text stringlengths 16 3.88k | source stringlengths 60 201 |
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0n
1
0 1 0ff 0n
1
1 0 0ff 0n
0ff 0
1 1 0n
How do you build a 2-input NOR Gate?
L2: 6.111 Spring 2006
Introductory Digital Systems Laboratory
11
Theorems of Boolean Algebra (I)
Theorems of Boolean Algebra (I)
(cid:132) Elementary
1. X + 0 = X
2. X + 1 = 1
3. X + X = X
4. (X) = X
5. X + X = 1
(cid:132) Commutativity:
6.... | https://ocw.mit.edu/courses/6-111-introductory-digital-systems-laboratory-spring-2006/340fab3bd54502c96ef177100c99b0f8_l2_combi_logic.pdf |
Z) =
X • (Y + Z)
(cid:132) Consensus:
12D. (X + Y) • (X + Z) =
X + (Y • Z)
13. (X • Y) + (Y • Z) + (X • Z) =
X • Y + X • Z
13D. (X + Y) • (Y + Z) • (X + Z) =
(X + Y) • (X + Z)
(cid:132) De Morgan's:
14. (X + Y + ...) = X • Y • ...
14D. (X • Y • ...) = X + Y + ...
(cid:132) Generalized De Morgan's:
15. f(X1,X2,...,Xn,... | https://ocw.mit.edu/courses/6-111-introductory-digital-systems-laboratory-spring-2006/340fab3bd54502c96ef177100c99b0f8_l2_combi_logic.pdf |
A B Cin + A B Cin + A B Cin
(cid:132) Product term (or minterm)
(cid:134) ANDed product of literals – input combination for which output
is true
(cid:134) Each variable appears exactly once, in true or inverted form (but
not both)
L2: 6.111 Spring 2006
Introductory Digital Systems Laboratory
14
Simplify Boolean Expr... | https://ocw.mit.edu/courses/6-111-introductory-digital-systems-laboratory-spring-2006/340fab3bd54502c96ef177100c99b0f8_l2_combi_logic.pdf |
B, C)
= Σm(1,3,5,6,7)
= m1 + m3 + m5 + m6 + m7
F =
canonical form ≠ minimal form
A B C
+ A B C+ A B C + A B C + ABC
F(A, B, C)
= A B C + A B C + AB C + ABC + ABC
= (A B + A B + AB + AB)C + ABC
= ((A + A)(B + B))C + ABC
= C + ABC = ABC + C = AB + C
(cid:132) Sum term (or maxterm) - ORed sum of literals – input combin... | https://ocw.mit.edu/courses/6-111-introductory-digital-systems-laboratory-spring-2006/340fab3bd54502c96ef177100c99b0f8_l2_combi_logic.pdf |
maxterm shorthand
replace minterm indices with the indices not already used
E.g., F(A,B,C) = Σm(3,4,5,6,7) = ΠM(0,1,2)
2. Maxterm to Minterm conversion:
rewrite maxterm shorthand using minterm shorthand
replace maxterm indices with the indices not already used
E.g., F(A,B,C) = ΠM(0,1,2) = Σm(3,4,5,6,7)
3. Minterm expan... | https://ocw.mit.edu/courses/6-111-introductory-digital-systems-laboratory-spring-2006/340fab3bd54502c96ef177100c99b0f8_l2_combi_logic.pdf |
A has a different value in the two rows
– A is eliminated
L2: 6.111 Spring 2006
Introductory Digital Systems Laboratory
18
Boolean Cubes
Boolean Cubes
(cid:132) Just another way to represent truth table
(cid:132) Visual technique for identifying when the uniting theorem
can be applied
(cid:132) n input variables = n-... | https://ocw.mit.edu/courses/6-111-introductory-digital-systems-laboratory-spring-2006/340fab3bd54502c96ef177100c99b0f8_l2_combi_logic.pdf |
A
The on-set is completely covered by the combination (OR) of the subcubes of
lower dimensionality - note that “111” is covered three times
L2: 6.111 Spring 2006
Introductory Digital Systems Laboratory
20
Higher Dimension Cubes
Higher Dimension Cubes
011
111
010
B C
110
001
101
000
A
100
F(A,B,C) = Σm(4,5,6,7)
on-set... | https://ocw.mit.edu/courses/6-111-introductory-digital-systems-laboratory-spring-2006/340fab3bd54502c96ef177100c99b0f8_l2_combi_logic.pdf |
, 11, 10 (only a single bit changes in code for adjacent map cells)
2-variable
K-map
0
1
0
1
2
3
B
A
0
1
AB
A
C
00
01
11
10
3-variable
K-map
0
1
0
1
6
7
4
5
2
3
B
AB
CD
00
01
11
10
C
A
00
01
11
10
0
1
3
2
4
5
7
6
B
12
13
8
9
15
11
14
10
D
4-variable
K-map
L2: 6.... | https://ocw.mit.edu/courses/6-111-introductory-digital-systems-laboratory-spring-2006/340fab3bd54502c96ef177100c99b0f8_l2_combi_logic.pdf |
1
1
0
0
1
1
B
F(A,B,C,D) = Σm(0,2,3,5,6,7,8,10,11,14,15)
F = C + A B D + B D
Find the smallest number
of the largest possible
subcubes that cover the
ON-set
0011
0010
C
0001
0000
D
A
B
0111
1010
0110
0101
0100
1000
1011
1111
1110
K-map Corner Adjacency
Illustrated in the 4-Cube
1001
1101
1100... | https://ocw.mit.edu/courses/6-111-introductory-digital-systems-laboratory-spring-2006/340fab3bd54502c96ef177100c99b0f8_l2_combi_logic.pdf |
B
L2: 6.111 Spring 2006
Introductory Digital Systems Laboratory
25
Hazards
Hazards
AB
C
00 01 11 10
0
1
0
0
0
1
1
1
1
0
F
A
C
B
Implemented with MSI gates:
'00
'00
'00
1
'00
F
2
Static hazards: Consider this function:
F = A * C + B * C
A
C
B
A = B = 1
C
1
2
F
Gate delay
Glitch
L2: 6.111 Spring 2006
Introductory Digit... | https://ocw.mit.edu/courses/6-111-introductory-digital-systems-laboratory-spring-2006/340fab3bd54502c96ef177100c99b0f8_l2_combi_logic.pdf |
2.3.2 Swollen (coil) polymers in good solvents
Most of the terms in the trial free energy of Eq. (2.49) have definite sign. The exception
is the term proportional to N 2(a/R)3 which has opposing contributions from the repulsive
1/2). The sign of this term
and attractive parts of the potential, and is proportional to (χ
... | https://ocw.mit.edu/courses/8-592j-statistical-physics-in-biology-spring-2011/341e304796bde8c040895e83d45706ea_MIT8_592JS11_lec11.pdf |
ization group theory is to estimate the exact value of ν = 0.591 . . . , remarkably
close to the Flory approximation of 3/5. While not directly relevant to real polymers, it is
possible to inquire about the exponent ν for self-avoding walks in d-spatial dimensions– e.g.,
for polymers confined to a d = 2 dimensional surf... | https://ocw.mit.edu/courses/8-592j-statistical-physics-in-biology-spring-2011/341e304796bde8c040895e83d45706ea_MIT8_592JS11_lec11.pdf |
−
2χ
1
−
2
ρ +
ρ2
6
+ higher order terms.
(2.54)
The optimal density for T < θ is obtained by minimizing the above free energy,
leading to
1
N
d ln Z
dρ
−
=
1
2 −
(cid:18)
χ
+
(cid:19)
ρ
3
+
,
· · ·
ρ = 3
χ
(cid:18)
+
,
· · ·
(cid:19)
1
2
−
42
(2.55)
i.e. a density that vanishes linearly on approaching the θ-temperatu... | https://ocw.mit.edu/courses/8-592j-statistical-physics-in-biology-spring-2011/341e304796bde8c040895e83d45706ea_MIT8_592JS11_lec11.pdf |
can be visualized as the
collection of all maximally compact configurations. In a lattice version, these are self-avoiding
walks that visit all sites, leaving no empty ones, and are referred to as Hamiltonian walks. The
number of Hamiltonian walks also grows exponentially with the number of steps as g′N , but is
much sm... | https://ocw.mit.edu/courses/8-592j-statistical-physics-in-biology-spring-2011/341e304796bde8c040895e83d45706ea_MIT8_592JS11_lec11.pdf |
αic = NBh
V 2
abic ≡
Nσ2 ,
(2.60)
h
,
where noting that NB = (z
have folded the proportionality constant into the definitions of ε0 and σ2.
−
44
For large N, the probability distribution for the energy will take the Gaussian form
p(E) =
1
√2πNσ2
exp
(E
Nε0)2
−
2Nσ2
.
(cid:21)
−
(cid:20)
(2.61)
Since the total number of... | https://ocw.mit.edu/courses/8-592j-statistical-physics-in-biology-spring-2011/341e304796bde8c040895e83d45706ea_MIT8_592JS11_lec11.pdf |
−
Ec
(cid:18)
Nε0
Ec −
Nσ2
= kB
(cid:19)
√2 ln g′
σ ⇒
kBTC =
σ
√2 ln g′
.
(2.64)
There are presumably a few low energy states with energy close to Ec, and the system freezes
into one of these for T
Tc.
(cid:12)
(cid:12)
(cid:12)
(cid:12)
≤
45
2.3.5 Designed REM for protein folding
It is tempting to equate the freezing... | https://ocw.mit.edu/courses/8-592j-statistical-physics-in-biology-spring-2011/341e304796bde8c040895e83d45706ea_MIT8_592JS11_lec11.pdf |
Nε0)2/(2Nσ2)
En
.
(2.65)
As depicted in the figure, the above result equates the slope of the tangent line from the
point at En computed in two different ways.
To justify the above result, note that in the canonical ensemble, the probability of finding
the system in the native state is
pn =
e−βEn
Z(β)
, with Z(β) = e−βEn ... | https://ocw.mit.edu/courses/8-592j-statistical-physics-in-biology-spring-2011/341e304796bde8c040895e83d45706ea_MIT8_592JS11_lec11.pdf |
Noise in Cascaded Amplifiers
F1,G1
F2,G2
1
2
≡
3
1
F1+2,G
1+2
3
F
∆+
2 1
NS
1
1
NS
3
3
where
S
3
=
S G G
121
N
2
=
kT
G F
1 o
1
⎛
⎜
Recall
⎜
⎝
F
1
=
N
3
=
G
kT
2
G F
+
1
1 o
(
FG
22
⎞
⎟
⎟
⎠
NS
1
1
NS
2
2
) o
1
kT
−
amplified from “2”
excess from “2”
∴
NS
3
3
=
S G G
1 ... | https://ocw.mit.edu/courses/6-661-receivers-antennas-and-signals-spring-2003/34219e23d858375d1c802669f2969dfc_lecture06.pdf |
…
cascade
noise
formula
A
B
vs.
B
A
The better choice is not obvious.
F1,2
also depends on interstage impedance mismatches
and gain of the first amplifier, not just on F1
L2
Noise in superheterodyne receivers
0
f
fo
Gc,Fc
S1,N1 S2,N2
×
0
fi.f.
f
L.O.
“lower (rf) sideband”
Gi.f.,Fi.f.
∆
“i.f.”... | https://ocw.mit.edu/courses/6-661-receivers-antennas-and-signals-spring-2003/34219e23d858375d1c802669f2969dfc_lecture06.pdf |
amp.
2
−≅
(
3 ~ 8
−
9
)
dB
L4
Basic receiver types-Amplification
1) Simple detector
B
f
( )2
h(t)
vo(t)
2) RF Amplifier
3) Superheterodyne
OR
OR
×
N1
1)
2)
3)
4)
Basic receiver types-Combinors
vo(t) Total power radiometer
vo(t)
Dicke radiometer
×
vo(t) Multiplication (correlation)
receiver ... | https://ocw.mit.edu/courses/6-661-receivers-antennas-and-signals-spring-2003/34219e23d858375d1c802669f2969dfc_lecture06.pdf |
) The scattering matrix
S
is defined only when the n-port
is imbedded in a network
2) The phases of a
i and bi can be defined as some linear
combinations of those for voltage and current
.g.e
a
i
V
= +
2Y
o
=
(
V
+
)
I Z
o
, Z 8
o
b
i
=
(
V
−
)
I Z
o
Z 8
o
P1
Gain definition for N-port networks
“Trans... | https://ocw.mit.edu/courses/6-661-receivers-antennas-and-signals-spring-2003/34219e23d858375d1c802669f2969dfc_lecture06.pdf |
=
k(T1 +T2)
∗
t
a
=•⇒
bba
∗
t
00
⎡
⎢
00
⎢
⎢
⎣
α
βαα
α
⎤
⎥
⎥
⎥
⎦
∗
t
b
b
=•
∗
(
aS
t
)
(
aS
•
)
•
=
∗
t
∗
t
Sa
aS
Therefore
∗
t
ISS
==
001
⎤
⎡
⎥
⎢
010
⎥
⎢
100
⎥
⎢
⎦
⎣
; if the combiner is lossless
P4
Example of constrained N-port networks
; if the container is lossless
Therefore
∗
t
I... | https://ocw.mit.edu/courses/6-661-receivers-antennas-and-signals-spring-2003/34219e23d858375d1c802669f2969dfc_lecture06.pdf |
atisfied simultaneously for this system
Ideal matched 2-input-port combiners are impossible
P5
Another N-port network example
Can we match all 3 ports simultaneously?
1
2
3
(Note:
S
is defined only when imbedded in network)
For 3 matched ports:
S
=
0
⎡
⎢
α
⎢
β
⎢
⎣
βα
0
⎤
⎥
γ
⎥
0
⎥
⎦
γ
Does
∗
t ... | https://ocw.mit.edu/courses/6-661-receivers-antennas-and-signals-spring-2003/34219e23d858375d1c802669f2969dfc_lecture06.pdf |
matched
3
4
We can show ∆φ for
1
3
versus
1
4
paths is unique
using losslessness, reciprocity, and symmetry
P8
signal
+
S(t
)
+
(cid:65)
(t) V
=
local
oscillator
Th
-
Mixer
vTh
-
+
vo(t)
Rs
i(t)
RL
i
o
=
V
Th
(
R R
+
S
L
)
>>
i(t)
i
diode:
sin
+ω
o
vt
s
sin
ω
s
t
i ≅
v
2
d
for
)t(v
... | https://ocw.mit.edu/courses/6-661-receivers-antennas-and-signals-spring-2003/34219e23d858375d1c802669f2969dfc_lecture06.pdf |
(Note:
4
v ⇒
d
. f.f 2
i
, can be in i.f. passband)
Normally:
a 4-port model suffices
to find F,T
" ,
: " T
R
SSB
signal
image
1
2
4
3
output
internal
noise
R2
Mixer Noise Figure Using 4-port Model
T
R
=
T
SSB
=
T)1F(
−
o
=
T
o
(
t L
rc
−
)1
signal
for single sideband (SSB) operation
image ... | https://ocw.mit.edu/courses/6-661-receivers-antennas-and-signals-spring-2003/34219e23d858375d1c802669f2969dfc_lecture06.pdf |
2
signal
S
43
2
S
41
image
Both ports 1 and 2 are signal, so
S
4
=
kT
o
(
2
+
S
S
41
2
) (
S1
−
42
output
1
2
4
3
internal
noise
)2
44
It follows that
T
DSB
=
T
3
S
43
2
[
S
2
41
+
S
2
42
]
Often suggested:
SSBF
≅
F2
DSB
R4
Double-sideband Receiver
Therefore
T
SSB
=
T
2
S... | https://ocw.mit.edu/courses/6-661-receivers-antennas-and-signals-spring-2003/34219e23d858375d1c802669f2969dfc_lecture06.pdf |
6.825 Techniques in Artificial Intelligence
Resolution Theorem Proving:
Propositional Logic
• Propositional resolution
• Propositional theorem proving
• Unification
Lecture 7 • 1
Today we’re going to talk about resolution, which is a proof strategy. First, we’ll
look at it in the propositional case, then in the... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
that that one rule is all you need to prove things. At least, to prove that a
set of sentences is not satisfiable. So, let's see how this is going to work. There's
a proof strategy called Resolution Refutation, with three steps. And it goes like
this.
4
Propositional Resolution
• Resolution rule:
α v β
β v γ
¬... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
right? If you assert that the negation of the thing that you're interested
in is true, and then you prove for a while and you manage to prove false, then
you've succeeded in a proof by contradiction of the thing that you were trying to
prove in the first place. So you run the resolution rule until you derive false o... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
So let's just do a proof. Let's say I'm given “P or Q”, “P implies R” and “Q implies
R”. I would like to conclude R from these three axioms. I'll use the word
"axiom" just to mean things that are given to me right at the moment.
9
Propositional Resolution Example
Step
Formula
Derivation
1
P v Q
Given
Prove R
1 ... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
.
We'd like to prove that R follows from these things. But what we're going to do
instead is say not R, and now we're trying to prove false. And if we manage to
prove false, then we will have a proof that R is entailed by the assumptions.
13
Propositional Resolution Example
Prove R
1
P v Q
2
R
P
→
Q
3
R
→... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
16
Propositional Resolution Example
Prove R
1
P v Q
2
R
P
→
Q
3
R
→
Step
Formula
Derivation
1
2
3
4
5
6
7
P v Q
P v R
Q v R
R
¬
¬
¬
Q v R
P
¬
Q
¬
Given
Given
Given
Negated
conclusion
1,2
2,4
3,4
Similarly, we can take lines 3 and 4, resolve away R, and get “not Q”.
Lecture 7 • 17
17
Proposit... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
desired contradiction.
19
Propositional Resolution Example
Step
Formula
Derivation
Prove R
1
P v Q
2
R
P
→
Q
3
R
→
false v R
R v false
¬
false v false
1
2
3
4
5
6
7
8
9
P v Q
P v R
Q v R
R
¬
¬
¬
Q v R
P
¬
Q
¬
R
•
Given
Given
Given
Negated
conclusion
1,2
2,4
3,4
5,7
4,8
Lecture 7 • 20
Ho... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
Lecture 7 • 22
22
The Power of False
Step
Formula
Derivation
Prove Z
P1
P
2
¬
It does, and it’s easy to prove using resolution refutation.
Lecture 7 • 23
23
The Power of False
Prove Z
P1
P
2
¬
Step
Formula
Derivation
1
2
3
P
¬
¬
P
Z
Given
Given
Negated
conclusion
We start by writing down the ass... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
are very brittle.
Lecture 7 • 27
So, we see, again, that any conclusion follows from a contradiction. This is the
property that can make logical systems quite brittle; they’re not robust in the
face of noise. We’ll address this problem when we move to probabilistic
inference.
27
Example Problem
Convert to CNF
... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
R
¬
R) Æ (
¬
P v R)
Lecture 7 • 30
The second turns into (“P or R” and “not P or R”). We probably should have
simplified it into “False or R” at the second step, which reduces just to R. But
we’ll leave it as is, for now.
30
Example Problem
Convert to CNF
Prove R
1
2
3
(P
(P
→
→
Q)
→
Q
P)
→
R
(... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
Q
P)
→
R
(R
3
→
S)
(S
→
¬
→
Q)
1
2
3
4
5
6
7
P v Q
P v R
¬
R v S
R v
P v R
Q
¬
S v
Q
¬
R
¬
¬
Neg
Lecture 7 • 32
Now we can almost start the proof. We copy each of the clauses over here, and we
add the negation of the query.
Please stop and do this proof yourself before going on.
32
Resolut... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
a shorter
clause. And the shorter a clause is, the closer it is to false.
35
Proof Strategies
• Unit preference: prefer a resolution step involving
an unit clause (clause with one literal).
• Produces a shorter clause – which is good since we are
trying to produce a zero-length clause, that is, a
contradiction.... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
to move to the first-order case.
And there's going to be a resolution rule in the first-order case, and it's
essentially the same inference rule, but the trick is what to do with the variables.
And what to do with the variables is pretty hard and pretty complicated. We’ll
spend the rest of this lecture understandin... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
)
Q(A)
Syllogism:
All men are mortal
Socrates is a man
Socrates is mortal
uppercase letters:
constants
lowercase letters:
variables
x.
¬
∀
P(x) v Q(x)
P(A)
Q(A)
Equivalent by
definition of
implication
P(A) v Q(A)
¬
Substitute A for
x, still true
P(A)
Q(A)
then
Propositional
resolution
Lectur... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
y. We can think about
different ways that we can substitute terms into this expression, right? Those are
called substitution instances of that expression.
45
Substitutions
P(x, F(y), B) : an atomic sentence
Substitution
instances
Substitution
{v1 /t1,…, vn /tn}
Comment
Lecture 7 • 46
A substitution is a set ... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
phabetic
variant
To get P(G(z), F(A), B), we substitute the term G(z) in for x and the constant A for
y.
Lecture 7 • 49
49
Substitutions
P(x, F(y), B) : an atomic sentence
Substitution
instances
P(z, F(w), B)
Substitution
{v1 /t1,…, vn /tn}
{x/z, y/w}
P(x, F(A), B)
{y/A}
P(G(z), F(A), B)
{x/G(z), y/A}
Co... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
B) : an atomic sentence
Substitution
instances
P(z, F(w), B)
Substitution
{v1 /t1,…, vn /tn}
{x/z, y/w}
P(x, F(A), B)
{y/A}
P(G(z), F(A), B)
{x/G(z), y/A}
Comment
Alphabetic
variant
P(C, F(A), B)
{x/C, y/A}
Ground instance
Applying a substitution:
subst({A/y},P(x, F(y), B)) = P(x, F(A), B)
P(x, F(y), B) {A... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
ifiers
s
ω1 s
ω2 s
So, let’s look at some unifiers of the expressions x and y. Since x and y are both
variables, there are lots of things you can do to make them match.
Lecture 7 • 55
55
Unification
• Expressions ω1 and ω2 are unifiable iff there exists a
substitution s such that ω1 s = ω2 s
• Let ω1 = x and ω... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
= ω2 s
• Let ω1 = x and ω2 = y, the following are unifiers
s
{y/x}
{x/y}
ω1 s
ω2 s
x
y
x
y
{x/F(F(A)), y/F(F(A))}
F(F(A))
F(F(A))
{x/A, y/A}
A
A
Or, you could substitute some constant, like A, in for both x and y.
Lecture 7 • 59
59
Most General Unifier
Lecture 7 • 59
Of the unifiers we considered on the pre... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
such that ω1 s = (ω1 g) s0
ω1
P(x)
ω2
P(A)
MGU
{x/A}
So, let's do a few more examples together, and then you can do some as recitation
problems. So, what’s a most general unifier of P(x) and P(A)? A for x.
Lecture 7 • 61
61
Most General Unifier
g is a most general unifier of ω1 and ω2 iff for all
unifiers s... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
y have to be the same as well
(to make the middle argument match), they all have to be the same variable.
Might as well make it x (though it could be any other variable).
63
Most General Unifier
g is a most general unifier of ω1 and ω2 iff for all
unifiers s, there exists s0
and ω2 s = (ω2 g) s0
such that ω1... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
))
and G(u). How can we make that work? By substituting F(v) in for u.
Lecture 7 • 65
65
Most General Unifier
g is a most general unifier of ω1 and ω2 iff for all
unifiers s, there exists s0
and ω2 s = (ω2 g) s0
such that ω1 s = (ω1 g) s0
ω1
P(x)
ω2
P(A)
MGU
{x/A}
P(F(x), y, G(x))
P(F(x), x, G(x))
{y/x} o... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
(v)), G(u))
P(x, x)
{x/G(F(v)), u/F(v)}
P(x, F(x))
P(x, x)
No MGU!
Lecture 7 • 67
The last time I explained this to a class, someone asked me what would happen if F
were the identity function. Then, couldn’t we unify these two expressions?
That’s a great question, and it illustrates a point I should have made befo... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
we go to a special subroutine that’s shown on the
next slide.
Lecture 7 • 71
71
Unification Algorithm
unify(Expr x, Expr y, Subst s){
if s = fail, return fail
else if x = y, return s
else if x is a variable, return unify-var(x, y, s)
else if y is a variable, return unify-var(y, x, s)
else if x is a predicate ... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
)
else if x is a predicate or function application,
if y has the same operator,
return unify(args(x), args(y), s)
return fail
else
else
; x and y have to be lists
return unify(rest(x), rest(y),
unify(first(x), first(y), s))
}
Lecture 7 • 75
Finally, (if we get to this case, then x and y have to be lists, or... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
, s)
else if x is bound to val in s,
return unify-var(var, val, s)
Lecture 7 • 78
Similarly, if x is a variable, and it is bound to val in s, then we have to unify var
with val in s. (We call unify-var directly, because we know that var is still a
var).
78
Unify-var subroutine
Substitute in for var and x as lo... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
careful. The last line of this algorithm in the book is incorrect.
Lecture 7 • 81
81
Unification Problems
For each pair of sentences, give an MGU.
Color(Tweety, Yellow)
Color(Tweety, Yellow)
Color(Hat(John), Blue)
R(F(x), B)
R(F(y), x)
R(F(y), y, x)
Loves(x, y)
F(G(w), H(w, J(x, y)))
F(G(w), H(w, J(x, u)))... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
Massachusetts Institute of Technology
Department of Materials Science and Engineering
77 Massachusetts Avenue, Cambridge MA 02139-4307
3.21 Kinetics of Materials—Spring 2006
February 21, 2006
Lecture 4: Interdiffusion. Effects of electical potential, capillarity, and stress on diffusion potential.
References
1. ... | https://ocw.mit.edu/courses/3-21-kinetic-processes-in-materials-spring-2006/3453d577b53c280708513740833c4932_ls4.pdf |
cient numbers of vacancy sources
and sinks. However, it is not unusual to find Kirkendall porosity on one side of the interdiffusion zone.
• The mobility Mi that relates the diffusive flux of i to its chemical potential gradient is always posi
tive. Diffusivities are usually positive also, but not always. For instance... | https://ocw.mit.edu/courses/3-21-kinetic-processes-in-materials-spring-2006/3453d577b53c280708513740833c4932_ls4.pdf |
1..
-2.
-"~~i~;e~a~i·-
30
ESL-, MSL4
. I
-
.. �
.e
A
�F,
El- bm-r-0
tj\ -J~(ti, Z o
wrr~s 14%
-,L
kt,1-6
(C§(2
) .
same
'S%;SA
Sq
Not
~or
sl,
c
-----------
�I
�---�--�------
----
·-
M~
·---
;~~
-.
^
AS-
4~~~~~~~~~~~~F
Yo -. ..------ -.
.P.~~~~~~~N~A
''-
~
~~ŽL.
Tt~
zc
j
ki
~Y
i
o
... | https://ocw.mit.edu/courses/8-322-quantum-theory-ii-spring-2003/3469fc3f8ad9f1562fbc172aaab9efcd_83228HigherMultiple.pdf |
~
roASA4
\ex*n
Ccmaair-0o-j
r 2
A
'-T o
44,1pP-
re)AL
Cof
\C>\^
(
e &~
f\ C","\ C.
7\ Lis ( r, 9, >) 9
(\
\
"b
ce4 v A.
Co
e'AT Mt s %,,,4vrr3
'5ene 6r,,
\(,
(ft, 6)
6 ;vR- Wa,,v~jrC*
J pj
C(:j
,,m
I es rc-t4~za-
1WxhI
C2p-k2lc6 | https://ocw.mit.edu/courses/8-322-quantum-theory-ii-spring-2003/3469fc3f8ad9f1562fbc172aaab9efcd_83228HigherMultiple.pdf |
LECTURE 19
Brauer Groups
In this lecture, we present an overview of Brauer groups. Our presentation
will be short on proofs, but we will give precise constructions and formulations of
claims. For complete proofs, see [Mil13, Ser79, Boy07]. Our motivating question
is: “what was all that stuff about Hamiltonian algebras?”... | https://ocw.mit.edu/courses/18-786-number-theory-ii-class-field-theory-spring-2016/346c269899e7aef6cd2da8960a965a61_MIT18_786S16_lec19.pdf |
have a short exact sequence
0 → Gal(L2/L1) → G2 → G1 → 0,
and maps
Brcoh(L1/K) = H 2(G1, L×
1 ) → H 2(G2, L×
1 ) → H 2(G2, L×
2 ) = Brcoh(L2/K)
since invariance with respect to G2 implies impvariance with respect to G1, and via
the embedding L×
2 . This motivates the following definition.
1 (cid:44)→ L×
Definition 19.3.... | https://ocw.mit.edu/courses/18-786-number-theory-ii-class-field-theory-spring-2016/346c269899e7aef6cd2da8960a965a61_MIT18_786S16_lec19.pdf |
-
algebra with center K. Then the following are equivalent:
(1) A is simple, that is, it has no non-trivial 2-sided ideals.
(2) A ⊗K L (cid:39) Mn(L) (i.e., an n × n matrix algebra) for some separable
extension L/K.
(3) A (cid:39) Mn(D) for some central (i.e., with center K) division (i.e., multi-
plicative inverses ex... | https://ocw.mit.edu/courses/18-786-number-theory-ii-class-field-theory-spring-2016/346c269899e7aef6cd2da8960a965a61_MIT18_786S16_lec19.pdf |
B] ∈ Brcsa(K), so Brcsa(K)
forms an abelian group.
The proof is omitted; showing that A ⊗K B is simple is rather annoying. We
note, however, that matrix algebras over K represent the identity element (or “0
equivalence class”) of Brcsa(K). The inverse of A is A, but with opposite multipli-
cation (i.e., x · y := yx), w... | https://ocw.mit.edu/courses/18-786-number-theory-ii-class-field-theory-spring-2016/346c269899e7aef6cd2da8960a965a61_MIT18_786S16_lec19.pdf |
claim, L is very non-
unique. Note that, unlike in this case, such an L need not be Galois over
K; examples are hard to find, but were discovered in the 1970s.
a) and K(
In fact, we can take K(
√
√
Claim 19.15. An n2-dimensional central division algebra D/K splits over a
degree-n extension L/K if and only if K ⊆ L ⊆ D a... | https://ocw.mit.edu/courses/18-786-number-theory-ii-class-field-theory-spring-2016/346c269899e7aef6cd2da8960a965a61_MIT18_786S16_lec19.pdf |
39) Brcsa(K/L).
Note that the cohomological Brauer group is defined only for Galois extensions
L/K, whereas the csa Brauer group is defined for all extensions L/K. This gives
meaning to the cohomological definition of the Brauer group. We provide the
following (incomplete) proof sketch:
Proof. For a G-Galois extension L/K... | https://ocw.mit.edu/courses/18-786-number-theory-ii-class-field-theory-spring-2016/346c269899e7aef6cd2da8960a965a61_MIT18_786S16_lec19.pdf |
the equivalence class of this
csa only depends on ϕ, up to coboundaries, and that it splits over L. Moreover,
(cid:3)
our map (19.1) is a group isomorphism.
This is not an especially deep theorem, despite being far from obvious; the
complete proof uses a lot of structure theory that is not particularly memorable.
We no... | https://ocw.mit.edu/courses/18-786-number-theory-ii-class-field-theory-spring-2016/346c269899e7aef6cd2da8960a965a61_MIT18_786S16_lec19.pdf |
This gives ϕ(n): the division algebras come from
classes in Z/nZ which do not arise from any smaller Z/mZ. Another upshot is the
following:
Corollary 19.20. Any degree-n2 central division algebra over K contains
every degree-n field extension of L.
Proof. Any n-torsion class in Br(K) maps to zero in the Brauer group Br(... | https://ocw.mit.edu/courses/18-786-number-theory-ii-class-field-theory-spring-2016/346c269899e7aef6cd2da8960a965a61_MIT18_786S16_lec19.pdf |
Asymptotic Expansions of Integrals
Lecture 12
The Laplace Method
We begin with integrals of the form
IΩRæ : X
b
a
e?RvΩxæhΩxædx,
(8.8)
where vΩxæ and hΩxæ are independent of the parameter R. The variable of integration x is real. We
shall show how to find the asymptotic form of IΩRæ as R î K.
We note that the ... | https://ocw.mit.edu/courses/18-305-advanced-analytic-methods-in-science-and-engineering-fall-2004/3477ac43d18b5f901fe9bae124e85ce6_twelve.pdf |
�Ræ u
1 X
R
K
?K
e?[2 d[ :
Z
R
.
(8.10)
In the above, we have made use of the fact that the Gaussian integral XK
given in Appendix A of this chapter.
?K
e?[2 d[ is equal to Z , as is
If one makes a small mistake in an approximation, the answer one gets can be far off the mark.
Thus it is often useful to ha... | https://ocw.mit.edu/courses/18-305-advanced-analytic-methods-in-science-and-engineering-fall-2004/3477ac43d18b5f901fe9bae124e85ce6_twelve.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-2010/34a75af33a916b46c2ad007e5c110502_MIT18_014F10_ChInotes.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
18.102 Introduction to Functional Analysis
Spring 2009
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
8
LECTURE NOTES FOR 18.102, SPRING 2009
Lecture 3. Tuesday, 10 Feb.
Recalled the proof from last time that the bounded o... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/34aa7af3453590196e4c834ad62b8129_MIT18_102s09_lec03.pdf |
through it all you will understand, before we get to it, the
main sorts of arguments needed to prove most of the integrability results we will
encounter later.
Let V be a normed space with norm � · �V . A completion of V is a Banach space
B with the following properties:
(1) There is an injective (1-1)linear map I... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/34aa7af3453590196e4c834ad62b8129_MIT18_102s09_lec03.pdf |
}.
This always gives an absolutely summable series by the triangle inequality:
(3.4)
�
�t1uk + t2u�
k� ≤ |t1|
�uk� + |t2|
k
k
k
�
�
�u�
k�.
LECTURE NOTES FOR 18.102, SPRING 2009
9
Within �V
(3.5)
consider the linear subspace
�
S = {uk};
�
�uk� < ∞,
�
�
uk = 0
of those which converge to 0. As always ... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/34aa7af3453590196e4c834ad62b8129_MIT18_102s09_lec03.pdf |
– namely the sequence of norms
is itself Cauchy but now in R. Moreover, adding an element of S to {uk} does not
change the norm of the sequence of partial sums, since the addtional term tends
to zero in norm. Thus �b�B is well-defined for each element b ∈ B and �b�B = 0
means exactly that the sequence {uk} used to d... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/34aa7af3453590196e4c834ad62b8129_MIT18_102s09_lec03.pdf |
0, 0, · · · , is the limit of the norms of the
sequence of partial sums and hence is �v�V so �I(v)�B = �v�V and I(v) = 0
therefore implies v = 0 and hence I is also injective.
So, we need to check that B is complete, and also that I(V ) is dense. Here is
an extended discussion of the difficulty – of course maybe you ... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/34aa7af3453590196e4c834ad62b8129_MIT18_102s09_lec03.pdf |
some sense – because it is supposed
u(
k
to represent the sum of the bn’s. Now, it would be very nice if we had the estimate
n k
(3.11)
� �
n)�V < ∞
�u(
k
n
k
since this should allow us to break up the double sum in some nice way so as to get
an absolutely summable series out of the whole thing. The trouble is ... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/34aa7af3453590196e4c834ad62b8129_MIT18_102s09_lec03.pdf |
j
Now, ‘resum the series’ defining instead v1
N1
(n) = �
k=1
u(
k
n), vj
(n) =
Nj
�
k=Nj−1
n) and do
u(
k
this setting � = 2−n for the nth series. Check that now
�
�
(3.14)
� �
(n)�V < ∞.
�vk
Of course, you should also check that bn = {vk
series work just as well as the old ones.
(n)} + S so that these ... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/34aa7af3453590196e4c834ad62b8129_MIT18_102s09_lec03.pdf |
102, SPRING 2009
Problem set 2, Due 11AM Tuesday 24 Feb.
I was originally going to make this problem set longer, since there is a missing
Tuesday. However, I would prefer you to concentrate on getting all four of these
questions really right!
Problem 2.1 Finish the proof of the completeness of the space B construc... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/34aa7af3453590196e4c834ad62b8129_MIT18_102s09_lec03.pdf |
Check that this is an absolutely summable series.
(2) For which x ∈ [0, 1) does
|fk(x)| converge?
�
k
(3) Describe a function on [0, 1) which is shown to be Lebesgue integrable
(as defined in Lecture 4) by the existence of this series and compute its
Lebesgue integral.
(4) Is this function Riemann integrable (th... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/34aa7af3453590196e4c834ad62b8129_MIT18_102s09_lec03.pdf |
the areas is less than or equal to that of the
containing rectangle.
(4) Show that if a finite collection of rectangles has union containing a given
rectange then the sum of the areas of the rectangles is at least as large of
that of the rectangle contained in the union.
(5) Prove the extension of the preceeding re... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/34aa7af3453590196e4c834ad62b8129_MIT18_102s09_lec03.pdf |
a good faith attempt to answer
each question. The first four problems concern the ‘little L p’ spaces lp. Note that
you have the choice of doing everything for p = 2 or for all 1 ≤ p < ∞.
Everyone who handed in a script received full marks.
Problem 1.1 Write out a proof (you can steal it from one of many places but ... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/34aa7af3453590196e4c834ad62b8129_MIT18_102s09_lec03.pdf |
p = 2p max(|a|p
i , |bi|p) ≤ 2p(|ai| + |bi|)
(3.19)
�a + b�p
p =
|ai + bi|p ≤ 2p(�a�p + �b�p),
�
where we use the fact that tp is an increasing function of t ≥ 0.
j
Now, to see that lp is a normed space we need to check that �a�p is indeed a
norm. It is non-negative and �a�p = 0 implies ai = 0 for all i which i... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/34aa7af3453590196e4c834ad62b8129_MIT18_102s09_lec03.pdf |
xp grows faster
than xβ. Moreover, it is differentiable and the derivative only vanishes at xp−1 =
β, where it must have a global minimum in x > 0. At this point f (x) = 0 so
Young’s inequality follows. Now, applying this with α = |ai|/�a�p and β = |bi|/�b�q
(assuming both are non-zero) and summing over i gives H¨ol... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/34aa7af3453590196e4c834ad62b8129_MIT18_102s09_lec03.pdf |
a normed space.
i
I did not necessarily expect you to go through the proof of Young-H¨older-
Minkowksi, but I think you should do so at some point since I will not do it in
class.
Problem 1.2 The ‘tricky’ part in Problem 1.1 is the triangle inequality. Suppose
you knew – meaning I tell you – that for each N
⎛
⎝ ... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/34aa7af3453590196e4c834ad62b8129_MIT18_102s09_lec03.pdf |
finite, but arbitrary, N.
Problem 1.3
Prove directly that each lp as defined in Problem 1.1 – or just l2 – is complete,
i.e. it is a Banach space. At the risk of offending some, let me say that this means
showing that each Cauchy sequence converges. The problem here is to find the limit
of a given Cauchy sequence. Sho... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/34aa7af3453590196e4c834ad62b8129_MIT18_102s09_lec03.pdf |
. Since C is complete
(n) − ai
(m)| ≤ �a(n) − a(m)�p so each of the sequences ai
(n) must
(3.29)
lim ai
n→∞
(n) = ai exists for each i = 1, 2, . . . .
So, our putative limit is a, the sequence {ai}∞ The boundedness of the norms
shows that
i=1.
(3.30)
N
�
(n)|p ≤ Ap
|ai
and we can pass to the limit here a... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/34aa7af3453590196e4c834ad62b8129_MIT18_102s09_lec03.pdf |
of inter
changing limits.
Problem 1.4
Consider the ‘unit sphere’ in lp – where if you want you can set p = 2. This is
the set of vectors of length 1 :
S = {a ∈ lp; �a�p = 1}.
(1) Show that S is closed.
(2) Recall the sequential (so not the open covering definition) characterization
of compactness of a set in a m... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/34aa7af3453590196e4c834ad62b8129_MIT18_102s09_lec03.pdf |
the unit sphere cannot be
compact whereas in the former it is.
Problem 1.5 Show that the norm on any normed space is continuous.
Solution:- Right, so I should have put this problem earlier!
The triangle inequality shows that for any u, v in a normed space
(3.37)
�u� ≤ �u − v� + �v�, �v� ≤ �u − v� + �u�
which imp... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/34aa7af3453590196e4c834ad62b8129_MIT18_102s09_lec03.pdf |
18.311:
Principles
of
Applied
Mathematics
Lecture
5
Rodolfo
Ros
ales
Spring
2014
servation
laws
and
pde:
igher
order
(TRANSPORT)
effects
beyond
quasi
Time
it
takes
to
reach
R
=
5
cm
[salt
in
water].
Con
H
Not
fl
ux
appe
•
•
e
be... | https://ocw.mit.edu/courses/18-311-principles-of-applied-mathematics-spring-2014/34ccec51ce5d47f7bd3e6fd81ae192f7_MIT18_311S14_Lecture5.pdf |
in
a
silo.
ow.
xample
of
a
higher
order
transport
effect
is
viscosity:
Forces
proportional
Another
e
to
the
flow
velocity
gradient.
REASON:
the
flow
velocity
is
a
macroscopic
"average"
variable.
The
same
phenomena
that
... | https://ocw.mit.edu/courses/18-311-principles-of-applied-mathematics-spring-2014/34ccec51ce5d47f7bd3e6fd81ae192f7_MIT18_311S14_Lecture5.pdf |
1st
order
(scalar)
quasilinear
equations
by
characteristics.
Examples:
traffic
flow
and
river
waves.
Tra
ffic
Flow
equations.
Traffic
density,
flow
and
car
velocity.
r
a
co
nea
ati
ion
t
a
equ
par
e
by
s
Linea
Do
Sho
s
b
ork
This
w
... | https://ocw.mit.edu/courses/18-311-principles-of-applied-mathematics-spring-2014/34ccec51ce5d47f7bd3e6fd81ae192f7_MIT18_311S14_Lecture5.pdf |
for
traffi
Density
waves
reach
cars
from
ahead.
River
waves
move
faster
than
the
flow.
c
flow
u
>
c
and
for
river
flows
u
<
c.
Interpret
equation
as
a
statement
about
a
directional
derivative
of
the
solution
in... | https://ocw.mit.edu/courses/18-311-principles-of-applied-mathematics-spring-2014/34ccec51ce5d47f7bd3e6fd81ae192f7_MIT18_311S14_Lecture5.pdf |
18.311:
Principles
of
Applied
Mathematics
Lecture
5
Rodolfo
Ros
ales
Spring
2014
Scalar
the
cor
hyper
respo
bolic
equations
can
be
r
nding
directional
derivati
ves.
educed
to
o.d.e.'s
along
the
cur
ves
integrating
2
MIT... | https://ocw.mit.edu/courses/18-311-principles-of-applied-mathematics-spring-2014/34ccec51ce5d47f7bd3e6fd81ae192f7_MIT18_311S14_Lecture5.pdf |
18.782 Introduction to Arithmetic Geometry
Lecture #17
Fall 2013
11/05/2013
Throughout this lecture k denotes an algebraically closed field.
17.1 Tangent spaces and hypersurfaces
For any polynomial f ∈ k[x1, . . . , xn] and point P = (a1, . . . , an) ∈ An we define the affine
linear form
fP (x1, . . . , xn) :=
(P )(xi
− ai... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/34da4eafb6fbe4d016d08aa7914a4be7_MIT18_782F13_lec17.pdf |
= 0 yields
gP =
(cid:88)(cid:0)hi(P )fi,P + hi,P fi(P )
(cid:1) =
(cid:88)
hi(P )fi,P ,
i
i
(1)
which is an element of the ideal (f1,P , . . . , fm,P ). Thus I(Tp(V )) = (f1,P , . . . , fm,P ).
When considering the tangent space of a variety at a particular point P , we may assume
without loss of generality that P = (0... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/34da4eafb6fbe4d016d08aa7914a4be7_MIT18_782F13_lec17.pdf |
I(TP (V )) is precisely the set of linear forms that vanish at every point in Tp(V ), which, by
∨. Moreover, we see from (1) that every linear
definition, is the orthogonal complement TP
form in I(TP (V )) is a k-linear combination of f1,P . . . , fm,P .
The vector space TP (V )⊥ is called the cotangent space of V at P ... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/34da4eafb6fbe4d016d08aa7914a4be7_MIT18_782F13_lec17.pdf |
of generators for I(V ) and P
is a smooth point of V if and only if dim TP = dim V .
Remark 17.5. For projective varieties V we defined smooth points P as points that are
smooth in all (equivalently, any) affine part containing P . One can also define tangent spaces
and Jacobian matrices for projective varieties directly u... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/34da4eafb6fbe4d016d08aa7914a4be7_MIT18_782F13_lec17.pdf |
. . , φ(xn) must be algebraically dependent as elements of k(V ). Thus there exists
g ∈ k[x2, . . . , xn] such that g(φ(x2), . . . , φ(xn)) = 0. But then φ(g) = 0, so g ∈ ker φ = (f ).
⊆ Pn, then one of its
But this is a contradiction, since degx1 g = 0. So dim V = n − 1. If V
affine parts Vi is a hypersurface in A , and... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/34da4eafb6fbe4d016d08aa7914a4be7_MIT18_782F13_lec17.pdf |
1) = 0;
such an f exists since γ1, . . . , γn+1 are algebraically dependent. We must have ∂f /∂xi = 0
for some xi;
if not than we must have char(k) = p > 0 and f = g(x1, . . . , xn+1) =
gp(x1, . . . , xn+1) for some g ∈ k[x1, . . . , xn+1], but this is impossible since f is irreducible. It
follows that γi is algebraic,... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/34da4eafb6fbe4d016d08aa7914a4be7_MIT18_782F13_lec17.pdf |
of a hypersurface: if k(V ) = k(γ1, . . . , γn+1)
with γ1, . . . , γn) algebraically independent, then there exists an irreducible polynomial f in
k[x1, . . . , xn+1] for which f (γ1, . . . , γn+1) = 0, and then V is birationally equivalent to the
zero locus of f in An+1.
Corollary 17.11. For any point P on an affine var... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/34da4eafb6fbe4d016d08aa7914a4be7_MIT18_782F13_lec17.pdf |
k[An].
⊆ An is an affine variety, then
the maximal ideals mP of its coordinate ring k[V ] = k[A ]/I(V ) are in one-to-one corre-
n
spondence with the maximal ideals MP of k[An] that contain I(V ); these are precisely the
maximal ideals MP for which P ∈ V .
If V
P as a subspace, and the quotient space MP /M 2
If we choose... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/34da4eafb6fbe4d016d08aa7914a4be7_MIT18_782F13_lec17.pdf |
surjective linear transformation
d : MP → TP (V )∨
whose kernel we claim is equal to M 2
P + I(V ) (this is a sum of ideals in k[x1, . . . , xn] that
is clearly a subset of MP ). A polynomial f ∈ MP lies in ker d if and only if the restriction
of fP to TP (V ) is the zero function, which occurs if and only if fP = gP f... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/34da4eafb6fbe4d016d08aa7914a4be7_MIT18_782F13_lec17.pdf |
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