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Softmax Outputs for Multiclass Networks
Softmax
pj =
1
S
ewj where S =
ewk
n
Xk=1
Softmax produces the probabilities in teachyourmachine.com
The World Championship at the Game of Go
Residual Networks (ResNets)
“skip connections”
29/32
Softmax Outputs for Multiclass Networks
Softmax
pj =
1
S
ewj where S =
ewk
n
Xk... | https://ocw.mit.edu/courses/18-085-computational-science-and-engineering-i-summer-2020/22a453f2f41f9c34ad274b7d7da9a0aa_MIT18_085Summer20_lec_GS.pdf |
i v + bi = 0, in other words ReLU at N neurons. F has r(N, m) linear
pieces :
r(N, m) =
P
m
i=0
N
i
(cid:18)
=
(cid:19)
(cid:18)
N
0
+
(cid:19)
(cid:18)
N
1
(cid:19)
+
· · ·
+
N
m
(cid:18)
(cid:19)
r(N, m) = r(N
1, m) + r(N
1, m
1)
−
−
−
4
Start with 2 planes
1a
3a
1b
2a
2b
r(2, 2) = 4
←
Add new plane H
H
3b
r(2, 1) = ... | https://ocw.mit.edu/courses/18-085-computational-science-and-engineering-i-summer-2020/22a453f2f41f9c34ad274b7d7da9a0aa_MIT18_085Summer20_lec_GS.pdf |
equations
aT
i v + bi = 0, in other words ReLU at N neurons. F has r(N, m) linear
pieces :
r(N, m) =
P
m
i=0
N
i
(cid:18)
=
(cid:19)
(cid:18)
N
0
+
(cid:19)
(cid:18)
N
1
(cid:19)
+
· · ·
+
N
m
(cid:18)
(cid:19)
r(N, m) = r(N
1, m) + r(N
1, m
1)
−
−
−
4
Start with 2 planes
1a
3a
1b
2a
2b
r(2, 2) = 4
←
Add new plane H
H
... | https://ocw.mit.edu/courses/18-085-computational-science-and-engineering-i-summer-2020/22a453f2f41f9c34ad274b7d7da9a0aa_MIT18_085Summer20_lec_GS.pdf |
OpencourseWare
2.12 Uncertainty in labelling
9 August 2006
While many stretches of speech will be straightforward to label with ToBI, many others
may prove more challenging to the labeller. Naturally-produced speech contains an
enormous amount of variation, and there are cases where a labeller may be uncertain
wh... | https://ocw.mit.edu/courses/6-911-transcribing-prosodic-structure-of-spoken-utterances-with-tobi-january-iap-2006/22ab414fc96a65b7bce85d1e08bc31b7_chap2_12.pdf |
to indicate places
where more than one label was seriously considered. There may be times when a labeller
spends a comparatively long time determining which of two possible labels to use for a
particular tone or break. The alternatives tier allows an outlet for the labeller to include
both labels, so that she can m... | https://ocw.mit.edu/courses/6-911-transcribing-prosodic-structure-of-spoken-utterances-with-tobi-january-iap-2006/22ab414fc96a65b7bce85d1e08bc31b7_chap2_12.pdf |
file, on the monosyllabic word
clouds. In the first full intonational phrase of the file (shown in Figure 2.12.1) with the
words and indeed, there is a strong prominence that is marked by a peak in the pitch track
on the syllable –deed of
indeed; the syllable is clearly pitch-accented. However, the
labeller was no... | https://ocw.mit.edu/courses/6-911-transcribing-prosodic-structure-of-spoken-utterances-with-tobi-january-iap-2006/22ab414fc96a65b7bce85d1e08bc31b7_chap2_12.pdf |
+!H* (where the bitonal has a downstepped High). In
this case, the labeller perceived the pitch height of the peak of the pitch-accented word
clouds as being about the same as the pitch height reached in the preceding pitch accent
on the –lec– of electron. However, the pitch track shows that the f0 of the peak on cl... | https://ocw.mit.edu/courses/6-911-transcribing-prosodic-structure-of-spoken-utterances-with-tobi-january-iap-2006/22ab414fc96a65b7bce85d1e08bc31b7_chap2_12.pdf |
pairs of pitch accents
H* vs L+H*:
The pitch accented syllable clearly has a High prominence, but the rise may be more
gradual than expected for the bitonal L+H*, but steeper than would be expected for a
single-tone H*, and there may be varying degrees of evidence for a preceeding L target.
L* vs H* (in a compress... | https://ocw.mit.edu/courses/6-911-transcribing-prosodic-structure-of-spoken-utterances-with-tobi-january-iap-2006/22ab414fc96a65b7bce85d1e08bc31b7_chap2_12.pdf |
, and its use will be discussed below.
Levels of prominence:
As discussed in section 2.3, syllables in spoken language can be produced with a range of
degrees of prominence: it is not the case, for example, that all syllables which do not bear
a pitch accent are equally weak. Some syllables can be prominent compare... | https://ocw.mit.edu/courses/6-911-transcribing-prosodic-structure-of-spoken-utterances-with-tobi-january-iap-2006/22ab414fc96a65b7bce85d1e08bc31b7_chap2_12.pdf |
as prominent as long, the word seems to be fairly prominent. The
pitch track also shows some ambiguity: there is a slight rise in f0 at the beginning of the
word how. While this is potentially the result of a pitch-tracking error, the labeler may
have been uncertain whether the rise in pitch, combined with the sense... | https://ocw.mit.edu/courses/6-911-transcribing-prosodic-structure-of-spoken-utterances-with-tobi-january-iap-2006/22ab414fc96a65b7bce85d1e08bc31b7_chap2_12.pdf |
accent: the signal is ambiguous
here. The labeller has captured this ambiguity, and resulting uncertainty, by using the *?
label in the tones tier, and L* in the alt tier.
Figure 2.12.4 Using *? to mark uncertainty on made
<marmalade7>
The example <anna3>, shown in Figure 2.12.5, shows a parallel case to the on... | https://ocw.mit.edu/courses/6-911-transcribing-prosodic-structure-of-spoken-utterances-with-tobi-january-iap-2006/22ab414fc96a65b7bce85d1e08bc31b7_chap2_12.pdf |
ertainty regarding level of disjuncture
Labellers may also encounter times when they are uncertain about the appropriate break
index. There are two ways which the labeller may signal uncertainty about level of
disjuncture, or type of break index, which will be discussed in the following sections: 1)
using the “minu... | https://ocw.mit.edu/courses/6-911-transcribing-prosodic-structure-of-spoken-utterances-with-tobi-january-iap-2006/22ab414fc96a65b7bce85d1e08bc31b7_chap2_12.pdf |
, and the ambiguity in the signal,
by using the label 4- in the breaks tier. The labeller also put the corresponding phrase
accent-boundary tone combination of L-L% in the tones tier, corresponding to a 4 break.
The use of the 4- break index with L-L% in the tones tier indicates that the labeller
considered that a ... | https://ocw.mit.edu/courses/6-911-transcribing-prosodic-structure-of-spoken-utterances-with-tobi-january-iap-2006/22ab414fc96a65b7bce85d1e08bc31b7_chap2_12.pdf |
and the corresponding phrase accent-
boundary tone combination with the “&” (“ampersand”) symbol between them: 4&L-
L%.
Figure 2.12.16 Using the alt tier and the ? symbol to indicate uncertainty about the
choice of break index on sure at the end of the first intonational phrase
<sure2> (version 2)
2.12.6 Uncertain... | https://ocw.mit.edu/courses/6-911-transcribing-prosodic-structure-of-spoken-utterances-with-tobi-january-iap-2006/22ab414fc96a65b7bce85d1e08bc31b7_chap2_12.pdf |
choice of phrase accent and boundary tone at the end of the intonational phrase
<diagonal>
In cases where there is uncertainty about the phrase accent portion of a phrase accent-
boundary tone combination, the labeller should put the “?” diacritic following the
boundary tone in the tones tier, and list the alternati... | https://ocw.mit.edu/courses/6-911-transcribing-prosodic-structure-of-spoken-utterances-with-tobi-january-iap-2006/22ab414fc96a65b7bce85d1e08bc31b7_chap2_12.pdf |
-: high phrase accent
L-: low phrase accent
!H-: downstepped high phrase accent
Break indices:
0: word boundary erased
1: typical inter-word disjuncture within a phrase
3: end of an intermediate phrase
4: end of an intonational phrase
Optional labels:
<: late High Tonal peak
Uncertainty markers:
*?
uncertai... | https://ocw.mit.edu/courses/6-911-transcribing-prosodic-structure-of-spoken-utterances-with-tobi-january-iap-2006/22ab414fc96a65b7bce85d1e08bc31b7_chap2_12.pdf |
%
unable to determine whether L-L%, H-L% or !H-L%
H-X%? H-L% or H-H%
L-L% or L-H%
L-X%?
!H-L% or !H-H%
!H-X%? | https://ocw.mit.edu/courses/6-911-transcribing-prosodic-structure-of-spoken-utterances-with-tobi-january-iap-2006/22ab414fc96a65b7bce85d1e08bc31b7_chap2_12.pdf |
The 1-D Wave Equation
18.303 Linear Partial Differential Equations
Matthew J. Hancock
Fall 2006
1 1-D Wave Equation : Physical derivation
Reference: Guenther & Lee
1.2, Myint-U & Debnath
2.1-2.4
§
§
[Oct. 3, 2006]
We consider a string of length l with ends fixed, and rest state coinciding with
x-axis. The string... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
, t) be the angle wrt the horizontal x-axis. Note that
tan θ (x, t) = slope of tangent at (x, t) in ux-plane =
∂u
∂x
(x, t) .
(1)
1
Newton’s Second Law (F = ma) states that
F = (ρΔx)
∂2u
∂t2
(2)
where ρ is the linear density of the string (M L−1) and Δx is the length of the segment.
The force comes from t... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
.
(cid:18)
Substituting F from (2) into Eq. (4) and dividing by Δx gives
(cid:19)
ρ
∂2
u
∂t2
(ξ, t) = τ ∂x (x + Δx, t)
Δx
∂u
∂u
− ∂x (x, t)
for ξ
∈
[x, x + Δx]. Letting Δx
→
0 gives the 1-D Wave Equation
∂2u
∂t2
= c
2 ∂2u
,
∂x2
τ
2
c = > 0.
ρ
Force
Note that c has units [c] = Density
i
1.1 Boundar... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
balances the mass of the cylinder.
But τ = T cos θ = const and tan θ = ux, so that (6) becomes
Rearranging yields
τ ux = T cos θ tan θ = mg
ux =
mg
,
τ
x = 0, 1
These are Type II BCs. If the string is really tight and the cylinders are very light,
then mg/τ
0 at x = 0, 1, and the BCs become Type
II homogeneou... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
gives
utt (x, 0) = c 2 uxx (x, 0) = c 2f ′′ (x) ,
∂3u
∂t3
(x, 0) = c utxx (x, 0) = c g (x) .
2 ′′
2
Higher order terms can be found similarly. Therefore, the two initial conditions for
u (x, 0) and ut (x, 0) are sufficient to determine u (x, t) near t = 0.
To summarize, the dimensional basic 1-D Wave Problem with... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
(ˆ =
ˆ x)
T∗g (x)
L∗
.
∂u
∂x
= L∗
∂uˆ ∂xˆ
∂xˆ ∂x
=
∂uˆ
∂xˆ
,
∂u
∂t
= L∗
∂uˆ ∂tˆ
∂tˆ ∂t
=
L∗ ∂uˆ
T∗ ∂tˆ
and similarly for higher derivatives. Substituting the dimensionless variables into 1-D
Wave Equation (7) gives
uˆtˆtˆ =
T 2c2
∗
L2
∗
uˆxˆxˆ
This suggests choosing T∗ = L∗/c = l/c, so that
uˆx,
u... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
IC :
u (0, t) = 0 = u (1, t) ,
t > 0,
u (x, 0) = f (x) ,
ut (x, 0) = g (x) ,
0 < x < 1
2 Separation of variables solution
(10)
(11)
(12)
(13)
(14)
(15)
Ref: Guenther & Lee
4.2, Myint-U & Debnath
6.2, and
7.1 – 7.3
§
§
§
Substituting u (x, t) = X (x) T (t) into the PDE (13) and dividing by X (x) T (t)
g... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
0
and hence each eigenvalue λn corresponds to a solution Tn (t)
Tn (t) = αn cos (nπt) + βn sin (nπt) .
Thus, a solution to the PDE and BCs is
un (x, t) = (αn cos (nπt) + βn sin (nπt)) sin (nπx)
where we have absorbed the constant bn into αn, βn.
In general, the individual un (x, t)’s will not satisfy the ICs. Thu... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
and an IC, the wave equation has a unique solution (Myint-U
−n
& Debnath
6.3).
§
3
Interpretation - Normal modes of vibration
[Oct 5, 2006]
Ref: Guenther & Lee p. 100 problem 5
The terms
un (x, t) = (αn cos (nπt) + βn sin (nπt)) sin (nπx)
for n = 1, 2, 3, ... are called the normal modes of vibration. The solut... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
dimensions of time,
so does T . Note that each normal mode un (x, t) has period 2/n, and in physical
variables, un
′ (x ′ , t ′) has period 2l/ (nc).
The analog to period for spatial coordinates is the wavelength. The wavelength of
a function g (x) (x is a scaled or physical spatial coordinate) is defined as the sm... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
cos (2πfnt ′ ) + βn sin (2πfnt ′ )) sin
(cid:16)
(cid:16)
(cid:16)
sin
= (αn cos (ωnt ′ ) + βn sin (ωnt ′ )) sin
(cid:18)
nπx′
l
(cid:18)
nπx′
l
(cid:19)
(cid:19)
(cid:18)
nπx′
l
(cid:19)
The first harmonic is the normal mode of lowest frequency, u1 (x, t) or in physical
variables, u1
′ (x ′ , t ′).
The fund... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
sin nπ x +
(cid:18) (cid:18)
sin nπ t +
(cid:18) (cid:18)
2m
n
2m
n
(cid:19)(cid:19)
(cid:19)(cid:19)
=
sin nπx,
=
sin nπt,
cos nπ x +
(cid:18) (cid:18)
cos nπ t +
(cid:18) (cid:18)
2m
n
2m
n
(cid:19)(cid:19)
(cid:19)(cid:19)
= cos nπx
= cos nπt
for all m, n = 1, 2, 3, .. Therefore, the dimensionless so... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
whether
or not we are working with physical x, t (a length and time) or with dimensionless x,
t (dimensions 1). However, in general you should always ask the question, “What are
the dimensions?” The quantity c/l has dimensions of 1/time since I have defined c, l
to be a speed and a length. The argument of any mathem... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
An (x) sin (nπt + ψn)
An (x) = γn sin (nπx) .
The mode un (x, t) vibrates sinusoidally in time between the two limits
An (x). We
call A (x) the time amplitude of the mode un (x, t), since this sets the bounds on the
oscillations in time. Locations where An (x) = 0 are called nodes and locations where
An (x)
|
|
=... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
resistive forces. Once
plucked, the string vibrates/oscillates forever, with period 2l/c in physical coordi
nates. We call this type of system conservative: energy is conserved. In addition, the
system comes back to its initial condition periodically - i.e.
it maintains a memory
of its initial state.
We define the... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
+
2
1 Δu
2 Δx
(cid:18)
(cid:19)
−
1 + O
Δu
Δx
(cid:19)
(cid:18)
4
!!
and hence the potential energy is
2
Δx Δu
≈ 2 Δx
(cid:18)
(cid:19)
P E ′ (x ′ , t ′ ) = τ Δl
2
τ Δu
2 Δx
(cid:18)
(cid:19)
≈
Δx
≈
2
τ
2
∂u
∂x
(cid:18) (cid:19)
Δx
9
as Δx
→
0. Thus, the total energy is, in dime... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
the form (19)
into Eq. (21) for the energy,
1
Z
1
2
0
(cid:18)
(nπγn)2
2
(nπγn)2
2
En (t) =
=
=
Note that
∂un
∂t
1
2
+
(cid:19)
(cid:18)
∂un
∂x
2
!
(cid:19)
dx
sin2 (nπx) cos2 (nπt + ψn) + cos 2 (nπx) sin2 (nπt + ψn)
dx
(cid:1)
0
Z
(cid:0)
cos 2 (nπt + ψn)
1
(cid:18)
Z
+ sin2 (nπt + ψn)
sin2... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
conserved, we proceed as follows. Dif
ferentiating the dimensionless energy equation (21) gives
dE
dt
=
1
0
Z
(ututt + uxuxt) dx
10
Replacing utt = uxx (the wave PDE) gives
(utuxx + uxuxt) dx
1
Z
0
1
dE
dt
=
=
(utux)x dx
Z
0
= [utux]x=0
1
Differentiating... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
1 ∂un ∂um
∂t ∂t
0
n=1 m=1
X
X
Z
(cid:18)
∂
∂x
un
n=1
X
∂un ∂um
∂x ∂x
+
2
dx
dx
!
!
+
∂un ∂um
∂x ∂x
dx
(cid:19)
Substituting 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 ... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
nπt + ψn) + sin2 (nπt + ψn)
(cid:1)
which is constant for all t.
5 D’Alembert’s Solution for the wave equation
Ref: Guenther & Lee
§
4.1, Myint-U & Debnath
4.3
§
5.1 Motivation
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 (n... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
2v
∂η2
=
∂2v
∂ξ2 −
2
∂2v
∂ξ∂η
+
∂2v
∂η2
Simplifying and dividing by 4 gives a new form of the wave equation,
∂2v (ξ, η)
∂ξ∂η
= 0
5.3 Forward and backward waves
We can write the new form (25) of the wave equation in two ways:
∂
∂ξ
(cid:18)
∂v (ξ, η)
∂η
(cid:19)
= 0,
∂
∂η
(cid:18)
∂v (ξ, η)
∂ξ
(cid:... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
,
we can write
v (ξ, η) = P (ξ) + Q (η)
Substituting for η and ξ from (24) and recalling that u (x, t) = v (ξ, η) gives
u (x, t) = v (ξ, η) = P (x
−
t) + Q (x + t)
(27)
5.3.1 Forward wave
The forward wave is the function P (x
t) which represents a wave travelling in
−
the positive x-direction with scaled velo... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
along the lines x + t = const (in physical variables,
x ′ + ct ′ = const and speed is c).
5.4 Characteristics
Ref: Myint-U & Debnath
3.2(A)
§
The solution to the wave equation is the superposition of a forward wave P (x
t)
t = const are called
−
and a backward wave Q (x + t), both with speed c. The lines x
cha... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
)
(Q (0)
P (0)) =
−
Eqs. (28) and (30) can be solved for P (x) and Q (x):
−
−
x
0
Z
g (s) ds
x
f (x) +
g (s) ds + Q (0)
−
P (0)
(cid:19)
Q (x) =
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
[Oc... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
t = t0 is
u (x0, t0) =
1
2
[f (x0 −
t0) + f (x0 + t0)] +
1
2
x0+t0
x0−t0
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 initia... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
down D’Alembert’s solution,
u (x, t) =
1
2
[f (x
−
t) + f (x + t)] +
1
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... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
3
x+t=−1
(t)
x
A
R
5
−1.5
−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).... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
1
}
(35)
The regions determine 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, w... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
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.
1 so that
x+t ... | 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 sets Rn wit... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
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) + F (x + 0)) +
1
2
x+0
x−0
Z
G (s) ds = F (x)
Similarly, we can check that in the other regions,... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
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 the intersection of Rn with the line t = 2:
R5 = (
3],
−∞
,
−
R3 = [
R6 = [3,
(41)
At this point, you usually want ... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
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
1
1
G (s) ds,
2 x−t
R
+t G (s) ds,
x
1
2 −
R
1
1
1
G (s) ds,
2 −1
R
0
R
(x, t)
(x, t)
(x, t)
(x, t)... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
+
−
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)) ,
1 ,
2
0
4π
4
(x, t)
(x, t)
(x, t)
(x, t)
R1
∈
R2
∈
R3
∈
R4
∈
R5, R6
(42)
(x, t)
∈
u (x, t) =
Step 4. We consi... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
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
0
x
1
2
3
4
Figure 2: The profiles of the displacement u(x, t) for the times t = 0, 1/2, 1, 2.
At t = 1, the regions Rn are given by (40) and (42) ... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
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
We now consider D’Alembert’s solution for a finite string. The dimensionless
4.5
§
§
21
problem with homogeneous Type I BCs (ends of string fix... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
+ t)) + βn cos (nπ (x + t)))
6.1 Zero initial velocity
The simplest ICs involve the case of zero initial velocity ut (x, 0) = 0 and a specified
initial displacement u (x, 0) = f (x). In this case, βn = 0 for all n and
u (x, t) =
∞
n=1
X
where
αn cos (nπt) sin (nπx) = P (x
−
t) + Q (x + t)
P (s) = Q (s) =
fˆ(s... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
fˆ(1 + t)
−
Thus,
u (0, t) =
1
fˆ(
2
(cid:16)
fˆ(1
t) + fˆ(t) = 0
−
(cid:17)
t) + fˆ(1 + t) = 0
u (1, t) =
1
2
−
Therefore, (45) satisfies the Type I BCs (fixed ends). One can also check directly
that (45) satisfies the ICs and PDE.
(cid:17)
(cid:16)
To summarize, we have written the displacement of the string... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
peak of the forward wave
(at x = 1/2) first reaches the end of the string x = 1 when
t =
1
−
1
2
t =
=
⇒
1
2
At the same time, the peak of the backward wave reaches the end x = 0. Then what
happens? On the first half of the string 0
1/2 and times 1/2
x
1,
t
≤
≤
≤
≤
and thus
1
−
≤
x
−
0
t
≤
fˆ(x
t) =
fˆ... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/22ead9d70b36836a68d13c7393e19649_waveeqni.pdf |
end x = 1 (also at t = 1/2) and upside down (negative sign). At t = 1, the
waves meet in the center,
−
−
−
u (x, 1) =
1
2
fˆ(x
(cid:16)
−
1) + fˆ(x + 1) =
1
2
fˆ(x)
−
−
fˆ(x) =
(cid:17)
(cid:17)
(cid:16)
fˆ(x) =
−
f (x)
−
and the displacement is the upside-down version of the initial displacement. The
t criss... | 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 |
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 dimensions will be macroscopic.
2. We will work in the limit of strong coupling. The mo... | https://ocw.mit.edu/courses/8-821-string-theory-fall-2008/22f010a84df39d4448ec5ff59382413d_lecture04.pdf |
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-SUSY theories,
if one takes the strong-coupling limit, it tends not to exist. The examples include QED
(where one hits the landau Pole for strong couplings) an... | https://ocw.mit.edu/courses/8-821-string-theory-fall-2008/22f010a84df39d4448ec5ff59382413d_lecture04.pdf |
al anal
z . The most general
ysis suggests that this will scale under the scale transformation, so z
λ
five-dimensional metric (one extra dimension) with this symmetry and Poincare invariance is
of the following form:
→
→
z˜
L˜
dz˜2
L˜ )2ηµν dxµdxν + z˜2 L2 . Let ˜z = L z˜
˜
ds2 = (
L = L We can now bring it into t... | https://ocw.mit.edu/courses/8-821-string-theory-fall-2008/22f010a84df39d4448ec5ff59382413d_lecture04.pdf |
defines a theory of gravity on spaces that are asymptotically AdS5.
S5 .
Why is AdS5 a solution?
1. Check on PSet
2. Use effective field theory (which in this case just means dimensional reduction) to elaborate
on my previous statement that the reason why this is a solution is that ”the flux holds it
up”. To do this ... | https://ocw.mit.edu/courses/8-821-string-theory-fall-2008/22f010a84df39d4448ec5ff59382413d_lecture04.pdf |
Fg)n1...nq
We should then think of R(x) as a ”moduli field”. Now we have to evaluate the following
integral Sq √G
�
We will now go to something called the FRAME GAME
To make the D-dimensional Einstein-Hilbert term canonical, we can do a Weyl-rescaling of
the metric, meaning to define a new metric gE =
So the new met... | https://ocw.mit.edu/courses/8-821-string-theory-fall-2008/22f010a84df39d4448ec5ff59382413d_lecture04.pdf |
ective 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
Rmin .
value of this we take 0 = V
The fact that the potential goes to zero at large R is unavoidable. It is because when R is
large, the curvature is small and fl... | https://ocw.mit.edu/courses/8-821-string-theory-fall-2008/22f010a84df39d4448ec5ff59382413d_lecture04.pdf |
−
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 way to compute curvatures, which is to use what is called “tetrad” or
“vielbein” or “Cartan-Weyl” method.
There are three steps
Rewrite the metric as ds2 = (erˆ)2 + eµˆeνˆηµν ,where I define e... | https://ocw.mit.edu/courses/8-821-string-theory-fall-2008/22f010a84df39d4448ec5ff59382413d_lecture04.pdf |
ˆνˆ = ωµˆrˆωrˆνˆ =
L
−
eνˆ
∧
Rµˆrˆ = dωµˆrˆ = L
1
2 erˆ
Rµν
ρσ =
−
1
2 (δρ
L
µδν
σ −
eµˆ
∧
δµ
σ δν
ρ )
Rµr
ρr =
µ
1
− L2 δν
−d
Rµν = L2 gµν
2Λ
d
2−d = L2
⇒
⇒
Λ =
−d(d+1)
2L2
Counting of degrees of freedom [Susskind-Witten, hep-th/9805114]
The holographic principle tells us that the area of the bounda... | https://ocw.mit.edu/courses/8-821-string-theory-fall-2008/22f010a84df39d4448ec5ff59382413d_lecture04.pdf |
on a lattice, introducing a low distance
cut-off δ and we put it also in a box of size R. The number of degrees of freedom is the number of
3
cells times N 2, so Nd = R
N 2 . To regulate the integral we write before, we stop not at r = 0 but
δ3
we cut it off at r = δ. Aδ =
�
L3
3
�
�
3
d3x L
r3
|r=δ = R
3
δ
Now w... | 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 |
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. Compute (a, n): if it is > 1, then n is composite.
3. Otherwise compute an−1: if (cid:54)≡ 1 mod a then done, n is co... | https://ocw.mit.edu/courses/18-781-theory-of-numbers-spring-2012/22f790db07c43592369000a37c31d487_MIT18_781S12_lec6.pdf |
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 (Agrawal, Kayal, Saxena). The main idea is that n > 2 is prime if and
only if
(x − a)n ≡ xn − a (mod n) as polynomials
Check different values of a, but there are n possible choices of a and expa... | https://ocw.mit.edu/courses/18-781-theory-of-numbers-spring-2012/22f790db07c43592369000a37c31d487_MIT18_781S12_lec6.pdf |
A and B have some shared
key. With a message m, A can send m + k, or m ⊕ k (where ⊕ is the bitwise
exclusive OR). B can decrypt the message by subtracting k: (m + k) − k = m,
(m ⊕ k) ⊕ k = m. This is not so good if we want to send multiple messages - if C
sees m1 + k and m2 + k, then C can figure out m1 − m2, which give... | https://ocw.mit.edu/courses/18-781-theory-of-numbers-spring-2012/22f790db07c43592369000a37c31d487_MIT18_781S12_lec6.pdf |
� 0 mod pj,
and f (cid:48)(a) (cid:54)≡ 0 mod p. Then there’s a unique t mod p such that f (a + tpj)
0
mod pj+1. That is, there’s a unique solution b mod pj+1
which is congruent to a mod
pj, (ie., b reduces to a mod pj, a lifts to b mod pj+1).
≡
Proof. We’re looking for solutions b = a + tpj where t ∈ {0, 1, . . . p − ... | 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 |
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: John
A variable symbol: ?x
The Application of a function symbol to set of terms: (Brother (Cousin Joh... | https://ocw.mit.edu/courses/6-871-knowledge-based-applications-systems-spring-2005/233e2e8021787d40f9c4e73e1f06283d_lect08_logic.pdf |
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:
A Sequence of statements, such that each element is either:
An Axiom
An Assumption warranted by a proof rule
Or the results of applying a deduc... | https://ocw.mit.edu/courses/6-871-knowledge-based-applications-systems-spring-2005/233e2e8021787d40f9c4e73e1f06283d_lect08_logic.pdf |
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 [a/x](P x).
It means that every free occurrence of x is replaced by a in the statement (P x).
The substitution is only valid if no occurrence of a is “captured” (... | https://ocw.mit.edu/courses/6-871-knowledge-based-applications-systems-spring-2005/233e2e8021787d40f9c4e73e1f06283d_lect08_logic.pdf |
Rules
GIGO
i (not A)
j A
k C
(GIGO i j)
Indirect Proof
i (Show A)
i+1
J
k
l A
| (not A)
| B
| (not B)
(IP (j k) (i+1))
Cut
i (Or A B)
j (not A)
k B
(Cut i j)
Modus Tolens
i (Implies A B)
j (not B)
k (not A)
(MT i j)
DeMorgan's Rules
(not (or A B)) = (and (not A) (not B))
(not (And A B)) = (or (not ... | https://ocw.mit.edu/courses/6-871-knowledge-based-applications-systems-spring-2005/233e2e8021787d40f9c4e73e1f06283d_lect08_logic.pdf |
)
13. | (Implies R
(Or (And P Q) (And P R)))
14. | (Or (And P Q) (And P R))
15. (Implies (And P (Or Q R))
(Or (And P Q) (And P R)))
Assumption motivated by 10
And Introduction 3,11
Conditional Proof 13 (11)
Or Elimination 4,9,13
Conditional Proof 14 (2)
Logic : Page 12
Logic : Page 12
Another Proof
1. Show... | https://ocw.mit.edu/courses/6-871-knowledge-based-applications-systems-spring-2005/233e2e8021787d40f9c4e73e1f06283d_lect08_logic.pdf |
) (Thereis (?y) (P ?x ?y)))
3. | (Thereis (?y) (P b ?y))
4. | (P b a)
5. | (Forall (?w) (P ?w a))
6. | (Forall (?x) (P ?x a))
7. | (Thereis (?z) (Forall (?x) (P ?x ?z)))
8. | (Thereis (?y) (Forall (?x) ( P ?x ?y)))
9.
(Implies (Forall (?x) (Thereis (?y) ( P ?x ?y)))
Assumption motivated by 1
Universal Instant... | https://ocw.mit.edu/courses/6-871-knowledge-based-applications-systems-spring-2005/233e2e8021787d40f9c4e73e1f06283d_lect08_logic.pdf |
) (Or (F .. x ..) (H .. x ..)))
(5) (Or (Q1 (x) (F .. x ..)) (Q2 (x) (H .. x ..))) = (Q1 (x) (Q2 (z) (Or (F .. x ..) (H .. z ..))))
(6) (And (Q1 (x) (F .. x ..)) (Q2 (x) (H .. x ..))) = (Q1 (x) (Q2 (z) (And (F .. x ..) (H .. z ..))))
Negation Rules
(1) (Not (Not A)) = A
(2) (Not (There-is (x) (F .. x ..))) = (For-... | https://ocw.mit.edu/courses/6-871-knowledge-based-applications-systems-spring-2005/233e2e8021787d40f9c4e73e1f06283d_lect08_logic.pdf |
Drop all the Thereis Q’s replacing their variables by functions of the universally
quantified variables.
Drop all the Forall Q’s leaving the variables free.
Get a set of quantifier free disjunctions.
Logic : Page 16
Logic : Page 16
Resolution Rule
G)
(Or P
(OR (Not Q) H)
If P and Q can be unified (with unifying ... | https://ocw.mit.edu/courses/6-871-knowledge-based-applications-systems-spring-2005/233e2e8021787d40f9c4e73e1f06283d_lect08_logic.pdf |
?x ?y)
Logic : Page 19
Logic : Page 19
Examples
• Forward Chaining:
Assert (Parent Abe Ike)
Assert (Gender Abe male)
Assert (Parent Ike Jake)
Deduce (Grandparent Abe Jake)
Deduce (Grandfather Abe Jake)
• Backward Chaining:
Same Facts, plus (Father Abe Ishmael).
Goal (Grandfather Abe Jake)
Goal (Grandparent ... | https://ocw.mit.edu/courses/6-871-knowledge-based-applications-systems-spring-2005/233e2e8021787d40f9c4e73e1f06283d_lect08_logic.pdf |
theorem is a valid statement
– The system is called consistent.
– You can make only sound deductions.
•
• A logicians day is made when a proof of soundness and
completeness is obtained.
Logic : Page 22
Logic : Page 22
What Formal Properties Hold?
Decidability
•
Is there an algorithm for deciding whether or not... | https://ocw.mit.edu/courses/6-871-knowledge-based-applications-systems-spring-2005/233e2e8021787d40f9c4e73e1f06283d_lect08_logic.pdf |
p x))
Logic : Page 25
Logic : Page 25
Using Logic in Practical Applications:
• Weaker logical systems seem to be strong enough in practice to be
useful and yet still controllable.
• Logical Rule Languages are examples:
– A1 & A2 & ... & An --> C1 & ...
– The A's are called antecedents
– The B's are called c... | https://ocw.mit.edu/courses/6-871-knowledge-based-applications-systems-spring-2005/233e2e8021787d40f9c4e73e1f06283d_lect08_logic.pdf |
more complicated than needed.
Logic : Page 28
Logic : Page 28
What are the "Logic Glasses" ?
• The statement is the appropriate unit of modularity.
•
Information is correctly captured in the form of single independent
statements and single independent inferences.
• Think about what's true and what follows from it... | 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 |
" + 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
var get_json_status = function () {
var url = 'http://localhost:3000/status.json';
$.ajax({ url : url,
success : func... | 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 |
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] and H(z) in OSB Figure 4.31(a). The bottom subfigure illustrates the ... | https://ocw.mit.edu/courses/6-341-discrete-time-signal-processing-fall-2005/23887e05bb83d893874e27d49f7135ba_lec11.pdf |
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 noble identity give an equivalent system which requires fewer
multiplic... | 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 |
≤
(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 n is specified by a sequence �0 , �1 , . . . , �n of vertices
72
|
|
|
|
|
−
of Y . We will ... | https://ocw.mit.edu/courses/18-318-topics-in-algebraic-combinatorics-spring-2006/23b76f41bf5e26f7221734e3499e73fb_notes2.pdf |
(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, . . . , �m) as �1 · · ·
of type UU DDUUUU DUU = U 2D2U 4DU 2 .
n−1 · · ·
A2A1.
|
There is a nice combinatorial interpretation of walks of type U n which
b... | https://ocw.mit.edu/courses/18-318-topics-in-algebraic-combinatorics-spring-2006/23b76f41bf5e26f7221734e3499e73fb_notes2.pdf |
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 � denote the number of SYT of shape �, so for instance f (2,2,1) = 5.
The numbers f � have many interesting properties; for instance, there is a
famous explicit formula for them known as the Frame-Robinson-Thrall hook
formula. We will be concerned with ... | https://ocw.mit.edu/courses/18-318-topics-in-algebraic-combinatorics-spring-2006/23b76f41bf5e26f7221734e3499e73fb_notes2.pdf |
will obtain by
such a formula should exist.) In particular, since f
setting � = Ø a simple formula for the number of (closed) Hasse walks of
type w from Ø to Ø (thus including a simple formula for (40)).
There is an easy condition for the existence of any Hasse walks of type
w from Ø to �, given by the next lemma. ... | https://ocw.mit.edu/courses/18-318-topics-in-algebraic-combinatorics-spring-2006/23b76f41bf5e26f7221734e3499e73fb_notes2.pdf |
to the reader. �
If w is a word in U and D satisfying the conditions of Lemma 8.1, then
we say that w is a valid �-word. (Note that the condition of being a valid
�-word depends only on � .)
|
|
The proof of our formula for �(w, �) will be based on linear transforma
tions analogous to those defined by (18) and (19).... | https://ocw.mit.edu/courses/18-318-topics-in-algebraic-combinatorics-spring-2006/23b76f41bf5e26f7221734e3499e73fb_notes2.pdf |
. Apply the left-hand side of (41) to a partition � of i, expand
in terms of the basis Yi, and consider the coefficient of a partition µ. If
µ = � and µ can be obtained from � by adding one square s to (the Young
∪diagram of) � and then removing a (necessarily different) square t, then there
is exactly one choice of s... | https://ocw.mit.edu/courses/18-318-topics-in-algebraic-combinatorics-spring-2006/23b76f41bf5e26f7221734e3499e73fb_notes2.pdf |
it
back in. Hence when we apply the left-hand side of (41) to �, the coefficient
of � is equal to 1.
Combining the conclusions of the three cases just considered shows that
the left-hand side of (41) is just Ii, as was to be proved. �
We come to one of the main results of this section.
Let � be a partition and w = ... | https://ocw.mit.edu/courses/18-318-topics-in-algebraic-combinatorics-spring-2006/23b76f41bf5e26f7221734e3499e73fb_notes2.pdf |
RYi. We illustrate the proof for the special case w = DU λ DU � DU �, where
�, λ, β
0, from which the general case will be clear. By the definition of w
≡
we have
�(w, �) = [�]w(Ø)
= [�]DU λ DU � DU �(Ø).
We will use the identity (easily proved by induction on i)
DU i = U iD + iU i−1 .
(43)
Thus
w(Ø) = DU λ D... | https://ocw.mit.edu/courses/18-318-topics-in-algebraic-combinatorics-spring-2006/23b76f41bf5e26f7221734e3499e73fb_notes2.pdf |
(f �)2 = n!
�
��n
Proof. When w = DnU n in Theorem 8.3 we have Sw =
i, and bi = n, from which the proof is immediate. �
n + 1, n +
{
2, . . . , 2n
, ai = n
}
−
√
Note (for those familiar with the representation theory of finite groups).
It can be shown that the numbers f �, for �
n, are the degrees of the irre... | https://ocw.mit.edu/courses/18-318-topics-in-algebraic-combinatorics-spring-2006/23b76f41bf5e26f7221734e3499e73fb_notes2.pdf |
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