text stringlengths 30 4k | source stringlengths 60 201 |
|---|---|
s 2(˙1
2
2 + ˙2
2) .
o
Using the inversion property of transforms, we conclude that X + Y = N (µ1 +
µ2, ˙1
2), thus corroborating a result we first obtained using convolutions.
2 + ˙2
d
2
SUM OF A RANDOM NUMBER OF INDEPENDENT RANDOM VARI-
ABLES
Let X1, X2, . . . be a sequence of i.i.d. random variables, with mean... | https://ocw.mit.edu/courses/6-436j-fundamentals-of-probability-fall-2018/1a592ed184fb4c444547f67c9bcdd8ec_MIT6_436JF18_lec13.pdf |
expression, and
evaluate it at s = 0, to recover the formula E[Y ] = E[N ]E[X].
Example : Suppose that each Xi is exponentially distributed, with parameter , and
that N is geometrically distributed, with parameter p
(0, 1). We find that
∈
MY (s) =
log MX (s)
e
p
elog MX (s)(1
−
1
−
=
p)
p/(
s)
−
p)... | https://ocw.mit.edu/courses/6-436j-fundamentals-of-probability-fall-2018/1a592ed184fb4c444547f67c9bcdd8ec_MIT6_436JF18_lec13.pdf |
contained in a multivariate transform, which we now define.
Consider n random variables X1, . . . , Xn related to the same experiment.
Let s1, . . . , sn be real parameters. The associated multivariate transform is a
function of these n parameters and is defined by
MX1,...,Xn (s1, . . . , sn) = E e
s1X1+···+snXn
.
... | https://ocw.mit.edu/courses/6-436j-fundamentals-of-probability-fall-2018/1a592ed184fb4c444547f67c9bcdd8ec_MIT6_436JF18_lec13.pdf |
random variables.
(b) If X and Y are independent, then MX,Y (s, t) = MX (s)MY (t).
8
MIT OpenCourseWare
https://ocw.mit.edu
6.436J / 15.085J Fundamentals of Probability
Fall 2018
For information about citing these materials or our Terms of Use, visit: https://ocw.mit.edu/terms | https://ocw.mit.edu/courses/6-436j-fundamentals-of-probability-fall-2018/1a592ed184fb4c444547f67c9bcdd8ec_MIT6_436JF18_lec13.pdf |
13.811
Advanced Structural
Dynamics and Acoustics
Acoustics
Lecture 5
13.811
ADVANCED STRUCTURAL DYNAMICS AND ACOUSTICS
Lecture 5
Ewald Sphere Construction
Baffled Piston
Directivity Function
f = ω/2π=kc/2π
Radiating Spectrum
Evanescent
Spectrum
13.811
ADVANCED STRUCTURAL DYNAMICS AND ACOUSTICS
Lecture 5
circ.m
%
%... | https://ocw.mit.edu/courses/2-067-advanced-structural-dynamics-and-acoustics-13-811-spring-2004/1a6a081bf8c514c5a812db5d121071b2_lect_5_4.pdf |
2)
nphi=361.
dphi=2*pi/(nphi-1)
nth=181;
dth=0.5*pi/(nth-0.5);
phi=[0:dphi:(nphi-1)*dphi]' * ones(1,nth);
th=([dth/2:dth:pi/2]'*ones(1,nphi))';
kx=ka*sin(th).*cos(phi);
ky=ka*sin(th).*sin(phi);
kr=ka*sin(th);
ss=rho*k*c*a^2*besselj(1,kr)./kr;
ss=dba(ss);
sm=max((max(ss))');
for i=1:size(ss,1)
for j=1:size(ss,2)
ss(i,j)... | https://ocw.mit.edu/courses/2-067-advanced-structural-dynamics-and-acoustics-13-811-spring-2004/1a6a081bf8c514c5a812db5d121071b2_lect_5_4.pdf |
a
y
x
Directivity Function
One Baffled
Piston
Array of Point Sources
13.811
ADVANCED STRUCTURAL DYNAMICS AND ACOUSTICS
Lecture 5
circ_arr.m
%
% MATLAB script for plotting the directivity function for
% an array of circular, baffled pistons
%
% Parameters:
Wavenumber
% k
% rho
Density
% c Speed of Sound
% a Radius of ... | https://ocw.mit.edu/courses/2-067-advanced-structural-dynamics-and-acoustics-13-811-spring-2004/1a6a081bf8c514c5a812db5d121071b2_lect_5_4.pdf |
=' num2str(d)
', ' num2str(nd) ]
b=title(tit);
set(b,'FontSize',20);nphi=361;
dphi=2*pi/(nphi-1);
phi=[0:dphi:2*pi];
xx=k*a*cos(phi);
yy=k*a*sin(phi);
hold on
b=plot(xx,yy,'b');
set(b,'LineWidth',3);
figure(2)
nphi=361.
dphi=2*pi/(nphi-1)
nth=181;
dth=0.5*pi/(nth-0.5);
phi=[0:dphi:(nphi-1)*dphi]' * ones(1,nth);
th=([d... | https://ocw.mit.edu/courses/2-067-advanced-structural-dynamics-and-acoustics-13-811-spring-2004/1a6a081bf8c514c5a812db5d121071b2_lect_5_4.pdf |
d
a
y
x
Directivity Function
13.811
ADVANCED STRUCTURAL DYNAMICS AND ACOUSTICS
Lecture 5
Radiated Power
Radiating Spectrum
k Real
z
Evanescent
Spectrum
k Imaginary
z
Radiated Power
Fourier Transforms
13.811
ADVANCED STRUCTURAL DYNAMICS AND ACOUSTICS
Lecture 5
Point-Driven Plate Radiation
Plate Bending Equation
z
... | https://ocw.mit.edu/courses/2-067-advanced-structural-dynamics-and-acoustics-13-811-spring-2004/1a6a081bf8c514c5a812db5d121071b2_lect_5_4.pdf |
k
f
Image removed due to copyright considerations.
Image removed due to copyright considerations.
See Figure 2.24 in [Williams].
See Figure 2.24 in [Williams].
Subsonic
Evanescent
Supersonic
Radiating
Coincidence Frequency
Water
13.811
ADVANCED STRUCTURAL DYNAMICS AND ACOUSTICS
Lecture 5 | https://ocw.mit.edu/courses/2-067-advanced-structural-dynamics-and-acoustics-13-811-spring-2004/1a6a081bf8c514c5a812db5d121071b2_lect_5_4.pdf |
6.087 Lecture 6 – January 19, 2010
Review
User defined datatype
Structures
Unions
Bitfields
Data structure
Memory allocation
Linked lists
Binary trees
1
Review: pointers
• Pointers: memory address of variables
•
’&’ (address of) operator.
• Declaring: int x=10; int ∗ px= &x;
• Dereferencing: ∗px=20;
•
Po... | https://ocw.mit.edu/courses/6-087-practical-programming-in-c-january-iap-2010/1aa34decd30f336826cd087d2f52035b_MIT6_087IAP10_lec06.pdf |
1 0 0 ] ;
i n t age ;
} ;
/ ∗ members o f d i f f e r e n t
t y p e ∗ /
4
Structure
• struct defines a new datatype.
• The name of the structure is optional.
struct {...} x,y,z;
•
The variables declared within a structure are called its
members
• Variables can be declared like any other built in data-type.
s... | https://ocw.mit.edu/courses/6-087-practical-programming-in-c-january-iap-2010/1aa34decd30f336826cd087d2f52035b_MIT6_087IAP10_lec06.pdf |
; / ∗ t o p
l e f t ∗ /
} ;
s t r u c t r e c t a n g l e r e c t ;
t l x = r e c t . t l . x ;
i n t
t l y = r e c t . t l . y ;
i n t
/ ∗ nested ∗ /
7
Structure pointers
• Structures are copied element wise.
• For large structures it is more efficient to pass pointers.
void foo(struct point ∗ pp); struct point... | https://ocw.mit.edu/courses/6-087-practical-programming-in-c-january-iap-2010/1aa34decd30f336826cd087d2f52035b_MIT6_087IAP10_lec06.pdf |
p [3]={0,1,10,20,30,12};
struct point p [3]={{0,1},{10,20},{30,12}};
9
Size of structures
• The size of a structure is greater than or equal to the sum
of the sizes of its members.
• Alignment
s t r u c t {
char c ;
/ ∗ padding ∗ /
i ;
i n t
• Why is this an important issue? libraries, precompiled files,
SIMD... | https://ocw.mit.edu/courses/6-087-practical-programming-in-c-january-iap-2010/1aa34decd30f336826cd087d2f52035b_MIT6_087IAP10_lec06.pdf |
set of adjacent bits within a single
’word’. Example:
f l a g {
s t r u c t
unsigned i n t
unsigned i n t has_sound : 1 ;
unsigned i n t
} ;
i s _ c o l o r : 1 ;
i s _ n t s c : 1 ;
• the number after the colons specifies the width in bits.
• each variables should be declared as unsigned int
Bit fields vs. masks ... | https://ocw.mit.edu/courses/6-087-practical-programming-in-c-january-iap-2010/1aa34decd30f336826cd087d2f52035b_MIT6_087IAP10_lec06.pdf |
records where each element contains a link to the
next record in the sequence.
• Linked lists can be singly linked, doubly linked or circular.
For now, we will focus on singly linked list.
• Every node has a payload and a link to the next node in
the list.
• The start (head) of the list is maintained in a separat... | https://ocw.mit.edu/courses/6-087-practical-programming-in-c-january-iap-2010/1aa34decd30f336826cd087d2f52035b_MIT6_087IAP10_lec06.pdf |
;
r e t u r n p ;
18
Linked list
Iterating:
f o r ( p=head ; p ! =NULL ; p=p−>n e x t )
/ ∗ do something ∗ /
f o r ( p=head ; p−>n e x t ! =NULL ; p=p−>n e x t )
/ ∗ do something ∗ /
19
Binary trees
• A binary tree is a dynamic data structure where each node
has at most two children. A binary search tree is a... | https://ocw.mit.edu/courses/6-087-practical-programming-in-c-january-iap-2010/1aa34decd30f336826cd087d2f52035b_MIT6_087IAP10_lec06.pdf |
node ∗ /
/ ∗ r e t u r n new r o o t ∗ /
}
/ ∗ r e c u r s i v e c a l l ∗ /
else
i f ( data < r o o t −>data )
r o o t −> l e f t =addnode ( r o o t −> l e f t , data )
else
r o o t −> r i g h t =addnode ( r o o t −> r i g h t , data )
}
22
For information about citing these materials or our Terms of Use, ... | https://ocw.mit.edu/courses/6-087-practical-programming-in-c-january-iap-2010/1aa34decd30f336826cd087d2f52035b_MIT6_087IAP10_lec06.pdf |
Lecture 8
MOSFET(I)
MOSFET I-V CHARACTERISTICS
Outline
1. MOSFET: cross-section, layout,
symbols
2. Qualitative operation
I-V characteristics
3.
Reading Assignment:
Howe and Sodini, Chapter 4, Sections 4.1-4.3
6.012 Spring 2009
Lecture 8
1
1. MOSFET: layout, cross-section, symbols
active area (thin
int... | https://ocw.mit.edu/courses/6-012-microelectronic-devices-and-circuits-spring-2009/1aadf127cb5e71ec1577aa244e2b3773_MIT6_012S09_lec08.pdf |
to
source
Gate-Source Voltage (VG& controls amount of
inversion charge that carries the current
Drain-Source Voltage (V'J :controls the electric
field that drifts the inversion charge fiom the source
to drain
Want to understand the relationship between the drain
current in the MOSFET as a function of gate-to-so... | https://ocw.mit.edu/courses/6-012-microelectronic-devices-and-circuits-spring-2009/1aadf127cb5e71ec1577aa244e2b3773_MIT6_012S09_lec08.pdf |
location y, V(y):
• If electric field is not too high:
vy (y) = −µµµµn • Ey (y) = µµµµn •
dV
dy
• For QN(y), use charge-control relation at location y:
]
[
QN (y) = −C ox VGS − V (y) − VT
for VGS – V(y) ≥ VT. .
Note that we assumed that VT is independent of y.
See discussion on body effect in Section 4.4 of t... | https://ocw.mit.edu/courses/6-012-microelectronic-devices-and-circuits-spring-2009/1aadf127cb5e71ec1577aa244e2b3773_MIT6_012S09_lec08.pdf |
µµµµnCox VGS −
VDS
2
• VDS
− VT
6.012 Spring 2009
Lecture 8
10
I-V Characteristics (Contd…)
ID =
W
L
• µµµµnCox
VGS −
VDS − VT
• VDS
2
for VDS < VGS − VT
Key dependencies:
• VDS↑ → ID↑ (higher lateral electric field)
• VGS↑ → ID↑ (higher electron concentration)
This is th... | https://ocw.mit.edu/courses/6-012-microelectronic-devices-and-circuits-spring-2009/1aadf127cb5e71ec1577aa244e2b3773_MIT6_012S09_lec08.pdf |
|
⇒ ID saturates when |QN| equals 0 at drain end.
Value of drain saturation current:
IDsat = I Dlin(VDS = VDSsat = VGS − VT )
Then
W
IDsat =
L
•µµµµnCox VGS −
VDS
2
−VT
•VDS
VDS =VGS−VT
IDsat
=
1 W
2 L
µµµµnCox[VGS − VT ]2
Will talk more about saturation region next tim... | https://ocw.mit.edu/courses/6-012-microelectronic-devices-and-circuits-spring-2009/1aadf127cb5e71ec1577aa244e2b3773_MIT6_012S09_lec08.pdf |
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/6-012-microelectronic-devices-and-circuits-spring-2009/1aadf127cb5e71ec1577aa244e2b3773_MIT6_012S09_lec08.pdf |
Lecture Six: More General Operators
1 The Weak definition of a harmonic function
It is sometimes useful to weaken the notion of a harmonic function somewhat. One way
of doing this is with the notion of a weakly harmonic function. Let u be a differentiable
function on some set Ω. We say that u is weakly harmonic on Ω ... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2005/1ac3c08d20609c08ab021ef33149291f_lecture6.pdf |
symmetric n by n matrix with entries aij (not necessarily constant). An
operator written like this is said to be in divergence form since Lu = div(A�u). Note
that if A is the identity matrix then L is simply the laplacian. We will often be interested
in functions satisfying Lu = 0. Such functions are called Lharmon... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2005/1ac3c08d20609c08ab021ef33149291f_lecture6.pdf |
0 on B2r then
| |
�
Br
2 u
|� | ≤
4Λ2
λ2r
2
�
B2r \Br
2
u .
Proof Again we start off by introducing a test function φ with φ ≥ 0 and φ = 0 on the
boundary of B2r . Calculate
�
B2r
�
0 ≤
≤
φ2uLu
φ2 u(� · A�
u)
B2r
�
≤ −
B2r
�
≤ −2
< �(φ2 u), A�u >
φu < �φ, A�u > −
�
B2r
B2r
2
φ2 < �u, A�u >
by S... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2005/1ac3c08d20609c08ab021ef33149291f_lecture6.pdf |
φ >
�1/2 ��
B2r
(8)
�1/2
< φ�u, φA�u >
(9)
.
�
λ
B2 r
�
λ
B2r
Applying uniform ellipticity and rearranging gives
φ2 < �u, �u > ≤ 2Λ
��
2
u < �φ, �φ >
�1/2 ��
Divide and square to get
B2r
B2r
�
λ2
B2 r
φ2
|�u|
2 ≤ 4Λ2
�
B2r
2u
|�φ|2
.
�
1
2r−|x|
r
if |x| ≤ r;
if r < x ≤ 2r
.
φ(x) =
Then
�
2
|... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2005/1ac3c08d20609c08ab021ef33149291f_lecture6.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 3
LECTURES: ABHINAV KUMAR
1. Birational maps continued
Recall that the blowup ... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/1ac4b82daa4485674ad67c21b5a49e7a_lect3.pdf |
1, 0) = 0 and each fk a homogeneous polynomial of degree k. In a
˜
U P1, we have local coordinates x
neighborhood of (p, = [1 : 0]) U
and t = x and π∗f = f (x, tx) = xmfm(1, t) + xm+1fm+1(1, t) +
), giving the
�
desired formula.
⊂ ×
· · ·
∞
∈
y
˜
˜
·
·
→
π∗ : Pic X →
= Pic X⊕Z.
Pic X and Z
Theorem 1. We ... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/1ac4b82daa4485674ad67c21b5a49e7a_lect3.pdf |
X
·
1
2
LECTURES: ABHINAV KUMAR
·
·
·
as desired. Moreover, since C (possibly after moving) does not pass through
p, (π∗C) E = 0. Next, taking a curve passing through p with multiplicity
1, its strict transform meets E transversely at one point which corresponds to
the tangent direction of p ∈ C, i.e. C˜ E =... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/1ac4b82daa4485674ad67c21b5a49e7a_lect3.pdf |
the same cohomology.
Proof. π is an isomorphism away from E, π : X˜ � E → X � {p
}, so it is clear
i
that OX → π∗O ˜ is an isomorphism except possibly at p, and R π∗OX˜ can only
be supported at p. By the theorem on formal functions, the completion at p
of this sheaf is R� = lim H i(E ,
iπ∗OX˜
n OEn
is the closed ... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/1ac4b82daa4485674ad67c21b5a49e7a_lect3.pdf |
→ OEn → 0 with I/I = OE (1) = ⇒
), where E
n
I n/I n+1
∼
2
This implies that the irregularity qX = h1(X, OX ) = q ˜ and geometric genus
X
pg(X) = h2(X, OX ) = pg(X˜ ) are invariant under blowup.
Let X, Y be varieties, X irreducible.
2. Rational maps
Definition 1. A rational map X ��� Y is a morphism φ from an ... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/1ac4b82daa4485674ad67c21b5a49e7a_lect3.pdf |
rational map is
defined on all but finitely many points F (those lying on the set of zeroes and
poles of φ). If C is an irreducible curve on X, φ is defined on C �(C ∩F ), and we
can set φ(C) = φ(C � (C ∩ F )) (and similarly φ(X) = φ(X � F )). Restriction
gives us an isomorphism between Pic (X) and Pic (X � F ), so we... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/1ac4b82daa4485674ad67c21b5a49e7a_lect3.pdf |
∈ P has no fixed
part. A point p of X is a base point of P if every divisor of P contains p. If p
has no fixed part, then it has finitely many base points (at most (D2)).
4. Properties of Birational Maps between Surfaces
(1) Elimination of indeterminacy
(2) Universal property of blowing up
(3) Factoring birational m... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/1ac4b82daa4485674ad67c21b5a49e7a_lect3.pdf |
→
◦
4
LECTURES: ABHINAV KUMAR
curves in P passing through p). Then P1 ⊂ |π∗D − kE| obtained by subtracting
kE from elements of π∗P has no fixed component, and defines a rational map
φ1 : X1 ��� Pm which coincides with φ π. If φ1 is a morphism, we are done;
◦
∗ Dn−1 −
otherwise, repeat the process. We obtain a seque... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/1ac4b82daa4485674ad67c21b5a49e7a_lect3.pdf |
empty, so there is an
f −1 : S ��� An is given by rational functions
embedding j : S
g1, g2, . . . , gn and at least one of them is undefined at p, say g1 ∈/ OS�,p. Let
g1 = u
S ,p are coprime and v(p) = 0. Let D be defined on S by
f ∗v = 0. On S we have f ∗u = (f ∗v)x1 (where x1 is the first coordinate function
on ... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/1ac4b82daa4485674ad67c21b5a49e7a_lect3.pdf |
MIT 6.972 Algebraic techniques and semidefinite optimization
March 2, 2006
Lecturer: Pablo A. Parrilo
Scribe: ???
Lecture 7
In this lecture we introduce a special class of multivariate polynomials, called hyperbolic. These
polynomials were originally studied in the context of partial differential equations. As we w... | https://ocw.mit.edu/courses/6-972-algebraic-techniques-and-semidefinite-optimization-spring-2006/1ae3402ea8aa2ab1edf9c607eeb0ea07_lecture_07.pdf |
with respect to e if
p(e) > 0 and, for all vectors x ∈ Rn, the univariate polynomial t �→ p(x − te) has only real roots.
A natural geometric interpretation is the following: consider the hypersurface in Rn given by p(x) = 0.
Then, hyperbolicity is equivalent to the condition that every line in Rn parallel to e inter... | https://ocw.mit.edu/courses/6-972-algebraic-techniques-and-semidefinite-optimization-spring-2006/1ae3402ea8aa2ab1edf9c607eeb0ea07_lecture_07.pdf |
shortly, it turns out that these cones are actually convex cones. We prove this following the arguments
in Renegar [Ren]; the original results are due to G˚
It is immediate from homogeneity and the definition above that λ > 0, x ∈ Λ++
arding [G˚
ar59].
Lemma 3. The hyperbolicity cone Λ++ is the connected component of ... | https://ocw.mit.edu/courses/6-972-algebraic-techniques-and-semidefinite-optimization-spring-2006/1ae3402ea8aa2ab1edf9c607eeb0ea07_lecture_07.pdf |
ie + βv + γx), where i is the imaginary unit.
We claim that if γ ≥ 0, this polynomial has only roots in the lower halfplane. Let’s look at the γ = 0
case first. It is clear that β �→ p(αie + βv) cannot have a root at β = 0, since p(αie) = (αi)dp(e) = 0. If
β = 0, we can write
p(αie + βv) = 0
⇔
p(αβ−1ie + v) = 0
⇒... | https://ocw.mit.edu/courses/6-972-algebraic-techniques-and-semidefinite-optimization-spring-2006/1ae3402ea8aa2ab1edf9c607eeb0ea07_lecture_07.pdf |
all x, or equivalently, p is hyperbolic in the direction v.
The following result shows that this set is actually convex:
Theorem 6 ([G˚ar59]). The hyperbolicity cone Λ++ is convex.
Proof. We want to show that u, v ∈ Λ++, β, γ > 0 implies that βu + γv ∈ Λ++. The previous result
implies that we can always assume v = ... | https://ocw.mit.edu/courses/6-972-algebraic-techniques-and-semidefinite-optimization-spring-2006/1ae3402ea8aa2ab1edf9c607eeb0ea07_lecture_07.pdf |
�
�
2
,
xk
n
�
�
2
4xn+1 − 4 xn+1 −
2
n
�
�
2
xk = 4
n
�
2
xk ,
which is always nonnegative, so the polynomial t �→ p(x − te) has only real roots. The corresponding
hyperbolicity cone is the Lorentz or second order cone given by
k=1
k=1
�
Λ+ =
x ∈ Rn+1
| xn+1 ≥ 0,
n
�
2
xk ≤ xn+1
2
�
.
Example 8 ... | https://ocw.mit.edu/courses/6-972-algebraic-techniques-and-semidefinite-optimization-spring-2006/1ae3402ea8aa2ab1edf9c607eeb0ea07_lecture_07.pdf |
++.
Λ++ = {x ∈ Rn
| x1A1 + · · ·
+ xnAn � 0}.
Based on the results discussed earlier regarding the number of real roots of a univariate polynomial,
we have the following lemma.
Lemma 9. The polynomial p(x) is hyperbolic with respect to e if and only if the Hermite matrix H1(p) ∈
S
n[x] is positive semidefinite for a... | https://ocw.mit.edu/courses/6-972-algebraic-techniques-and-semidefinite-optimization-spring-2006/1ae3402ea8aa2ab1edf9c607eeb0ea07_lecture_07.pdf |
: Rn
→
1
|
2
2
f ���(x) 3
| .
73
One of the main open issues regarding hyperbolic cones is about their generality. As Example 8 shows,
the cone associated with a semidefinite program is a hyperbolic cone. An open question (known as the
generalized Lax conjecture) is whether the converse holds, more specifically,... | https://ocw.mit.edu/courses/6-972-algebraic-techniques-and-semidefinite-optimization-spring-2006/1ae3402ea8aa2ab1edf9c607eeb0ea07_lecture_07.pdf |
. An inequality for hyperbolic polynomials. J. Math. Mech., 8:957–965, 1959.
[G¨ul97] O. G¨uler. Hyperbolic polynomials and interior point methods for convex programming. Math.
Oper. Res., 22(2):350–377, 1997.
[NN94] Y. E. Nesterov and A. Nemirovski. Interior point polynomial methods in convex programming,
volume 1... | https://ocw.mit.edu/courses/6-972-algebraic-techniques-and-semidefinite-optimization-spring-2006/1ae3402ea8aa2ab1edf9c607eeb0ea07_lecture_07.pdf |
Massachusetts Institute of Technology
Department of Electrical Engineering and Computer Science
6.341: Discrete-Time Signal Processing
OpenCourseWare 2006
Lecture 5
Sampling Rate Conversion
Reading: Section 4.6 in Oppenheim, Schafer & Buck (OSB).
It is often necessary to change the sampling rate of a discrete-tim... | https://ocw.mit.edu/courses/6-341-discrete-time-signal-processing-fall-2005/1b110cd38e3cbd3efd56f43c9ddf8b2a_lec05.pdf |
with
r = i + kM
− ∞ < k < ∞, 0 ≤ i ≤ M − 1
= ⇒ Xd(ejω) =
�
1
T
1
M
M −1
�
i=0
� �
Xc j
−
2πk
T
−
���
2πi
M T
ω
M T
∞
�
k=−∞
= X(e
j(ω−2πi)/M )
= Xd(ejω) =
⇒
M −1
1 �
M
i=0
X(ej(ω−2πi)/M ) .
As an example, the following figure illustrates decimation by M = 2 in the time domain. We
see that re... | https://ocw.mit.edu/courses/6-341-discrete-time-signal-processing-fall-2005/1b110cd38e3cbd3efd56f43c9ddf8b2a_lec05.pdf |
ing an anti-aliasing filter with a decimator gives a downsampler. OSB Figure 4.22 illustrates
downsampling with and without aliasing.
M
2π
1
2
Sampling Rate Expansion by an Integer Factor
A typical system for increasing the sampling rate of a discrete sequence by an integer factor is
illustrated in OSB Figure 4.2... | https://ocw.mit.edu/courses/6-341-discrete-time-signal-processing-fall-2005/1b110cd38e3cbd3efd56f43c9ddf8b2a_lec05.pdf |
they are placed, as long as the rates M and L are
mutually prime.
L
3 | https://ocw.mit.edu/courses/6-341-discrete-time-signal-processing-fall-2005/1b110cd38e3cbd3efd56f43c9ddf8b2a_lec05.pdf |
Topic 12 Notes
Jeremy Orloff
12 Laplace transform
12.1 Introduction
The Laplace transform takes a function of time and transforms it to a function of a complex variable
. Because the transform is invertible, no information is lost and it is reasonable to think of a function
() and its Laplace transform () as two v... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1b1cbc6540b338b58b7bc1b1cc1f280b_MIT18_04S18_topic12.pdf |
. Consider the constant coefficient differential equation
3 ′′ + 8 ′ + 7 = ()
This equation models a damped harmonic oscillator, say a mass on a spring with a damper, where
() is the force on the mass and () is its displacement from equilibrium. If we consider to be
the input and the output, then this is a linear tim... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1b1cbc6540b338b58b7bc1b1cc1f280b_MIT18_04S18_topic12.pdf |
ee
− = ∫
0
∞
e(−) =
(cid:243)
e(−) ∞
(cid:243)
(cid:243)
− (cid:243)
0
=
1
−
divergent otherwise
if Re() > Re()
The last formula comes from plugging ∞ into the exponential. This is 0 if Re( − ) < 0 and
undefined otherwise.
Example 12.4. Let () = . Compute () = ( ; ) directly. Give the region in the compl... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1b1cbc6540b338b58b7bc1b1cc1f280b_MIT18_04S18_topic12.pdf |
) =
1∕( − ) + 1∕( + )
2
=
.
2 + 2
12.3.2 Connection to Fourier transform
The Laplace and Fourier transforms are intimately connected. In fact, the Laplace transform is often
called the Fourier-Laplace transform. To see the connection we’ll start with the Fourier transform
of a function ().
̂() = ∫
If we assum... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1b1cbc6540b338b58b7bc1b1cc1f280b_MIT18_04S18_topic12.pdf |
4
The key here is that Re() > implies Re( − ) < 0. So, we can write
∞
∞
(cid:240)e(−)(cid:240) = ∫
The last integral clearly converges when Re( − ) < 0. QED
(cid:240) ()e−(cid:240) ≤
∫
0
∫
0
0
∞
eRe(−)
Example 12.7. Here is a list of some functions of exponential type.
(cid:240)
< 2eRe()
() = e ∶
() = 1 ... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1b1cbc6540b338b58b7bc1b1cc1f280b_MIT18_04S18_topic12.pdf |
if and are constants and and are
functions then
( + ) = ( ) + ().
(3)
(The proof is trivial –integration is linear.)
Property 2. A key property of the Laplace transform is that, with some technical details,
Laplace transform transforms derivatives in to multiplication by (plus some details).
This is proved in... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1b1cbc6540b338b58b7bc1b1cc1f280b_MIT18_04S18_topic12.pdf |
consider functions that are discontinuous at the
origin or if we want to allow () to be a generalized function like (). In these cases (0) is not
defined, so our formulas are undefined. The technical fix is to replace 0 by 0− in the definition and
all of the formulas for Laplace transform. You can learn more about this b... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1b1cbc6540b338b58b7bc1b1cc1f280b_MIT18_04S18_topic12.pdf |
(; ) = ( ; ) = (−1) () =
2
3
!
+1
Property 4. -shift rule. As usual, assume () = 0 for < 0. Suppose > 0. Then,
( ( − ); ) = e− ()
(10)
12 LAPLACE TRANSFORM
6
Proof. We go back to the definition of the Laplace transform and make the change of variables
= − .
( ( − ); ) = ∫
0
∞
∞
= ∫
0
( − )e− = ∫
... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1b1cbc6540b338b58b7bc1b1cc1f280b_MIT18_04S18_topic12.pdf |
1
− 1
Use partial fractions to write
+ 1
The coverup method gives = 1∕3, = −1∕2, = 1∕6.
− 1
=
− 2
+
+
+
1
− 1
.
We recognize
as the Laplace transform of e, so
1
−
1
e2 −
3
Example 12.10. Solve ′′ − = 1, (0) = 0, ′(0) = 0.
() = e2 + e + e− + e =
1
2
e +
1
e− + e.
6
Solution: The rest (zero) i... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1b1cbc6540b338b58b7bc1b1cc1f280b_MIT18_04S18_topic12.pdf |
() we have
=
.
We read as ‘ applied to .’
Example 12.11. If () = 3 + 2 then = 32, 2 = 6.
2. If () is a polynomial then () is called a polynomial differential operator.
Example 12.12. Suppose () = 2 + 8 + 7. What is ()? Compute () applied to () =
3 + 2 + 5. Compute () applied to () = e2.
Solution: () = 2 + 8 + 7... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1b1cbc6540b338b58b7bc1b1cc1f280b_MIT18_04S18_topic12.pdf |
ACE TRANSFORM
8
Let’s continue to work from this specific example. From it we’ll be able to remind you of the general
approach to solving constant coefficient differential equations.
Example 12.14. Suppose () = 2+8+7. Find the exponential modes of the equation () = 0.
. Using the substititution rule
Solution: The expo... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1b1cbc6540b338b58b7bc1b1cc1f280b_MIT18_04S18_topic12.pdf |
− + e−7
.
That is, () is a linear combination of the exponential modes.
You should notice that the denominator in the expression for () is none other than the characteristic
polynomial ().
12.7.2 The system function
Example 12.16. With the same () as in Example 12.12 solve the inhomogeneous DE with rest
initial ... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1b1cbc6540b338b58b7bc1b1cc1f280b_MIT18_04S18_topic12.pdf |
=
⋅ ()
()
()
Using the formulation
output = system function × input,
we see that the system function is
()
.
() = ()
Note that when () = 0 the differential equation becomes () = 0. If we make the assumption
that the ()∕ () is in reduced form, i.e. and have no common zeros, then the modes of the
system (whic... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1b1cbc6540b338b58b7bc1b1cc1f280b_MIT18_04S18_topic12.pdf |
. Then,
since the integrals won’t notice the difference at one point, () = () = 1∕( − ). In this sense
it is impossible to define −1( ) uniquely.
The good news is that the inverse exists as long as we consider two functions that only differ on a
negligible set of points the same. In particular, we can make the followi... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1b1cbc6540b338b58b7bc1b1cc1f280b_MIT18_04S18_topic12.pdf |
−∞
e
+ −
.
12 LAPLACE TRANSFORM
11
The (conditional) convergence of this integral follows using exactly the same argument as in the
example near the end of Topic 9 on the Fourier inversion formula for () =
. That is, the
integrand is a decaying oscillation, around 0, so its integral is also a decaying oscill... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1b1cbc6540b338b58b7bc1b1cc1f280b_MIT18_04S18_topic12.pdf |
243)
(cid:243)
e(+)
(cid:243)
(cid:243)
− (cid:243) + − (cid:243)
= ∫
e
=
e − e−
.
≤
∫
−
Since and are fixed, it’s clear this goes to 0 as goes to infinity.
The bottom
4 is handled in exactly the same manner as the top
3 is parametrized by = () = − + , with − ≤ ≤ . So,
3:
(cid:243)
(cid:243)
(cid:243)
(ci... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1b1cbc6540b338b58b7bc1b1cc1f280b_MIT18_04S18_topic12.pdf |
0. Laplace inversion 1. Assume is continuous and of exponential type . Then for
> we have
()e .
(15)
As usual, this formula holds for > 0.
() =
+∞
1
2 ∫−∞
Proof. The proof uses the Fourier inversion formula. We will just accept this theorem for now.
Example 12.18 above illustrates the theorem.
Theorem 12.... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1b1cbc6540b338b58b7bc1b1cc1f280b_MIT18_04S18_topic12.pdf |
=
1 + e−
Note even if you start with a rational function the system function of the closed loop with delay is
not rational. Usually it has an infinite number of poles.
Example 12.22. Suppose () = 1, = 1 and = 1 find the poles of ().
Solution:
() =
1
.
1 + e−
So the poles occur where e− = −1, i.e. at , where is a... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1b1cbc6540b338b58b7bc1b1cc1f280b_MIT18_04S18_topic12.pdf |
2 −2 − 3 −3 + …
e
e
() = () () = () − e
− () + 2e−2 () − 3e−3 () + …
Using the shift formula Equation 10, we have
() = () − ( − ) + 2 ( − 2) − 3 ( − 3) + …
(This is not really an infinite series because () = 0 for < 0.) If the input is bounded and < 1
then even for large the series is bounded. So bounded input p... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1b1cbc6540b338b58b7bc1b1cc1f280b_MIT18_04S18_topic12.pdf |
2
(sin() + cos())
1
2
e
1ø
Transform
1∕
1∕( − )
1∕2
!∕+1
∕(2 + 2)
∕(2 + 2)
( − )∕(( − )2 + 2)
∕(( − )2 + 2)
1
−
e
∕(2 − 2)
∕(2 − 2)
1
(2 + 2)2
(2 + 2)2
2
(2 + 2)2
!∕( − )+1
1ø
Γ( + 1)
+1
Region of convergence
Re() > 0
Re() > Re()
Re() > 0
Re() > 0
Re() > 0
Re() > 0
Re() > Re()
Re() ... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1b1cbc6540b338b58b7bc1b1cc1f280b_MIT18_04S18_topic12.pdf |
MATH 18.152 COURSE NOTES - CLASS MEETING # 7
18.152 Introduction to PDEs, Fall 2011
Professor: Jared Speck
Class Meeting # 7: The Fundamental Solution and Green Functions
1. The Fundamental Solution for ∆ in Rn
Here is a situation that often arises in physics. We are given a function f x on Rn representing
( )
the spat... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/1b2023e54a5d0c4023418a27274787ef_MIT18_152F11_lec_07.pdf |
to the operator ∆ is
(1.0.3)
def
Φ x
( )
1
2π ln ∣x∣
2,
1
ωn∣x∣n−2 n 3,
i=1(xi)2 and ωn is the surface area of a unit ball in Rn (e.g. ω3
= { −
n =
≥
n
√
def
=
where as usual ∣x∣
Remark 1.0.1. Some people prefer to define their Φ to be the negative of our Φ.
∑
4π).
=
that this holds for now. We then claim that the solut... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/1b2023e54a5d0c4023418a27274787ef_MIT18_152F11_lec_07.pdf |
ready to state and prove a rigorous
using the fact that ∆ ∂2
r
r Φ +
r
4
def
n = 3. Note that Φ(x) = Φ(r) (r = ∣x∣) is spherically symmetric.
0 for spherically symmetric functions, we have that
(cid:3)
>
version of the aforementioned heuristic results.
Theorem 1.1 (Solution to Poisson’s equation in Rn
∞
C0
). Let f x
o... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/1b2023e54a5d0c4023418a27274787ef_MIT18_152F11_lec_07.pdf |
n
Remark 1.0.2. As we alluded to above, Theorem 1.1 shows that ∆Φ x
“delta distribution.” For on the one hand, as we have previously discussed, we have that f
On the other hand, our proof of Theorem 1.1 below will show that f ∆
(
Thus, for any f, we have δ
δ x , where δ is the
(
δ f.
= ∗
∆Φ f.
) ∗
f ∆Φ f, and so ∆Φ δ.
... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/1b2023e54a5d0c4023418a27274787ef_MIT18_152F11_lec_07.pdf |
0 y
∣
We first show that I goes to 0 as (cid:15) →
∆xu(x) = −
∆
yf (x − y) d3
y
−
1
∫
4π Bc 0
(cid:15)
( ) ∣
1
y
∣
+
0
. To this end, let
∆
yf (x − y) d y = I
+ I.
I
3
def
(1.0.8)
def
M = sup
y R3
∈
∣f (y)∣ + ∣∇f (y)∣ + ∣∆yf (y)∣.
Then using spherical coordinates r, ω for the y variable, and recalling that d3y
ω ∈ ∂B
th... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/1b2023e54a5d0c4023418a27274787ef_MIT18_152F11_lec_07.pdf |
x
show this, we can combine
e
w
)
(
f x as desired.
To show (1.0.10), we will use integration by parts via Green’s identity and simple
estimates to
control the boundary terms. Recall that Green’s identity for two functions u, v is
(1.0.7),
(1.0.9), and (1.0.10) and let (cid:15)
+
0 to deduce
→
(1.0.11)
( ) −
Ω
∫ v x ∆u... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/1b2023e54a5d0c4023418a27274787ef_MIT18_152F11_lec_07.pdf |
dσ (cid:15)2dω on ∂Bc
c
(cid:15)
=
1
∂B
(cid:15)
0 . This corresponds to the
( )
(cid:15) 0 ,
)
(
on the right-hand side of
,
0
( )
Bc
in the standard
form
ulation of Green’s identity for
(1.0.13)
− ∫
1
(cid:15) 0 y
Bc
( ) ∣
∣
∆yf (x − y) d3y = − ∫
( )
∂B1 0
(cid:15)ω ⋅ (∇f )(x − (cid:15)ω) dω + ∫
( )
∂B1 0
f (x (cid:1... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/1b2023e54a5d0c4023418a27274787ef_MIT18_152F11_lec_07.pdf |
that
∣ >
Using this inequalit
and
R.
2
∣ >
∣ ≤
∣
(1.0.14)
∣u(x)∣ =
1
4π
∣ ∫
1
f (
y d3y
)
∣ ≤
M
∣ ∫ ( )
2π x BR 0
∣
1 d3y
=
2R3M
3 x
∣
∣
,
and we have shown (1.0.5) in the case n
BR 0 x y
( ) ∣ − ∣
= 3
.
4
MATH 18.152 COURSE NOTES - CLASS MEETING # 7
To prove uniqueness, we will make use of Corollary 4.0.4,
are two so... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/1b2023e54a5d0c4023418a27274787ef_MIT18_152F11_lec_07.pdf |
Ω)
Proof. This proof is a bit beyond this course.
). Let g be
the PDE (2.0.16) has a unique solution u
theorem
Then
.
a bounde
C 2
Ω ∩
( )
d Lipschitz domain, and let
C(Ω).
∈
g
∈
(cid:3)
Definition 2.0.2. Let Ω ⊂
x, y Ω Ω verifying the following conditions for each fixed x Ω
(
Rn be a domain. A Green function in Ω is defi... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/1b2023e54a5d0c4023418a27274787ef_MIT18_152F11_lec_07.pdf |
)
)
Therefore, Φ x y
( − ) −
Furthermore, using (2.0.21), w
) − (
σ
φ
( − ) − (
φ x, y also verifies the boundary condition (2.0.18).
φ x, y verifies equation (2.0.17).
ve
e ha
that
Φ x
y
)
)
(
x, σ
) =
0
whenev
er σ ∈
∂Ω. Thus, Φ x
(
−
(cid:3)
MATH 18.152 COURSE NOTES - CLASS MEETING # 7
5
The following technical propo... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/1b2023e54a5d0c4023418a27274787ef_MIT18_152F11_lec_07.pdf |
„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„‚„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„¶
σ
‚„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„
( )∇
+ ∫
u σ
∂Ω
∂Ω
·„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„
single layer potential
double layer potential
Proof. We’ll do the proof for... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/1b2023e54a5d0c4023418a27274787ef_MIT18_152F11_lec_07.pdf |
( ) − ( )∇ (
u σ
ˆ
N
1
∣x − σ∣
dσ
)
1
∂Ω x − σ∣
= ∫ ∣
∇ ˆN u(σ) dσ
− ∫ u σ
∂Ω
( ∇ (
)
ˆN
1
x σ
∣ −
∣
) dσ
1
∣x − σ
In the last two integrals above, N σ denotes the radially outward unit
normal to the oundary of
b
( )
the ball B(cid:15) x . This corresponds to the “opposite” choice of normal that appears in the standard... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/1b2023e54a5d0c4023418a27274787ef_MIT18_152F11_lec_07.pdf |
.26)
1
Ω x − y∣
∣
∣∫
∆u y d3y
( )
1
− ∫Ω(cid:15) ∣x − y∣
∆u y d y
( )
3
∣ ≤ ∫B(cid:15)(x) ∣x
M
≤
∫ ( )
B(cid:15) x
∣
1
− y
∣
1
x y
−
∣
∆u y d3y
)∣
(
d
3y
→
0 as (cid:15) 0.
↓
∣
This shows that L converges to
Ω x y ∆u y d3y as (cid:15) 0.
∫ ∣ − ∣
The limits for R1 and R2 are obvious since these terms do not dep
We now a... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/1b2023e54a5d0c4023418a27274787ef_MIT18_152F11_lec_07.pdf |
φ
, π
0
× [
) ×
)
∞
∫θ∈[0,π] 1
π]
∈[
0,2
0, 2π
)
[
dφ
dθ
=
centered at
4π. We now
∫
∂
x, we ha
estimate
(2.0.28)
1
∣ R4 − [ − u(x)]∣ = ∣u(x) +
4π
1
π
4
∫ ( )
B(cid:15) x
∂
u σ
( )∇ ˆ
1
4π
1
=
∣
∫∂ (cid:15)( )
B
x
u
( ( ) − ( ))
(
u
σ
x
≤ 4π ∫∂B(cid:15)(x)
u x
∣ ( ) − ( )∣(
u σ
1
) dσ
∣
N (σ)( x σ
− ∣
∣
1
∣x − σ
1
∣x − ... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/1b2023e54a5d0c4023418a27274787ef_MIT18_152F11_lec_07.pdf |
−
Proof. Applying Proposition 2.0.3, we have that
∫ g
∂
Ω
x,
(σ) ∇ ˆN G(
·
σ
)
¶
„„„„„„„„„„„„„„„„„„„„„„„‚„„„„„„„„„„„„„„„„„„„„„„„
kernel
Poisson
(2.0.30) u(x) = − ∫ Φ
Recall also that
Ω
(x − y)f y
( ) dny + ∫ Φ
∂Ω
(x − σ)∇ ˆN (σ)u(σ) dσ − ∫ g
∂Ω
dσ.
(σ)∇ ˆN (σ)Φ(x − σ) dσ.
(2.0.31)
(2.0.32)
Applying the Green
G(x, y) = ... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/1b2023e54a5d0c4023418a27274787ef_MIT18_152F11_lec_07.pdf |
x, σ dσ.
∇ (
)
∫
∂Ω
(2.0.31),
MATH 18.152 COURSE NOTES - CLASS MEETING # 7
7
(cid:3)
3. Poisson’s Formula
Let’s compute the
def
=
Green
in the case that Ω BR 0
( )
called the method of images that works for special domains.
G x, y and Poisson kernel P x, σ
)
the
(
)
of radius R centered at
(
ball
function
R3 is a
⊂
de... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/1b2023e54a5d0c4023418a27274787ef_MIT18_152F11_lec_07.pdf |
x∗
2
− y
∣
= q2
∣x − y∣
2.
(3.0.37)
2
∣x∣
− 2x ⋅ y + R2
2
∣
= q2
∣x∗
2
− y∣
= q2
(∣x∗
∣
2
− 2x∗
⋅ y + R2
).
Then
performing simple algebra, we ha
= ∣x y
−
ve
(3.0.38)
2
∣x∗
∣
+ R2
− q2
(R2
+ ∣x∣
2
) = 2y ⋅ (x∗
− q2x).
Now since the left-hand side of (3.0.38) does not
term on the right-hand side vanishes. This implies t... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/1b2023e54a5d0c4023418a27274787ef_MIT18_152F11_lec_07.pdf |
3.0.42) to
→
G(x, y
) =
G 0, y
(
) =
(3.0.44)
(3.0.45)
(3.0.46)
1
− 4π
R
R2
x∣2 x − y∣
∣
,
∣x
∣∣
x ≠ 0,
1
4π x y∣
∣
−
1
1
R
4π
−
y
∣
∣
.
yG x, y
∇ (
) =
x
y
−
x y
4π
∣ −
∗
1 R x
y
∣3 − 4π ∣x∣ ∣x∗ − y∣3
−
Now when σ ∂B 0 , (3.0.36) and (3.0.40)
imply
that
∈
R
( )
(3.0.47)
∣
Therefore, using (3.0.46) and (3.0.47), w
x∗ σ... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/1b2023e54a5d0c4023418a27274787ef_MIT18_152F11_lec_07.pdf |
σ
∣ − ∣
(3.0.49)
∇ ˆN (σ)G(
x, σ
)
def
= ∇
ˆ
σG x, σ N
)
⋅
(
Remark 3.0.3. If the ball were cen
(3.0.49)
would be replaced with
tered
at
the p
t
oin
p ∈ R
R
σ) =
(
1
2
− ∣x∣2
4πR ∣x
of
3 instead
.
origin, then the formula
σ 3
∣
−
the
(3.0.50)
ˆ G x, σ
(
N (σ)
∇
)
def
= ∇ (
σG x, σ N σ
) ⋅
ˆ
( ) =
Theorem
3.1 (Poisson’s... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/1b2023e54a5d0c4023418a27274787ef_MIT18_152F11_lec_07.pdf |
)
∣
gets
Remark 3.0.4. In n dimensions, the
for
m
ula
(3.0.
52)
with
(3.0.53)
u(x) =
R2 − ∣x − p∣2
ωnR
∫
where as usual, ω is the surface area of the unit ball in
n
f σ
)
(
− σ
∣
dσ,
n
∂BR p x
( )
∣
Rn.
Proof. The identity (3.0.52) follows immediately from Theorem 2.2 and (3.0.50).
4. Harnack’s inequality
Theorem 4.1 (... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/1b2023e54a5d0c4023418a27274787ef_MIT18_152F11_lec_07.pdf |
)
∣ − R ≤ ∣x − σ∣ ≤ ∣x∣ + R.
f, we deduce that
∣ =
σ
∣
∈
(4.0.56)
No
w recall that y the
b
(x) R2
u
≤
prop
1
R x
∣
+ ∣
x∣2 4πR ∂BR 0
( )
∣
−
e that
, we hav
y
ert
∫
mean value
f σ dσ.
( )
(4.0.57)
u(0) =
1
4πR2 ∫
∂BR
)
(0
f σ dσ.
( )
Th
us, combining (4.0.56) and (4.0.57), we have that
(4.0.58)
u x
( ) ≤
Rn−2(R x
∣
+
n... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/1b2023e54a5d0c4023418a27274787ef_MIT18_152F11_lec_07.pdf |
0.59)
Allowing R
→ ∞
(and therefore u is too).
Rn−2(R − ∣x∣)
R
( + ∣
x n− u 0
∣)
in (4.0.59), we conclude that v x
u
( ) ≤ ( ) ≤
x
(R +
∣)
∣
−1
n
R − ∣x∣)
(
v 0 . Th
( ) = ( )
R
x
−2
n
1
u
0
.
( )
us, v is a constant-valued function
(
v x , and we argue as ab
o handle the case u x M, we simply consider the function w x... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/1b2023e54a5d0c4023418a27274787ef_MIT18_152F11_lec_07.pdf |
18.405J/6.841J: Advanced Complexity Theory
Spring 2016
Prof. Dana Moshkovitz
(cid:47)(cid:72)(cid:70)(cid:87)(cid:88)(cid:85)(cid:72)(cid:3)(cid:20)(cid:23)(cid:29)(cid:3)(cid:44)(cid:81)(cid:87)(cid:85)(cid:82)(cid:3)(cid:87)(cid:82)(cid:3)(cid:51)(cid:38)(cid:51)
Scribe: Dana Moshkovitz
Scribe Date: Spring 2015
1 Ch... | https://ocw.mit.edu/courses/18-405j-advanced-complexity-theory-spring-2016/1b2c1d36df1e2a9cfa79d6531ffb8d29_MIT18_405JS16_IntroPCP.pdf |
reductions, meaning that a witness/proof for the new problem can be efficiently
transformed to a witness/proof for the initial problem.
Examples: Some NP-complete problems that will be important for the course are:
1. \Does [your favorite math conjecture here] have a proof of length n?"
2. Max-3Sat: Given clauses C1; : :... | https://ocw.mit.edu/courses/18-405j-advanced-complexity-theory-spring-2016/1b2c1d36df1e2a9cfa79d6531ffb8d29_MIT18_405JS16_IntroPCP.pdf |
P -hard problems arise are also diverse: they range from biology and
chemistry to medicine and technology.
So all these N P -hard problems were not going anywhere { in real life people had to solve them.
It became clear that researchers must (cid:12)nd ways to cope with N P -hardness. Such ways included
identifying eas... | https://ocw.mit.edu/courses/18-405j-advanced-complexity-theory-spring-2016/1b2c1d36df1e2a9cfa79d6531ffb8d29_MIT18_405JS16_IntroPCP.pdf |
running time depends polynomially on 1=ϵ, we say that it is an FPTAS: \a fully polynomial-time
approximation scheme".
The approach of designing efficient approximation algorithms has been extremely successful.
It
resulted in the development of many new algorithmic methods, such as the Markov chain Monte-
Carlo method (MC... | https://ocw.mit.edu/courses/18-405j-advanced-complexity-theory-spring-2016/1b2c1d36df1e2a9cfa79d6531ffb8d29_MIT18_405JS16_IntroPCP.pdf |
lemma follows from the
linearity of expectation.
8
Note that Lemma 2.1 immediately implies that there exists an assignment that satis(cid:12)es at least
7 fraction of the clauses. This can be seen as an instance of the "Probabilistic Method" [AS92],
8
where the existence of an object with a certain property is proved b... | https://ocw.mit.edu/courses/18-405j-advanced-complexity-theory-spring-2016/1b2c1d36df1e2a9cfa79d6531ffb8d29_MIT18_405JS16_IntroPCP.pdf |
distinguish whether (cid:5)(x) (cid:21) c
or (cid:5)(x) (cid:20) s, then it is N P -hard to approximate (cid:5) to within c=s for a minimization problem, or
s=c for a maximization problem.
The problem of distinguishing whether (cid:5)(x) (cid:21) c or (cid:5)(x) (cid:20) s is called a gap problem, and we denote
it by (... | https://ocw.mit.edu/courses/18-405j-advanced-complexity-theory-spring-2016/1b2c1d36df1e2a9cfa79d6531ffb8d29_MIT18_405JS16_IntroPCP.pdf |
the randomized veri(cid:12)er makes
polynomially-many rounds of interaction with the prover, and multi-prover interactive proofs (MIP),
proof systems where the randomized veri(cid:12)er interacts with more than one prover, and the provers
cannot communicate after seeing the veri(cid:12)er’s questions.
The new notion wa... | https://ocw.mit.edu/courses/18-405j-advanced-complexity-theory-spring-2016/1b2c1d36df1e2a9cfa79d6531ffb8d29_MIT18_405JS16_IntroPCP.pdf |
6) if (cid:6) is the binary alphabet.
Notice that a probabilistic veri(cid:12)er with randomness r and running time t can be simulated by a
deterministic veri(cid:12)er than runs in time 2r (cid:1) t. The randomness allows lower running time and lower
query complexity.
Note that a veri(cid:12)er with randomness r and q... | https://ocw.mit.edu/courses/18-405j-advanced-complexity-theory-spring-2016/1b2c1d36df1e2a9cfa79d6531ffb8d29_MIT18_405JS16_IntroPCP.pdf |
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