text stringlengths 30 4k | source stringlengths 60 201 |
|---|---|
used separate keys: K_{c-‐>s},
K_{s-‐>c}
What
if users connect to wrong
server (analogue of MITM
/ phishing
attack)?
If server is intercepting
packets,
learns what service
user connects to.
•
• What if user accidentally types ssh malicious.server?
o Server learns user's principal name.
o Serve... | https://ocw.mit.edu/courses/6-858-computer-systems-security-fall-2014/1812c0fe15b67c5b73b983475243e19d_MIT6_858F14_lec13.pdf |
match.
Using Kerberos
in an application.
• Paper suggests using special functions to seal messages, 3 security
levels.
• Requires moderate changes to an application.
o Good for flexibility, performance.
o Bad for ease of adoption.
o Hard
for developers
to
understand
subtle
security
guarantees.
•... | https://ocw.mit.edu/courses/6-858-computer-systems-security-fall-2014/1812c0fe15b67c5b73b983475243e19d_MIT6_858F14_lec13.pdf |
) is fast -‐-‐ O(100MB/sec) on
current
hardware.
o Tickets are small, O(100 bytes), so can support 1M tickets/second.
o Easy
to scale
by adding
slaves.
• Potential problem: password changes take a while to propagate.
• Adversary can still use a stolen password for a while after user changes it.
5
... | https://ocw.mit.edu/courses/6-858-computer-systems-security-fall-2014/1812c0fe15b67c5b73b983475243e19d_MIT6_858F14_lec13.pdf |
time.
o Better alternatives are available: SRP, PAKE.
• What
can
adversary do with a stolen
ticket?
• What
can
adversary do with a stolen
K_c?
• What
can
adversary do with a stolen
K_s?
o Remember: two parties share each key (and rely on it) in Kerberos!
• What happens after a password change... | https://ocw.mit.edu/courses/6-858-computer-systems-security-fall-2014/1812c0fe15b67c5b73b983475243e19d_MIT6_858F14_lec13.pdf |
(a and b).
o Alice and Bob send (g^a mod p) and (g^b mod p) to each other.
o Each party
computes (g^(ab) mod p) = (g^a^b mod p) = (g^b^a mod p).
o Use (g^(ab) mod p) as secret key.
o Assume discrete log (recovering a from (g^a mod p)) is hard.
Cross-‐realm
in Kerberos.
•
• Kerberos v4 only supported pairwise cro... | https://ocw.mit.edu/courses/6-858-computer-systems-security-fall-2014/1812c0fe15b67c5b73b983475243e19d_MIT6_858F14_lec13.pdf |
HTTP
requests
,
.
7
MIT OpenCourseWare
http://ocw.mit.edu
6.858 Computer Systems Security
Fall 2014
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/6-858-computer-systems-security-fall-2014/1812c0fe15b67c5b73b983475243e19d_MIT6_858F14_lec13.pdf |
LECTURE 15
LECTURE OUTLINE
Subgradient methods
Calculation of subgradients
•
•
Convergence
•
***********************************************
Steepest descent at a point requires knowledge
•
of the entire subdifferential at a point
Convergence failure of steepest descent
•
3
2
1
0
-1
-2
2
x
-3
-3
-2
-1
0
x1
1
2
3
60
40
z... | https://ocw.mit.edu/courses/6-253-convex-analysis-and-optimization-spring-2012/1814d845fc72f1ad285be44cf8cb3663_MIT6_253S12_lec15.pdf |
⌦
⌦
X
x
f (x) +µ �g(x)
⇤
⌅
or minµ
⇧
0 F (µ), where F (
µ)
−
q(µ).
⌃ −
2 ALGORITHMS: SUBGRADIENT METHOD
Problem: Minimize convex function f :
over a closed convex set X.
n
�
→
Subgradient method:
•
�
•
xk+1 = PX (xk −
αkgk),
where gk is any subgradient of f at xk, αk is a
positive stepsize, and PX ( ) is projection on ... | https://ocw.mit.edu/courses/6-253-convex-analysis-and-optimization-spring-2012/1814d845fc72f1ad285be44cf8cb3663_MIT6_253S12_lec15.pdf |
for all αk such that
�
y
�
<
xk −
�
y
,
�
0 < αk <
2
�
f (y)
f (xk)
−
2
gk�
�
⇥
.
4 PROOF
Proof of nonexpansive property
•
PX (x)
�
PX (y)
x
� ⌥ �
y
,
�
−
−
x, y
n.
⌘ �
Use the projection theorem to write
PX (x)
�
x
z
−
PX (x)
−
0,
⌥
z
⌘
X
�
�
⇥
PX (y)
from which
Similarly, PX (x)
�
Adding and using the
⇥
�
−
�
x
P... | https://ocw.mit.edu/courses/6-253-convex-analysis-and-optimization-spring-2012/1814d845fc72f1ad285be44cf8cb3663_MIT6_253S12_lec15.pdf |
−
�
y
2
y
⌃
⌃
) +α 2
gk�
k�
+ α2
k�
2
2,
gk�
where the last inequality follows from the subgra-
�
dient inequality. Q.E.D.
f (y)
⇥
5 CONVERGENCE MECHANISM
Assume constant stepsize: αk ⌃
If
c for some constant c and all k,
α
•
•
�
gk� ⌥
x⇤�
2
�
x
−
k+1
x⇤�
so the distance to the optimum decreases if
f (x⇤)
f (x )
k
⌥ �
... | https://ocw.mit.edu/courses/6-253-convex-analysis-and-optimization-spring-2012/1814d845fc72f1ad285be44cf8cb3663_MIT6_253S12_lec15.pdf |
imum.
fk}
level set
fk can be adjusted based on the progress of
the method.
f (x)
⌥
x
{
|
.
Example of dynamic stepsize rule:
•
fk = min f (xj)
j
⌅
⌅
k
0
⌅k,
−
and ⌅k (the “aspiration level of cost reduction”) is
updated according to
⌅k+1 =
⌦⌅k
max
�
⇥⌅k, ⌅
fk,
if f (xk+1)
if f (xk+1) > fk,
⌥
⌅
where ⌅ > 0, ⇥ < 1, and ... | https://ocw.mit.edu/courses/6-253-convex-analysis-and-optimization-spring-2012/1814d845fc72f1ad285be44cf8cb3663_MIT6_253S12_lec15.pdf |
6.001 Structure and Interpretation of Computer Programs. Copyright © 2004 by Massachusetts Institute of Technology.
6.001 Notes: Section 3.1
Slide 3.1.1
In this lecture, we are going to put together the basic pieces of
Scheme that we introduced in the previous lecture, in order to
start capturing computational pro... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/182629e35d886325280dbc1bb4b5643c_lecture3webhand.pdf |
by Massachusetts Institute of Technology.
Slide 3.1.3
So here are the rules for the substitution model. Given an
expression, we want to determine its value. If the expression is
a simple expression there are several possibilities. If it is a self-
evaluating expression, like a number, we just return that value.
If... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/182629e35d886325280dbc1bb4b5643c_lecture3webhand.pdf |
just a name, and we look up its value, getting back the
procedure we created when we evaluated the lambda. The
second subexpression is just a number, so that is its value. Now
the rule says to substitute 4 everywhere we see the formal
parameter x in the body of the procedure and replace the original expression with... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/182629e35d886325280dbc1bb4b5643c_lecture3webhand.pdf |
. This is called an
applicative order evaluation model.
Slide 3.1.7
Notice in particular that under this model I need to get the value
of the operands first, which caused me to evaluate that interior
compound expression before I got around to applying average.
Once I have simple values, I can continue the substit... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/182629e35d886325280dbc1bb4b5643c_lecture3webhand.pdf |
recognize that this latter computation is just factorial of n-1.
So notice what this does, it reduces a computation to a simpler
operation, multiplication, and a simpler version of the same
problem, in this case factorial of a smaller number. So I have
reduced one version of a problem to a simpler version of the
s... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/182629e35d886325280dbc1bb4b5643c_lecture3webhand.pdf |
of rules recursively on this
expression. It does this before it ever looks at either of the two
other expressions. Now, if the predicate evaluates to a true
value, then the if expression takes the consequent, and evaluates
it, returning that value as the value of the overall if expression.
On the other hand, if th... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/182629e35d886325280dbc1bb4b5643c_lecture3webhand.pdf |
to a simple answer.
Note the form shown in red, in which an expression is reduced to this pattern of a simpler operation and a simpler
version of the same problem. This unwrapping continues until we are left with a nested expression whose
innermost subexpression only involves simple expressions and operations, at wh... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/182629e35d886325280dbc1bb4b5643c_lecture3webhand.pdf |
solution to the full version of factorial.
Notice that the second step here requires some ingenuity. One
should be careful to think through this strategy, before
beginning to code up a solution. In the case of factorial, we saw
how we did this decomposition into a multiplication and a
smaller version of factorial.... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/182629e35d886325280dbc1bb4b5643c_lecture3webhand.pdf |
do that
decomposition and identify the smallest version of the problem
that will stop the unwinding of the computation into simpler
versions of the same problem. As a consequence, algorithms
associated with this kind of approach have a test, a base case,
and a recursive case to control the order of evaluation in o... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/182629e35d886325280dbc1bb4b5643c_lecture3webhand.pdf |
computation does not require keeping
track of any deferred operations, but rather can perform its
work without using up any additional memory to keep track of
things to do.
6.001 Structure and Interpretation of Computer Programs. Copyright © 2004 by Massachusetts Institute of Technology.
Slide 3.4.4
So that's ou... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/182629e35d886325280dbc1bb4b5643c_lecture3webhand.pdf |
current product and the next
multiplier.
6.001 Structure and Interpretation of Computer Programs. Copyright © 2004 by Massachusetts Institute of Technology.
Slide 3.4.8
The reason I can do this is because multiplication satisfies two
nice properties from mathematics, that in essence say I can do
the multiplicati... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/182629e35d886325280dbc1bb4b5643c_lecture3webhand.pdf |
bigger than n.
Slide 3.4.14
Ah -- and that indicates one more thing. I will need another
column to keep track of n, since each column specifies a
needed piece of information and that is one more. The fact that
the value doesn't change is irrelevant, what matters is that I
need to keep track of this information.
... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/182629e35d886325280dbc1bb4b5643c_lecture3webhand.pdf |
value, an answer to give when the test case holds. The iterative case, the thing to
do if the test case is not true, is slightly different from our recursive case. Here, the call to the procedure has new
arguments, one for each parameter. Notice how the new argument in each case captures the update rule from the
tab... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/182629e35d886325280dbc1bb4b5643c_lecture3webhand.pdf |
special form, and this means we first evaluate
the predicate. Since in this case it returns a false value, the
value of the whole if expression reduces to the value of the
alternative clause.
6.001 Structure and Interpretation of Computer Programs. Copyright © 2004 by Massachusetts Institute of Technology.
Slide ... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/182629e35d886325280dbc1bb4b5643c_lecture3webhand.pdf |
The value of one call to the procedure is the same as
the value of a different call to the same procedure, that is,
different arguments but nothing deferred. As a consequence,
this procedure has a fixed size, or constant, evolution.
So we have seen two different implementations of the same
computation, with differ... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/182629e35d886325280dbc1bb4b5643c_lecture3webhand.pdf |
or logical proposition is
a chain or sequence of logical deductions that starts with a base
set of axioms and leads to the proposition we are attempting to
prove. Now, we need to fill in the details on these pieces.
First, a proposition is basically a statement that is either true or
false. Typically, in propositi... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/182629e35d886325280dbc1bb4b5643c_lecture3webhand.pdf |
ling. The proposition is defined to
be true if either P is false or Q is true. To see this, take a simple
example. Suppose we consider the proposition: "If n is greater
than 2, then n squared is greater than 4". Then for example, if n equals 4, both propositions are true, and our truth
table also states that the im... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/182629e35d886325280dbc1bb4b5643c_lecture3webhand.pdf |
prove statements
of fact. Unfortunately, the things we want to prove about
programs (e.g. does this code run for all correct inputs) require
potentially infinite propositions, since we would have to take a
conjunction of propositions, one for each input value. So we
need to extend our propositional logic to predic... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/182629e35d886325280dbc1bb4b5643c_lecture3webhand.pdf |
2004 by Massachusetts Institute of Technology.
essence, you can "bootstrap" a proof that the statement is always true.
Slide 3.5.7
Let's make this a bit less murky with an example. Suppose I
want to prove the predicate shown here for all non-negative
values of n, that is, that the sum of the powers of 2 can be
ca... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/182629e35d886325280dbc1bb4b5643c_lecture3webhand.pdf |
3.5.9
So finally we can relate this back to programs. Suppose we
want to prove that our code for factorial is correct, that is, that it
will always run correctly. Note that there is a hidden
assumption here, namely that we give it a correct input. Note
that here, we have made the unstated assumption that factorial... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/182629e35d886325280dbc1bb4b5643c_lecture3webhand.pdf |
!
Slide 3.5.12
Now, let's pull this together. The message to take away from
this exercise is that induction provides a basis for
understanding, analyzing and proving correctness of recursive
procedure definitions.
Moreover, exactly this kind of thinking can be used when we
design programs, not just when we analy... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/182629e35d886325280dbc1bb4b5643c_lecture3webhand.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
18.014 Calculus with Theory
Fall 2010
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/18-014-calculus-with-theory-fall-2010/182970ab73d4377688a774a8db9db7f3_MIT18_014F10_ChDnotes.pdf |
Queueing Systems: Lecture 6
Amedeo R. Odoni
November 6, 2006
Lecture Outline
• Congestion pricing in transportation: the
fundamental ideas
• Congestion pricing and queueing theory
• Numerical examples
• A real example from LaGuardia Airport
• Practical complications
Reference: Handout on “Congestion Pricing
an... | https://ocw.mit.edu/courses/1-203j-logistical-and-transportation-planning-methods-fall-2006/18342098bf20f7b29102666916eba761_lec10.pdf |
C = cLq = cλWq
and the marginal delay cost, MC , imposed by an
additional (“marginal”) user is given by:
dW q
dC
dλ
dλ
= c W + cλ
q
MC =
Marginal
cost
Internal
cost
External
cost
Numerical Example
• Three types of aircraft; Poisson; FIFO service
_ Non-jets: λ1 = 40 per hour; c1 = $600 per hour
... | https://ocw.mit.edu/courses/1-203j-logistical-and-transportation-planning-methods-fall-2006/18342098bf20f7b29102666916eba761_lec10.pdf |
= (c1 ⋅λ1 + c2 ⋅λ2 + c3 ⋅λ3 ) ⋅Wq = c ⋅Wq
Or: C = c ⋅Wq = ($138,000) ⋅ (0.0464) = $6,400
dWq
dλ
=
2
E 2[S] +σ S
2 ⋅ (1 − ρ)
+
2 ]
λ⋅[E 2[S] +σ S
⋅
1
2 ⋅ (1 − ρ)2 μ
≈ 5.1556 ×10−6 hours ≈ 18.6 sec
Numerical Example [3]
dC
dλ1
= c1 ⋅Wq + c ⋅
dWq
dλ
≈ $28 + $711 = $739
internal
cost
external cost=
co... | https://ocw.mit.edu/courses/1-203j-logistical-and-transportation-planning-methods-fall-2006/18342098bf20f7b29102666916eba761_lec10.pdf |
λ
μ i = 1
= ∑ρi = ∑ λi
i =1 μi
c = ∑
m ⎛ λi ci
⎞
i= 1 ⎝ λ ⎠
Generalization (continued)
• As before: C = cL = cλ W
q
q
giving: MC(i) =
dC
dλ i
= ciW q + cλ
dW q
dλ i
• When we have explicit expressions for Wq, we
can also compute explicitly the total marginal
delay cost MC(i), the internal (or private)... | https://ocw.mit.edu/courses/1-203j-logistical-and-transportation-planning-methods-fall-2006/18342098bf20f7b29102666916eba761_lec10.pdf |
ultaneously the m equations:
xi = i
q[ ˆ( ˆ)]
c ⋅W λ x +
⎛ m
⎜ ∑ j
⎜
⎝ j =1
where λˆ(xˆ) = {λ1(x1),λ2 (x2 ),...,λm (xm )} .
dW [λˆ(xˆ)]
)⎞
q
⎟
j x j ⎟
⋅
(
dλ
i
⎠
c ⋅λ (
)x
i
+ K
i
∀i
The missing piece: Demand functions can
only be roughly estimated, at best!
An illustrative example from airports
Servi... | https://ocw.mit.edu/courses/1-203j-logistical-and-transportation-planning-methods-fall-2006/18342098bf20f7b29102666916eba761_lec10.pdf |
40
60
0,001
0,01
Demand Functions for three types of users
0,00001
0,00008
lambda 1 lambda 2 lambda 3
60
58,2
54,8
49,8
43,2
35
25,2
13,8
0,8
13,8
-
-
30
-
47,8
67,2
-
-
88,2
110,8
-
-
135
-
160,8
-
188,2
217,2
-
247,8
-
280
-
Total cost ($)
50
49,5
48,6
47,3
45,6
43,5
41
38,1
34,8
31,1
27
22,5
17,6
12,3
6,6
0,5
-
6
... | https://ocw.mit.edu/courses/1-203j-logistical-and-transportation-planning-methods-fall-2006/18342098bf20f7b29102666916eba761_lec10.pdf |
0
1600
1700
0
1800
1900
2000
Ty pe 1
Ty pe 2
Ty pe 3
Case 1: No Congestion Fee
No Congestion Fee
(1) Delay cost (IC) per aircraft
(2) Congestion fee
(3) Total cost of access
[=(1)+(2)]
(4) Demand (no. of movements
per hour)
(5) Total demand (no. of
movements per hour)
(6) Expected delay per aircraft
(7) U... | https://ocw.mit.edu/courses/1-203j-logistical-and-transportation-planning-methods-fall-2006/18342098bf20f7b29102666916eba761_lec10.pdf |
U
(
e
t
a
r
l
a
v
i
r
r
A
70
60
50
40
30
20
10
0
o
+
+
+
0
200
400
600
800
1000
1200
1400
1600
1800
Total cost ($)
Type 1
Type 2
Type 3
o No Fee
+ With Fee
o
2000
Important to note…
• The external costs computed in the absence
of congestion pricing give only an upper
bound on the magnitu... | https://ocw.mit.edu/courses/1-203j-logistical-and-transportation-planning-methods-fall-2006/18342098bf20f7b29102666916eba761_lec10.pdf |
to
longer-term policy that would use “market-based” mechanisms
• Process stopped after September 11, 2001; re-opened in 2004
Scheduled aircraft movements at LGA
before and after slot lottery
120
Scheduled
movements
100
per hour
80
60
40
20
0
Nov, 00
Aug, 01
81 flights/hour
5
7
9
11
13
15
17
19... | https://ocw.mit.edu/courses/1-203j-logistical-and-transportation-planning-methods-fall-2006/18342098bf20f7b29102666916eba761_lec10.pdf |
0
($ per
movt)
$3,500
$3,000
$2,500
$2,000
$1,500
$1,000
$500
$0
5
7
9
11
13
19
15
17
21
Time of day
23
1
3
Issues that arise in practice
-- Toll may vary in time and by location
-- Facility users may be driven by “network”
considerations
-- “Social benefit” considerations
-- Political ... | https://ocw.mit.edu/courses/1-203j-logistical-and-transportation-planning-methods-fall-2006/18342098bf20f7b29102666916eba761_lec10.pdf |
PDE Examples Sheet
Problem 1. Prove that a Harmonic function with an interior maximum is constant.
Problem 2. Write out the laplacian in planepolar coordinates.
Problem 3. A Green’s function on Rn is a harmonic function on Rn \ {0} which
depends only on the radius (for example log r on R2). Find nontrivial Green’s f... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2005/184a3abfb02fb6d9a8d27996023feb00_pset.pdf |
Lecture 3
8.251 Spring 2007
Lecture 3 - Topics
• Relativistic electrodynamics.
•
Gauss’ law
• Gravitation and Planck’s length
Reading: Zwiebach, Sections: 3.1 - 3.6
Electromagnetism and Relativity
Maxwell’s Equations
Source-Free Equations:
� × E� = −
1 ∂B�
c ∂t
� · B� = 0
With Sources (Charge, Current):
... | https://ocw.mit.edu/courses/8-251-string-theory-for-undergraduates-spring-2007/184b1fe03c3d32c9ac0a860af62b877c_lec3.pdf |
= −�Φ
(Φ scalar)
E� = −�Φ −
1 ∂A�
c ∂t
( �
�
E, B� ) encoded as (Φ, A)
Φ, A are the fundamental quantities we’ll use
Gauge Transformations
A� → A�� = A� + �
B� � = � × A� = � × (A + �) = B�
function of �x,t. � function = vector.
Φ → Φ� = Φ −
1 ∂
c ∂t
�
E� � = −�(Φ�) = −� Φ −
�
1 ∂
c ∂t
−
1 ∂
c ∂t
(A ... | https://ocw.mit.edu/courses/8-251-string-theory-for-undergraduates-spring-2007/184b1fe03c3d32c9ac0a860af62b877c_lec3.pdf |
= −Fνµ
Foi =
1 ∂Ai
c ∂t
−
∂
∂xi
(−Φ) = −Ei
F12 = ∂xAy − ∂yAx = Bz
⎛
Fµν = ⎜
⎜
⎝
0
0 −Ex −Ey −Ez
Bz −By
Ex
Bx
0
Ey −Bz
0
−Bx
Ez
By
⎞
⎟
⎟
⎠
What happens under gauge transformation?
Aµ → Aµ
� = Aµ + ∂µ
Then get:
F � = ∂µA�
µν
ν − ∂ν A�
µ
= ∂µ(Aν + ∂ν ) − ∂ν (Aµ + ∂µ)
= Fµν + ∂µ∂ν − ∂ν ∂µ
= Fµν
Defi... | https://ocw.mit.edu/courses/8-251-string-theory-for-undergraduates-spring-2007/184b1fe03c3d32c9ac0a860af62b877c_lec3.pdf |
conserved and a Lorentz invar, (cρ, J�) form
a 4-vector J µ
Now let’s do what a typical theoretical physicist does for a living: guess the
equation!
F µν ≈ J µ
No, derivatives not right.
∂F µν /∂xν ≈ J µ
No, constants not right.
F µν /∂xµ = J µ
1
c
Correct, amazingly! (even sign)
µ = 0:
∂F 0ν /∂xν = ρ
∂F ... | https://ocw.mit.edu/courses/8-251-string-theory-for-undergraduates-spring-2007/184b1fe03c3d32c9ac0a860af62b877c_lec3.pdf |
3D spacial world E and B are both vectors.
Let’s look at � · E� = ρ in all dimensions.
Notation: Circle S� is a 1D manifold, the boundary of a ball B2
Sphere S2(R) : x2
1
Ball B3(R) : x2
2
+ x + x
1
2
2
+ x + x
2
2
3 = R2
3 ≤ R2
2
5
Lecture 3
8.251 Spring 2007
When talking about S2(R), call it S2 (R = 1 implied)
... | https://ocw.mit.edu/courses/8-251-string-theory-for-undergraduates-spring-2007/184b1fe03c3d32c9ac0a860af62b877c_lec3.pdf |
d(r)
Bd(r)
This represents the flux of E� through Sd−1(r)
E(r) vol(Sd−1(r)) = q
·
E(r) =
Γ(d/2) q
2πd/2 rd−1
Electric field of a point charge in d dimensions.
If there are extra dimensions, then would see larger E at very small distances.
7 | https://ocw.mit.edu/courses/8-251-string-theory-for-undergraduates-spring-2007/184b1fe03c3d32c9ac0a860af62b877c_lec3.pdf |
Introduction to Algorithms: 6.006
Massachusetts Institute of Technology
Instructors: Erik Demaine, Jason Ku, and Justin Solomon
Recitation 3
Recitation 3
Recall that in Recitation 2 we reduced the Set interface to the Sequence Interface (we simulated
one with the other). This directly provides a Set data structur... | https://ocw.mit.edu/courses/6-006-introduction-to-algorithms-spring-2020/1869dbf640ded6b31f1bd369d2001ef5_MIT6_006S20_r03.pdf |
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
Recitation 3
Sorting
3
Sorting an array A of comparable items into increasing order is a common subtask of many com-
putational problems. Insertion sort and selection sort are common sorting algorithms for sorting
small numbers of ... | https://ocw.mit.edu/courses/6-006-introduction-to-algorithms-spring-2020/1869dbf640ded6b31f1bd369d2001ef5_MIT6_006S20_r03.pdf |
O(i) search for max in A[:i]
# O(1) check for larger value
# O(1) new max found
# O(1) swap
Insertion Sort
Here is a Python implementation of insertion sort. Having already sorted sub-array A[:i], the
algorithm repeatedly swaps item A[i] with the item to its left until the left item is no larger than
A[i]. As ca... | https://ocw.mit.edu/courses/6-006-introduction-to-algorithms-spring-2020/1869dbf640ded6b31f1bd369d2001ef5_MIT6_006S20_r03.pdf |
for sorting large numbers
of items. The algorithm recursively sorts the left and right half of the array, and then merges the
two halves in linear time. The recurrence relation for merge sort is then T (n) = 2T (n/2) + Θ(n),
which solves to T (n) = Θ(n log n). An Θ(n log n) asymptotic growth rate is much closer to
... | https://ocw.mit.edu/courses/6-006-introduction-to-algorithms-spring-2020/1869dbf640ded6b31f1bd369d2001ef5_MIT6_006S20_r03.pdf |
i = i + 1
else:
A[a] = R[j]
j = j + 1
a = a + 1
# O(1) merge from left
# O(1) decrement left pointer
# O(1) merge from right
# O(1) decrement right pointer
# O(1) decrement merge pointer
Merge sort uses a linear amount of temporary storage (temp) when combining the two halves, so
it is not in-place. While th... | https://ocw.mit.edu/courses/6-006-introduction-to-algorithms-spring-2020/1869dbf640ded6b31f1bd369d2001ef5_MIT6_006S20_r03.pdf |
recurrences:
• Substitution: Guess a solution and substitute to show the recurrence holds.
• Recursion Tree: Draw a tree representing the recurrence and sum computation at nodes.
This is a very general method, and is the one we’ve used in lecture so far.
• Master Theorem: A general formula to solve a large class of... | https://ocw.mit.edu/courses/6-006-introduction-to-algorithms-spring-2020/1869dbf640ded6b31f1bd369d2001ef5_MIT6_006S20_r03.pdf |
some constant k ≥ 0
T (n) = Θ(f (n))
f (n) = Ω(nlogb a+ε) for some constant ε > 0
and af (n/b) < cf (n) for some constant 0 < c < 1
The Master Theorem takes on a simpler form when f (n) is a polynomial, such that the recurrence
has the from T (n) = aT (n/b) + Θ(nc) for some constant c ≥ 0.
Recitation 3
6
case ... | https://ocw.mit.edu/courses/6-006-introduction-to-algorithms-spring-2020/1869dbf640ded6b31f1bd369d2001ef5_MIT6_006S20_r03.pdf |
) + O(1)
Solution: T (n) = O(2n), height n binary tree, O(1) work per node.
5. T (n) = T (2n/3) + O(1)
Solution: T (n) = O(log n), length log3/2(n) chain, O(1) work per node.
6. T (n) = 2T (n/2) + O(1)
Solution: T (n) = O(n), height log2 n binary tree, O(1) work per node.
7. T (n) = T (n/2) + O(n)
Solution: T (n... | https://ocw.mit.edu/courses/6-006-introduction-to-algorithms-spring-2020/1869dbf640ded6b31f1bd369d2001ef5_MIT6_006S20_r03.pdf |
8.701
0. Introduction
0.2 Course Organization
Introduction to Nuclear
and Particle Physics
Markus Klute - MIT
1
Course Setup
Inverted classroom in online setting (see additional video A.02)
Lectures are organized in short video discussing specific
concepts or methods. Each video is supplemented with slides,
text, ... | https://ocw.mit.edu/courses/8-701-introduction-to-nuclear-and-particle-physics-fall-2020/18834425462a7fb4f77dd5b94bedfd54_MIT8_701f20_lec0.2.pdf |
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-1
Lecture 5 - PN Junction and MOS
Electrostatics (II)
pn Junction in Thermal Equilibrium
September 22, 2005
Contents:
1. Introduction to pn junction
2. Electrostatics of pn junction in thermal equilibrium
3. The depletion approximation
4. Contact potent... | https://ocw.mit.edu/courses/6-012-microelectronic-devices-and-circuits-fall-2005/188f6340d4d10b0e923d1c0166c8fba1_lec5.pdf |
)
to p side
x = 0
x
(cid:0)(cid:0)(cid:0)(cid:0)
(Nd)
n
(cid:0)(cid:0)(cid:0)(cid:0)
(b)
metal contact to(cid:13)
n side
Doping distribution of abrupt p-n junction:
p-region
Na
N
0
n-region
Nd
x
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-5
What is the carrier concentration distribution in the... | https://ocw.mit.edu/courses/6-012-microelectronic-devices-and-circuits-fall-2005/188f6340d4d10b0e923d1c0166c8fba1_lec5.pdf |
regions:
• two quasi-neutral n- and p-regions (QNR’s)
• one space charge region (SCR)
Now, want to know no(x), po(x), ρ(x), E(x), and φ(x).
Solve electrostatics using simple, powerful approximation.
(cid:13)
(cid:13)
(cid:13)
(cid:13)
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-8
3. The depletio... | https://ocw.mit.edu/courses/6-012-microelectronic-devices-and-circuits-fall-2005/188f6340d4d10b0e923d1c0166c8fba1_lec5.pdf |
φB
-xpo
0
xno
x
x
φn
x
0
0
0
φp
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-13
φ
φB
-xpo
0
xno
0
φp
φn
x
• x < −xpo
φ(x) = φp
• − xpo < x < 0
φ(x) − φ(−xpo)
= − R
= qNa
2(cid:15)s
−qNa
x
−xpo
(cid:15)s
(x + xpo)2
(x + xpo)dx
φ(x) = φp + qNa
2(cid:15)s
(x + xpo)2
• 0 < x < xno
φ(x) = φn − qNd
2(c... | https://ocw.mit.edu/courses/6-012-microelectronic-devices-and-circuits-fall-2005/188f6340d4d10b0e923d1c0166c8fba1_lec5.pdf |
15
Other results:
Total width of space charge region:
xdo = xno + xpo =
v
u
u
u
u
u
t
2(cid:15)sφB(Na + Nd)
qNaNd
Field at metallurgical junction:
|Eo| =
v
u
u
u
u
u
t
2qφBNaNd
(cid:15)s(Na + Nd)
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-16
Three cases:
• Symmetric junction: Na = Nd ⇒ xpo = xn... | https://ocw.mit.edu/courses/6-012-microelectronic-devices-and-circuits-fall-2005/188f6340d4d10b0e923d1c0166c8fba1_lec5.pdf |
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-18
We are missing contact potential at metal-semiconductor
contacts:
p
-
+
n
p-QNR
SCR
φ
n-QNR
φmp
φB
-xpo
0
xno
φmn
x
Metal-semiconductor contacts: junctions of dissimilar ma-
terials
⇒ built-in potentials: φmn, φmp
Potential difference across structure... | https://ocw.mit.edu/courses/6-012-microelectronic-devices-and-circuits-fall-2005/188f6340d4d10b0e923d1c0166c8fba1_lec5.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
18.01 Single Variable Calculus
Fall 2006
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
Lecture 5
18.01 Fall 2006
Implicit
5
Lecture
Differentiation
and
Inverses
Implicit Differentiation
d
dx
(x a) = ax a−1 .
Example 1.... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2006/18d6a86a30a4bc046c5f7034d47587f1_lec5.pdf |
of 1: x2 +y2 = 1 which we can write as y2 = 1−x2 .
So y = ±
1 − x2. Let us look at the positive case:
√
1
y = + 1 − x2 = (1 − x 2) 2
dy
dx
=
1
−
(1 − x 2)
2
(−2x) =
√
−x
1 − x2
=
−x
y
�
� �
1
2
1
Lecture 5
18.01 Fall 2006
Now, let’s do the same thing, using implicit differentiation.
d ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2006/18d6a86a30a4bc046c5f7034d47587f1_lec5.pdf |
f , f −1:
x = g(y) = f −1(y)
Now, let us use implicit differentiation to find the derivative of the inverse function.
y = f (x)
f −1(y) = x
d
dx
(f −1(y)) =
d
dx
(x) = 1
By the chain rule:
= 1
d
dy
(f −1(y))
dy
dx
and
(f −1(y)) =
d
dy
1
dy
dx
2
Lecture 5
18.01 Fall 2006
So, implicit differentiation ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2006/18d6a86a30a4bc046c5f7034d47587f1_lec5.pdf |
dx
=
1
1 + x2
d
dx
arctan(x) =
1
1 + x2
18.01 Fall 2006
Graphing an Inverse Function.
Suppose y = f (x) and g(y) = f −1(y) = x. To graph g and f together we need to write g as a
function of the variable x. If g(x) = y, then x = f (y), and what we have done is to trade the
variables x and y. This is illustra... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2006/18d6a86a30a4bc046c5f7034d47587f1_lec5.pdf |
§ 18. Lattice codes (by O. Ordentlich)
Consider the n-dimensional additive white Gaussian noise (AWGN) channel
Y X Z
+
=
where Z ∼ N (0, In×n) is statistically independen
t
reliably over this channel, under the power constraint
of the input X. Our goal is to communicate
∥ ∥2 ≤ SNR
X
1
n
where SNR is the signal-to-noise... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/18e413df88c6303554cc45a6b3b876b6_MIT6_441S16_chapter_18.pdf |
.1)
where B(r) is an n-dimensional ball of radius r.
18.1 Lattice Definitions
A lattice Λ is a discrete subgroup of
lattice Λ in R
R
n is spanned by some n n matrix G such that
×
n which is closed under reflection and real addition. Any
Λ = {t = Ga ∶ a ∈ Z
n}.
185
We will assume G is full-rank. Denote the nearest neighb... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/18e413df88c6303554cc45a6b3b876b6_MIT6_441S16_chapter_18.pdf |
Dt ≜ V ∩ (t + S).
Vol
(V)
. Furthermore
Note that
Thus
Moreo
ver
and therefore
∩
+ V) S] + t
Dt = [(−t
= A− +
t
t.
Vol(S) =
∑
Λ
t
∈
Vol
(A ) =
t
∑
∈Λ
t
Vol
(A− + ) =
t
t
Vol
(Dt) = Vol(V).
∑
∈Λ
t
S =
⊍
∈Λ
t
At =
⊍
∈Λ
t
A−t =
⊍
Λ
t
∈
Dt − t,
[S] mod Λ =
Dt = V.
⊍
∈Λ
t
186
Corollary
(
∣
det G
)∣. In Particular, Vol
(V
)... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/18e413df88c6303554cc45a6b3b876b6_MIT6_441S16_chapter_18.pdf |
n
synd
for some g
mod 1 operation above is to be understood as componentwise
modulo reduction. Thus, a lattice decoder is indeed much more “structured” than ML decoder for a
random code.
R , where
the
Note that for an additive channel Y X Z, if X Λ we have that
=
+
∈
Pe = Pr
(QΛ(Y
)
≠ X) = Pr(Z ∉ V).
(18.4)
We therefor... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/18e413df88c6303554cc45a6b3b876b6_MIT6_441S16_chapter_18.pdf |
ly speaking, the Voronoi region of a lattice
norm-ergodic noise as a ball with the same volume.
that is
go
od for
coding is as resilient to semi
187
(a)
Figure 18.1: (a) shows a lattice in R
effective ball.
(b)
2, and (b) shows its Voronoi region and the corresponding
18.2 First Attempt at AWGN Capacity
√
n with reff(Λ)... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/18e413df88c6303554cc45a6b3b876b6_MIT6_441S16_chapter_18.pdf |
error probability is
t + Z for some t
v =
( − )
Y v
on the shifted
output. Since
Pe = Pr Q
( Λ(Y − v) ≠ t
) = Pr(Z ∉ V).
188
reff heme over
with n. Taking δ → 0 we see that any rate
log(SNR) can be achieved reliably. Note that for this coding scheme (encoder+decoder) the
∥
Since Λ is good for coding and
additive semi no... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/18e413df88c6303554cc45a6b3b876b6_MIT6_441S16_chapter_18.pdf |
ol(Vc) 1 n
V
)
Vol(Vf
)
(18.7)
A nested
⊂
pair Λc Λf is defined as [CS83, For89, EZ04]
lattice code (sometimes also called “Voronoi constellation”)
based on the nested lattice
(18.8)
Proposition 18.2.
L ≜ Λf ∩ Vc.
∣L∣ =
Vol(Vc)
Vol(Vf )
.
Thus, the codebook L has rate R = 1 log ∣L∣ = log Γ
n
(Λf , Λc).
Proof. First note... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/18e413df88c6303554cc45a6b3b876b6_MIT6_441S16_chapter_18.pdf |
related to a quantity called the second
n E∥U∥2.
moment of a lattice. Let U Uniform
Let W ∼ Uniform(B(reff(Λ)). By the isoperimetric inequality [Zam14]
The average transmission
∼
L
L
V
σ2(Λ) ≥
1
n
E∥W∥2 =
r2
eff(Λ)
+
n 2
.
exhibits
( )
(
B
reff Λ .
a good tradeoff between average power and volume if its second moment is cl... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/18e413df88c6303554cc45a6b3b876b6_MIT6_441S16_chapter_18.pdf |
EZ04]. Let Λc Λf
) =
) = n 1+SNR (1 + (cid:15)). The rate is
be a
nested
SNR − (cid:15) , whereas the fine lattice is good for coding and has r2
(
)
1
therefore
(
eff Λf
SNR
(
1
n
1
2
log
(
log (
log
)
ol(Vc)
V
Vol(Vf )
r2
eff(Λc)
r2
eff(Λf )
SNR(1 − (cid:15))
SNR
1+SNR (1 + (cid:15))
)
⎛
⎝
R =
=
→
→
1
2
1
2
log
(1 + SNR) ... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/18e413df88c6303554cc45a6b3b876b6_MIT6_441S16_chapter_18.pdf |
distributed over
e in R
lattic
V
n, let U Uniform
∼
vector X
(V)
= [V + U
and let V be a
] mod Λ is
Proof.
[
that v
or any v
F
V]
+
mod Λ
∈ R
n the set v
= V
and Vol
+ V
( +
v
fundamen
a
is
(V)
=
V)
Vol
tal cell of Λ. Th
us, by Proposition 18.1 we have
∈
. Thus, for any v R
n
X∣V = v ∼ [v + U] mod Λ ∼ Uniform(V).
The C... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/18e413df88c6303554cc45a6b3b876b6_MIT6_441S16_chapter_18.pdf |
ffective noise, that is statistically independen
t of t, with effective variance
σ2
eff(α) ≜
1
n
E
∥Zeff∥2 α2
<
+ ( −
1 α 2SNR.
)
(18.11)
(18.12)
(18.13)
(18.14)
dic,
Since Z is semi norm-ergo
is good for MSE
α . Setting α SNR 1 SNR , such as to minimize the upper bound on σ2
variance σ2
( )
eff
ariance σ2
in effective v
er ... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/18e413df88c6303554cc45a6b3b876b6_MIT6_441S16_chapter_18.pdf |
scheme the error probability does not depend on the chosen message,
such that Pe,max Pe,avg. However, this required common randomness in the form of the dither U.
By a standard averaging argument it follows that there exist some fixed shift u that achieves the
same, or better, Pe,avg. However, for a fixed shift the error... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/18e413df88c6303554cc45a6b3b876b6_MIT6_441S16_chapter_18.pdf |
known to achieve the interference free capacity. There are many more scenarios where lattice
codes can reliably achieve better
than the best known random coding schemes.
rates
( +
)
2
18.5 Construction of Good Nested Lattice Pairs
We now briefly describe a method for constructing nested lattice pairs. Our construction i... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/18e413df88c6303554cc45a6b3b876b6_MIT6_441S16_chapter_18.pdf |
dew
a lattice. Moreover, if F is full-rank over Zp, then
pk. Consequently, the number of lattice points
ev
=
corresponding Voronoi region must satisfy Vol
C( )
ctors in
F , and a
F due to the linearity of
)
( )
it
holds that x Λ F , and that if all
that for any x Λ F
minimum distance. Thus, Λ F is indeed
F are distinct... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/18e413df88c6303554cc45a6b3b876b6_MIT6_441S16_chapter_18.pdf |
∩
b
y
′
′
w˜ T
= [
k
0 ⋯ 0 wT ] ∈ Z
p,
†
k′ zeros
(18.16)
and taking t = t(w
the (finite field) generating
) = [[
w˜ T F mod p mod Λ F
]
( ′
is needed.
F
matrix
]
). Also, in order to specify the codebook , only
L
we take the elements of F to be i.i.d. and
If
nested
lattice
p O n 1 (cid:15) 2
( + )/ )
= (
coarse lattice ... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/18e413df88c6303554cc45a6b3b876b6_MIT6_441S16_chapter_18.pdf |
Massachusetts Institute of Technology
6.042J/18.062J, Fall ’05: Mathematics for Computer Science
Prof. Albert R. Meyer and Prof. Ronitt Rubinfeld
September 16
revised September 20, 2005, 1273 minutes
Solutions to InClass Problems Week 2, Fri.
Problem 1. Subset takeaway1 is a two player game involving a fixed finit... | https://ocw.mit.edu/courses/6-042j-mathematics-for-computer-science-fall-2005/194a6380ce2c26852fde38af8e2daa39_cp2fsol.pdf |
2nd player should choose the size 1 subset con
taining the remaining element. This results in a position corresponding to a size 3 game, and is
therefore a win for the 2nd player.
Case 3:[1st player chooses a size 2 subset {a, b}] Then 2nd player should choose the complemen
tary size 2 subset {c, d}. Here a, b, c, ... | https://ocw.mit.edu/courses/6-042j-mathematics-for-computer-science-fall-2005/194a6380ce2c26852fde38af8e2daa39_cp2fsol.pdf |
as the game on the
complement of {x, y}, which Player 2 can win.
Subcase 3.3: At some point in further play, Player 1 picks a size 1 subset and no size 2
subset remains as a legal move. Then some other size 1 subset will also available as a
possible move, and Player 2 should choose any such size 1 subset. This eith... | https://ocw.mit.edu/courses/6-042j-mathematics-for-computer-science-fall-2005/194a6380ce2c26852fde38af8e2daa39_cp2fsol.pdf |
18.782 Introduction to Arithmetic Geometry
Lecture #15
Fall 2013
10/29/2013
As usual, k is a perfect field and k is a fixed algebraic closure of k. Recall that an affine
(resp. projective) variety is an irreducible alebraic set in An = An(k) (resp. Pn = Pn(k)).
¯
¯
¯
15.1 Rational maps of affine varieties
Before defining rati... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/194e47df6e1f016ee3fb2719abe7e8d3_MIT18_782F13_lec15.pdf |
not defined for any P (cid:54)∈ X).
One advantage of this approach is that there is then a one-to-one correspondence between
morphisms and the functions they define. If φ = (φ1, . . . , φn) and ψ = (ψ1, . . . , ψn) define
the same function from X to Y then each of the polynomials φi−ψi in k[x1, . . . , xm] contains
X in i... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/194e47df6e1f016ee3fb2719abe7e8d3_MIT18_782F13_lec15.pdf |
(P ) = 0 at
points where g(P ) = 0 (and vice versa).
¯
1 2 − x3x4 in A (which is in fact a
Example 15.1. Consider the the zero locus X of x x
variety) and the rational function r = x1/x3 = x4/x2. At the point P = (0, 1, 0, 0) ∈ X the
value x1(P )/x3(P ) is not defined, but x4(P )/x2(P ) = 0 is defined, and the reverse oc... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/194e47df6e1f016ee3fb2719abe7e8d3_MIT18_782F13_lec15.pdf |
the closed
subset of X defined by the denominator ideal {g ∈ k[X] : gr ∈ k[X]}, which we note is not
the zero ideal, since r = f /g for some nonzero g.3
¯
¯
¯
We now associate to r the function from dom(r) to k that maps P to
¯
r(P ) = (f /g)(P ) = f (P )/g(P ),
where g is chosen so that g(P ) = 0 and gr = f
k[X]. Now i... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/194e47df6e1f016ee3fb2719abe7e8d3_MIT18_782F13_lec15.pdf |
a complete function from X to Y precisely when dom(φ) = X, that
is, when φ is regular. This occurs precisely when φ is a morphism.
Theorem 15.5. A rational map of affine varieties is a morphism if and only if it is regular.
Proof. A morphism is clearly a regular rational map. For the converse, apply Lemma 15.3
to each co... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/194e47df6e1f016ee3fb2719abe7e8d3_MIT18_782F13_lec15.pdf |
xm]/I(X) rather than k[x1, . . . , xm]. The one-to-one corrsepondence between radical ideals and
closed sets still holds, as does the correspondence between prime ideals and (sub-) varieties.
¯
¯
2
2(cid:54)
(cid:54)
To fix this problem we want to restrict our attention to rational maps whose image is
dense in its codo... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/194e47df6e1f016ee3fb2719abe7e8d3_MIT18_782F13_lec15.pdf |
is algebraic
then F = k; this corresponds to the function field of a zero-dimensional variety (a point).
Given a function field F generated by elements α1, . . . , αn over k, let R be the k-algebra
generated by α1, . . . , αn in F ; this means that R is equal to the set of all polynomial
expressions in α1, . . . , αn, bu... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/194e47df6e1f016ee3fb2719abe7e8d3_MIT18_782F13_lec15.pdf |
pose/reduce,
that is, pick representatives for φ1, . . . , φn that are rational functions in k(x1, . . . , xm),
pick a representative of r in k(y1, . . . , yn), then compose and reduce the numerator and
denominator modulo I(X). The fact that φ is dominant ensures that the denominator of
the composition does not lie in ... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/194e47df6e1f016ee3fb2719abe7e8d3_MIT18_782F13_lec15.pdf |
0, so certainly
f (θ∗(P )) = 0 for all P ∈ dom(θ∗). Thus θ∗ is a rational map from X to Y .
But we also need to check that θ∗ is dominant (this is the only new part of the proof).
This is equivalent to showing that the only element of k[Y ] that vanishes on the image of θ∗
is the zero element, which is in turn equivale... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/194e47df6e1f016ee3fb2719abe7e8d3_MIT18_782F13_lec15.pdf |
φ = (φ1, . . . , φn) is defined over k if the φi all lie in k(X).
Corollary 15.12. Let X and Y be affine varieties defined over k.
If φ : X → Y is a
dominant rational map defined over k then φ∗ : k(Y ) → ¯k(X) restricts to a morphism
k(Y ) → k(X).
¯
15.2 Morphisms and rational maps of projective varieties
We now want to gen... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/194e47df6e1f016ee3fb2719abe7e8d3_MIT18_782F13_lec15.pdf |
: . . . : φn) with φi ∈ k(X) not all zero
such that at any point P ∈ X where all the φi are regular and at least one is nonzero, the
point (φ0(P ) : . . . : φn(P )) lies in Y . The equivalence relation is given by
¯
(φ0 : . . . : φn) = (λφ0 : . . . : λφn)
for any λ ∈ ¯k(X)×. We say that φ is regular at P if there is a ... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/194e47df6e1f016ee3fb2719abe7e8d3_MIT18_782F13_lec15.pdf |
(and the choice may vary with P ), but note that in this case Remark 15.15 no longer applies.
¯
Now that we have defined rational maps for projective varieties we can define morphisms
and dominant rational maps; the definitions are exactly the same as in the affine case, so
we can now state them generically.
Definition 15.17... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/194e47df6e1f016ee3fb2719abe7e8d3_MIT18_782F13_lec15.pdf |
) =
(cid:18) y
x + 1
(cid:19) (cid:18)
=
(cid:19)
1 − x
y
sends each point Q = (x, y) ∈ X different from P to the slope of the line P Q. The map φ
is not regular (hence not a morphism), because it is not regular at P , but it is dominant
(even surjective). The rational map φ−1 : A1 → X defined by
φ−1(t) =
(cid:18) − t2
1... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/194e47df6e1f016ee3fb2719abe7e8d3_MIT18_782F13_lec15.pdf |
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