text stringlengths 30 4k | source stringlengths 60 201 |
|---|---|
b
aT x ≤ b
D
C
a
the hyperplane {x | aT x = b} separates C and D
strict separation requires additional assumptions (e.g., C is closed, D is a
singleton)
Convex sets
2–19
Supporting hyperplane theorem
supporting hyperplane to set C at boundary point x0:
{x | a T x = a T x0}
where a
= 0 and a
T x ≤ T x0 for... | https://ocw.mit.edu/courses/6-079-introduction-to-convex-optimization-fall-2009/26c4c530c9db63a12b898d720dd89a44_MIT6_079F09_lec02.pdf |
all x �K 0
Convex sets
2–21
Minimum and minimal elements via dual inequalities
minimum element w.r.t. �K
x is minimum element of S iff for all
λ ≻K∗ 0, x is the unique minimizer
of λT z over S
minimal element w.r.t. �K
S
x
•
if x minimizes λT z over S for some λ ≻K∗ 0, then x is minimal
λ1
x1
S
x2
λ2
• ... | https://ocw.mit.edu/courses/6-079-introduction-to-convex-optimization-fall-2009/26c4c530c9db63a12b898d720dd89a44_MIT6_079F09_lec02.pdf |
Terms of Use, visit: http://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/6-079-introduction-to-convex-optimization-fall-2009/26c4c530c9db63a12b898d720dd89a44_MIT6_079F09_lec02.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
6.006 Introduction to Algorithms
Spring 2008
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
Lecture 3 Ver 2.0
Scheduling and Binary Search Trees
6.006 Spring 2008
Lecture 3: Scheduling and Binary Search Trees
Lecture Overvie... | https://ocw.mit.edu/courses/6-006-introduction-to-algorithms-spring-2008/26f81dc853b4980d8c0873c0f8875268_lec3.pdf |
error" %\Theta (n)
R.append(t)
R = sorted(R)
land: t = R[0]
if (t != now) return error
R = R[1: ] (drop R[0] from R)
Can we do better?
• Sorted list: A 3 minute check can be done in O(1). It is possible to insert new
time/plane rather than append and sort but insertion takes Θ(n) time.
• Sorted array: It is possible... | https://ocw.mit.edu/courses/6-006-introduction-to-algorithms-spring-2008/26f81dc853b4980d8c0873c0f8875268_lec3.pdf |
3
4949797949467949464164insert 49insert 79insert 46insert 41insert 64BSTBSTBSTBSTrootall elements > 49 off to the right, in right subtreeall elements < 49, go into left subtreeBSTNIL79494179494646 Lecture 3 Ver 2.0
Scheduling and Binary Search Trees
6.006 Spring 2008
Finding the next larger element
next-larger(x)
i... | https://ocw.mit.edu/courses/6-006-introduction-to-algorithms-spring-2008/26f81dc853b4980d8c0873c0f8875268_lec3.pdf |
7: Insert into BST in sorted order
The tree in Fig. 7 looks like a linked list. We have achieved O(n) not O(log(n)!!
Balanced BSTs to the rescue...more on that in the next lecture!
5
49461 + 2 + 1 + 1 = 57964subtreesubtree46434955..| | https://ocw.mit.edu/courses/6-006-introduction-to-algorithms-spring-2008/26f81dc853b4980d8c0873c0f8875268_lec3.pdf |
§ 10. Binary hypothesis testing
10.1 Binary Hypothesis Testing
Two possible distributions on a space X
H0 ∶ X ∼ P
H1 X Q
∼
∶
Where under hypothesis H0 (the null hypothesis)
is distributed according to P , and under H1
(the alternative hypothesis) X is distributed according to Q. A test between two distributions
chooses... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/26fd180f40b6773bf19b659a4c5e8656_MIT6_441S16_chapter_10.pdf |
many oth
{
er pairs from π , π , π , π
}
∼
.
0∣0
0∣1
1∣0
1 1
∣
So for any test P
test”
Z∣X there is an associated
(α, β). There are a few ways to determine the “best
• Bayesian: Assume prior distributions P[H0
] = π0 and P[H1
] = π1, minimize the expected error
∗ =
Pb min
tests
+
π0π1 0 π1π0 1
∣
∣
112
• Minimax: Assum... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/26fd180f40b6773bf19b659a4c5e8656_MIT6_441S16_chapter_10.pdf |
) =
Theorem 10.1 (Prop
erties of
R(
P, Q ).
)
R(
P, Q (HW).
)
1. R(P, Q) is a closed, convex
subset
of [0, 1]2.
2. R(P, Q) contains the diagonal.
1Recall that P is mutually singular w.r.t. Q, denoted by P ⊥ Q, if P [E] = 0 and Q[E] = 1 for some E.
113
R(P,Q)ββα(P,Q)α 3. Symmetry: (α, β
(
) ∈ R P, Q
) ⇔
(
1 α, 1 β
−
−
)... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/26fd180f40b6773bf19b659a4c5e8656_MIT6_441S16_chapter_10.pdf |
ipping a coin, i.e., let Z ∼ Bern
)
( −
1 α X. This achieves the point α, α .
⊥⊥
(
)
3. If (α, β
) ∈ R(P, Q), then form the test that
P
whenev
er PZ∣X choses Q, and chooses
Q whenever PZ∣X choses P , which gives 1 α, 1 β) ∈ R(
(
P, Q).
c
ho
−
oses
−
The region R(P, Q) consists of the operating points of all randomized ... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/26fd180f40b6773bf19b659a4c5e8656_MIT6_441S16_chapter_10.pdf |
”: Comparing (10.1) and (10.2), by definition,
oof. “⊃
P,
Q is closed convex, and we are done with the
⊂
“ ”: Given any randomized test PZ∣X , put g
Moreover,
⊃
R(
P, Q
direction.
) ⊃ R (
det P, Q . By Theorem 10.1,
))
(x) = PZ=0∣X=x. Then g is a
measurable function.
P [Z = 0] = ∑ g(x
)P
x
(x) = EP
[g
(X
)] =
∫
0
Q
[ = ... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/26fd180f40b6773bf19b659a4c5e8656_MIT6_441S16_chapter_10.pdf |
. The log likelihood ratio (LLR) is F = log dP
dQ
ratio test (LRT) with threshold τ
dQ = q(x)dµ (one can take µ = P + Q, for example) and set
∈ R is 1{log dP
∶ X → ∪ {±∞}
. The likelihood
R
dQ ≤ τ }. Formally, we assume that dP = p(x)dµ and
F (x) ≜
⎧⎪⎪⎪⎪⎪⎪⎪⎪
⎪⎪⎪⎪⎪⎪⎪⎪⎩
⎨
log p(x)
q(x)
+∞
,
−∞
,
/
n a,
>
, p(x) > 0, x
( ... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/26fd180f40b6773bf19b659a4c5e8656_MIT6_441S16_chapter_10.pdf |
≥ τ
EP [
1{
X)
g(
≤ }] ≥
{
)
(
[
EQ X 1 F τ
g
}] ≥
F
{ } ⋅
[ (
) { ≥ }]
exp τ EQ g X 1 F τ
exp{−τ } ⋅ EP [g(X)1{F ≤ τ }]
(10.3)
(10.4)
115
Below, these and similar inequalities are only checked for the cases of F not taking extended
values, but from this remark it should be clear how to treat the general case.
• Anoth... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/26fd180f40b6773bf19b659a4c5e8656_MIT6_441S16_chapter_10.pdf |
(F (x)) + g(−∞)Q[F = −∞]
= ∫
= EP
x
{−∞< ( )<∞}
−F }g(F )] + g(−∞)Q[F
{
F
[exp
= −∞]
,
dµ p(x) exp{−F (x)}g(F (x)) + g(−∞)Q F
[ = −∞]
(10.8)
(10.9)
(10.10)
where we used (10.5) to justify
restriction to finite values of F .
(1) To show F
is
a s.s, w
e need to show PX∣F = Q
PX∣
F (x∣f ) =
( )
( ∣ )
PX x PF X f x
∣
PF (f ... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/26fd180f40b6773bf19b659a4c5e8656_MIT6_441S16_chapter_10.pdf |
belong to R(P, Q)...
116
10.4 Converse bounds on
(
R P, Q
)
Theorem 10.4 (Weak Converse). ∀( , β
α
) ∈ R(P, Q),
d(α∥β) ≤ D(P
( ∥ ) ≤
( ∥
d β α D Q
∥ )
Q
P )
where d
(⋅∥⋅)
is the binary divergence.
Proof. Use
data processing with PZ∣X .
Lemma 10.1 (Deterministic tests). ∀E, ∀γ > 0 ∶ P [E] − γQ[E] ≤ P [ log dP
dQ >
log
... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/26fd180f40b6773bf19b659a4c5e8656_MIT6_441S16_chapter_10.pdf |
P
dµ
dQ
dµ
= dP .dQ
[So we see that the only difference between the discrete and the general case is that the counting
measure is replaced by some other measure µ.]
Note: In this case, we do not need P Q, since
log likelihood ratio.
≪
±∞
is a reasonable and meaningful value for the
Lemma 10.2 (Randomized tests). P
[Z = ... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/26fd180f40b6773bf19b659a4c5e8656_MIT6_441S16_chapter_10.pdf |
xes γ
∗
there is going to be a maximal c,
t the line touches the lower boundary of the
region. Then (10.11) says that c cannot exceed P log dP
R
log γ . Hence must lie to the left
dQ
of the line.
arly, (10.12) provides bounds for the upper boundary. Altogether Theorem 10.5
)
states that
strong converse Theorem 10.5, we... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/26fd180f40b6773bf19b659a4c5e8656_MIT6_441S16_chapter_10.pdf |
X (0∣x) = 1 {
log
(x)
P
Q(x)
≥
log
}
t
.
Thus, we have shown that all supporting hyperplanes are parameterized
R(
completely recovers the region
pieces) of the region. To be precise,
by LLR-tests. This
P, Q except for the points corresponding to the faces (linear
)
state the following result.
we
Theorem 10.6 (Neyman-Pe... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/26fd180f40b6773bf19b659a4c5e8656_MIT6_441S16_chapter_10.pdf |
„„„„„
dP
dQ >t}
{
„„„„„„„„„„„„„„„„„„¶
= ] ) +
t
[ (
Q g X 1
)
E
]
dP
dQ >t}
{
≥ EQ[(1 − g(X))1
{
= Q[
dP
dQ
> ] + λQ[
t
dP
dQ >t}
P
d
dQ
= t].
] + λQ[
dP
dQ
= t] + EQ[g(X)1
]
dP
dQ >t}
{
a consequence of the Neyman-Pearson lemma, all the points on the boundary of
Remark 10.2.
the region
R(
As
)
P, Q are attainable. The... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/26fd180f40b6773bf19b659a4c5e8656_MIT6_441S16_chapter_10.pdf |
. The test (10.13) is related to LRT2 as follows:
1. Left figure: If α = P [log dP
dQ > ] for some τ , then λ = 0, and (10.13) becomes the LRT
τ
=
Z 1
{log dP
dQ ≤τ }
.
2. Right figure: If α ≠ P [log dP
to randomize over tests: Z = 1
dQ ≤ }
2Note that it so happens that in Definition 10.3 the LRT is defined with an ≤ inste... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/26fd180f40b6773bf19b659a4c5e8656_MIT6_441S16_chapter_10.pdf |
the
∣
∣
ponential.
types of tests, both which the convergence rate to zero error is ex
error probabilities
π0 1 and π1 0. There are two main
1. Stein
≤
(cid:15)?
Regime: What is the best exponential rate of convergence for π0∣1 when π1∣0 has to be
⎧⎪⎪
π1∣0 ≤ (cid:15)
⎨
⎪⎪
→
0
⎩π0∣1
2. Chernoff Regime: What is the trade ... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/26fd180f40b6773bf19b659a4c5e8656_MIT6_441S16_chapter_10.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-2005/27035dfb2ffa2460b38ef54ac5238d69_lec11.pdf |
n(z)
= −Γ(z)
=
1
Zn(z)
= Yn(z) =
Y (z)
Y0
=
ˆi(z)
Y0vˆ(z)
D. If line is matched, ZL = Z0, ΓL = 0, Zn(z) = 1
II. Load Impedance Reflected Back to the Source
From Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987. Used with permission.
Zn(z = 0) =
�
Zn z = −
=
�
λ
4
ZL =
Z0
RL... | https://ocw.mit.edu/courses/6-013-electromagnetics-and-applications-fall-2005/27035dfb2ffa2460b38ef54ac5238d69_lec11.pdf |
avelength Matching
From Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987. Used with permission.
To match Z1 to RL ⇒ Z1 = RL/Z2
Z2
=
Z2
2
RL
, Z2
=
√
Z1RL.
3
V0coswtYRs IV. Smith Chart
Zn(z) = r + jx
(z) = 1+Γ(z)
Zn
1−Γ2
Γ(z) = Γr + jΓi
1+Γr +jΓi
1−Γ(z) ⇒ r + jx = 1−Γr −jΓi
r −Γ... | https://ocw.mit.edu/courses/6-013-electromagnetics-and-applications-fall-2005/27035dfb2ffa2460b38ef54ac5238d69_lec11.pdf |
|Iˆ| =
V
0
(z
= −l)|
|50 + Z
�
�
Im [Z(z = −l)]
50 + Re [Z(z = −l)]
φ = tan−1
6
V. Standing Wave Parameters
From Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987. Used with permission.
vˆ(z) = Vˆ+e−jkz [1 + Γ(z)]
ˆi(z) = Y0Vˆ+e−jkz [1 − Γ(z)]
⇒ |
⇒ |
vˆ(z) = Vˆ+||1 + Γ(z)|
ˆi(z... | https://ocw.mit.edu/courses/6-013-electromagnetics-and-applications-fall-2005/27035dfb2ffa2460b38ef54ac5238d69_lec11.pdf |
�
= |1 + ΓL|;
�
�
Y0Vˆ+
= 1 − |ΓL|
= |1 − ΓL|
E. If ZL = RL (real), then ΓL real. If ZL > Z0, VSWR = ZL . If ZL < Z0, VSWR = Z0
ZL
Z0
7
From Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987. Used with permission.
Load Impedance: ZL = Z0
1 + ΓL ejφ
|
|
ejφ
1
ΓL
− |
|
�
VSWR + 1... | https://ocw.mit.edu/courses/6-013-electromagnetics-and-applications-fall-2005/27035dfb2ffa2460b38ef54ac5238d69_lec11.pdf |
Introduction to C++
Massachusetts Institute of Technology
January 19, 2011
6.096
Lecture 7 Notes: Object-Oriented Programming
(OOP) and Inheritance
We’ve already seen how to define composite datatypes using classes. Now we’ll take a step
back and consider the programming philosophy underlying classes, known as ob... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/270def7b1f68535b7c3846c606b220eb_MIT6_096IAP11_lec07.pdf |
the world in terms of interacting objects: we’d talk about interactions between
the steering wheel, the pedals, the wheels, etc. OOP allows programmers to pack away details
into neat, self-contained boxes (objects) so that they can think of the objects more abstractly
and focus on the interactions between them.
The... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/270def7b1f68535b7c3846c606b220eb_MIT6_096IAP11_lec07.pdf |
ish your goal.
If you remember the analogy from Lecture 6 about objects being
boxes with buttons you can push, you can also think of the interface of a class as the set
of buttons each instance of that class makes available. Interfaces abstract away the details
of how all the operations are actually performed, allo... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/270def7b1f68535b7c3846c606b220eb_MIT6_096IAP11_lec07.pdf |
to duplicate the functionality that all vehicles have in common. Instead, C++ allows
us to specify the common code in a Vehicle class, and then specify that the Car and Truck
classes share this code.
The Vehicle class will be much the same as what we’ve seen before:
1
2
3
4
class Vehicle {
protected :
string ... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/270def7b1f68535b7c3846c606b220eb_MIT6_096IAP11_lec07.pdf |
0. This line assumes the existence of some function stringify for converting numbers to
strings.
Now we want to specify that Car will inherit the Vehicle code, but with some additions.
This is accomplished in line 1 below:
string style ;
1 class Car : public Vehicle { // Makes Car inherit from Vehicle
2
3
4 pub... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/270def7b1f68535b7c3846c606b220eb_MIT6_096IAP11_lec07.pdf |
and shares its code. This
would give a class hierarchy like the following:
Vehicle
Truck
C a r
Class hierarchies are generally drawn with arrows pointing from derived classes to base
classes.
3.1
Is-a vs. Has-a
There are two ways we could describe some class A as depending on some other class B:
1. Every A ob... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/270def7b1f68535b7c3846c606b220eb_MIT6_096IAP11_lec07.pdf |
2
3
4
4 public :
5
Car ( const string & myLicense , const int myYear , const string
& myStyle )
: Vehicle ( myLicense , myYear ) , style ( myStyle ) {}
const string getDesc () // Overriding this member function
{ return stringify ( year ) + ’ ’ + style + " : " + license
;}
const string & getStyle () { retur... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/270def7b1f68535b7c3846c606b220eb_MIT6_096IAP11_lec07.pdf |
are still public in the derived class. Specifying protected would make inherited methods,
even those declared public, have at most protected visibility. For a full table of the effects
of different inheritance access specifiers, see
http://en.wikibooks.org/wiki/C++ Programming/Classes/Inheritance.
4 Polymorphism
Poly... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/270def7b1f68535b7c3846c606b220eb_MIT6_096IAP11_lec07.pdf |
2
3
4 };
...
virtual const string getDesc () {...}
With this definition, the code above would correctly select the Car version of getDesc.
Selecting the correct function at runtime is called dynamic dispatch. This matches the whole
OOP idea – we’re sending a message to the object and letting it figure out for itself ... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/270def7b1f68535b7c3846c606b220eb_MIT6_096IAP11_lec07.pdf |
Vehicle by making the function pure virtual
via the following odd syntax:
1 class Vehicle {
2
3
4 };
...
virtual const string getDesc () = 0; // Pure virtual
The = 0 indicates that no definition will be given. This implies that one can no longer create
an instance of Vehicle; one can only create instances of Cars, ... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/270def7b1f68535b7c3846c606b220eb_MIT6_096IAP11_lec07.pdf |
In general, avoid multiple inheritance unless you know exactly what you’re doing.
7
MIT OpenCourseWare
http://ocw.mit.edu
6.096 Introduction to C++
January (IAP) 2011
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/270def7b1f68535b7c3846c606b220eb_MIT6_096IAP11_lec07.pdf |
LECTURE 17
LECTURE OUTLINE
Review of cutting plane method
Simplicial decomposition
•
•
Duality between cutting plane and simplicial
•
decomposition
All figures are courtesy of Athena Scientific, and are used with permission.1 CUTTING PLANE METHOD
Start with any x0
X. For k
0, set
≥
⌘
•
xk+1 ⌘
where
arg min Fk(x),
X
x
⌦... | https://ocw.mit.edu/courses/6-253-convex-analysis-and-optimization-spring-2012/2710d17aff6b2989d7c1b2eac2af2c84_MIT6_253S12_lec17.pdf |
is much
simpler than minimizing f over X.
−
3◆
SIMPLICIAL DECOMPOSITION METHOD
f (x0)
x0
f (x1)
x1
X
˜x2
f (x2)
x2
f (x3)
x3
˜x4
x4 = x
˜x1
˜x3
Level sets of f
•
(initially x0
Given current iterate xk, and finite set Xk ⌦
x0
{
Let x˜k+1 be extreme point of X that solves
X, X0 =
⌘
).
}
•
X
minimize
∇
subject to x
f (xk)... | https://ocw.mit.edu/courses/6-253-convex-analysis-and-optimization-spring-2012/2710d17aff6b2989d7c1b2eac2af2c84_MIT6_253S12_lec17.pdf |
eventually occur.
The method will find a minimizer of f over X
•
in a finite number of iterations.
5 COMMENTS ON SIMPLICIAL DECOMP.
Important specialized applications
•
Variant to enhance e⌅ciency. Discard some of
•
the extreme points that seem unlikely to “partici-
pate” in the optimal solution, i.e., all x˜ such that
f... | https://ocw.mit.edu/courses/6-253-convex-analysis-and-optimization-spring-2012/2710d17aff6b2989d7c1b2eac2af2c84_MIT6_253S12_lec17.pdf |
(x) = max
j=1,...,⌫
y
j� x
−
f (yj)
⇤
[this follows using x�jyj = f (xj) +f (yj), which is
◆f (xj) – the Conjugate Subgra-
implied by yj
⌘
dient Theorem]
⌅
7
◆
INNER LINEARIZATION OF FNS
f (x)
F (y)
f (y)
x0
x1
0
x2
x
Slope = y0
O
F (x)
Slope = y1
f
Outer Linearization of f
Slope = y2
f
y0
y1
f
0
y2
y
Inner Lineariza... | https://ocw.mit.edu/courses/6-253-convex-analysis-and-optimization-spring-2012/2710d17aff6b2989d7c1b2eac2af2c84_MIT6_253S12_lec17.pdf |
Const.
f (x)
−
xk
xk+1
˜xk+1
x
Given Ck: inner linearization of c, obtain
xk ⌘
arg min f
⌦�
x
n
⇤
Obtain x˜k+1 such that
(x) +C k(x)
⌅
•
�
•
•
•
f (xk)
◆c(˜xk+1),
−∇
⌘
x˜k+1}
and form Xk+1 = Xk ∪ {
9 NONDIFFERENTIABLE CASE
Given Ck: inner linearization of c, obtain
arg min f (x) +C k(x)
xk ⌘
n
x
⌦�
Obtain a subgradient... | https://ocw.mit.edu/courses/6-253-convex-analysis-and-optimization-spring-2012/2710d17aff6b2989d7c1b2eac2af2c84_MIT6_253S12_lec17.pdf |
•
2,k(x)
1(x) +F
min f
n
x
⌦�
F2,k and F
•
tions of f and f
2
2
f
min 1 (⌃) +F 2,k(
⌅
⌦�
n
⌃)
−
2,k are inner and outer approxima-
x˜i+1 and gi are solutions of the primal or the
•
dual approximating problem (and corresponding
subgradients)
11
MIT OpenCourseWare
http://ocw.mit.edu
6.253 Convex Analysis and Optimiz... | https://ocw.mit.edu/courses/6-253-convex-analysis-and-optimization-spring-2012/2710d17aff6b2989d7c1b2eac2af2c84_MIT6_253S12_lec17.pdf |
í9ëà B†ô9_¢2 ëo ov_Äí ^â ëíâ _m 9v† oI†9ôv†o Aí PA8w¢†
1b; t_9† 9vë9 Äv†í 9v† õvëo† B†à_ôA9â ëo ëí wõÄë¢f ô_Uõ_í†í92 9v† 8¢_wõ B†à_ôA9â vëo
ë f_ÄíÄë¢f ô_Uõ_í†í92 ëíf BAô† B†¢oë; t_Ä à†9 wo ô_íoAf†¢ †í†¢8â 9¢ëíoõ_¢9; m¢_U
ïù;1Ea Ć 8†9
v†íô† 9v† fâíëUAô õ¢†oow¢† Ao
\ F ρψcM F ρωβψi
ûf
−
Tïα\iβc
ωMa
−
Tïα\iβc
ωMa
−... | https://ocw.mit.edu/courses/2-062j-wave-propagation-spring-2017/27220e77673d0f1dff71a461b1574259_MIT2_062J_S17_Chap4.pdf |
XE
t
MIT OpenCourseWare
https://ocw.mit.edu
2.062J / 1.138J / 18.376J Wave Propagation
Spring 2017
For information about citing these materials or our Terms of Use, visit: https://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/2-062j-wave-propagation-spring-2017/27220e77673d0f1dff71a461b1574259_MIT2_062J_S17_Chap4.pdf |
Spring 2010
Semantic Analysis
Semantic Analysis
Saman Amarasinghe
Massachusetts Institute of Technology
Massachusetts Institute of Technology
Symbol Table Summary
• Program Symbol Table (Class Descriptors)
• Class Descriptors
D
Cl
– Field Symbol Table (Field Descriptors)
i t
Pointer to Field Symbol Table for... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/275980aad79ddb35c3872f07019d3278_MIT6_035S10_lec06.pdf |
Flat Lists
x = a*b + c
x = a*b + c
– Need to represent
intermediate values
intermediate values
• In a stack
push a; push b; mul; push c; add; pop x
• In single use temporary
registers
i t
t1 = mul a, b
x add t1, c
x = add t1 c
=
• Trees
– Intermediate values are
implicit in the edges
St x
St x
add
m... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/275980aad79ddb35c3872f07019d3278_MIT6_035S10_lec06.pdf |
What not to do!
What not to do!
• Keep data in the abstract (in descriptors)
• Keep data in the abstract (in descriptors)
– Don’t try to do register allocation!
• No optimizations!
Even when they seem sooo easy
– Even when they seem sooo easy
• Theme:
– take small stepsp
– don’t try to do too many at once
– don... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/275980aad79ddb35c3872f07019d3278_MIT6_035S10_lec06.pdf |
• Make sure the program confirms to the
programming language definition
• Provide meaningful error messages to the user
Don t need to do additional work, will discover
• Don’t need to do additional work will discover
•
in the process of intermediate representation
generation
ti
Saman Amarasinghe
14
6.035 ©MIT ... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/275980aad79ddb35c3872f07019d3278_MIT6_035S10_lec06.pdf |
– Identifiers in an expression should be “evaluatable”
– LHS of an assignment should be “assignable”
– In an expression all the types of variables, method return
I
h
types and operators should be “compatible”
bl
h
d
ll
f
i
i
Saman Amarasinghe
18
6.035 ©MIT Fall 2006
Dynamic checks
Dynamic ch... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/275980aad79ddb35c3872f07019d3278_MIT6_035S10_lec06.pdf |
type error
– special type that’ll signal an error
• void
id
– basic type denoting “the absence of a value”
Saman Amarasinghe
23
6.035 ©MIT Fall 2006
Type Expressions: Names
Type Expressions: Names
• Since type expressions maybe be named, a
• Since type expressions maybe be named a
type name is a type expressi... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/275980aad79ddb35c3872f07019d3278_MIT6_035S10_lec06.pdf |
structures and classes
– Example
class { int i; int j;}
integer integer
Functional Languages
• Functional Languages
– functions that take functions and return
functions
functions
– Example
(integer integer) integer (integer integer)
(i
)
(i
)
i
i
i
Saman Amarasinghe
28
6.035 ©MIT Fall 2006
... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/275980aad79ddb35c3872f07019d3278_MIT6_035S10_lec06.pdf |
6.035 ©MIT Fall 2006
Parser actions
Parser actions
23
P D; E
P D; E
D D; D
D id : T
D id : T
T char
T integer
T integer
T array [ num ] of T1
{ addtype(id.entry, T.type); }
{ addtype(id entry T type); }
{ T.type = char; }
{ T.type = integer; }
{ T type = integer; }
{ T.type ... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/275980aad79ddb35c3872f07019d3278_MIT6_035S10_lec06.pdf |
==
E1 .type == array(s, t) then
s
E type = s
E.type
=
else
E.type = type_error
E type = type error
}
Saman Amarasinghe
34
6.035 ©MIT Fall 2006
Type Equivalence
Type Equivalence
25
• How do we know if two types are equal?
• How do we know if two types are equal?
– Same type entry
– Example:
int A[128]; ... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/275980aad79ddb35c3872f07019d3278_MIT6_035S10_lec06.pdf |
to chars
– longs to shorts
longs to shorts
• Rare in languages
Saman Amarasinghe
38
6.035 ©MIT Fall 2006
Widening conversions
Widening conversions
• Conversions without loss of information
• Conversions without loss of information
• Examples:
– integers to floats
shorts to longs
– shorts to longs
• What is... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/275980aad79ddb35c3872f07019d3278_MIT6_035S10_lec06.pdf |
one
•
type.
• Example
l
E
int A, B, C;
float X, Y, Z;
A = A + B
X = X + Y
X
X
+
Y
• Complicates the type system
– Example
A = A + X
• What is the type of + ?
Saman Amarasinghe
43
6.035 ©MIT Fall 2006
Outline
Outline
• Practical Issues in Intermediate
t
di
I
t
I
l
ti
P
i
Representation
Repr... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/275980aad79ddb35c3872f07019d3278_MIT6_035S10_lec06.pdf |
When building the local symbol table, have
• When building the local symbol table have
a list of local descriptors
• What to check for?
duplicate variable names
– duplicate variable names
– shadowed variable names
• When to check?
k?
h
h
– when insert descriptor into local symbol table
p
• Parameter and field ... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/275980aad79ddb35c3872f07019d3278_MIT6_035S10_lec06.pdf |
5 ©MIT Fall 2006
Load Array Instruction
Load Array Instruction
What else can/should be checked?
What else can/should be checked?
Saman Amarasinghe
51
6.035 ©MIT Fall 2006
Add Operations
Add Operations
• What does compiler have?
• What does compiler have?
– two expressions
• What can go wrong?
?
Wh t
– exp... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/275980aad79ddb35c3872f07019d3278_MIT6_035S10_lec06.pdf |
expression
• If variable type not compatible with expression
type, error
t
Saman Amarasinghe
54
6.035 ©MIT Fall 2006
Store Array Instruction
Store Array Instruction
What does compiler have?
• What does compiler have?
– Variable name, array index expression
– Expression
• What does it do?
– Look up variable ... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/275980aad79ddb35c3872f07019d3278_MIT6_035S10_lec06.pdf |
does compiler have?
•
– Expression for the if-condition and the
statement list of then (and else) blocks
t li t f th ( d l ) bl k
t t
• Checks:
If the conditional expression producing a
– If the conditional expression producing a
Boolean value?
Saman Amarasinghe
58
6.035 ©MIT Fall 2006
Semantic Check Summary ... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/275980aad79ddb35c3872f07019d3278_MIT6_035S10_lec06.pdf |
18.354J Nonlinear Dynamics II: Continuum Systems
Lecture 1
Spring 2015
1 Math basics
1.1 Derivatives and differential equations
In this course, we will mostly deal with ordinary differential equations (ODEs) and partial
differential equations (PDEs) real-valued scalar or vector fields. Usually, non-bold symbols
will be... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/27c776ae2d0b07942bb58afcdf7f2029_MIT18_354JS15_Ch1.pdf |
as application of ∇ to a tensorial quantity depends on the choice of the product: For
instance, in the case of a 3D vector field v(t, x), we can obtain a scalar field called divergence
another (pseudo-)vector field called curl
∇ · v ≡ ∂ivi,
and the gradient matrix
∇ ∧ v ≡ ((cid:15)ijk∂jvk)
∇v ≡ (∂ivj).
The (scalar) Laplac... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/27c776ae2d0b07942bb58afcdf7f2029_MIT18_354JS15_Ch1.pdf |
60)z = x ∈ R and imaginary part (cid:61)z = y ∈ R. The complex conjugate of a
real number is given by
z¯ = x − iy
(11)
and corresponds to a reflection at the real axis or, equivalently, at the line (cid:61)(z) = 0.
Addition of complex numbers is linear
z = z1 + z2 = (x1 + iy1) + (x2 + iy2) = (x1 + x2) + i(y1 + y2) = x +... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/27c776ae2d0b07942bb58afcdf7f2029_MIT18_354JS15_Ch1.pdf |
zn
(cid:88)
k!
k=0
= 1 + z +
z
2!
+ . . .
eiφ = cos φ + i sin φ ,
φ ∈ R
(16)
(17)
relates exp to the trigonometric sin-and cos-functions.
When dealing with axisymmetric problems it is often advantageous to use the polar
representation of a complex number
√
r = |z| = zz¯ ∈ R+
z = reiφ ,
0 , ,
φ = arctan 2(y, x) ∈ [0, 2π... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/27c776ae2d0b07942bb58afcdf7f2029_MIT18_354JS15_Ch1.pdf |
2J / 12.207J Nonlinear Dynamics II: Continuum Systems
Spring 2015
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/27c776ae2d0b07942bb58afcdf7f2029_MIT18_354JS15_Ch1.pdf |
Introduction to Simulation - Lecture 6
Krylov-Subspace Matrix Solution Methods
Jacob White
Thanks to Deepak Ramaswamy, Michal Rewienski,
and Karen Veroy
Outline
• General Subspace Minimization Algorithm
– Review orthogonalization and projection formulas
• Generalized Conjugate Residual Algorithm
– Krylov-subspace
– ... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/27d0ff96c912a862ba47b64f1ce0cc7c_lec6.pdf |
i sα
'
Residual Minimizing idea: pick
b Mx
= −
=
b
k
k
1
−
to minimize
k
r
≡
(
k
r
T
) (
k
r
)
=
2
2
b
−
⎛
⎜
⎝
SMA-HPC ©2003 MIT
k
1
−
∑
i
=
0
α
i
(cid:71)
Mw
i
T
⎞ ⎛
⎟ ⎜
⎠ ⎝
b
−
k
1
−
∑
i
=
0
α
i
(cid:71)
Mw
i
⎞
⎟
⎠
Arbitrary Subspace
methods
Residual Minimization
Computational Approach
Minimizing
k
r
2
2
=
b
(c... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/27d0ff96c912a862ba47b64f1ce0cc7c_lec6.pdf |
T
) (
Mp
) (
r
) (
T
T
j
j
j
)
)
pMp
j
)
M
j
Mp
j
Mp
(
+
(
r
−
Normalize
Update Solution
Update Residual
j
p
j
SMA-HPC ©2003 MIT
Arbitrary Subspace
methods
Subspace Selection
Criteria
,...,
All that matters is the
'
k
1
w
0
Criteria for selecting
w −
{
}
span w
,...,
w −
0
k
= − ∑ (cid:71)
b
Mw
i
i
0
=
}
(cid:19)
... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/27d0ff96c912a862ba47b64f1ce0cc7c_lec6.pdf |
SMA-HPC ©2003 MIT
Arbitrary Subspace
methods
Subspace Selection
Krylov Subspace
Note:
span
{
∇
(cid:71)
{
span w
0
x
)
0
(
f x
(cid:71)
w
k
,...,
,...,
∇
}
− =
1
x
k
1
−
(
f x
}
)
{
span r
0
=
0
{
span r
}
k
1
r −
,...,
k
1
−
,...,
r
}
If:
i
Mrα
i
k
1
−
− ∑
i
,...,
0
=
r
k
1
−
}
=
0
,
}
{
0
1 0
k
−
span r Mr
M r
(... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/27d0ff96c912a862ba47b64f1ce0cc7c_lec6.pdf |
p
j
Compute the new
orthogonalized
search direction
Krylov Methods
The Generalized Conjugate
Residual Algorithm
Algorithm Cost for iter k
r
(
Mp
k
Tk
) (
) (
T
Mp
k
)
Mp
k
)
α =
k
(
1k
+ =
x
1k
+ =
r
k
k
x
r
+
−
p
k
1
+
=
r
k
1
+
pα
k
k
Mpα
k
(
Mr
(
Mp
0
k
j
k
− ∑
j
=
Vector inner products, O(n)
Matrix-vector produc... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/27d0ff96c912a862ba47b64f1ce0cc7c_lec6.pdf |
98) n, then symmetric,
sparse, GCR is O(n2 )
Better Converge Fast!
SMA-HPC ©2003 MIT
Krylov Methods
Nodal Formulation
“No-leak Example”
Insulated bar and Matrix
Incoming Heat
(0)T
Near End
Temperature
m
SMA-HPC ©2003 MIT
(1)T
Far End
Temperature
Discretization
2
1
−
⎡
⎤
⎢
⎥
1 2
−⎢
⎥
⎢
⎥−
1
⎥
⎢
⎣
⎦
(cid:8)(cid:11)(... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/27d0ff96c912a862ba47b64f1ce0cc7c_lec6.pdf |
1)(cid:11)(cid:11)(cid:11)(cid:9)(cid:11)(cid:11)(cid:11)(cid:11)(cid:10)
M
(cid:37)
−
Nodal
Equation
Form
SMA-HPC ©2003 MIT
Krylov Methods
Nodal Formulation
“leaky” Example
Circuit and Matrix
1
2
3
4
1m −
m
m
SMA-HPC ©2003 MIT
1
−
2.01
2.01
⎤
⎡
⎥
⎢
1
−⎢
⎥
⎥−
⎢
1
⎥
⎢
1 2.01
⎣
⎦
(cid:8)(cid:11)(cid:11)(cid:11)(cid:1... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/27d0ff96c912a862ba47b64f1ce0cc7c_lec6.pdf |
(residual) versus Iteration
SMA-HPC ©2003 MIT
Krylov Subspace
Methods
I
f
span
{
w
0
w
k
}
=
span
k
1
+
x
=
α
i
i
M r
0
=
,...,
k
∑
i
=
0
Convergence Analysis
Polynomial Approach
0
r Mr
,
0
,...,
k
M r
0
}
(
)
M r
0
{
ξ
k
k
1
+
r
=
0
r
Note: for any
{
1
r
span
r
,
0
k
−
i
α
i
∑
0
=
0α ≠
0
0
r
−
α=
0
SMA-HPC ©2003... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/27d0ff96c912a862ba47b64f1ce0cc7c_lec6.pdf |
21
2
+℘
1
k
rder
kr +
3)
o
r
1
+
0
0
0
k
k
21
2
SMA-HPC ©2003 MIT
Krylov Methods
Convergence Analysis
Optimality of GCR poly
GCR Optimality Property
k
1
+
r
2
2
(cid:4)
≤ ℘
k+1
(
M r
)
2
0
where
(cid:4)
℘
2
polynomial such that
is
k
ny
a
( )
0 =1
k+1
k+1
℘(cid:4)
th
order
Therefore
Any polynomial ... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/27d0ff96c912a862ba47b64f1ce0cc7c_lec6.pdf |
7)
(cid:37)
M
NN
SMA-HPC ©2003 MIT
0
M−
1
NN
1
−
1
0
0
0
(cid:35)
0
0
1
1
−
0
⎤
⎥
0
⎥
⎥−
1
⎥
2
⎦
Eigenvalues?
Eigenvectors?
⎤
⎥
⎥
⎥
⎥
⎥
⎦
What about a lower
triangular matrix
Eigenvalues and
Vectors Review
A Simplifying
Assumption
Almost all NxN matrices have N linearly
independent Eigenvectors
↑
(cid:71)
u
2
↓
↑
... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/27d0ff96c912a862ba47b64f1ce0cc7c_lec6.pdf |
=
U U
λ
−
1
Does NOT imply distinct eigenvalues,
M
Does NOT imply
is nonsingular
λ
i
can equal
λ
j
SMA-HPC ©2003 MIT
Eigenvalues and
Vectors Review
Spectral Radius
)
(
Im λ
iλ
)
(
Re λ
The spectral Radius of M is the radius of the smallest
circle, centered at the origin, which encloses all of
M’s eigenvalu... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/27d0ff96c912a862ba47b64f1ce0cc7c_lec6.pdf |
−
U U
λ
MM U U U U
λ
λ
=
p
p
1
M U Uλ −
Apply to the polynomial of the matrix
1
+
)
(
f M a UU
=
+…
=
+
1
−
−
1
0
Factoring
(
f M U a I
=
0
)
1
p
−
−
p
a U U
a U U
λ
λ
1
)
(
p
λ −
U
+
p
(cid:8)(cid:11)(cid:11)(cid:11)(cid:11)(cid:9)(cid:11)(cid:11)(cid:11)(cid:11)(cid:10)
Diagonal
a
λ
1
+…
+
a
1
(
f M U U a I
=
0
)
(
+... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/27d0ff96c912a862ba47b64f1ce0cc7c_lec6.pdf |
MA-HPC ©2003 MIT
Krylov Methods
Convergence Analysis
Important Observations
1) The GCR Algorithm converges to the exact solution
in at most n steps
(
( )
(cid:4)
x
x
=
℘
−
−
λ
n
1
(
where
M
λ λ∈
Proof: Let
)(
x
)
.
λ
n
λ
2
...
−
x
)
(
)
i
(cid:4)
⇒ ℘
(
) 0
n M r
=
0 and therefore
n
r
=
0
2) If M has only q dist... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/27d0ff96c912a862ba47b64f1ce0cc7c_lec6.pdf |
4
Convolution
In Lecture 3 we introduced and defined a variety of system properties to
which we will make frequent reference throughout the course. Of particular
importance are the properties of linearity and time invariance, both because
systems with these properties represent a very broad and useful class and be-
cau... | https://ocw.mit.edu/courses/res-6-007-signals-and-systems-spring-2011/27da9a018e92fc06d2cf1fbce5cd3e71_MITRES_6_007S11_lec04.pdf |
exponentials.
In this lecture we develop in detail the representation of both continuous-
time and discrete-time signals as a linear combination of delayed impulses
and the consequences for representing linear, time-invariant systems. The re-
sulting representation is referred to as convolution. Later in this series of... | https://ocw.mit.edu/courses/res-6-007-signals-and-systems-spring-2011/27da9a018e92fc06d2cf1fbce5cd3e71_MITRES_6_007S11_lec04.pdf |
, time-in-
variant system as a linear combination of delayed impulse responses also be-
comes an integral. The resulting integral is referred to as the convolution in-
tegral and is similar in its properties to the convolution sum for discrete-time
signals and systems. A number of the important properties of convolutio... | https://ocw.mit.edu/courses/res-6-007-signals-and-systems-spring-2011/27da9a018e92fc06d2cf1fbce5cd3e71_MITRES_6_007S11_lec04.pdf |
x[0] 1
x[2]
-l IOJ
fr 2
x[0]
x[O]8a[n]
-.- e--.-0-
-1 0 I2
n
X[1] x[1]8[n-1]
-1 0 I 2
0-0
--
*-
x[-I]8[n+1]
n
-1 0
I 2
X[-] x [-2]8[n +2]
0--0-0
-1 0 1 2
n
x[o]8[n]+x(I] 8[n -1]
+ x [-I]8[n+ ]+.--
+X kr
=2 x[k]8[n-k]
k= -c
Signals and Systems
TRANSPARENCY
4.2
The convolution sum
for linear, time-
invariant disc... | https://ocw.mit.edu/courses/res-6-007-signals-and-systems-spring-2011/27da9a018e92fc06d2cf1fbce5cd3e71_MITRES_6_007S11_lec04.pdf |
added
to form the total
output.
x[n] =E
x[k] S[n-k]
Linear System:
k = -o
+0n
y [n] =E
x[k] hk [n]
k = - 010
5 [n - k] - h [n]
If Time-invariant:
hk [n] = h [n-k]
+ o0
LTI: y[n]
=E x[k] h[n - k]I
Convolution Sum
Convolution
4-5
TRANSPARENCY
4.4
Approximation of a
continuous-time signal
as a linear
combination of
weigh... | https://ocw.mit.edu/courses/res-6-007-signals-and-systems-spring-2011/27da9a018e92fc06d2cf1fbce5cd3e71_MITRES_6_007S11_lec04.pdf |
-6
TRANSPARENCY
4.6
Derivation of the
convolution integral
representation for
continuous-time LTI
systems.
TRANSPARENCY
4.7
Interpretation of the
convolution integral as
a superposition of the
responses from each
of the rectangular
pulses in the
representation of the
input.
x(t) =
(
Eim
'L+0 k=-o
x(k A)
56(t - kA) A... | https://ocw.mit.edu/courses/res-6-007-signals-and-systems-spring-2011/27da9a018e92fc06d2cf1fbce5cd3e71_MITRES_6_007S11_lec04.pdf |
]
k= -o00
Convolution Integral:
+00
x(t)
=f x(-) 6(t-r) dr
+fd
y(t)
= X(r) h(t-,r) dr- = x(t) -* h(t)
y [n] Z x[k]h [n-k]
x [n]= u [n]
h[n]=an u[n]
x [n]
h [n]
x [k]
h [n-k]
n
k
k
0
t
n
Signals and Systems
TRANSPARENCY
4.10
Evaluation of the
convolution integral
for an input that is a
unit step and a system
impulse... | https://ocw.mit.edu/courses/res-6-007-signals-and-systems-spring-2011/27da9a018e92fc06d2cf1fbce5cd3e71_MITRES_6_007S11_lec04.pdf |
LECTURE 2
Hilbert Symbols
Let K be a local field over Qp (though any local field suffices) with char(K) (cid:54)= 2.
Note that this includes fields over Q2, since it is the characteristic of the field, and
not the residue field, with which we are concerned. Recall from the previous lecture
the duality
(2.1) Gal2(K) := Galab(K... | https://ocw.mit.edu/courses/18-786-number-theory-ii-class-field-theory-spring-2016/27f59437d11cbc1a6039b250e4c325ca_MIT18_786S16_lec2.pdf |
and b, as these
can be absorbed in x and y.
Proposition 2.3. The Hilbert symbol satisfies the following properties:
(1) Bimultiplicativity. For all a, b, c ∈ K ×,
(a, bc) = (a, b) · (a, c).
(2) Non-degeneracy. For all a ∈ K ×, if (a, b) = 1 for all b ∈ K ×, then
a ∈ (K ×)2.
Note that (a, b) = (b, a) trivially. Bimultipl... | https://ocw.mit.edu/courses/18-786-number-theory-ii-class-field-theory-spring-2016/27f59437d11cbc1a6039b250e4c325ca_MIT18_786S16_lec2.pdf |
and π ∈ p a uniformizer, that is, π /∈ p2.
Claim 2.5. Let x ∈ K ×, and write x = πv(x)y, where y ∈ O×
K. Then the
following are equivalent:
(1) x is a square;
(2) v(x) is even and y is a square;
(3) y mod p is a square in K ×.
Note that we may reduce to x ∈ O×
K. We offer two proofs:
Proof (via Hensel’s Lemma). All expl... | https://ocw.mit.edu/courses/18-786-number-theory-ii-class-field-theory-spring-2016/27f59437d11cbc1a6039b250e4c325ca_MIT18_786S16_lec2.pdf |
+ p2) (cid:39) k as for any 1 + aπ, 1 + bπ ∈ 1 + p, where a, b ∈ OK/p,
we have (1 + aπ)(1 + bπ) = 1 + (a + b)π + abπ2, and since abπ2 ∈ p2, we are left with
1 + (a + b)π in the associated graded term, hence multiplication simply corresponds
to addition in k. Similarly, for each n ≥ 1, we have (1 + pn)/(1 + pn+1) (cid:3... | https://ocw.mit.edu/courses/18-786-number-theory-ii-class-field-theory-spring-2016/27f59437d11cbc1a6039b250e4c325ca_MIT18_786S16_lec2.pdf |
.9, the map O×
K
σ−→ S is surjective, as desired.
Remark 2.6. In general, the tools we have to deal with O×
K are the p-adic
exponential map, and this filtration, which, though an abstract formalism, has the
advantage of being simpler than O×
K, as the quotients are all isomorphic to finite
fields. As a general principle,... | https://ocw.mit.edu/courses/18-786-number-theory-ii-class-field-theory-spring-2016/27f59437d11cbc1a6039b250e4c325ca_MIT18_786S16_lec2.pdf |
1B. Then, since
the associated graded map is surjective by assumption, we can solve the equation
f ((cid:15)1) ≡ x − f (y0) mod F2B, where (cid:15)1 ∈ F1A describes an “error term” lifted from
Gr1A. Observe that, since f is a homomorphism, we have
f (y1) = f (y0 + (cid:15)1) = f (y0) + f ((cid:15)1) ≡ x mod F2B,
where ... | https://ocw.mit.edu/courses/18-786-number-theory-ii-class-field-theory-spring-2016/27f59437d11cbc1a6039b250e4c325ca_MIT18_786S16_lec2.pdf |
Z, so
have [K × : (K ×)2] = 4 since [O×
K ×/(K ×)2 is isomorphic to Z/2Z × Z/2Z as it has a basis {π, r} ⊂ O×
K, where π
is a uniformizer and r is not a square modulo p (so certainly π and r don’t differ
by a square).
K : (O×
We now reformulate the Hilbert symbol in terms of norms over extension fields;
in contrast to th... | https://ocw.mit.edu/courses/18-786-number-theory-ii-class-field-theory-spring-2016/27f59437d11cbc1a6039b250e4c325ca_MIT18_786S16_lec2.pdf |
again (a, b) = 1.
·
(a + 1)2
a2
+ (−β2a) ·
1
4β2 ·
(a − 1)2
a2
=
(a + 1)2 − (a − 1)2
4a
= 1,
The forward implication is a trivial reversal of the argument for nonzero α. (cid:3)
We state, without proof, the main result about Hilbert Symbols. It’s important
that that the image of L× under the norm is not too big (not ev... | https://ocw.mit.edu/courses/18-786-number-theory-ii-class-field-theory-spring-2016/27f59437d11cbc1a6039b250e4c325ca_MIT18_786S16_lec2.pdf |
Sufficiency
Sufficiency
MIT 18.443
Dr. Kempthorne
Spring 2015
MIT 18.443
Sufficiency
1 Sufficiency
Definition
Example
Theorems
Outline
1 Sufficiency
Definition
Example
Theorems
MIT 18.443
Sufficiency
2 Sufficiency
Definition
Example
Theorems
Sufficient Statistics
Definition: Sufficiency
X1, X2, . . . , Xn iid with ... | https://ocw.mit.edu/courses/18-443-statistics-for-applications-spring-2015/27f9a8f9aaacfe7feb9475af1ec13a79_MIT18_443S15_LEC7.pdf |
ition
Example
Theorems
Sufficiency: Example
Example 8.8.1A Bernoulli Trials Let X = (X1, . . . , Xn) be the
outcome of n i.i.d Bernoulli(θ) random variables
The pmf function of X is:
p(X | θ) = P(X1 = x1 | θ) × · · · × P(Xn = xn | θ)
= θx1 (1 − θ)1−x1 × θx2 (1 − θ)1−x2 × · · · θxn (1 − θ)1−xn
= θ
xi (1 − θ)(n− x... | https://ocw.mit.edu/courses/18-443-statistics-for-applications-spring-2015/27f9a8f9aaacfe7feb9475af1ec13a79_MIT18_443S15_LEC7.pdf |
To make statistical inferences concerning θ, we should only
need the information of T (X) = t, since the value of X given
t reflects only the order information in X which is
independent of θ.
MIT 18.443
Sufficiency
6 Sufficiency
Definition
Example
Theorems
Outline
1 Sufficiency
Definition
Example
Theorems
MIT 18.... | https://ocw.mit.edu/courses/18-443-statistics-for-applications-spring-2015/27f9a8f9aaacfe7feb9475af1ec13a79_MIT18_443S15_LEC7.pdf |
).
Corollary A
If T is sufficient for θ, then the maximum likelihood estimate is a
function of T .
Rao-Blackwell Theorem
Let θˆ be an estimator of θ with E [θˆ2] < ∞ for all θ.
Suppose that T is sufficient for θ
Define θ˜ = E [θˆ | T ].
Then for all θ,
E [(θ˜ − θ)2] ≤ E [(θˆ − θ)2].
The inequality is strict unless... | https://ocw.mit.edu/courses/18-443-statistics-for-applications-spring-2015/27f9a8f9aaacfe7feb9475af1ec13a79_MIT18_443S15_LEC7.pdf |
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
6.265/15.070J
Lecture 5
Fall 2013
9/16/2013
Extension of LD to Rd and dependent process. G¨artner-Ellis Theorem
Content.
1. Large Deviations in may dimensions
2. G¨artner-Ellis Theorem
3. Large Deviations for Markov chains
1
Large Deviations in Rd
Most of the developmen... | https://ocw.mit.edu/courses/15-070j-advanced-stochastic-processes-fall-2013/27fd98c78d35c40fd821548867809ed5_MIT15_070JF13_Lec5.pdf |
, and the addi
tional condition such as M (θ) < ∞ for all θ is needed. Known counterex
amples are somewhat involved and can be found in a paper by Dinwoodie [2]
which builds on an earlier work of Slaby [5]. The difficulty arises that there is
no longer the notion of monotonicity of I(x) as a function of the vector x... | https://ocw.mit.edu/courses/15-070j-advanced-stochastic-processes-fall-2013/27fd98c78d35c40fd821548867809ed5_MIT15_070JF13_Lec5.pdf |
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