text stringlengths 30 4k | source stringlengths 60 201 |
|---|---|
define the period
of a function as the smallest number T such that f (t + T ) = f (t). We say that f (t)
has period T or, equivalently, that f (t) is T -periodic. The period T has the same
dimensions as t. If t is dimensionless then so is T ; if t has the dimensions of time,
so does T . Note that each normal mode un... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
fined as ω = 2πf where f is the frequency. Each mode
has angular frequency ωn = 2πfn = nπc/l. In terms of the frequency, we can write
the normal mode as
u ′ (x ′ , t ′ ) =
n
αn cos
πnc
l
nπc
t ′ + βn sin
t ′
l
(cid:17)
(cid:17)(cid:17)
= (αn cos (2πfnt ′ ) + βn sin (2πfnt ′ )) sin
(cid:16)
(cid:16)
(cid:16)
... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
sin (πx) is 2-periodic i.e. has period 2, sin (x)
is 2π-periodic, sin (nπt) is 2/n-periodic, sin (nπct/l) has period 2l/ (nc), sin (nπx/l)
has period 2l/n. Note that if a function has period T , then f (t + mT ) = f (t) for all
m = 1, 2, 3, .... In particular,
sin (nπx) , cos (nπx) , sin (nπt) , cos (nπt)
are all ... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
)
n=1
X
n=1
X
and for each normal mode, un (x, t) = un (x, t + 2) (check for yourself). Thus, the
period of u (x, t) is the same as that for the first harmonic u1 (x, t).
In physical
′ =
variables, the period of u ′ (x ′ , t ′ ) is T1
1/T 1
′ = 2l/c = 2l/ τ /ρ and the frequency is f1
′ = c/ (2l).
p
7
... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
of 1/time, c is a speed, and l is a length.
Then t must have dimensions of time, so that the arguments ωt, 2πf t, or πct/l of
cosine are dimensionless.
πct
l
(cid:1)
(cid:0)
3.3 Amplitude
Note that un (x, t) can be written
un (x, t) = γn sin (nπx) sin (nπt + ψn)
(19)
where γn =
sinusoidally according to
p
n ... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
solution to the Heat Problem with a homogeneous PDE (i.e. no sources,
§
sinks) Homogenous Type I BCs (u = 0 at x = 0, 1) is
u (x, t) =
∞
n=1
X
Bn sin (nπx) e −n2π2t
8
and as t
of the initial state, or initial condition.
, u (x, t)
→ ∞
→
0. Thus, the rod looses heat energy and with it the memory
In c... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
(x ′ , t ′ ) =
1
2
2 mv =
1
2
(ρΔx)
2
∂u
∂t
(cid:18) (cid:19)
and the local potential energy is
P E ′ (x ′ , t ′ ) = work done getting to displacement u
= Force
×
change in length of string
Making a displacement Δu of a segment of string from x to x+Δx results in a change
in the string segment length, initi... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
is, in dimensional form,
l
l
2
2
E ′ (t ′ ) =
1
2
0
ρ
∂u
∂t′
+ τ
∂u
∂x′
dx ′ =
!
ρ
2
0
(cid:18)
(cid:19)
(cid:18)
Z
(cid:0)
t ′ = (∂u/∂t′ )2 . Substituting c2 = τ /ρ and the
where we use the shorthand notation u2
dimensionless variables into (20) gives the dimensionless total energy
E′ (t ′)
τ l ... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
1)
0
Z
(cid:0)
cos 2 (nπt + ψn)
1
(cid:18)
Z
+ sin2 (nπt + ψn)
sin2 (nπx) dx
1
0
cos 2 (nπx) dx
(cid:19)
0
Z
sin2 (nπx) =
1
1
2 − 2
cos (2nπx) ,
cos 2 (nπx) =
1
2
+
1
2
cos (2nπx)
so that
1
sin2 (nπx) dx =
1
2
,
0
Z
1
0
cos 2 (nπx) dx =
1
2
Z
Substituting these integrals into the energy En (t) give... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
ux)x dx
Z
0
= [utux]x=0
1
Differentiating the BCs in time t gives
Therefore, (22) becomes
d
dt
u (0, t) = 0,
d
dt
u (1, t) = 0
dE
dt
= 0
(22)
(23)
dx.
Thus
E(t) = const = E (0) =
1
′
2
(f (x)) + (g (x))
2
1
2
0
Z
Thus the energy of a multi-mode wave system remains constant in time. This is
not... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
for the normal modes un (x, t) = γn sin (nπx) sin (nπt + ψn) gives
E (t) =
1
2
n=1 m=1
X
X
+
1
2
n=1 m=1
X
X
γnγmnmπ2 cos (nπt + ψn) cos (mπt + ψm)
1
×
0
Z
sin (nπx) sin (mπx) dx
γnγmnmπ2 sin (nπt + ψn) sin (mπt + ψm)
cos (nπx) cos (mπx) dx
1
×
0
Z
11
Subst... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
ivation
Note that each normal mode can be written in alternative form:
un (x, t) =
=
(αn cos (nπt) + βn sin (nπt)) sin (nπx)
1
2
(αn sin (nπ (x
βn cos (nπ (x
t))
−
−
−
t)))
+
(αn sin (nπ (x + t)) + βn cos (nπ (x + t)))
1
2
where we used the trig identities
sin (a + b) + sin (a
cos (a + b)
−
cos (a
−
−
... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
η
∂η ∂t
=
=
∂v
∂ξ
+
∂v
∂η
∂v
− ∂ξ
+
∂v
∂η
Similarly,
∂2u
∂x2
=
=
+
∂ ∂v
∂x ∂ξ
∂2v
∂ξ2
+ 2
∂ ∂v
∂x ∂η
∂2v
∂ξ∂η
+
∂2v
∂ξ2
+
∂2v
∂ξ∂η
+
(cid:19) (cid:18)
∂2v
∂ξ∂η
+
∂2v
∂η2
(cid:19)
=
(cid:18)
∂2v
∂η2
∂2
u
∂t2 =
=
v
∂2
− − ∂ξ2 +
(cid:18)
∂2v
∂2v
∂ξ2 −
∂ξ∂η
2
+
∂2v
∂η2
∂2
v
... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
:
∂
∂ξ
(cid:18)
∂v (ξ, η)
∂η
(cid:19)
= 0,
∂
∂η
(cid:18)
∂v (ξ, η)
∂ξ
(cid:19)
= 0
Integrating the first equation in ξ gives
∂v (ξ, η)
∂η
= G (η)
(25)
(26)
for some function G (η) (due to partial integration with respect to ξ). Define the
antiderivative of G (η) as Q (η), i.e. such that Q′ (η) = G (η). ... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
which represents a wave travelling in
−
the positive x-direction with scaled velocity 1. In physical coordinates, the function
depends on x
τ /ρ. The shape of the wave is
t = const.
determined by the function P (x) and the motion is governed by the line x
The wave (same shape) moves forward in time along the strin... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
speed c. The lines x
characteristics.
±
14
5.5 Determining the shape functions
The shapes of the forward and backward waves, P (x) and Q (x), are determined from
the initial conditions,
u (x, 0) = f (x) = P (x) + Q (x) ,
P ′ (x) + Q ′ (x)
ut (x, 0) = g (x) =
−
(28)
(29)
To obtain the second equation, the c... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
P (x) =
1
2
1
2
(cid:18)
(cid:18)
f (x)
Z
0
x
−
0
Z
g (s) ds
−
Q (0) + P (0)
(cid:19)
(30)
(31)
(32)
5.6 D’Alembert’s solution to the wave equation
[Oct 19, 2006]
Ref: Guenther & Lee
Summarizing our results: from (27), (31) and (32), the wave equation and initial
4.1, Myint-U & Debnath
4.3
§
§
condi... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
0
Z
g (s) ds.
−
t = x0 −
In other words, the solution is found by tracing backwards in time along the charac
t0 and x + t = x0 + t0 to the initial state (f (x), g (x)), then
teristics x
applying (33) to compute u (x0, t0) from the initial state. Information from the initial
x0 + t0 is all that is needed to find u ... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
2
x+t
x−t
Z
g (s) ds
Step 2.
Identify the regions.
In general, the function f (x) and g (x) are case
functions. You need to determine various regions by plotting the salient characteristics
t and x + t are relative to the cases
x
for the functions f (x) and g (x) and tells us what part of the case functions ... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
.5
−1
−0.5
R
2
x−t=1
x (t)
B
x (t)
C
(t)
x
D
R
1
0
x
R
6
0.5
1
1.5
2
Figure 1: Regions of interest separated by four characteristics.
Step 2. Identify the regions. The functions f (x) and g (x) are equal to functions
1 and are zero for x > 1. Thus, the regions
t =
1 (Figure 1). The
F (x) and G (x), resp... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
where x
1, which tells us what
part of the case functions f (x) and g (x) should be used. It is helpful to define the
lines
t and x + t are relative to
±
−
xA (t) =
t
−
−
1,
xB (t) = t
−
1,
xC (t) =
t + 1,
−
xD (t) = t + 1.
Step 3. Consider the solution in each region. In R1, we have
(34) implies
x
|
±
t
| ... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
t) =
In region R3, we have
1
−
≤
f (x + t) = F (x + t) ,
and hence
x + t
≤
f (x
In region R4, x
−
t
≤ −
1 and x + t
≥
x+t
Z
x−t
and hence
g (s) ds =
g (s) ds +
1
x−t
Z
1
−1
Z
x+t
u (x, t) =
1
2
x−t
Z
hence u = 0. To summarize,
t)
F (x
−
2
1 and x
+
1
1
2
Z
t
≤ −
x−t
−
t) = 0,
−
G (s) ds.
... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
x+t
1
t) + F (x + t)) +
2 x−t
1
F (x−t)
1
G (s) ds,
2 x−t
R
2
+t G (s) ds,
x
F (x
1
−
R
2
1
G (s) ds,
R
0
+
+t) + 2
1
1
2 −1
R
G (s) ds,
(x, t)
(x, t)
(x, t)
(x, t)
(x, t)
∈
Step 4. For each specific time t = t0, write the x-intervals corresponding to the
intersection of the... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
1
1
18
(37)
At t = 0, we use Table (37) and Figure 1 to find the x intervals Rn
′ corresponding
to the intersection of Rn with the line t = 0:
′ = (
R5
1],
,
−
−∞
′ = [
R1
1, 1] ,
−
′ = [1,
R6
).
∞
In R1, we have (recall that t = 0),
u (x, 0) =
1
2
(F (x
−
0... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
,
R2 = [0.5, 1.5] .
−
At t = 1, we use Table (37) and Figure 1 to find the x intervals Rn
−
′ corresponding
to the intersection of Rn with the line t = 1:
R5 = (
2],
,
−
−∞
R3 = [
2, 0] ,
−
R2 = [0, 2] ,
R6 = [2,
).
∞
(40)
At t = 2, we use Table (37) and Figure 1 to find the x intervals Rn
′ corresponding
to th... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
G (x) = cos2
19
π
x .
2
(cid:0)
(cid:1)
Step 1. D’Alembert’s solution (33) becomes
x+t
u (x, t) =
1
2
g (s) ds
x−t
Z
Step 2. The regions are the same as those in (35) and are plotted in Figure 1
above.
Step 3. Determine u (x, t) in each region. From (36), we have
x+t
1
G (s) ds,
2 x−t
... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
t
Z
x+t
−1
Z
1
G (s) ds =
t +
sin (π (x + t))
sin (π (x
−
t))
= t +
1
π
cos (πx) sin (πt)
G (s) ds =
G (s) ds =
t)
1
−
−
(x
2
1 + x + t
2
+
−
2π
sin (π (x
2π
−
sin (π (x + t))
2π
t))
−
G (s) ds = 1
−1
Z
Thus
t
+
2
−
1
2π
1−(x−t)
4
sin (πt) cos (πx) ,
sin(π(x−t))
,
4π
1+x+t + sin(π(x+t)) ... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
�
(cid:19) (cid:0)
(cid:0)
1 + cos πx ,
2π
4
,
+
2 −
3 + x + cos πx ,
2
cos πx
4π
x
4π
1
2
− ≤
1
2 ≤
3
2
− ≤
x
|
x
x
x
1
≤ 2
3
≤ 2
≤ −
3
2
| ≥
1
2
(cid:1)
0
(cid:1)
20
t=1/2
1
2
0.5
)
0
t
,
x
(
u
0
−4
−3
−2
−1
t=0... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
3+
x
4
πx)
sin(
,
4π
− sin(πx)
,
+
4π
1
,
2
0
3
1
≤
3
≤
−
1
≤
−
x
| ≥
|
x
≤
x
1
≤ −
x
1
≤
3
In Figure 2, the profiles of the displacement u(x, t) are plotted for the times t =
0, 1/2, 1, 2.
6 Waves on a finite string
[Oct 24, 2006]
Ref: Myint-U & Debnath
4.4 – 4.6, Guenther & Lee... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
) dx
1
g (x) sin (nπx) dx
0
Z
2
nπ 0
Z
βn =
We wrote, equivalently, that
u (x, t) = P (x
−
t) + Q (x + t)
where
P (x
−
t) =
Q (x + t) =
1
2
1
2
∞
n=1
X
∞
n=1
X
(αn sin (nπ (x
−
t))
−
βn cos (nπ (x
−
t)))
(αn sin (nπ (x + t)) + βn cos (nπ (x + t)))
6.1 Zero initial velocity
The simplest ICs invo... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
)
(cid:17)
.
(45)
22
This looks like D’Alembert’s solution for the infinite string - but fˆ replaces f (x)
so we’ll see that some properties of the solution are different.
To see directly that (45) satisfies the BCs, note that since fˆ is odd, fˆ(
t) =
fˆ(t),
−
−
and since fˆ is 2-periodic and od... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
to 1, but time t
). Thus
). We use the 2-periodic and odd properties of fˆ to
∞
∈
[0,
[0,
x + t
−
see what u (x, t) looks like for 0 < x < 1.
) and x
−∞
[0,
∞
∈
∈
t
Example. Suppose f (x) is a thin pulse symmetric about x = 1/2 and with a
maximum at x = 1/2. The initial pulse breaks into forward and backward... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
the end x = 1 and reappeared at x = 0, UPSIDE DOWN (i.e. with negative
sign)! On the second half of the string 1/2
1 and times 1/2
x
−
−
1,
t
≤
≤
≤
≤
1
≤
x + t
2
≤
23
and thus
fˆ(x + t) =
fˆ(x + t
1) =
f (x + t
1)
−
Thus the backward wave went off the end x = 0 at t = 1/2 and reappeared at the
... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
, 1]
x + t
range
[0, 1]
1/2
t
≤
≤
1
[0, 1/2]
1, 0]
[
−
[1/2, 1]
[1, 2]
left half string
right half string
bkwd wave
fwd wave,
upside down
fwd wave
bkwd wave,
upside down
24 | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
8.821 String Theory
Fall 2008
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
8.821 F2008 Lecture 0
4
Lecturer: McGreevy
September 22, 2008
Today
1. Finish hindsight derivation
2. What holds up the throat?
3. Initial che... | https://ocw.mit.edu/courses/8-821-string-theory-fall-2008/22f010a84df39d4448ec5ff59382413d_lecture04.pdf |
we will start with a special case.
1. There are many colors in our non-Abelian gauge theory. The motivation for this is best given
by the quote:
“You can hide a lot in a large-N matrix.”
1
– Shenker
The idea is that at large N the QFT has many degrees of freedom, thus corresponding to the
limit where the extra di... | https://ocw.mit.edu/courses/8-821-string-theory-fall-2008/22f010a84df39d4448ec5ff59382413d_lecture04.pdf |
of the interactions. This means that there are fewer candidates
for the dual.
(b) Supersymmetric theories have more adiabatic invariants, meaning observables that are
independent of the coupling. So there are more ways to check the duality.
(c) It controls the strong-coupling behavior. The argument is that in non-S... | https://ocw.mit.edu/courses/8-821-string-theory-fall-2008/22f010a84df39d4448ec5ff59382413d_lecture04.pdf |
is soft gluons.
Scale invariance applies to both space and time, so X µ
λX µ ,where µ = 0, 1, 2, 3. As we
said, the extra dimension coordinate is to be thought of as an energy scale. Dimensional anal
z . The most general
ysis suggests that this will scale under the scale transformation, so z
λ
five-dimensional metric... | https://ocw.mit.edu/courses/8-821-string-theory-fall-2008/22f010a84df39d4448ec5ff59382413d_lecture04.pdf |
S5
A specific example is N=4 SYM which is related to IIB Supergravity on AdS5 ×
Note: On a theory of gravity space-time is a dynamical variable, where you specify asymp
totics. This CFT defines a theory of gravity on spaces that are asymptotically AdS5.
S5 .
Why is AdS5 a solution?
1. Check on PSet
2. Use effective ... | https://ocw.mit.edu/courses/8-821-string-theory-fall-2008/22f010a84df39d4448ec5ff59382413d_lecture04.pdf |
q-sphere we have Sq Fq = N
To find the form of this part of the metric as a result of the flux, we integrate over the q-sphere
and get the action in D-dimensions, which will contain a term which is
Sq √GGm1n1 ...Gmq nq (Fg)m1...mq (Fg)n1...nq
We should then think of R(x) as a ”moduli field”. Now we have to evaluate th... | https://ocw.mit.edu/courses/8-821-string-theory-fall-2008/22f010a84df39d4448ec5ff59382413d_lecture04.pdf |
λ(x)gµν .
E E
R
.
R
gE(
R
�
�
(D)
R2
R
R
R
N
=
(d)
(d)
Sq
F
µν
g
�
|
N 2
Rα −
∼
1
β . where α > β > 0
R
So Vf lux(R)
At small R the first term in the effective potential dominates, while for R large, the effective
potential approaches zero from below. In between there is some minimum. To find the actual
α−β
Rm... | https://ocw.mit.edu/courses/8-821-string-theory-fall-2008/22f010a84df39d4448ec5ff59382413d_lecture04.pdf |
∂S
The equation of motion is the following: 0 = ∂g
µν ⇒
−2Λ )
2−d
V ′′(Rmin), which
2Λ)R . The cosmological constant is determined as
Rµν = gµν ( 2
ddx√g e(RE
R = 2d Λ
R ∼
Rµν (
1 R
Λ)
−
−
1
2
�
d−2 ⇒
⇒
We know the solution for AdSd, which has the metric ds2 =
L2 dr2+ηµν dxµdxν
r2
Now I will describe the right... | https://ocw.mit.edu/courses/8-821-string-theory-fall-2008/22f010a84df39d4448ec5ff59382413d_lecture04.pdf |
? Here’s why:
Cartan 2 Rab =
dωab + ωacωcb Rab
=
µν dxµ
dxν
∧
From that you can get the Ricci tensor, which appears in the equation of motion, by contracting
some indices Rν
µ = Rρν
ρµ
Let’s do some examples:
1
2 eµˆ
Rµˆνˆ = ωµˆrˆωrˆνˆ =
L
−
eνˆ
∧
Rµˆrˆ = dωµˆrˆ = L
1
2 erˆ
Rµν
ρσ =
−
1
2 (δρ
L
µδν
σ −
e... | https://ocw.mit.edu/courses/8-821-string-theory-fall-2008/22f010a84df39d4448ec5ff59382413d_lecture04.pdf |
boundary is the locus where they get really big. The area is defined as A = R3 √gd3x = R3 d3x L
r3
�
�
This is infinite for two reasons: from the integral over x and from the fact that r is going to zero.
Let’s regulate the Field Theory first. We put the thing on a lattice, introducing a low distance
cut-off δ and we put... | https://ocw.mit.edu/courses/8-821-string-theory-fall-2008/22f010a84df39d4448ec5ff59382413d_lecture04.pdf |
Lecture 6
Primality, Factoring, RSA, Hensel's Lemma
CRT and the number of solutions - we have a congruence
akxk + ak 1xk−1 + · · · + a
−
0 ≡
0
(mod n),
ai ∈ Z
2 . . . per
We want to know all solutions mod n, and in particular the number of solutions.
1 pe2
Write n = pe1
r . Then solving the congruence mod m reduces to ... | https://ocw.mit.edu/courses/18-781-theory-of-numbers-spring-2012/22f790db07c43592369000a37c31d487_MIT18_781S12_lec6.pdf |
polynomial in input
- in this case, poly(log n) steps. Obvious algorithm is to divide by every prime
starting from 2 to
√
(cid:98) n(cid:99), which is O(
n) steps, or exp( 1 log n)
.
√
2
Test using Fermat’s Little Theorem - if n is prime and n (cid:45) a, then an−1 ≡ 1
mod n.
1. Pick an integer a ∈ {2 . . . n − 1}.
2. ... | https://ocw.mit.edu/courses/18-781-theory-of-numbers-spring-2012/22f790db07c43592369000a37c31d487_MIT18_781S12_lec6.pdf |
, which decreases exponentially with k. And so if you
do c log n trials, the at most probability goes like 1
n . So if n passes c log n trials
(for some large enough c ≈ 100), then probability that n is prime is very close to
1.
4
∗
This is poly(log) steps, but we want a deterministic algorithm. Solved in 2002
by AKS (... | https://ocw.mit.edu/courses/18-781-theory-of-numbers-spring-2012/22f790db07c43592369000a37c31d487_MIT18_781S12_lec6.pdf |
ryptography - RSA Alice and Bob (A and B) want to pass messages, and Carol
is eavesdropper. A can send a message to B, converted into bits - equivalent to
sending an integer m. Wants to encrypt the number so C can’t understand it.
2 Obvious method is to use a shared key model, where A and B have some shared
key. With a... | https://ocw.mit.edu/courses/18-781-theory-of-numbers-spring-2012/22f790db07c43592369000a37c31d487_MIT18_781S12_lec6.pdf |
C has no way to compute m from me - this relies on the hardness
of factoring.
Hensel’s Lemma - this is a way to solve congruences mod pe if we know solu-
tions mod p (analog to Newton’s Method for finding roots of polynomials).
Theorem 26 (Hensel’s Lemma). Suppose that f (x) ∈ Z[x], f (a) ≡ 0 mod pj,
and f (cid:48)(a) (... | https://ocw.mit.edu/courses/18-781-theory-of-numbers-spring-2012/22f790db07c43592369000a37c31d487_MIT18_781S12_lec6.pdf |
Representations for KBS:
Logic: When Sound Deduction is Required
Spring 2005
6.871 Knowledge Based Systems
Howard Shrobe and Kimberle Koile
Syntax
Proofs
Semantics
Sound Inference and Complete Inference
What Properties hold?
The Language as a Representation
Comprehensiveness
Ambiguity
Lack of Commitment
Compro... | https://ocw.mit.edu/courses/6-871-knowledge-based-applications-systems-spring-2005/233e2e8021787d40f9c4e73e1f06283d_lect08_logic.pdf |
3
Logic : Page 3
Predicate Logic Syntax Solves These Problems
• The unit of representation is the Statement which is the application
of a predicate to a set of arguments: John Loves Mary
• Building blocks:
Constant Symbols, Variable Symbols, Function Symbols, Predicate Symbols
• A Term is:
A Constant symbol: Joh... | https://ocw.mit.edu/courses/6-871-knowledge-based-applications-systems-spring-2005/233e2e8021787d40f9c4e73e1f06283d_lect08_logic.pdf |
connective: An introduction rule and an elimination rule
– And Elimination: From (And A B) you can deduce A
– Modus Ponens:
• From (IMPLIES P Q) and P you can deduce Q
– Universal Instantiation:
• (FORALL (X) (P X)) you can deduce (P A) for any A
• Axioms: Statements that are given as a priori true
• A Proof is:... | https://ocw.mit.edu/courses/6-871-knowledge-based-applications-systems-spring-2005/233e2e8021787d40f9c4e73e1f06283d_lect08_logic.pdf |
assumption j)
(OE (i l n) (k m))
Double Negation Elimination
i
j A
(not (not A))
(DNE i)
Negation Introduction
i Show (Not A)
i+1 | A
| B
k
| (not B)
l
m (not A)
(assumption)
(NI (k l) (i+1))
Logic : Page 8
Logic : Page 8
Quantifier Rules
A Substitution of a for x in the statement (P x) is written... | https://ocw.mit.edu/courses/6-871-knowledge-based-applications-systems-spring-2005/233e2e8021787d40f9c4e73e1f06283d_lect08_logic.pdf |
j (Thereis (x)[x/a](P ... ))
where a is any term at all
(EG i)
Alphabetic Variance
i (Q (x)(P ... ))
j (Q (z)[z/x](P ... ))
where Q is either quantifier and [z/x] is a valid substitution.
(AV i)
Logic : Page 10
Logic : Page 10
Derived Rules
GIGO
i (not A)
j A
k C
(GIGO i j)
Indirect Proof
i (Show A)
i+1
J ... | https://ocw.mit.edu/courses/6-871-knowledge-based-applications-systems-spring-2005/233e2e8021787d40f9c4e73e1f06283d_lect08_logic.pdf |
plies Q
(Or (And P Q) (And P R)))
Assumption motivated by 1
And Elimination 2
And Elimination 2
trying for Or Elimination from 4
Assumption motivated by 5
And Introduction 3,6
Or Introduction 7
Conditional Proof 8 (6)
10. | Show (Implies R
trying for Or Elimination from 4
(Or (And P Q) (And P R)))
11. | | R... | https://ocw.mit.edu/courses/6-871-knowledge-based-applications-systems-spring-2005/233e2e8021787d40f9c4e73e1f06283d_lect08_logic.pdf |
?y)))
Assumption motivated by 1
Existential Instantiation 2
Universal Instantiation 3
Existential Generalization 4
Alphabetic Variation 5
Universal Generalization 6
Alphabetic Variation 7
(Forall (?x) (Thereis (?y) ( P ?x ?y))))
Conditional Proof 8 (2)
Logic : Page 13
Logic : Page 13
Bogus Proof
1.
Show (Implies... | https://ocw.mit.edu/courses/6-871-knowledge-based-applications-systems-spring-2005/233e2e8021787d40f9c4e73e1f06283d_lect08_logic.pdf |
is some expression containing no free occurence of the variables of interest.
Quantifier Rules
(1) (Or (Q (x) (F .. x ..)) G) = (Q (x) (Or (F .. x ..) G))
(2) (And (Q (x) (F .. x ..)) G) = (Q (x) (And (F .. x ..) G))
(3) (And (For-all (x) (F .. x ..)) (For-all (x) (H .. x ..))) = (For-all (x) (And (F .. x ..) (H ..... | https://ocw.mit.edu/courses/6-871-knowledge-based-applications-systems-spring-2005/233e2e8021787d40f9c4e73e1f06283d_lect08_logic.pdf |
D)) = (And (Or A C) (Or A D) (Or B C) (Or B D))
(2) (And (Or A B) (Or C D)) = (Or (And A C) (And A D) (And B C) (And B D))
Implication
(Implies A B) = (Or (Not A) B)
Logic : Page 15
Logic : Page 15
Normal Form (2)
Order of Use of Identities:
1. Implication
2. Negations and deMorgan
3. Quantifers
4. Distributio... | https://ocw.mit.edu/courses/6-871-knowledge-based-applications-systems-spring-2005/233e2e8021787d40f9c4e73e1f06283d_lect08_logic.pdf |
)
Logic : Page 17
Logic : Page 17
Horn Clauses: A Restriction to Rule-like Form
Suppose that after normalization all clauses have at most 1 positive
statement
(Or p (not q) (not r) (not s) (not t))
where any of p q r s t may contain with free variables
Then by implication identity this is the same as:
(implies (... | https://ocw.mit.edu/courses/6-871-knowledge-based-applications-systems-spring-2005/233e2e8021787d40f9c4e73e1f06283d_lect08_logic.pdf |
A Model
– Domain (a mathematical set)
– Interpretation, a function that maps:
• Constant symbols of the syntax to elements of the domain.
• Function symbols of the syntax to functions over the domain.
• Predicate symbols of the syntax to predicates over the domain (note that this
is a mapping from a set of argume... | https://ocw.mit.edu/courses/6-871-knowledge-based-applications-systems-spring-2005/233e2e8021787d40f9c4e73e1f06283d_lect08_logic.pdf |
Turing machine.
– Does this seem contradictory?
Logic : Page 23
Logic : Page 23
What Formal Properties Hold?
Completeness and Consistency
Is the system consistent?
Yes (Hilbert and Ackerman).
Is the system complete?
Yes (Godel).
•
•
•
•
• What happens if you add some interesting axioms? For example,
those f... | https://ocw.mit.edu/courses/6-871-knowledge-based-applications-systems-spring-2005/233e2e8021787d40f9c4e73e1f06283d_lect08_logic.pdf |
Kee's rule system
–
–
ART
Joshua
Logic : Page 26
Logic : Page 26
When is Logic the Right Representation?
• Two Case Studies:
– American Express Authorizer's Assistant
– British Nationality Act
• Goal in both cases is Policy Distribution: The systematic and
reproducible interpretation of the intended meani... | https://ocw.mit.edu/courses/6-871-knowledge-based-applications-systems-spring-2005/233e2e8021787d40f9c4e73e1f06283d_lect08_logic.pdf |
(namely what inference rules they cause to
run).
• The policies become active interpreters of the data.
• Control is secondary or to be ignored.
Logic : Page 29
Logic : Page 29
How well does this work in practice?
• Authorizer's assistant:
• Often's people's purchasing patterns are idiosyncratic.
– Holidays, vac... | https://ocw.mit.edu/courses/6-871-knowledge-based-applications-systems-spring-2005/233e2e8021787d40f9c4e73e1f06283d_lect08_logic.pdf |
software
studio
asynchronous calls:
examples
Daniel Jackson
1 timers
var alert_timers = function () {
setTimeout(function () {alert("page about to expire!");}, 2000);
setInterval(function () {alert("take a typing break!");}, 4000);
};
› asynchronous event due to timeouts
› note that alert is modal (and syn... | https://ocw.mit.edu/courses/6-170-software-studio-spring-2013/23728628acd5486841f72e85b02ef24c_MIT6_170S13_48-asyn-exam.pdf |
get(url, data, function (d) {alert(d);});
};
# GET /welcome
def welcome
user = params[:user]
render :text => "Welcome, " + user
end
server side
client side
› client passes Javascript object
› because call is $.get, appended as query string on url
› server returns string
6
getting a JSON object
v... | https://ocw.mit.edu/courses/6-170-software-studio-spring-2013/23728628acd5486841f72e85b02ef24c_MIT6_170S13_48-asyn-exam.pdf |
Studio
Spring 2013
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/6-170-software-studio-spring-2013/23728628acd5486841f72e85b02ef24c_MIT6_170S13_48-asyn-exam.pdf |
Massachusetts Institute of Technology
Department of Electrical Engineering and Computer Science
6.341: Discrete-Time Signal Processing
OpenCourseWare 2006
Lecture 11
Multirate Systems and Polyphase Structures
Reading: Section 4.7 in Oppenheim, Schafer & Buck (OSB).
Consider the two systems depicted below.
The f... | https://ocw.mit.edu/courses/6-341-discrete-time-signal-processing-fall-2005/23887e05bb83d893874e27d49f7135ba_lec11.pdf |
ampling systems which is discussed in
that subsection and which is depicted in OSB Figure 4.31. Further insight about the upsampling
identity can be gained by considering its behavior in the frequency domain:
The top subfigure shows two DTFTs, a signal X(ej�) and filter H(ej�). These respectively cor
respond to x[n] ... | https://ocw.mit.edu/courses/6-341-discrete-time-signal-processing-fall-2005/23887e05bb83d893874e27d49f7135ba_lec11.pdf |
system
in OSB Figure 4.33, picking off the polyphase components ek [n], and forming an equivalent
realization of the filter h[n] as in OSB Figure 4.34. When this new structure is used to im
plement the filter h[n] in our downsampling system, a series of flow graph manipulations and
applications of the downsampling nobl... | https://ocw.mit.edu/courses/6-341-discrete-time-signal-processing-fall-2005/23887e05bb83d893874e27d49f7135ba_lec11.pdf |
8 A glimpse of Young tableaux.
We defined in Section 6 Young’s lattice Y , the poset of all partitions of all
nonnegative integers, ordered by containment of their Young diagrams.
� �
� �
�
11111
� �
��
��
���
��
��
�
�
�
2111
221
��
��
��
��
��
�
311
��
��
��
���
�
�
32
��
��
��
�
41
��
� �
� �
�
5
� �
�
�
� ... | https://ocw.mit.edu/courses/18-318-topics-in-algebraic-combinatorics-spring-2006/23b76f41bf5e26f7221734e3499e73fb_notes2.pdf |
of infinite rank), with Yi consisting of all
partitions of i. In other words, we have Y = Y0 ≤
(disjoint union),
where every maximal chain intersects each level Yi exactly once. We call Yi
the ith level of Y .
Y1 ≤ · · ·
Since the Hasse diagram of Y is a simple graph (no loops or multiple
edges), a walk of length ... | https://ocw.mit.edu/courses/18-318-topics-in-algebraic-combinatorics-spring-2006/23b76f41bf5e26f7221734e3499e73fb_notes2.pdf |
order to that of the
walk. This is because we will soon regard U and D as linear transfor
mations, and we multiply linear transformations right-to-left (opposite to
the usual left-to-right reading order). For instance (abbreviating a partition
�m), the walk Ø, 1, 2, 1, 11, 111, 211, 221, 22, 21, 31, 41 is
(�1, . .... | https://ocw.mit.edu/courses/18-318-topics-in-algebraic-combinatorics-spring-2006/23b76f41bf5e26f7221734e3499e73fb_notes2.pdf |
. . , n inserted into the squares of D, so that each number
appears exactly once, and every row and column is increasing. We call � the
shape of the SYT θ , denoted � = sh(θ ). For instance, there are five SYT of
73
shape (2, 2, 1), given by
2
4
1
3
5
2
5
1
3
4
3
4
1
2
5
3
5
1
2
4
4
5
1
2
3
Let f �... | https://ocw.mit.edu/courses/18-318-topics-in-algebraic-combinatorics-spring-2006/23b76f41bf5e26f7221734e3499e73fb_notes2.pdf |
An−1 · · ·
√
√
�(DnU n , Ø) =
(f �)2 .
�
��n
(40)
Our object is to find an explicit formula for �(w, �) of the form f �cw ,
where cw does not depend on �. (It is by no means a priori obvious that
Ø = 1, we will obtain by
such a formula should exist.) In particular, since f
setting � = Ø a simple formula for the n... | https://ocw.mit.edu/courses/18-318-topics-in-algebraic-combinatorics-spring-2006/23b76f41bf5e26f7221734e3499e73fb_notes2.pdf |
) = � = n.
|
|
After
j
i=1(ri +si) steps we will be at level
j
level is level 0, we must have
�
�
j
i=1(ri −
si)
≡
si). Since the lowest
i=1(ri −
j
�
k.
�
0 for 1
�
The easy proof that the two conditions of the lemma are sufficient for
the existence of a Hasse walk of type w from Ø to � is left to the reader... | https://ocw.mit.edu/courses/18-318-topics-in-algebraic-combinatorics-spring-2006/23b76f41bf5e26f7221734e3499e73fb_notes2.pdf |
4422221 + 544222111
D21(54422211) = 44422211 + 54322211 + 54422111 + 5442221.
It is clear [why?] that if r is the number of distinct (i.e., unequal) parts of �,
then Ui(�) is a sum of r + 1 terms and Di(�) is a sum of r terms. The next
lemma is an analogue for Y of the corresponding result for Bn (Lemma 4.6).
75
... | https://ocw.mit.edu/courses/18-318-topics-in-algebraic-combinatorics-spring-2006/23b76f41bf5e26f7221734e3499e73fb_notes2.pdf |
� is 0.
If now µ = � and we cannot obtain µ by adding a square and then deleting
a square from � (i.e., µ and � differ in more than two rows), then clearly
when we apply the left-hand side of (41) to �, the coefficient of µ will be 0.
Finally consider the case � = µ. Let r be the number of distinct (unequal)
parts of... | https://ocw.mit.edu/courses/18-318-topics-in-algebraic-combinatorics-spring-2006/23b76f41bf5e26f7221734e3499e73fb_notes2.pdf |
�
i�Sw
ai).
(42)
76
∪
Before proving Theorem 8.3, let us give an example. Suppose w =
U 3D2U 2DU 3 = UU UDDUU DUUU and � = (2, 2, 1). Then Sw =
}
and a4 = 0, b4 = 3, a7 = 1, b7 = 5, a8 = 2, b8 = 5. We have also seen earlier
that f 221 = 5. Thus
4, 7, 8
{
�(w, �) = 5(3
−
0)(5
−
1)(5
−
2) = 180.
Proof o... | https://ocw.mit.edu/courses/18-318-topics-in-algebraic-combinatorics-spring-2006/23b76f41bf5e26f7221734e3499e73fb_notes2.pdf |
+ λ
1)U �+�−2)(Ø)
= �(� + λ
= �(� + λ
= �(� + λ
−
−
−
−
1)DU �+�+λ−2(Ø)
1)(U �+�+λ−2D + (� + λ + β
1)(� + λ + β
2)U �+�+λ−3(Ø).
−
−
2)U �+�+λ−3)(Ø)
The coefficient of � in U �+�+λ−3(Ø) is f �, so we get
[�]DU λ DU � DU �(Ø) = �(� + λ
1)(� + λ + β
−
2)f � ,
−
which is equivalent to (42). �
77
An i... | https://ocw.mit.edu/courses/18-318-topics-in-algebraic-combinatorics-spring-2006/23b76f41bf5e26f7221734e3499e73fb_notes2.pdf |
G of
G. There are many other intimate connections between the representation
theory of
Sn, on the one hand, and the combinatorics of Young’s lattice and
Young tableaux, on the other. There is also an elegant combinatorial proof of
Corollary 8.4, known as the Robinson-Schensted correspondence, with many
fascinating... | https://ocw.mit.edu/courses/18-318-topics-in-algebraic-combinatorics-spring-2006/23b76f41bf5e26f7221734e3499e73fb_notes2.pdf |
.
(44)
�
i,j
Just as in the proof of Theorem 8.3, the numbers bij (φ) exist and are well-
defined.
8.5 Lemma. We have bij (φ) = 0 if φ
i
−
j is odd. If φ
−
j = 2m
i
−
−
then
bij (φ) =
φ!
.
2m i! j! m!
(45)
Proof. The assertion for φ
j odd is equivalent to (F1) above, so
j is even. The proof is by i... | https://ocw.mit.edu/courses/18-318-topics-in-algebraic-combinatorics-spring-2006/23b76f41bf5e26f7221734e3499e73fb_notes2.pdf |
is
unique. Hence equating coefficients of U iDj on both sides of (46) yields the
recurrence
bij (φ + 1) = bi,j−1(φ) + (i + 1)bi+1,j (φ) + bi−1,j (φ).
(47)
It is a routine matter to check that the function φ!/2mi!j!m! satisfies the same
recurrence (47) as bij (φ), with the same intial condition b00(0) = 1. From this ... | https://ocw.mit.edu/courses/18-318-topics-in-algebraic-combinatorics-spring-2006/23b76f41bf5e26f7221734e3499e73fb_notes2.pdf |
all j. We could have
b
carried out the proof so as only to compute bi0(φ), but the general value of
ij (φ) is so simple that we have included it too.
b
8.7 Corollary. The total number of Hasse walks in Y of length 2m
from Ø to Ø is given by
λ(2m, Ø) = 1
·
3 5
·
· · ·
(2m
−
1).
Proof. Simply substitute � = Ø (so... | https://ocw.mit.edu/courses/18-318-topics-in-algebraic-combinatorics-spring-2006/23b76f41bf5e26f7221734e3499e73fb_notes2.pdf |
for 1
p(j
−
≥s are eigenvalues of multiplicity p(j
−
s)
p(j
±
−
Proof. Let A denote the adjacency matrix of Yj−1,j . Since RYj−1,j =
RYj−1,j can be written
j−1 →
RYi. The matrix A acts on the vector
RYj (vector space direct sum), any vector v
RY
uniquely as v = vj−1 + vj , where vi ⊕
space RYj−1,j as fol... | https://ocw.mit.edu/courses/18-318-topics-in-algebraic-combinatorics-spring-2006/23b76f41bf5e26f7221734e3499e73fb_notes2.pdf |
=
j
�
±
RYj−1,j , with v� =
Note that v�
j−1
Using equation (43), we compute
−
⊕
≥j
±
−
sU j−1−s(v) and v� = U j−s(v).
j
�)
A(v �) = U (vj−1) + D(vj
�
=
=
j
±
±
�
j
�
sU j−s(v) + DU j−s(v)
sU j−s(v) + U j−sD(v) + (j
−
−
s)U j−s−1(v)
−
81
sU j−s(v) + (j
=
=
±
±
j
�
�
�
−
j
s v... | https://ocw.mit.edu/courses/18-318-topics-in-algebraic-combinatorics-spring-2006/23b76f41bf5e26f7221734e3499e73fb_notes2.pdf |
we have found all its eigenvalues. �
s are eigenvalues.) Since the graph Yj−1,j has p(j
s
1) + p(j)
≥j
−
−
−
−
An elegant combinatorial consequence of Theorem 8.8 is the following.
8.9 Corollary. Fix j
1. The number of ways to choose a partition �
of j, then delete a square from � (keeping it a partition), ... | https://ocw.mit.edu/courses/18-318-topics-in-algebraic-combinatorics-spring-2006/23b76f41bf5e26f7221734e3499e73fb_notes2.pdf |
Engineering Risk Benefit Analysis
1.155, 2.943, 3.577, 6.938, 10.816, 13.621, 16.862, 22.82,
ESD.72, ESD.721
DA 1. The Multistage Decision Model
George E. Apostolakis
Massachusetts Institute of Technology
Spring 2007
DA 1. The Multistage Decision Model
1
Why decision analysis?
A structured way for ranking decision op... | https://ocw.mit.edu/courses/esd-72-engineering-risk-benefit-analysis-spring-2007/23bb56e577f168fe62b4e4a37e7dd81c_da1.pdf |
] = 1.0
L1: $150,000, old product,
L2: $300,000, new product and the market is
P[s] = P[L2/N] = 0.3
strong,
L3: $100,000, new product and the market is mild,
P[m] = P[L3/N] = 0.5
L4: -$100,000, new product and the market is weak,
P[w] = P[L4/N] = 0.2
DA 1. The Multistage Decision Model
6
Building the decision tre... | https://ocw.mit.edu/courses/esd-72-engineering-risk-benefit-analysis-spring-2007/23bb56e577f168fe62b4e4a37e7dd81c_da1.pdf |
: Maximize the expected monetary
value (EMV) of the earnings (payoffs).
• In the decision tree, work from right to left and
compute expectations.
DA 1. The Multistage Decision Model
10
Calculation of the EMV
EMV[N] = 0.3x300 + 0.5x100 + 0.2x(-100) = $120K
EMV[O] = 1.0x150 = $150K
Option O has the largest EMV, theref... | https://ocw.mit.edu/courses/esd-72-engineering-risk-benefit-analysis-spring-2007/23bb56e577f168fe62b4e4a37e7dd81c_da1.pdf |
.0
P(w/L4) = 0.7
1.0
DA 1. The Multistage Decision Model
14
The new decision tree
*Entries in earnings column
do not yet take account
of the cost of the survey.
Purchase
survey
C
D
Survey
response
"strong"
Survey
response
"mild"
Survey
response
"weak"
D
D
D
Do not
purchase
survey
C
No new
information
is obtained
D
M... | https://ocw.mit.edu/courses/esd-72-engineering-risk-benefit-analysis-spring-2007/23bb56e577f168fe62b4e4a37e7dd81c_da1.pdf |
L4
• Evidence: “Survey result is strong”
)s/L(P
j
=
)L(P)L/s(P
j
j
4
∑
2
)L(P)L/s(P
j
j
4,3,2j
=
DA 1. The Multistage Decision Model
17
Calculations for “survey result is s”
Payoff Prior Likelihood
Product
Posterior
Probability
Prob.
0.3
0.5
0.2
1.0
L2
L3
L4
P(s/ L2)=0.8
P(s/ L3)=0.2
P(s/ L4)=0.0
0.24
0.10
0.00
0.34
P... | https://ocw.mit.edu/courses/esd-72-engineering-risk-benefit-analysis-spring-2007/23bb56e577f168fe62b4e4a37e7dd81c_da1.pdf |
.5
.2
L1
L2
L3
L4
L1
L2
L3
L4
L1
L2
L3
L4
L1
L2
L3
L4
130,000
280,000
80,000
-120,000
130,000
280,000
80,000
-120,000
130,000
280,000
80,000
-120,000
150,000
300,000
100,000
-100,000
DA 1. The Multistage Decision Model
20
Figure by MIT OCW.
Optimal terminal decisions
1. Solve “backwards in time.”
2. Determine the best... | https://ocw.mit.edu/courses/esd-72-engineering-risk-benefit-analysis-spring-2007/23bb56e577f168fe62b4e4a37e7dd81c_da1.pdf |
000
O
120,000
N
C
1
.706
.294
0
1
.143
.714
.143
1
0
.417
.583
1
.3
.5
.2
L1
L2
L3
L4
L1
L2
L3
L4
L1
L2
L3
L4
L1
L2
L3
L4
130,000
280,000
80,000
-120,000
130,000
280,000
80,000
-120,000
130,000
280,000
80,000
-120,000
150,000
300,000
100,000
-100,000
DA 1. The Multistage Decision Model
22
Figure by MIT OCW.
Best decis... | https://ocw.mit.edu/courses/esd-72-engineering-risk-benefit-analysis-spring-2007/23bb56e577f168fe62b4e4a37e7dd81c_da1.pdf |
General form of a decision tree
P(H1|C1)
H1
G1
G2
Gk
P(C1)
P(C2)
P(Cn)
C1
C2
Cn
D1
D2
D3
Dm
GENERAL FORM OF A DECISION TREE
DA 1. The Multistage Decision Model
25
The multistage decision model
1. Each stage consists of a decision node followed by a
chance node for each of the decision options available in
this stag... | https://ocw.mit.edu/courses/esd-72-engineering-risk-benefit-analysis-spring-2007/23bb56e577f168fe62b4e4a37e7dd81c_da1.pdf |
§ 13. Hypothesis testing asymptotics II
Setup:
H0 ∶ X n ∼ P
∶
test PZ
X n
∣
n
X
X
n
→
n
H1 ∶ X ∼ QX
}
{
0, 1
n
(i.i.d.
)
sp
ecification:
n
)
1 − α = π(
∣0
1
≤ −nE0
2
n
( )
β = π0
∣1
≤ 2−nE1
Bounds:
• achievability (Neyman Pearson)
α = 1 − π1∣0 = PX n[Fn > τ ],
β = π0∣1 = QX n[Fn > τ
]
• converse (strong)
where
∀(α, β) a... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/23be1265744c2e9e8962e79985cdda45_MIT6_441S16_chapter_13.pdf |
be finite and also T
≠ const since P ≠ Q):
ψP (λ) = log EP [eλT ]
= log ∑ P (x 1−λQ x λ
)
( ) =
x
θλ − ψP
(λ)
∗ ( )
ψP θ
= sup
λ
R
∈
138
log
∫
dP 1
)
(
−λ
(
dQ
λ
)
P
≪
≪
Q and Q
( )
Note that since ψP (0) = ψP (1) =
1. Furthermore,
λ
[
(
assuming
ψP λ continuous everywhere on 0, 1 (
)
(
arguments).
on 0, 1 it follows f... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/23be1265744c2e9e8962e79985cdda45_MIT6_441S16_chapter_13.pdf |
1.3
ψP θ . Note that
) =
is increasing, θ E1 θ is decreasing.
( ) =
0
)
(
ψP 1
ψ
P
(
↦
Remark 13.1 (R´enyi divergence). R´enyi defined a family of divergence indexed by λ ≠ 1
Dλ(P ∥Q) ≜
1
λ − 1
log EQ [(
λ
)
] ≥ 0.
dP
dQ
( ∥ ) = −
which generalizes Kullback-Leibler
( − )
λ 1 Dλ Q P
pro
( )
and 1, and the slope at endpoi... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/23be1265744c2e9e8962e79985cdda45_MIT6_441S16_chapter_13.pdf |
C
139
Proof of Theorem 13.1. The idea is to apply the large deviation theory to iid sum n
∑
k 1 Tk. Specifi-
=
cally, let’s rewrite the bounds in terms of T :
• Achievability (Neyman Pearson)
let τ = −
nθ,
(n)
π
0
1
∣
=
P [
n
∑
k
1
=
T
≥
k nθ
]
π
n
( )
∣1
0
= Q [
n
∑
1
k
=
Tk
< nθ]
• Converse (strong)
let γ = 2−nθ,
π1∣... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/23be1265744c2e9e8962e79985cdda45_MIT6_441S16_chapter_13.pdf |
ψP θ
( ) −
(
)
] =
( + )
ψP λ 1
)
(
thus E0, E1
verse:
Con
(
( ))
)
(
E0 θ , E1 θ
bound we have:
in (13.1) is achievabl
e.
We want to show that any achievable (
0, E1 pair must be below the curve
in the above Neyman-Pearson test with parameter θ. Apply the strong converse
E
)
2−nE0 + 2−nθ2−nE1 ≥ 2−nψ∗
1 + θ) ≤
⇒ min(E0... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/23be1265744c2e9e8962e79985cdda45_MIT6_441S16_chapter_13.pdf |
≤
0 E0 D Q P
≤
(
∥
)
(13.3)
Proof. The first part is verified trivially. Indeed,
we have
if
we
fix
λ and let
θ
(λ) ≜ E
P
λ
[T
], then from (11.13)
D(Pλ∥P
) = ψP
∗ (θ) ,
whereas
Also
from
)
( ∥
D Q P .
D(Pλ∥Q) = EPλ
[
log
dPλ
dQ
] = EPλ
[log
dPλ
dP
dP
dQ
] =
D Pλ P
∥ )
(
−
EPλ T
[ ] =
∗ θ
( ) −
ψP
θ .
(11.12) we know that ... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/23be1265744c2e9e8962e79985cdda45_MIT6_441S16_chapter_13.pdf |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.