-spring-2003/6ad96b72dd085923176112e82960ba8f_kimmolex.pdf |
why it fails:
>show rule 3
3 on Epenthesis Epenthesis. 0:e ==> S +:0____s#"
s:s ( s:s )
S:S ( x:x z:z )
+:0 ( +:0 )
0:e ( 0:e )
#:# ( #:# )
@:@ ( b:b d:d F:F g:g j:j k:k l:l m:m n:n p:p q:q r:r t:t
v:v w:w y:y a:a e:e i:i o:o u:u ':' -:- -:0 ':0 )
From this display, it is obvious that the column header S:S d... | https://ocw.mit.edu/courses/6-863j-natural-language-and-the-computer-representation-of-knowledge-spring-2003/6ad96b72dd085923176112e82960ba8f_kimmolex.pdf |
as finite-state machines, and
all the types of rule constraints can be translated
into finite-state tables. We then summarize the rule
semantics. This is followed by a detailed discussion
of rule conflicts; specficity and conflicts amongst
SUBSETS; and finally, and explanation of the rule
file format and the rule... | https://ocw.mit.edu/courses/6-863j-natural-language-and-the-computer-representation-of-knowledge-spring-2003/6ad96b72dd085923176112e82960ba8f_kimmolex.pdf |
. When the generator moves
on to the second character of the input word, it finds that it is a
lexical a, and thus R2 FAILS, so the generator must back up, undo
what it has done so far, and try to find a different path. Backing up
to the first character t, it now tries the DEFAULT correspondence t:t
(which is guar... | https://ocw.mit.edu/courses/6-863j-natural-language-and-the-computer-representation-of-knowledge-spring-2003/6ad96b72dd085923176112e82960ba8f_kimmolex.pdf |
generator is not yet
done. It will continue backtracking, trying to find alternative
realizations of the lexical form. First, it will undo the i:i
correspondence of the last character of the input word, then it will
consider the third character, lexical t. Having already tried the
correspondence t:c, it will try t... | https://ocw.mit.edu/courses/6-863j-natural-language-and-the-computer-representation-of-knowledge-spring-2003/6ad96b72dd085923176112e82960ba8f_kimmolex.pdf |
such that the entire set
of feasible pairs in the rule description is partition among the
column headers WITH NO OVERLAP (this is the source of MANY bugs in
Kimmo rule systems). T2 specifies the special correspondence t:c and
the environment in which it is allowed. (the machine goes to state 2
to anticipate that a... | https://ocw.mit.edu/courses/6-863j-natural-language-and-the-computer-representation-of-knowledge-spring-2003/6ad96b72dd085923176112e82960ba8f_kimmolex.pdf |
perhaps THE biggest source of difficulty in building
the FSTs.
In our case above, it is natural to think of R2 as saying that t:c
succeeds when it occurs preceding i:i. But T2 actually works because
it FAILS when ANYTHING BUT i:i follows t:c.
2.2 A <== rule.
Now consider R4.
R4 t:c <... | https://ocw.mit.edu/courses/6-863j-natural-language-and-the-computer-representation-of-knowledge-spring-2003/6ad96b72dd085923176112e82960ba8f_kimmolex.pdf |
:t that is disallowed before i, but rather the pair t:not-c
(lexical t and surface anything but c) Given that the more specific
correspondence t:c is already in the table, the more general
correspondence t:@ will take care of all the rest of the characters,
including t:t. (I'll leave the details of this to you..)
... | https://ocw.mit.edu/courses/6-863j-natural-language-and-the-computer-representation-of-knowledge-spring-2003/6ad96b72dd085923176112e82960ba8f_kimmolex.pdf |
Trends in III-V and II-VI Compounds
Larger atoms, weaker bonds, smaller U, smaller Eg, higher μ, more costly!
Energy Gap and Lattice Constants
)
V
e
(
p
a
g
y
g
r
e
n
E
2.6
2.4
2.2
2.0
1.8
1.6
1.4
1.2
1.0
0.8
0.6
0.4
0.2
0
0.54
AlP
GaP
AlAs
GaAs (D)
InP(D)
Si
Ge
Indirect band gap
Direct band gap (D)
AlSb
GaSb (D)
InAs... | https://ocw.mit.edu/courses/3-225-electronic-and-mechanical-properties-of-materials-fall-2007/6b110b909753ad4d0814478c2c0702a7_lecture_6.pdf |
� E
∂k 2
Note: These
electrons have
negative mass!
2
1-D Crystal Metals and Insulators
• How do band gaps affect properties of materials?
• Only electrons near EF participate in properties
•
•
• Need to find out where EF is!
If EF is in the middle of the band, free e- and metallic behavior
If EF is near th... | https://ocw.mit.edu/courses/3-225-electronic-and-mechanical-properties-of-materials-fall-2007/6b110b909753ad4d0814478c2c0702a7_lecture_6.pdf |
π/a
π/a
2-D
‘Fermi Surface’
Zone center
π/a
kx
−π/a
−π/a 0 π/a
3-D
5
Metals and Insulators
• Covalent bonds, weak U seen by e-, with EF
being in mid-band area: free e-, metallic
• Covalent or slightly ionic bonds, weak U to
medium U, with EF near band edge
– EF in or near kT of band edge: semimetal
– EF in... | https://ocw.mit.edu/courses/3-225-electronic-and-mechanical-properties-of-materials-fall-2007/6b110b909753ad4d0814478c2c0702a7_lecture_6.pdf |
Courtesy of Louis L. Whitcomb. Used with permission.
Notes on Kronecker Products
Louis L. Whitcomb ∗
Department of Mechanical Engineering
G.W.C. Whiting School of Engineering
The Johns Hopkins University
This note is a brief description of the matrix Kronecker product and matrix stack algebraic operators.
For a d... | https://ocw.mit.edu/courses/20-482j-foundations-of-algorithms-and-computational-techniques-in-systems-biology-spring-2006/6b6ebc6b4bd040a60b769f02fa4e730f_kronecker.pdf |
6.581J / 20.482JFoundations of Algorithms and Computational Techniques in Systems BiologyProfessor Bruce TidorProfessor Jacob K. White Courtesy of Louis L. Whitcomb. Used with permission.
3. trace(AB) = ((AT )S )T BS .
2
The Kronecker Product
The Kronecker product is a binary matrix operator that maps two arbitraril... | https://ocw.mit.edu/courses/20-482j-foundations-of-algorithms-and-computational-techniques-in-systems-biology-spring-2006/6b6ebc6b4bd040a60b769f02fa4e730f_kronecker.pdf |
6
3
6
12
0 −1 −2 −3
0 −4 −5 −6
⎤
⎥
⎥
⎦
.
4×6
2
6.581J / 20.482JFoundations of Algorithms and Computational Techniques in Systems BiologyProfessor Bruce TidorProfessor Jacob K. White Courtesy of Louis L. Whitcomb. Used with permission.
2.1 Properties of the Kronecker Product Operator
In the following it is assum... | https://ocw.mit.edu/courses/20-482j-foundations-of-algorithms-and-computational-techniques-in-systems-biology-spring-2006/6b6ebc6b4bd040a60b769f02fa4e730f_kronecker.pdf |
(A ⊗ B) = trace(A) · trace(B).
(9)
(10)
(11)
(12)
(13)
(14)
(15)
(16)
(17)
10. Stack of a matrix multiplication, when dimensions are appropriate for the product ABC to be well
defined, is
References
(ABC)S = (C T ⊗ A)BS .
(18)
[1] Alexander Graham. Kronecker Products and Matrix Calculus With Applications. Ha... | https://ocw.mit.edu/courses/20-482j-foundations-of-algorithms-and-computational-techniques-in-systems-biology-spring-2006/6b6ebc6b4bd040a60b769f02fa4e730f_kronecker.pdf |
Transient heat conduction solutions
Conduction into a semi-infinite medium from the surface
at a fixed temperature
Conduction into a plate, both surfaces
at a fixed temperature
2XerfX21n2cos41n2exp1n214122n MIT OpenCourseWare
http://ocw.mit.edu
3.044 Materials Processing
Sp... | https://ocw.mit.edu/courses/3-044-materials-processing-spring-2013/6b75f358c8da03e1a3ea70a6e2098dd5_MIT3_044S13_TranHeaCondSol.pdf |
1 Basic notions of representation theory
1.1 What is representation theory?
In technical terms, representation theory studies representations of associative algebras. Its general
content can be very briefly summarized as follows.
An associative algebra over a field k is a vector space A over k equipped with an associ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/6b7e3b7277dc86249ec40bb134186639_MIT18_712F10_ch1.pdf |
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