text stringlengths 30 4k | source stringlengths 60 201 |
|---|---|
� : P1 → X defined by
(cid:18) s2 − t2
s2 + t2
(cid:19) (cid:18)
: 1 = 1 :
2st
s2 + t2
ϕ−1(s : t) =
:
2st
s2 − t2
:
s2 + t2
s2 − t2
(cid:19)
which can also be written as
ϕ−1(s : t) = (s2 − t2 : 2st : s2 + t2).
1
◦ ϕ−1 and
The map ϕ−1 is regular everywhere, hence a morphism, and the compositions ϕ
ϕ− ◦ ϕ are both the ide... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/194e47df6e1f016ee3fb2719abe7e8d3_MIT18_782F13_lec15.pdf |
Topic 3 Notes
Jeremy Orloff
3 Line integrals and Cauchy’s theorem
3.1 Introduction
The basic theme here is that complex line integrals will mirror much of what we’ve seen for multi-
variable calculus line integrals. But, just like working with is easier than working with sine and
cosine, complex line integrals are ea... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1968cab6de7c96ead1c83001459a007b_MIT18_04S18_topic3.pdf |
TEGRALS AND CAUCHY’S THEOREM
2
so the right hand side of this equation is
That is, it is exactly the same as the expression in Equation 1a.
(()) ′() .
∫
2 along the straight line from 0 to 1 + .
Example 3.1. Compute ∫
Solution: We parametrize the curve as () = (1 + ) with 0 ≤ ≤ 1. So ′() = 1 + . The line
inte... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1968cab6de7c96ead1c83001459a007b_MIT18_04S18_topic3.pdf |
Theorem 3.5. (Fundamental theorem of complex line integrals) If () is a complex analytic func-
tion on an open region and is a curve in from
0 to
1 then
′() = (
1) − (
0).
∫
3 LINE INTEGRALS AND CAUCHY’S THEOREM
3
Proof. This is an application of the chain rule. We have
(())
= ′(()) ′().
So
′() = ∫
′... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1968cab6de7c96ead1c83001459a007b_MIT18_04S18_topic3.pdf |
. More precisely, if () is defined on a region then ∫
in , if it has the same value for any two paths in with the same endpoints.
() is path independent
The following theorem follows directly from the fundamental theorem. The proof uses the same
argument as Example 3.7.
Theorem 3.8. If () has an antiderivative in ... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1968cab6de7c96ead1c83001459a007b_MIT18_04S18_topic3.pdf |
= 0. So,
() = 0 for any closed curve. Consider the two curves
2 shown
1. By the assumption that integrals over closed paths are
0 and end at
1 and
() = ∫
2
∫
1
() .
That is, any two paths from
are path independent.
0 to
1 have the same line integral. This shows that the line integrals
Figure (i)
Figure... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1968cab6de7c96ead1c83001459a007b_MIT18_04S18_topic3.pdf |
we could use log() if we were careful to let the argument increase by 2 as it went
around the origin once.
Example 3.12. Compute ∫
(i) Using the fundamental theorem.
1
2
, where is the unit circle in two ways.
(ii) Directly from the definition.
Solution: (i) Let () = −1∕. Since ′() = 1∕2, the fundamental theorem... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1968cab6de7c96ead1c83001459a007b_MIT18_04S18_topic3.pdf |
flavor to the proof of Green’s theorem.
Let be the region inside the curve. And write = + . Now we write out the integral as follows
() = ∫
( + ) ( + ) = ∫
( − ) + ( + ).
∫
Let’s apply Green’s theorem to the real and imaginary pieces separately. First the real piece:
− = ∫
We get 0 because the Cauchy-Riemann e... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1968cab6de7c96ead1c83001459a007b_MIT18_04S18_topic3.pdf |
i′) we would need a more technically precise definition of simply connected so we could
say that all closed curves within can be continuously deformed to each other.
3 LINE INTEGRALS AND CAUCHY’S THEOREM
7
For reference, we note that using the path () = () + (), with (0) =
0 and () = we have
() = ∫... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1968cab6de7c96ead1c83001459a007b_MIT18_04S18_topic3.pdf |
Az0zz+hz+ihCCxCy
3 LINE INTEGRALS AND CAUCHY’S THEOREM
Similarly, we get (remember: = + , so = )
∫
1
= lim
ℎ→0
( + ℎ) − ()
ℎ
= lim
ℎ→0
()
ℎ
∫
0
ℎ (, + ) + (, + )
ℎ
= lim
ℎ→0
= (, ) + (, )
= ().
Together Equations 5 and 6 show
() =
1
=
By Equation 2a we have shown that is analyti... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1968cab6de7c96ead1c83001459a007b_MIT18_04S18_topic3.pdf |
and −
3 cancel, which leaves ∫
() = 0. QED
1−
2
Note. This clearly implies ∫
Example 3.15. Let () = 1∕. () is defined and analytic on the punctured plane.
() = ∫
() .
1
2
Punctured plane: − {0}
Question: What values can ∫
the plane?
Solution: We have two cases (i)
1 not around 0, and (ii)
2 around 0
() t... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1968cab6de7c96ead1c83001459a007b_MIT18_04S18_topic3.pdf |
3.16. A further extension: using the same trick of cutting the region by curves to make it
simply connected we can show that if is analytic in the region shown below then
∫
1−
2−
3−
4
() = 0.
That is,
1 −
2 −
3 −
4 is the boundary of the region .
Orientation. It is important to get the orientation of the curve... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1968cab6de7c96ead1c83001459a007b_MIT18_04S18_topic3.pdf |
18.782 Introduction to Arithmetic Geometry
Lecture #10
Fall 2013
10/8/2013
In this lecture we lay the groundwork needed to prove the Hasse-Minkowski theorem
for Q, which states that a quadratic form over Q represents 0 if and only if it represents 0
over every completion of Q (as proved by Minkowski). The statement sti... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/196f7a627ae6d75ab009bbef96e7c8c9_MIT18_782F13_lec10.pdf |
We say that a solution (x0, . . . , xn) to a homogeneous polynomial equation over Qp is
p . The following lemma
primitive if all of its elements lie in Zp and at least one lies in Z×
gives several equivalent definitions of the Hilbert symbol.
Lemma 10.2. For any a, b ∈ Q×
p , the following are equivalent:
(i) (a, b)p = ... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/196f7a627ae6d75ab009bbef96e7c8c9_MIT18_782F13_lec10.pdf |
10.3. For all a, b, c ∈ Q×
p , the following hold:
(i) (1, c)p = 1.
(ii) (−c, c)p = 1.
(iii) (a, c)p = 1 =⇒ (a, c)p(b, c)p = (ab, c)p.
(iv) (c, c)p = (−1, c)p.
c) to Qp. For (i) we have N (1) = 1. For (ii),
Proof. Let N denote the norm map from Qp(
−c = N (−c) for c ∈ Q×2 and −c = N (
c) otherwise. For (iii), If a and ... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/196f7a627ae6d75ab009bbef96e7c8c9_MIT18_782F13_lec10.pdf |
(u, v)p = 1 for all u, v ∈ Z×
p .
Proof. Recall from Lecture 3 (or the Chevalley-Warning theorem on problem set 2) that
every plane projective conic over Fp has a rational point, so we can find a non-trivial solution
to z2 − ux2 − vy2 = 0 modulo p. If we then fix two of x, y, z so that the third is nonzero,
Hensel’s lemm... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/196f7a627ae6d75ab009bbef96e7c8c9_MIT18_782F13_lec10.pdf |
If v is not a square then z2 − px2 − vy2 = 0 has no non-
trivial solutions modulo p, hence no primitive solutions. This implies
(cid:17) (cid:16)
= −1 = ( ).p
p−1
p
p
v
Case α = 1, β = 1: We must show (pu, pv)
p = (−1) 2
. Applying Corol-
(p, v)p
(cid:17)
v
p
(cid:16) u
p
lary 10.3 we have
(pu, pv)p = (pu, pv)p(−pv, pv... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/196f7a627ae6d75ab009bbef96e7c8c9_MIT18_782F13_lec10.pdf |
strong form of Hensel’s lemma proved in Problem Set 4. In both cases, if we
have a non-trivial solution (x0, y0, z0) modulo 8 we can fix two of x0, y0, z0 to obtain a
quadratic polynomial f (w) over Z2 and w0 ∈ Z×
2 that satisfies v2(f (w0)) = 3 > 2 =
2v2(f (cid:48)(w0)). In the case of the second equation, note that a p... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/196f7a627ae6d75ab009bbef96e7c8c9_MIT18_782F13_lec10.pdf |
where one of a, b lies in Z×
p . By Lemma 10.8, to compute (a, b)2 with one of a, b
in Z×, it suffices to check for primitive solutions to z2 − ax2
− by = 0 modulo 8, which
reduces the problem to a finite verification which performed by
(cid:58)(cid:82)(cid:85)(cid:78)(cid:86)(cid:75)(cid:72)(cid:72)(cid:87)(cid:17)
(cid:2... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/196f7a627ae6d75ab009bbef96e7c8c9_MIT18_782F13_lec10.pdf |
c)
p(b, c)p = ε
w
p
γ
(cid:19) (cid:18)
γ
(cid:18)
α u
p
(cid:18) uv
p
(cid:19)
(cid:19) (cid:18)
γ
(cid:19)
β
w
p
(cid:18)
v
p
α+β
(cid:19)
α
ε
β
(cid:19)
(cid:18) w
p
= εα+β
=
(ab, c) .
p
To verify non-degeneracy, we note that if c is not square than either γ = 1 or ( w ) =p
−1. If
γ = 1 we can choose b = v with ( v ... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/196f7a627ae6d75ab009bbef96e7c8c9_MIT18_782F13_lec10.pdf |
1(0) in (Z/4Z)× is {1}, a subgroup of index 2, and the
image of ω−1(0) in (Z/8Z)× is {±1}, which is again a subgroup of index 2.
We now verify non-degeneracy for p = 2. If c is not square then either γ = 1, or one
of (cid:15)(w) and ω(w) is nonzero. If γ = 1, then (5, c)2 = −1. If γ = 0 and ω(w) = 1, then
(2, c)2 = −1.... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/196f7a627ae6d75ab009bbef96e7c8c9_MIT18_782F13_lec10.pdf |
= (−1, 1)
− 2 = −1 and (−1, −1)p = 1 for p odd.
Case 2: a = −1 and b is prime. If b = 2 then (1, 1) is a solution to −x2 + 2y2 = 1
(cid:54)∈ {2, b}, while
p(−1, 2) = 1. If b is odd, then (
−1, b)p = 1 for p
(cid:81)
∞
and (−1, b)b = ( − ), both of which are equal to (
−1) − /2.
(b 1)
1
b
Case 3: a and b are
the same pr... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/196f7a627ae6d75ab009bbef96e7c8c9_MIT18_782F13_lec10.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
18.727 Topics in Algebraic Geometry: Algebraic Surfaces
Spring 2008
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
ALGEBRAIC SURFACES, LECTURE 2
LECTURES: ABHINAV KUMAR
Remark. In the definition of (L, M ) we wrote M = OX (... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/198274c0c471d31fc05d600e28e403db_lect2.pdf |
.t. E| = C and E| = D. We say C ∼alg D.
·
We say C is numerically equivalent to D, C ≡ D, if C E = D E for every
t1
t2
·
divisor E on X.
We have an intersection pairing Div X × Div X → Z which factors through
Pic X × Pic X → Z, which shows that linear equivalence = ⇒ num equivalence.
alg equivalence (map to P1 d... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/198274c0c471d31fc05d600e28e403db_lect2.pdf |
⊗ OC = OX (−C)|C ⊗ ΩC or KC = (KX + C)|C so deg KC =
2g(C) − 2 = C.(C + K) (genus formula). Note: C 2 = deg (OX (C) ⊗ OC ) by
definition. I/I 2 is the conormal bundle, and is ∼ O(−C) ⊗ OC , while NC/X is
the normal bundle ∼= O(C) ⊗ OC .
Theorem 1 (Riemann-Roch). χ(L) = χ(OX ) + 1
2 (L2 − L ωX ).
=
·
Proof. L−1 · L ... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/198274c0c471d31fc05d600e28e403db_lect2.pdf |
].
12
12
1
Remark. If H is ample on X, then for any curve C on X, we have C H > 0
(equals n (degree of C in embedding by nH) for larger n).
1.2. Hodge Index Theorem.
·
·
Lemma 1. Let D1, D2 be two divisors on X s.t. h0(X, D2) = 0
h0(X, D1 + D2).
. Then h0(X, D1) ≤
Proof. Let a = 0
H0(X, D1 + D2) is injective.... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/198274c0c471d31fc05d600e28e403db_lect2.pdf |
). Similarly,
(otherwise h0(nD) = 0 gives
h0(nD) can’t go to ∞ both as n → ∞ and as n → −∞. Similarly for h0(K −nD).
�
Finally, note that h0(nD) �= 0 implies (nD · H) > 0 and so D · H > 0.
Corollary 1. If D is a divisor on X and H is a hyperplane section on X s.t.
(D · H) = 0 then D2 ≤ 0 and D2 = 0 ⇔ D ≡ 0.
Proo... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/198274c0c471d31fc05d600e28e403db_lect2.pdf |
M = NumX ⊗Z R. This is a finite dimensional vector space
over R of dimension ρ (the Picard number) and signature (1, ρ − 1).
Proof. Embed this in �-adic cohomology H 2(X, Q�(1)) which is finite dimen
sional, and C.D equals C ∪ D under
(4)
H 2(X, Q�(1)) × H 2(X, Q�(1)) → H 4(X, Q�(2)) ∼= Q�
The map NumX � C → [C] ∈ ... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/198274c0c471d31fc05d600e28e403db_lect2.pdf |
number of L · M to be the coefficient of n1 · n2
in χ(Ln1 ⊗ Mn2 ) (check that this is bilinear, etc., and that it coincides with the
previous definition).
is easy. For the converse, χ(Ln) → ∞
⇒
Proof. Sketch when X is projective. =
as n → ∞ (Riemann-Roch, or by defn). Replace L by Ln to assume L =
OX (D), D effectiv... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/198274c0c471d31fc05d600e28e403db_lect2.pdf |
else ∃C ⊂ X with f (C) = � = ⇒ C · L = 0, a contradiction).
�
OPk (1) is ample =⇒ Ln is ample =⇒ L is ample.
=
4
LECTURES: ABHINAV KUMAR
→
=
1.4. Blowups. Let X be a smooth surface, p a point on X. The blowup X˜ π X→
at p is a smooth surface s.t. X˜ � π−1(p)
X � {p} is an isomorphism and
π−1(p) is a curve ∼ ... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/198274c0c471d31fc05d600e28e403db_lect2.pdf |
The normal bundle NE/ X˜ is OE (−1).
Note: this is a specific case of a more general fact (Hartshorne 8.24). For
Y ⊂ X a closed subscheme with corresponding ideal sheaf I, blow up X along
Y to get the projective bundle Y �
�
Y given by P(I/I 2), and overall blowup
→
I d , OX˜ (1) = I˜ = π−1IO X˜
X˜ = Proj
(7)
I... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/198274c0c471d31fc05d600e28e403db_lect2.pdf |
8.04: Quantum Mechanics
Massachusetts Institute of Technology
Professor Allan Adams
2013 February 12
Lecture 3
The Wavefunction
Assigned Reading:
E&R 16,7, 21,2,3,4,5, 3all NOT 4all!!!
1all, 23,5,6 NOT 2-4!!!
Li.
12,3,4 NOT 1-5!!!
Ga.
3
Sh.
In classical mechanics, the configuration or state of a system i... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2013/19862abcdf8e4f7cf1321d060cd6dd7c_MIT8_04S13_Lec03.pdf |
drawn for simplicity,
as complex-plane paper is hard to find. Furthermore, ψ must be singly-valued and not
“stupid”; the latter point will be elaborated later.
Let us examine this set of examples in further detail. The first wavefunction ψ1 is sharply
peaked at a particular value of x, and the probability density, be... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2013/19862abcdf8e4f7cf1321d060cd6dd7c_MIT8_04S13_Lec03.pdf |
lot of sense to think of a
sinusoidal wave as being localized in some place. Indeed, the positions for these two wave-
functions are ill-defined, so they are not well-localized, and the uncertainty in the position
is large in each case.
8.04: Lecture 3
3
The fifth wavefunction is multiply-valued, so it is considere... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2013/19862abcdf8e4f7cf1321d060cd6dd7c_MIT8_04S13_Lec03.pdf |
with small uncertainties
in their respective momenta, with ψ4 having a smaller wavelength and therefore a larger
momentum than ψ3. On the other hand, ψ1 and ψ2 do not look like sinusoidal waves, so it
is difficult to define a wavelength and therefore a momentum for each, and the respective
momentum uncertainties are l... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2013/19862abcdf8e4f7cf1321d060cd6dd7c_MIT8_04S13_Lec03.pdf |
and ψb, the system could also be in a superposition
ψ = αψa + βψb with α and β as arbitrary complex coefficients satisfying
normalization.
This forms the soul of quantum mechanics!
Note that for a superposition state
the probability density
ψ(x) = αψa(x) + βψb(x),
p(x) = |αψa(x) + βψb(x)|2 = |αψa(x)|2 + |βψb(x)|2 ... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2013/19862abcdf8e4f7cf1321d060cd6dd7c_MIT8_04S13_Lec03.pdf |
though,
the probability density will exhibit interference because the product of the two wavefunctions
is not always zero as they are both sinusoidal waves.
Note for the example of ψ5 that this superposition state has more spatial localization than
each of the component sinusoidal wavefunctions. This spatial locali... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2013/19862abcdf8e4f7cf1321d060cd6dd7c_MIT8_04S13_Lec03.pdf |
of the Fourier expansion coefficients
αj =
(cid:90)
π
1
2π −π
f (x)e −ikj x dx.
(0.7)
The physical interpretation of this is that any wavefunction ψ(x) can
be expressed as a superposition of states eikx with definite momenta
p = k as
ψ(x) =
√1
∞
2π −∞
˜
ψ(k)e
ikx dk.
(0.8)
Furthermore, ψ˜(k) gives the exact ... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2013/19862abcdf8e4f7cf1321d060cd6dd7c_MIT8_04S13_Lec03.pdf |
vein, aside from normalization, ψ5 and ψ6 are Fourier transforms of each other.
This means that a wavefunction that is well-localized around a given position has a Fourier
transform that looks like a sinusoidal function of k, and the frequency of oscillation as a
function of k is given by that position. Similarly, a... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2013/19862abcdf8e4f7cf1321d060cd6dd7c_MIT8_04S13_Lec03.pdf |
3.15: Transistors in ‘forward active’ mode
Common base circuit
E
B
C
Input
IE,VEB
Output
IE,VCB
IC,mA
5
0
_
IE = 5mA
3mA
1 mA
IE = 0
0
+
_VcB
Figure by MIT OCW.
This is the easiest to visualize, though of limited usefulness. We use one power supply to put EB
into forward bias (p side positive) and another one to put... | https://ocw.mit.edu/courses/3-15-electrical-optical-magnetic-materials-and-devices-fall-2006/1996e5617f952ee322c739d9e24f4ff9_lecture7b.pdf |
IB between the base and the emitter. The polarity is chosen so that EB is in
forward bias. We connect another power supply between E and C (this is chosen to make sure BC
is in reverse bias).
The base current IB is primarily composed of electrons that contribute to the forward current
through EB. In the EB junction, ... | https://ocw.mit.edu/courses/3-15-electrical-optical-magnetic-materials-and-devices-fall-2006/1996e5617f952ee322c739d9e24f4ff9_lecture7b.pdf |
LECTURE 11
LECTURE OUTLINE
Review of convex progr. duality/counterexamples
Fenchel Duality
•
•
Conic Duality
•
Reading: Sections 5.3.1-5.3.6
Line of analysis so far:
Convex analysis (rel. int., dir. of recession, hy-
•
perplanes, conjugacy)
MC/MC - Three general theorems: Strong dual-
•
ity, existence of dual optimal s... | https://ocw.mit.edu/courses/6-253-convex-analysis-and-optimization-spring-2012/19c72f4d87303765c3925112f5926282_MIT6_253S12_lec11.pdf |
there exists a dual optimal
solution.
⌘
⌘
(b) f ⇤ = q⇤, and (x⇤, µ⇤, ⌃⇤) are a primal and
dual optimal solution pair if and only if x⇤
is feasible, µ⇤ ≥
x⇤
arg min L(x, µ⇤, ⌃⇤), µ⇤j gj(x⇤) = 0,
0, and
⌘
x X
⌦
j
2 COUNTEREXAMPLE I
Strong Duality Counterexample: Consider
•
minimize f (x) = e−
subject to x1 = 0,
x1x2
... | https://ocw.mit.edu/courses/6-253-convex-analysis-and-optimization-spring-2012/19c72f4d87303765c3925112f5926282_MIT6_253S12_lec11.pdf |
1
0.8
0.6
0.4
0.2
0
0
√x1x2
e−
5
10
15
x1 = u
5
10
x2
20
0
20
15
Connection with counterexample for preserva-
•
tion of closedness under partial minimization.
4 COUNTEREXAMPLE II
Existence of Solutions Counterexample:
, f (x) = x, g(x) = x2. Then x⇤ = 0 is
•
Let X =
the only feasible/optimal solution, and we have
�
q(µ... | https://ocw.mit.edu/courses/6-253-convex-analysis-and-optimization-spring-2012/19c72f4d87303765c3925112f5926282_MIT6_253S12_lec11.pdf |
+ f2(x2) + ⌥⇧(x2
x1)
−
⌥⇧x1
⇤
+ inf
x2⌥
n
2(x2) + ⌥⇧x2
f
−
⇤
⌅
Dual problem: max
⌅
{−
min⌅{−
q(⌃)
or
}
•
−
f (⌃)
1
⇤
−
f
2 (
⌃)
}
−
⌅
=
1 (⌃) +f
f
2 (
minimize
n,
subject to ⌃
⌘ �
⌃)
−
where f
1 and f
2 are the conjugates.
6◆
◆
•
FENCHEL DUALITY THEOREM
Consider the Fenchel framework:
(a) If f ⇤ is finite and ri
... | https://ocw.mit.edu/courses/6-253-convex-analysis-and-optimization-spring-2012/19c72f4d87303765c3925112f5926282_MIT6_253S12_lec11.pdf |
✓
GEOMETRIC INTERPRETATION
f
1 ()
q()
f
2 (
)
f = q
f1(x)
Slope
Slope
f2(x)
x
x
When dom(f1) = dom(f2) =
n, and f1 and
•
f2 are differentiable, the optimality condition is
equivalent to
�
⌃⇤ =
∇
f1(x⇤) =
−∇
f2(x⇤)
By reversing the roles of the (symmetric) primal
•
and dual problems, we obtain alternative criteria
f... | https://ocw.mit.edu/courses/6-253-convex-analysis-and-optimization-spring-2012/19c72f4d87303765c3925112f5926282_MIT6_253S12_lec11.pdf |
x
⌥
⇤
where C ⇤ =
⇤⇧x
−
f (x)
⌅
, f ⌥
2 (⇤) = sup ⇤⇧x =
x
C
⌥
0
⇧
if ⇤
if ⇤
C⇥,
⌃
/ C ,
⇥
⌃
�
⌃
⌃�x
⌥
{
The dual problem is
|
0,
x
C
.
}
⌘
•
•
minimize
subject to ⌃
f (⌃)
ˆ
C,
⌘
where f is the conjugate of f and
Cˆ =
⌃
{
|
⌃�x
0,
x
C
.
}
⌘
≥
Cˆ and
−
ˆ
C are called the dual and polar cones.
10◆
CONIC DUALITY THEO... | https://ocw.mit.edu/courses/6-253-convex-analysis-and-optimization-spring-2012/19c72f4d87303765c3925112f5926282_MIT6_253S12_lec11.pdf |
b
c)�x = sup(⌃
c)�(y + b)
=
⌦
−
(⌃
�
⇣
c)�b
−
if ⌃
if ⌃
y S −
⌦
c
⌘
c /
⌘
S⊥,
S.
−
−
Dual problem is equivalent to
minimize
b�⌃
subject to ⌃
c S ,
− ⌘ ⊥
⌃ C.ˆ
⌘
If X
ri(C) =Ø, there is no duality gap an
d
•
there exists a dual optimal solution.
⌫
12 ANOTHER APPROACH TO DUALITY
Consider the problem
•
minimize f (x)
subj... | https://ocw.mit.edu/courses/6-253-convex-analysis-and-optimization-spring-2012/19c72f4d87303765c3925112f5926282_MIT6_253S12_lec11.pdf |
Lecture 8
Symmetries, conserved quantities, and the labeling of states
Angular Momentum
Today’s Program:
1. Symmetries and conserved quantities – labeling of states
2. Ehrenfest Theorem – the greatest theorem of all times (in Prof. Anikeeva’s opinion)
3. Angular momentum in QM
4. Finding the eigenfunctions of Lˆ... | https://ocw.mit.edu/courses/3-024-electronic-optical-and-magnetic-properties-of-materials-spring-2013/19da8ffb4044af736d595f30eebe310d_MIT3_024S13_2012lec8.pdf |
operation on it and then compare the result to the initial situation they
are indistinguishable.�When one speaks of a symmetry it is critical to state symmetric with
respect to which operation.
How do symmetries manifest themselves in equations?�Let us suppose that your system is
symmetric with respect to translati... | https://ocw.mit.edu/courses/3-024-electronic-optical-and-magnetic-properties-of-materials-spring-2013/19da8ffb4044af736d595f30eebe310d_MIT3_024S13_2012lec8.pdf |
�t
Aˆ, Hˆ
0
d
dt
ˆ
A 0
States are labeled by specific values of their properties, which do not change with time –
these properties are called constants of motion. We learned that in QM physical properties are
represented by operators and that the values of properties obtained in measuremen... | https://ocw.mit.edu/courses/3-024-electronic-optical-and-magnetic-properties-of-materials-spring-2013/19da8ffb4044af736d595f30eebe310d_MIT3_024S13_2012lec8.pdf |
t
This type of differential equation is separable, i.e. we can look for a solution in the following
r, t
form:
V
t . Let’s substitute it into the Schrodinger’s equation above:
r
r
t i
r t
t
r
2
2m
1
r
2
... | https://ocw.mit.edu/courses/3-024-electronic-optical-and-magnetic-properties-of-materials-spring-2013/19da8ffb4044af736d595f30eebe310d_MIT3_024S13_2012lec8.pdf |
�� t
E t t e
i
E
d
dt
II
2
2m
2 V
r
r E r uE
x , E
Then the solutions to time-dependent Schrodinger’s equation will have a form:
E
r, t
uE
r E
t uE
t
E
i
r e
In general, since the Hamiltonian may have many e... | https://ocw.mit.edu/courses/3-024-electronic-optical-and-magnetic-properties-of-materials-spring-2013/19da8ffb4044af736d595f30eebe310d_MIT3_024S13_2012lec8.pdf |
� at any time!
r, t
3
Conserved Quantities Example I: Particle in free space
Labeling of states is particularly important when the energy eigenvalues are degenerate such as
in the case of the particle in free space:
Hˆ x E x
2
2
2m x2
E uE
x
... | https://ocw.mit.edu/courses/3-024-electronic-optical-and-magnetic-properties-of-materials-spring-2013/19da8ffb4044af736d595f30eebe310d_MIT3_024S13_2012lec8.pdf |
uE,k eikx
x
2mE
2
In fact we represent all the eigenfunctions (eigenstates) of the free space Hamiltonian and the
momentum on the E vs k plot. Every point on this plot uniquely and completely specifies the
state.
4
Conserved Quantities Example II: Parity operator and symmetric potentials ... | https://ocw.mit.edu/courses/3-024-electronic-optical-and-magnetic-properties-of-materials-spring-2013/19da8ffb4044af736d595f30eebe310d_MIT3_024S13_2012lec8.pdf |
�2
2
x x 0
This means that one can always find a set of eigenfunctions common to Hˆ and ˆ . In fact, last
lecture we have shown that SHO eigenfunctions are always even or odd.
1
2
1
2
m2 x x
2
5
Conserved Quantities Example II: Angular momentum... | https://ocw.mit.edu/courses/3-024-electronic-optical-and-magnetic-properties-of-materials-spring-2013/19da8ffb4044af736d595f30eebe310d_MIT3_024S13_2012lec8.pdf |
r
p
iˆ
x
x
p
x
iˆ
y
y
p
y
z
iˆ
z iˆ yp zp iˆ zp xp iˆ xp yp iˆ L iˆ L iˆ L
p
y
x
x
z
y
y
y
x
x
y
x
z
z
z
z
z
In quantum mechanics we can define a corresponding observable (Hermitian operator):
ˆ
rˆ pˆ
L
iˆ
x
x
iˆ
y
y
iˆ
z
z
i
x
i
y
i
... | https://ocw.mit.edu/courses/3-024-electronic-optical-and-magnetic-properties-of-materials-spring-2013/19da8ffb4044af736d595f30eebe310d_MIT3_024S13_2012lec8.pdf |
y
x
While Descartes’s coordinate system allows us to intuitively derive the
components of the angular momentum operator, the spherical symmetry of
the problem makes it more logical to move to the spherical coordinate
system:
x r sin cos
y r sin sin
z r cos
6
In spherical co... | https://ocw.mit.edu/courses/3-024-electronic-optical-and-magnetic-properties-of-materials-spring-2013/19da8ffb4044af736d595f30eebe310d_MIT3_024S13_2012lec8.pdf |
since it has a classical equivalent.
If one looks at the commutation relations of this operator one finds:
Lˆ Lˆ
x, y iLˆ
Lˆ Lˆ
y, z iLˆ
Lˆ , Lˆ
iLˆ
z
x
x
y
z
Note: In QM there is another very important quantity, which is called spin. It turns out that this
quantity also obeys th... | https://ocw.mit.edu/courses/3-024-electronic-optical-and-magnetic-properties-of-materials-spring-2013/19da8ffb4044af736d595f30eebe310d_MIT3_024S13_2012lec8.pdf |
��!
Pl cos e
im
, m 0
2l 1 l m! m
4 l m!
Pl
cos e
im
, m 0
Since the wave function needs to be continuous at 2 , m is an integer. One can show that l also
needs to be an integer and that −l≤m≤l. l ≥ 0. Pl cos are Legendre polynomials and the
m
eigenfunctions Yl , are called spherica... | https://ocw.mit.edu/courses/3-024-electronic-optical-and-magnetic-properties-of-materials-spring-2013/19da8ffb4044af736d595f30eebe310d_MIT3_024S13_2012lec8.pdf |
Representation Invariants
6.170, Lecture 8
Fall 2005
8.1
Context
What you’ll learn: How to find representation invariants and avoid representation exposure.
Why you should learn this: An understanding of the theory of abstract types helps you avoid whole
classes of nasty, subtle bugs – or at minimum alerts you to thei... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/1a1f80fa85176ed3cae09ccd74f902d3_lec8.pdf |
will suffice to view it simply as a mathematical value.
The space of abstract values consists of the values that the type is designed to support. These
are a figment of our imagination. They’re platonic entities that don’t exist as described, but they
are the way we want to view the elements of the abstract type, as cl... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/1a1f80fa85176ed3cae09ccd74f902d3_lec8.pdf |
(!member(ch)) s.append(ch);
}
public void remove(char ch) {
int index = s.indexOf(String.valueOf(ch));
if (index>=0) {
s.deleteCharAt(index);
}
}
public boolean member(char ch) {
return s.indexOf(String.valueOf(ch))!=-1;
}
}
This works just fine. What are the rep values and the abstract values of this type? The abstr... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/1a1f80fa85176ed3cae09ccd74f902d3_lec8.pdf |
’t check for duplicates, so there is no
constraint on s. The remove method needs to be changed to dovetail with the new add behavior.
We can look at the rep (R) and abstract (A) value spaces for the two implementations graphi
cally, drawing arcs from each rep value to the abstract value it represents:
First impleme... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/1a1f80fa85176ed3cae09ccd74f902d3_lec8.pdf |
3
an interpretation for all rep values. A doubly-linked list representation, for example, can be
twisted into all kinds of pretzel configurations that won’t correspond to simple sequences,
and for which we won’t want to write special cases in the code. Or sometimes we will want
to impose certain properties on the re... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/1a1f80fa85176ed3cae09ccd74f902d3_lec8.pdf |
us
whether a given rep value is well-formed. Alternatively, you can think of RI as a set: it’s the
subset of rep values on which AF is defined.
We will look at abstraction functions next day. Today, let’s look at the rep invariant, RI. For our
first implementation of CharSet, it was:
RI(r) = r.s != null &&
r.s cont... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/1a1f80fa85176ed3cae09ccd74f902d3_lec8.pdf |
). We would like to write a rep invariante RI that is always true for our chosen representation:
all tree: BTree | RI(tree)
We can write RI recursively as:
RI(tree) = (tree.root!=null) =>
(tree.root.parent = null && RI(tree,tree.root))
RI(tree,node) = ((node.left!=null) => (node.left!=node.right)) &&
((node.left!=n... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/1a1f80fa85176ed3cae09ccd74f902d3_lec8.pdf |
produces one that also satisfies it. Now we can argue that every
object of the type satisfies the rep invariant, since it must have been produced by a constructor
and some sequence of mutator or producer applications.
Rep invariants can often be translated to code, and used to periodically check whether an object
is ... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/1a1f80fa85176ed3cae09ccd74f902d3_lec8.pdf |
(3) the representation of the type is never exposed, then the invariant is true of all instances of the
abstract data type.
Notice the caveat about the representation being exposed (called rep exposure). How can this
happen?
8.6
Rep Exposure
The notion of rep invariants that we have espoused offers a systematic me... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/1a1f80fa85176ed3cae09ccd74f902d3_lec8.pdf |
to a mutable object is passed in and made part of the rep, despite being
accessible as an existing reference from the outside.
• when a reference to a mutable object in the rep is returned as the result of an operation.
Consider the following class representing an interval of time between two Dates:
public class In... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/1a1f80fa85176ed3cae09ccd74f902d3_lec8.pdf |
example of rep exposure can occur when a method returns a collection. When
the representation already contains a collection object of the appropriate type, it is quite tempting
to return it directly. For example, in Java Lists must have a method toArray() that returns
an array of elements corresponding to the elemen... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/1a1f80fa85176ed3cae09ccd74f902d3_lec8.pdf |
{
return list.iterator();
}
Unfortunately, this too is a rep exposure! The Iterator interface in Java includes an optional
remove operation that allows the client to remove elements from the underlying collection. ArrayList
implements this operation, so the result of this method is an iterator object that can actual... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/1a1f80fa85176ed3cae09ccd74f902d3_lec8.pdf |
White blood cell (e.g., neutrophil) scavenging:
rolling, adhesion, and extravasation
© John Wiley & Sons, Inc. All rights reserved. This content is excluded from our Creative Commons
license. For more information, see http://ocw.mit.edu/help/faq-fair-use/.
Source: Man, Shumei, Eroboghene E. Ubogu, and Richard M. Ran... | https://ocw.mit.edu/courses/20-430j-fields-forces-and-flows-in-biological-systems-fall-2015/1a2b50cdc52990a6dac50a4591a53a61_MIT20_430JF15_Lecture19.pdf |
Biophysical Journal 56, no. 1 (1989): 151.
4
Chondrocytes are
more like elastic spheres
Hochmuth 2000
Courtesy of Elsevier, Inc., http://www.sciencedirect.com. Used with permission.
Source: Hochmuth, Robert M. "Micropipette aspiration of living cells." Journal of Biomechanics 33, no. 1 (2000): 15-22.
5
Com... | https://ocw.mit.edu/courses/20-430j-fields-forces-and-flows-in-biological-systems-fall-2015/1a2b50cdc52990a6dac50a4591a53a61_MIT20_430JF15_Lecture19.pdf |
19:45
© Education Development Center, Inc. All rights reserved. This content is excluded from our
Creative Commons license. For more information, see http://ocw.mit.edu/help/faq-fair-use/.
12 Drag on spherical versus slender bodies
© source unknown. All rights reserved. This content is excluded from our Creative
C... | https://ocw.mit.edu/courses/20-430j-fields-forces-and-flows-in-biological-systems-fall-2015/1a2b50cdc52990a6dac50a4591a53a61_MIT20_430JF15_Lecture19.pdf |
6.890: Algorithmic Lower Bounds: Fun With Hardness Proofs
Fall 2014
Prof. Erik Demaine
Lecture 5 Scribe Notes
1 Overview
We’ll start this week with a few more examples of classic Nintendo games, most of which are hard
via reductions from 3SAT or a variant. Then, we’ll look at a few more puzzles, including one who... | https://ocw.mit.edu/courses/6-890-algorithmic-lower-bounds-fun-with-hardness-proofs-fall-2014/1a2ff45b31e8e695010a55eb322a46f9_MIT6_890F14_Lec5.pdf |
It’s called Push-Once: not only can
you only push one block at a time, but you can only push a block once. The lock mechanism from
Push-1 can be modified for a Push-Once environment, so Push-Once is also hard.)
Figure 1: The modification of the Push-1 lock to the Push-Once lock prevents Link from pushing
a block mo... | https://ocw.mit.edu/courses/6-890-algorithmic-lower-bounds-fun-with-hardness-proofs-fall-2014/1a2ff45b31e8e695010a55eb322a46f9_MIT6_890F14_Lec5.pdf |
paper for more details!
Figure 3: The clause gadget allows Samus to come through one of the tunnels below the monsters
and shoot up to kill them. However, if she came from the top path, then there is nothing she could
do because she can’t shoot down. The cross over gadget allows her to either come from the left
or ... | https://ocw.mit.edu/courses/6-890-algorithmic-lower-bounds-fun-with-hardness-proofs-fall-2014/1a2ff45b31e8e695010a55eb322a46f9_MIT6_890F14_Lec5.pdf |
weak trainers and always
loses against strong trainers. Trainers you’ve beaten (all weak trainers, of course) act as obstacles
for strong trainers once they’re activated. Satisfying a clause corresponds to activating several
barrier trainers, which allow the player to exit all clause gadgets if and only if they’re a... | https://ocw.mit.edu/courses/6-890-algorithmic-lower-bounds-fun-with-hardness-proofs-fall-2014/1a2ff45b31e8e695010a55eb322a46f9_MIT6_890F14_Lec5.pdf |
Gastly moves
first and uses Self Destruct, causing the player to lose (even if it defeats the enemy Snorlax, the
opponent holds another one). This implementation only works in Generations I and II since TM36
exists only in Generation I and the Time Capsule feature in Generation II allows a Gastly with Self
Destruct... | https://ocw.mit.edu/courses/6-890-algorithmic-lower-bounds-fun-with-hardness-proofs-fall-2014/1a2ff45b31e8e695010a55eb322a46f9_MIT6_890F14_Lec5.pdf |
the top and bottom. The player
makes a series of long horizontal jumps to choose which variables to set to true, and is able to
traverse top to bottom if and only if none of the crucial squares have been cleared.
Figure 6: Here clauses are traversed by going top to bottom or bottom to top. Variable values are
set b... | https://ocw.mit.edu/courses/6-890-algorithmic-lower-bounds-fun-with-hardness-proofs-fall-2014/1a2ff45b31e8e695010a55eb322a46f9_MIT6_890F14_Lec5.pdf |
, with the rightmost three columns
reading:
0 p 0
0 p 0
1 q 0
These three columns force the letter “0” to have the value 0 and the letter “1” to have the value 1.
A variable va is considered true if va = 1 (mod 4) and false if va = 0 (mod 4)—the variable gadget
1For the puzzle in Table 1, S = 9, E = 5, N = 6, D ... | https://ocw.mit.edu/courses/6-890-algorithmic-lower-bounds-fun-with-hardness-proofs-fall-2014/1a2ff45b31e8e695010a55eb322a46f9_MIT6_890F14_Lec5.pdf |
0 hi
ti
ri 0 gi wi 0 fi 0
ri 0 gi wi 0 fi 0
0 gi 0
si 0 hi xi
Table 3: The clause gadget.
gi = 2fi
hi = 2gi + {0, 1}
= 4fi + {0, 1}
ti = hi + 1 + {0, 1}
= 4fi + 1 + {0, 1, 2}
= 4fi + {1, 2, 3}
va + vb + vc = ti = {1, 2, 3} (mod 4)
This construction can be simplified considerably by instead reducing from 1... | https://ocw.mit.edu/courses/6-890-algorithmic-lower-bounds-fun-with-hardness-proofs-fall-2014/1a2ff45b31e8e695010a55eb322a46f9_MIT6_890F14_Lec5.pdf |
+ vk + vl − vm − vp, for j, k, l, m, p < i. Since there are only n
5
such quintuples, as long as we have more than n5 numbers it will always be possible to select a vi
satisfying this condition.
Since the parameters remain polynomial, this construction proves that cryptarithms are strongly
NP-hard.
2.9 Star Tessell... | https://ocw.mit.edu/courses/6-890-algorithmic-lower-bounds-fun-with-hardness-proofs-fall-2014/1a2ff45b31e8e695010a55eb322a46f9_MIT6_890F14_Lec5.pdf |
are the same, we can produce a flat fold. However, if
they are all the same, then we can’t produce a flat fold because each fold of the paper is on top of
another.
Remark: A nice reason to hope for a reduction from Not-All-Equal-3SAT is that “true” and
“false” are totally symmetric in origami, and Not-All-Equal-3SA... | https://ocw.mit.edu/courses/6-890-algorithmic-lower-bounds-fun-with-hardness-proofs-fall-2014/1a2ff45b31e8e695010a55eb322a46f9_MIT6_890F14_Lec5.pdf |
clause pairs, each one corresponding
to a “satisfying” path.
11
Figure 13: Clauses are vertical and variables horizontal.
This construction can also be carried through even if we require that G be planar by designing
crossover gadgets for edges. Crossover gadgets force you to either take one path or the other.
... | https://ocw.mit.edu/courses/6-890-algorithmic-lower-bounds-fun-with-hardness-proofs-fall-2014/1a2ff45b31e8e695010a55eb322a46f9_MIT6_890F14_Lec5.pdf |
may be absent.
• Once the gadgets were constructed properly, there was the issue of fitting them together.
But the sizes of the variable gadgets and crossover gadgets were inconsistent—another parity
problem!
Figure 17: Gadget size parity.
13
• Once the full crossover versions were completed, it was necessary ... | https://ocw.mit.edu/courses/6-890-algorithmic-lower-bounds-fun-with-hardness-proofs-fall-2014/1a2ff45b31e8e695010a55eb322a46f9_MIT6_890F14_Lec5.pdf |
51. Springer, 2014.
[ADO+14] A Adcock, E D Demaine, M P O’Brien, F Reidl, F Sánchez Villaamil, and B D Sullivan.
Zig-Zag Numberlink is NP-complete. Under review, 2014.
[BH96] Marshall Bern and Barry Hayes. The complexity of flat origami. In Proceedings of the
seventh annual ACM-SIAM symposium on Discrete algorithms... | https://ocw.mit.edu/courses/6-890-algorithmic-lower-bounds-fun-with-hardness-proofs-fall-2014/1a2ff45b31e8e695010a55eb322a46f9_MIT6_890F14_Lec5.pdf |
] 古妻浩一 and 武永康彦 . ナンバーリンクの NP 完全性と問題の列挙 . 電子情報通信学
会技術研究報告. COMP, コンピュテーション, 109(465):1–7, 2010.
[Lyn75]
James F Lynch. The equivalence of theorem proving and the interconnection problem.
ACM SIGDA Newsletter, 5(3):31–36, 1975.
15
MIT OpenCourseWare
http://ocw.mit.edu
6.890 Algorithmic Lower Bounds: Fun with Hard... | https://ocw.mit.edu/courses/6-890-algorithmic-lower-bounds-fun-with-hardness-proofs-fall-2014/1a2ff45b31e8e695010a55eb322a46f9_MIT6_890F14_Lec5.pdf |
ESD.86
Random Incidence
A Major Source of Selection Bias
Richard C. Larson
February 21, 2007
Examples
(cid:139) Waiting for a bus at 77 Mass. Avenue.
– “Clumping”
(cid:139) Interview passengers disembarking
from an airplane.
Doctoral Exam Question
(cid:139) You arrive at a bus stop where busses arrive
according to ... | https://ocw.mit.edu/courses/esd-86-models-data-and-inference-for-socio-technical-systems-spring-2007/1a54fef489a72a6f19f64f10cabbff4a_lec5.pdf |
(x) = fY2
fY(x) = fY1
(x) = ...
•If the Yi’s are mutually independent then
we have a renewal process.
•But the Random Incidence results we are
about to obtain do not require that we
have a renewal process.
The Gap We Fall Into by Random Incidence
fW(w)dw=P{length of gap is between w and w+dw}
fW(w)dw is proportiona... | https://ocw.mit.edu/courses/esd-86-models-data-and-inference-for-socio-technical-systems-spring-2007/1a54fef489a72a6f19f64f10cabbff4a_lec5.pdf |
6
...
W=y4
Key result:
E[V] = E[Y]2(1+η2)/(2E[Y])
where
η = coefficient of variation of the R.V. Y
Let’s Visit Several Examples,
Including that Bus Stop Doctoral
Exam Question! | https://ocw.mit.edu/courses/esd-86-models-data-and-inference-for-socio-technical-systems-spring-2007/1a54fef489a72a6f19f64f10cabbff4a_lec5.pdf |
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
6.436J/15.085J
Lecture 13
Fall 2018
MOMENT GENERATING FUNCTIONS
Contents
1. Moment generating functions
2. Sum of a random number of random variables
3. Transforms associated with joint distributions
Moment generating functions, and their close relatives (probability gener... | https://ocw.mit.edu/courses/6-436j-fundamentals-of-probability-fall-2018/1a592ed184fb4c444547f67c9bcdd8ec_MIT6_436JF18_lec13.pdf |
analysis methods can be brought to bear on probability problems.
(h) They provide powerful tools for proving limit theorems, such as laws of
large numbers and the central limit theorem.
1
1 MOMENT GENERATING FUNCTIONS
1.1 Definition
Definition 1. The moment generating function associated with a random
variable X ... | https://ocw.mit.edu/courses/6-436j-fundamentals-of-probability-fall-2018/1a592ed184fb4c444547f67c9bcdd8ec_MIT6_436JF18_lec13.pdf |
the values 1, 2, and 3, with probabilities 1/2, 1/3, and
1/6, respectively, then
3s
MX (s) = e + e + e ,
2s
s
1
2
1
3
1
6
(1)
which is finite for every s
fX (x) = 1/(ˇ(1 + x2)), for all x, it is easily seen that MX (s) =
s
R. On the other hand, for the Cauchy distribution,
, for all
= 0.
∞
∈
In general... | https://ocw.mit.edu/courses/6-436j-fundamentals-of-probability-fall-2018/1a592ed184fb4c444547f67c9bcdd8ec_MIT6_436JF18_lec13.pdf |
The various powers esx indicate the possible
values of the random variable X, and the associated coefficients provide the
corresponding probabilities.
At the other extreme, if we are told that MX (s) =
for every s
is certainly not enough information to determine the distribution of X.
∞
= 0, this
On this subject... | https://ocw.mit.edu/courses/6-436j-fundamentals-of-probability-fall-2018/1a592ed184fb4c444547f67c9bcdd8ec_MIT6_436JF18_lec13.pdf |
Moment generating properties
There is a reason why MX is called a moment generating function. Let us con-
sider the derivatives of MX at zero. Assuming for a moment we can interchange
the order of integration and differentiation, we obtain
dMX (s)
|
ds
dmMX (s)
|
dsm
s=0
s=0
d
= E[e
ds
dm
dsm
=
|
sX ]
|
sX... | https://ocw.mit.edu/courses/6-436j-fundamentals-of-probability-fall-2018/1a592ed184fb4c444547f67c9bcdd8ec_MIT6_436JF18_lec13.pdf |
(d) Use the DCT to argue that
−
∞
E[X] = E lim
h#0
h
e
hX
−
h
1
i
= lim
h#0
E[e
hX ]
h
−
1
.
1.5 The probability generating function
For discrete random variables, the following probability generating function
is sometimes useful. It is defined by
gX (s) = E[s X ],
with s usually restricted to positi... | https://ocw.mit.edu/courses/6-436j-fundamentals-of-probability-fall-2018/1a592ed184fb4c444547f67c9bcdd8ec_MIT6_436JF18_lec13.pdf |
mpX (m) = E[X].
m1
X
4
1.6 Examples
d
Example : X = Exp(). Then,
MX (s) =
1
0
Z
sxe−x dx =
e
−s , s < ;
,
ˆ
∞
otherwise.
Example : X = Ge(p)
d
1
MX (s) =
sm
e
m=1
X
p(1
p)m−1
−
(
,
∞
In this case, we also find gX (s) = ps/(1
gX (s) =
, otherwise.
∞
se p
1−(1−p)es
s... | https://ocw.mit.edu/courses/6-436j-fundamentals-of-probability-fall-2018/1a592ed184fb4c444547f67c9bcdd8ec_MIT6_436JF18_lec13.pdf |
(c) Let X and Y be independent random variables. Let Z be equal to X, with
probability p, and equal to Y , with probability 1
−
p. Then,
MZ (s) = pMX (s) + (1
−
p)MY (s).
Proof: For part (a), we have
MX (aX + b) = E[exp(saX + sb)] = exp(sb)E[exp(saX)] = exp(sb)MX (as).
5
For part (b), we have
MX+Y (s) =... | https://ocw.mit.edu/courses/6-436j-fundamentals-of-probability-fall-2018/1a592ed184fb4c444547f67c9bcdd8ec_MIT6_436JF18_lec13.pdf |
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