text stringlengths 16 3.88k | source stringlengths 60 201 |
|---|---|
,j
Di−1,j−1 + �(S1(i), S2(j))
Vi−1,j − �
D
i−1,j − � − �
Multiple Alignment
S1, S2, S3, · · · , Sd are the sequences to be aligned.
�(S1, S2, · · ·) d-way score measures the distance for all possible pairwise alignments
as shown in the example below.
O(nd) because all possible pairs need to be evaluated.
Three ... | https://ocw.mit.edu/courses/18-417-introduction-to-computational-molecular-biology-fall-2004/4acde14888af72996f60ab92a9a430bf_lecture_05.pdf |
clustal (based on aligning to an alignment)
• does not use ( �
2 ) pairwise alignments, instead aligns first pair and then aligns
next sequence to the existing alignment and then continues until all sequences
have been aligned.
Common implementations of progressive alignment algorithms are called Clustal W
and Clu... | https://ocw.mit.edu/courses/18-417-introduction-to-computational-molecular-biology-fall-2004/4acde14888af72996f60ab92a9a430bf_lecture_05.pdf |
Computational Ocean
Acoustics
• Ray Tracing
• Wavenumber Integration
• Normal Modes
• Parabolic Equation
13.853
COMPUTATIONAL OCEAN ACOUSTICS
Lecture 19
Parabolic Equation
• Mathematical Derivation (6.2)
– Phase Errors and Angular Limitations (6.2.4)
• Starting Fields (6.4)
– Modal starter
– PE Self Starter
– Analyti... | https://ocw.mit.edu/courses/2-068-computational-ocean-acoustics-13-853-spring-2003/4afbf5c07b1676c30dbea0bb4ef2dda1_lect_19.pdf |
Energy Conserving
13.853
COMPUTATIONAL OCEAN ACOUSTICS
Lecture 19
Student Demos
Wavenumber Integration Models
13.853
COMPUTATIONAL OCEAN ACOUSTICS
Lecture 19 | https://ocw.mit.edu/courses/2-068-computational-ocean-acoustics-13-853-spring-2003/4afbf5c07b1676c30dbea0bb4ef2dda1_lect_19.pdf |
18.405J/6.841J:
Advanced Complexity Theory
Spring 2016
Prof. Dana Moshkovitz
Lecture 5: Toda's Theorem
Scribe: Ilya Razenshteyn
Scribe Date: Fall 2012
1 Overview
In the last lecture we covered Valiant–Vazirani reduction and investigated the complexity of ap-
proximate counting.
In this lecture we prove Toda’s Theorem:... | https://ocw.mit.edu/courses/18-405j-advanced-complexity-theory-spring-2016/4b169a1c4e9ce5a4e0e1ba2ace2b2eba_MIT18_405JS16_Todas.pdf |
), and a parameter m and
outputs a formula (cid:76)
y ψ(x, y) such that for every x
Pr[ϕ(x) =
(cid:77)
y
ψ(x, y)]
≥ 1 − 2−m.
The running time of the reduction is (nm)Oc(1).
1
To prove this theorem we first need to establish some properties of (cid:76)-formulae.
3.1 Properties of Parity-Quantified Formulae
Let ϕ, ψ : {0,... | https://ocw.mit.edu/courses/18-405j-advanced-complexity-theory-spring-2016/4b169a1c4e9ce5a4e0e1ba2ace2b2eba_MIT18_405JS16_Todas.pdf |
y
Now by union bound we have that with probability at least 1 − 2−(m+10) we have
ψ(x1, x) for every x, x1.
remove the outer quantifier. For this we use Valiant–Vazirani
somewhat “abstract” setting.
In this case we have ϕ(x) = ∃x1
y τ (x1, x, y) =
y τ (x1, x, y), and it is only left to
reduction. Let us first recall it in... | https://ocw.mit.edu/courses/18-405j-advanced-complexity-theory-spring-2016/4b169a1c4e9ce5a4e0e1ba2ace2b2eba_MIT18_405JS16_Todas.pdf |
failure is at most 2−(m+9) (cid:28) 2−m.
It is left to observe that we can (and should!) transform (1) to a ⊕SAT instance using tricks from
Section 3.1.
4 Derandomization
In this section we show how to derandomize Theorem 3 at a cost of switching from ⊕SAT to #SAT.
We can think about the reduction from the proof of The... | https://ocw.mit.edu/courses/18-405j-advanced-complexity-theory-spring-2016/4b169a1c4e9ce5a4e0e1ba2ace2b2eba_MIT18_405JS16_Todas.pdf |
:48)) ≡ 0 or −1 (mod 2t). We will show
3
how to switch from 2t to 22t. Namely, consider the formula γ = 4γ(cid:48)3 + 3γ(cid:48)4 (here by arithmetic
operations we mean the tricks from Section 3.1). It turns out that it gives exactly what we need!
If #(γ(cid:48)) ≡ 0 (mod 2t), then #(γ) ≡ 0 (mod 22t), but if #(γ(cid:4... | https://ocw.mit.edu/courses/18-405j-advanced-complexity-theory-spring-2016/4b169a1c4e9ce5a4e0e1ba2ace2b2eba_MIT18_405JS16_Todas.pdf |
Lecture Notes on Wave Optics (04/07/14)
2.71/2.710 Introduction to Optics –Nick Fang
Outline:
A. Imaging with coherent light
B. Optical Spatial Filtering
C. The significance of PSF and ATF, and effect of coherence
D. Phase Contrast Imaging: Zernike and Schlieren methods
A. Imaging with Coherent Light
Recap: a c... | https://ocw.mit.edu/courses/2-71-optics-spring-2014/4b53a5747f9b58b9b73b2bbcc39945f0_MIT2_71S14_lec16_notes.pdf |
Lecture Notes on Wave Optics (04/07/14)
2.71/2.710 Introduction to Optics –Nick Fang
By cascading two lenses together, we can reveal Abbe’s theory of imaging process:
Ideally, applying two forward Fourier transforms recovers the original function of
the object field, with a reversal in the coordinates:
𝐸𝑖𝑚𝑎𝑔�... | https://ocw.mit.edu/courses/2-71-optics-spring-2014/4b53a5747f9b58b9b73b2bbcc39945f0_MIT2_71S14_lec16_notes.pdf |
−
𝑓2
𝑓1
𝑥, −
𝑓2
𝑓1
𝑦) (6)
Potentially, the magnification ratio 𝑀 = 𝑓2/𝑓1can be arbitrarily large. This however
does not mean that the microscope is able to resolve arbitrarily small objects. The
finite size of the aperture stop, and the corresponding transmission 𝐴𝑆(𝑥′, 𝑦′) will
contribute to the above... | https://ocw.mit.edu/courses/2-71-optics-spring-2014/4b53a5747f9b58b9b73b2bbcc39945f0_MIT2_71S14_lec16_notes.pdf |
��
𝑓1
𝑘
, 𝑘𝑦
𝑓1
𝑘
)]
(8)
Note: In Goodman’s book, the term𝐴𝑆(𝑘𝑥
𝑓1
𝑘
, 𝑘𝑦
𝑓1
𝑘
) is called Amplitude Transfer
Function(ATF), and its Fourier transform, ℱ [𝐴𝑆(𝑘𝑥
Function(PSF) (since it is the spread of an ideal point source 𝛿(𝑥, 𝑦) at the image).
)] is called Point Spread
, 𝑘𝑦
𝑓1
𝑘
... | https://ocw.mit.edu/courses/2-71-optics-spring-2014/4b53a5747f9b58b9b73b2bbcc39945f0_MIT2_71S14_lec16_notes.pdf |
��𝑐 (
𝑦)]
=[𝑎𝑠𝑖𝑛𝑐 (−
𝑎𝑘
𝑓2
𝑥")] [𝑏𝑠𝑖𝑛𝑐 (−
𝑎𝑘
𝑓1
𝑏𝑘
𝑓2
𝑓1
𝑓2
𝑏𝑘
𝑓1
𝑦")]
(10)
3
a/λf1b/λf1ATFPSF
Lecture Notes on Wave Optics (04/07/14)
2.71/2.710 Introduction to Optics –Nick Fang
2) Circular apertures:
𝐴𝑇𝐹 = 𝑐𝑖𝑟𝑐 (
𝑓1
𝑅𝑘
√𝑘... | https://ocw.mit.edu/courses/2-71-optics-spring-2014/4b53a5747f9b58b9b73b2bbcc39945f0_MIT2_71S14_lec16_notes.pdf |
)
(12)
B. Optical Spatial Filtering
Spatial Filtering is a technique to process signals in an optical way, where the
irradiance content in the Fraunhofer plane is manipulated to control the irradiance
pattern in the image plane. A digital element to provide Spatial Filtering in a
dynamical way is coined as a spat... | https://ocw.mit.edu/courses/2-71-optics-spring-2014/4b53a5747f9b58b9b73b2bbcc39945f0_MIT2_71S14_lec16_notes.pdf |
filter in the
microscopy and used frequently in materials science, as it allows only light
scattered from sharp edges (such as grain boundaries) to pass through the filter.
are
𝑓1
𝑎
5
Low pass filterf1f1f2f2InputOutput ScreenHigh pass filterf1f1f2f2InputOutput Screen
Lecture Notes on Wav... | https://ocw.mit.edu/courses/2-71-optics-spring-2014/4b53a5747f9b58b9b73b2bbcc39945f0_MIT2_71S14_lec16_notes.pdf |
Lecture Notes on Wave Optics (04/07/14)
2.71/2.710 Introduction to Optics –Nick Fang
system is shift-invariant, then the observed image is a sum of all the field
E(x”, y”) contributed by the spread of each point (x, y) from the object.
Under (spatially) coherent illumination, the image field is a convolution of
... | https://ocw.mit.edu/courses/2-71-optics-spring-2014/4b53a5747f9b58b9b73b2bbcc39945f0_MIT2_71S14_lec16_notes.pdf |
𝑖𝑚𝑎𝑔𝑒(𝑥", 𝑦") = 𝐼𝑜𝑏𝑗𝑒𝑐𝑡 (−
𝑥, −
𝑦) ⨂|𝑃𝑆𝐹(𝑥, 𝑦)|2
𝑓2
𝑓1
𝑓2
𝑓1
𝑂𝑇𝐹(𝑘𝑥, 𝑘𝑦) = ∫ ∫|𝑃𝑆𝐹(𝑥, 𝑦)|2 exp(𝑖𝑘𝑥𝑥 + 𝑖𝑘𝑥𝑦) 𝑑𝑥𝑑𝑦 = 𝐴𝑇𝐹 ⊗ 𝐴𝑇𝐹
e.g. OTF of single square aperture:
|𝑂𝑇𝐹| = Λ (
𝑓1𝑘𝑥
𝑎𝑘
) Λ(
𝑓1𝑘𝑦
𝑏𝑘
)
(17)
(18)
(19)
7
E(x, y)E(x”, y”)
... | https://ocw.mit.edu/courses/2-71-optics-spring-2014/4b53a5747f9b58b9b73b2bbcc39945f0_MIT2_71S14_lec16_notes.pdf |
’s consider a transparent object with a small phase shift in the following form:
t(𝑥, 𝑦) = exp(𝑖𝑘(𝑛 − 1)ℎ(𝑥, 𝑦)) ≈ 1 + 𝑖𝑘(𝑛 − 1)ℎ(𝑥, 𝑦)
(20)
When the transparent object is uniformly illuminated by a plane wave, the
transmitted intensity is close to unity, leaving very low contrast. The idea behind
the ... | https://ocw.mit.edu/courses/2-71-optics-spring-2014/4b53a5747f9b58b9b73b2bbcc39945f0_MIT2_71S14_lec16_notes.pdf |
/2.710 Introduction to Optics –Nick Fang
𝑡(𝑥, 𝑦) = exp (𝑖
𝜋
2
) + 𝑖𝑘(𝑛 − 1)ℎ(𝑥, 𝑦) = 𝑖(1 + 𝑘(𝑛 − 1)ℎ(𝑥, 𝑦))
(21)
𝐼(𝑥, 𝑦) ∝ |t(𝑥, 𝑦)|2 = 1 + 2𝑘(𝑛 − 1)ℎ(𝑥, 𝑦) + 𝑂(ℎ2)
(22)
Now the transmitted intensity reflects the phase information. Actually, since the
intensity with phase change is nearly... | https://ocw.mit.edu/courses/2-71-optics-spring-2014/4b53a5747f9b58b9b73b2bbcc39945f0_MIT2_71S14_lec16_notes.pdf |
𝑦
𝑓1
𝑘
) × ℱ (𝐸𝑜𝑏𝑗𝑒𝑐𝑡(𝑥, 𝑦)))
𝐸𝑖𝑚𝑎𝑔𝑒(𝑥", 𝑦") ≈ ℱ ((1 + 𝑖(∆𝑛)𝑘𝑥
𝑓1
𝑎
) × ℱ (𝐸𝑜𝑏𝑗𝑒𝑐𝑡(𝑥, 𝑦)))
𝐸𝑖𝑚𝑎𝑔𝑒(𝑥", 𝑦") ≈ 𝐸𝑜𝑏𝑗𝑒𝑐𝑡(𝑥, 𝑦) + (∆𝑛)
𝑓1
𝑎
ℱ (ℱ (
𝜕
𝜕𝑥
𝐸𝑜𝑏𝑗𝑒𝑐𝑡(𝑥, 𝑦)))
(25)
9
Wedge Or spiral phase platef1f1f2f2Phase object𝜙(𝑥)Output Screen𝑡𝑥′≈1+𝑖𝑘... | https://ocw.mit.edu/courses/2-71-optics-spring-2014/4b53a5747f9b58b9b73b2bbcc39945f0_MIT2_71S14_lec16_notes.pdf |
�
𝜕
𝜕𝑥
𝐸𝑜𝑏𝑗𝑒𝑐𝑡(𝑥, 𝑦)
𝐸𝑖𝑚𝑎𝑔𝑒(𝑥", 𝑦") ≈ 𝐸𝑜𝑏𝑗𝑒𝑐𝑡(𝑥, 𝑦) [1 + 𝑖(∆𝑛)
𝑓1
𝑎
𝜕
𝜕𝑥
𝜙(𝑥, 𝑦)]
(26)
(27)
Note that using a mask with phase gradient, the intensity fringes of image are
connected to the index gradient of the fluid flow! Such effect was first reported
by Hooke and Huygens, ... | https://ocw.mit.edu/courses/2-71-optics-spring-2014/4b53a5747f9b58b9b73b2bbcc39945f0_MIT2_71S14_lec16_notes.pdf |
LECTURE 3
Matroids and geometric lattices
3.1. Matroids
A matroid is an abstraction of a set of vectors in a vector space (for us, the normals
to the hyperplanes in an arrangement). Many basic facts about arrangements
(especially linear arrangements) and their intersection posets are best understood
from the more... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/4b5dc2db132d782df7a9a5ca665c94da_lec3.pdf |
now
means affine independence.
≤
≤
It should be clear what is meant for two matroids M = (S, I) and M � = (S� , I�)
I
to be isomorphic, viz., there exists a bijection f : S
I� . Let M be a matroid and S a set of points in
if and only if
Rn, regarded as a matroid with independence meaning affine independence. If M
a... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/4b5dc2db132d782df7a9a5ca665c94da_lec3.pdf |
(where 123 is short for
and
{
3, 4, 5
1, 2, 3
}
(b) Write I = S1, . . . , Sk for the simplicial complex I generated by S1, . . . , Sk,
{
, etc.).
}
i.e.,
�
�
S1, . . . , Sk =
�
�
T : T
{
= 2S1
∗
∅ · · · ∅
Si for some i
}
2Sk
.
�
Then I = 13, 14, 23, 24 is the set of independent sets of a matroid M on [4]. Thi... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/4b5dc2db132d782df7a9a5ca665c94da_lec3.pdf |
111
100
101
001
Let us now define a number of important terms associated to a matroid M .
A basis of M is a maximal independent set. A circuit C is a minimal dependent
set, i.e., C is not independent but becomes independent when we remove any point
from it. For example, the circuits of the matroid of Figure 1 are... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/4b5dc2db132d782df7a9a5ca665c94da_lec3.pdf |
properties, such as T = T and T �
T
�
T
T .
⊆
∗
∗
3.2. The lattice of flats and geometric lattices
For a matroid M define L(M ) to be the poset of flats of M , ordered by inclusion.
Since the intersection of flats is a flat, L(M ) is a meet-semilattice; and since L(M )
has a top element S, it follows from Lemma 2.... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/4b5dc2db132d782df7a9a5ca665c94da_lec3.pdf |
4.
Let M be a matroid and x
¯
�
) = 0)
≤
}
then we call x a loop. Thus
M ,
is just the set of loops of M . Suppose that x, y
) = 1. We then call x and y parallel points.
neither x nor y are loops, and rk(
}
{
A matroid is simple if it has no loops or pairs of parallel points. It is clear that the
following thr... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/4b5dc2db132d782df7a9a5ca665c94da_lec3.pdf |
0.)
�
�
�
Example 3.8. Let S be any finite set and V a vector space. If f : S
define a matroid Mf on S by the condition that given I
S,
V , then
∃
I(M )
I
≤
√ {
f (x) : x
I
}
≤
is linearly independent.
∗
⇔
LECTURE 3. MATROIDS AND GEOMETRIC LATTICES
35
Then a loop is any element x satisfying f (x) = 0,... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/4b5dc2db132d782df7a9a5ca665c94da_lec3.pdf |
Proof. M has no loops, since every H
≤
nonparallel hyperplanes have linearly independent normals, so the points of M are
closed. Hence M is simple.
A, and set
Let B, B�
≤
∗
X =
H = XB, X � =
H = XB� .
H⊆B
⎦
Then X = X � if and only if
H⊆B�
⎦
nH : H
span
{
B
}
nH : H
= span
{
B� .
}
≤
≤
Now the closure ... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/4b5dc2db132d782df7a9a5ca665c94da_lec3.pdf |
�
Proof. Assume (1). Let x, y � x
rk(x
y) > rk(x) = rk(y). By (1),
⇒
y, so rk(x) = rk(y) = rk(x
y) + 1 and
∈
∈
rk(x) + rk(y)
rk(y)
y
x
⊆
⊆
(rk(x)
rk(x
1) + rk(x
1
−
y)
⇒
−
⊂
⊂
� x.
Similarly x
For (2)
⇒
⊆
⇒
y � y, proving (2).
(1), see [18, Prop. 3.3.2].
y)
⇒
�
36
R. STANLEY, HYPERPLANE ARRA... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/4b5dc2db132d782df7a9a5ca665c94da_lec3.pdf |
nor atomic.
We are now ready to characterize the lattice of flats of a matroid.
Theorem 3.8. Let L be a finite lattice. The following two conditions are equivalent.
(1) L is a geometric lattice.
(2) L ∪= L(M ) for some (simple) matroid M .
Proof. Assume that L is geometric, and let A be the set of atoms of L. If T
... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/4b5dc2db132d782df7a9a5ca665c94da_lec3.pdf |
we may assume is simple, we need to
show that L(M ) is a geometric lattice. Clearly L(M ) is atomic, since every flat is
the join of its elements. Let S, T
M . We will show that
�
�
(T
�
�
�
→
∗
⇔→
∅
≤
≤
≤
∅
I
{
∗
⇒
.
}
A and x
(24)
∗
rk(S) + rk(T )
rk(S
⊕
⊂
T ) + rk(S
T ).
∅
LEC... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/4b5dc2db132d782df7a9a5ca665c94da_lec3.pdf |
[x, y] is also a geometric lattice. (See Exercise 3.) In
general, however, an interval of an atomic lattice need not be atomic.
→
∅
⊕
For noncentral arrangements L(A) is not a lattice, but there is still a connection
with geometric lattices. For a stronger statement, see Exercise 4.
Proposition 3.8. Let A be an arr... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/4b5dc2db132d782df7a9a5ca665c94da_lec3.pdf |
1 = [n],
of [n], and let a =
{
viz., x1 = [n
0, x1] = Bn−1 and
[ˆ
0, x2] = Bn, we easily obtain µBn (ˆ
1] and x2 = [n]. Hence µ(x1) + µ(x2) = 0. Since [ˆ ⇒
. There are two elements x
}
1)n, agreeing with (4).
1) = (
−
≤
n
If x
y in a graded lattice L, write rk(x, y) = rk(y)
rk(x), the length of
every saturat... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/4b5dc2db132d782df7a9a5ca665c94da_lec3.pdf |
.
⇔
38
R. STANLEY, HYPERPLANE ARRANGEMENTS
Hence x a = ˆ
⇒
From Theorem 3.9 there follows
1 if and only if x = ˆ
1 or x is a coatom (i.e., x � ˆ
1) satisfying a
x.
⇔→
µ(ˆ ˆ
0, 1) =
−
µ(ˆ
0, x).
a∈⊇x�ˆ1
�
The sum on the right is nonempty since L is atomic, and by induction every x
�
indexing the sum satisfie... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/4b5dc2db132d782df7a9a5ca665c94da_lec3.pdf |
corresponding to A, as defined in Proposition 3.6, then
(26)
ψA(t) = tn−r ψM (t).
It follows from Corollary 3.5 and equation (26) that we can write
ψA(t) = bntn + bn−1tn−1 +
i
→ →
1)n−ibi > 0 for n
+ bn−r tn−r ,
where (
· · ·
n.
−
−
r | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/4b5dc2db132d782df7a9a5ca665c94da_lec3.pdf |
Transient Analysis of First Order RC and RL circuits
The circuit shown on Figure 1 with the switch open is characterized by a particular
operating condition.
Since the switch is open, no current flows in the circuit (i=0) and vR=0. The voltage
across the capacitor, vc, is not known and must be defined. It could be ... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/4b6a5fe626d65de20ceb97d8d23f38a4_transient1_rl_rc.pdf |
of time for t>0. Since the voltage across
the capacitor must be continuous the voltage at
is also Vo.
+=
0
t
Our first task is to determine the equation that describes the behavior of this circuit. This
is accomplished by using Kirchhoff’s laws. Here we use KLV which gives,
( )
v t
R
+
( )
v t
c
0
=
(0.1)
Using... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/4b6a5fe626d65de20ceb97d8d23f38a4_transient1_rl_rc.pdf |
is called the characteristic equation of the system. Therefore s is
And the solution is
s
= −
1
RC
vc t
( )
=
Ae
t
−
RC
t
−
Ae τ
=
(0.4)
(0.5)
(0.6)
(0.7)
(0.8)
The constant A may now be determined by applying the initial condition
gives
0tvc
V
= = 0
which
And the final solution is
A V=
0
vc t
( )
=
V e
0
t... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/4b6a5fe626d65de20ceb97d8d23f38a4_transient1_rl_rc.pdf |
=
(0.11)
Using the current voltage relationship of the resistor and the inductor, Equation (0.11)
becomes
L di t
( )
R dt
+
i t
( )
=
0
6.071/22.071 Spring 2006, Chaniotakis and Cory
(0.12)
4
The ratio
L
R
has the units of time as can be seen by simple dimensional analysis.
B... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/4b6a5fe626d65de20ceb97d8d23f38a4_transient1_rl_rc.pdf |
0.18)
R3
R1
C
+
vc
-
Figure 8
Therefore once the switch is closed, the equivalent circuit becomes
Req
C
+
vc
-
The characteristic time is now given by
Figure 9
And the evolution of the voltage vc is
τ =
eqR C
t
−
R C
eq
vc t
( )
=
V e
0
(0.19)
(0.20)
6.071/22.071 Spring 2006, Chaniotakis and Cory
6
... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/4b6a5fe626d65de20ceb97d8d23f38a4_transient1_rl_rc.pdf |
R
L
i(t)
0.5R
2R
2L
L
2R
Figure 13
By combining the resistors and the inductors the circuit reduces to
2R
5L/3
i(t)
Figure 14
With the initial condition for the current
i
t
= =
0
I
0
=
2
Vs
R
the solution for the current i(t)
becomes
( )
i t
=
2
6
R
−
L
5
t
e
Vs
R
(0.21)
For this example we have been able to co... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/4b6a5fe626d65de20ceb97d8d23f38a4_transient1_rl_rc.pdf |
.22)
(0.23)
(0.24)
The solution of this equation is the combination (superposition) of the homogeneous
solution
cpv
and the particular solution
chv
( )
t
t
( )
.
v
c
=
v
ch
+
v p
c
6.071/22.071 Spring 2006, Chaniotakis and Cory
(0.25)
9
The homogeneous solution ... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/4b6a5fe626d65de20ceb97d8d23f38a4_transient1_rl_rc.pdf |
thus there is no voltage across the capacitor plates at time t=0. At time
t=0 the witch is moved from position a to position b where it stays for time t1 and
subsequently returned to position a. This switch action corresponds to the rectangular
pulse shown on Figure 19.
b
a
R
+ vR -
C
+
vc
-
Figure 18
+
Vs
-
Vs
0
... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/4b6a5fe626d65de20ceb97d8d23f38a4_transient1_rl_rc.pdf |
known
τ =
R C
eq
eq
or
τ =
L
eq
R
eq
4. Calculate the initial value for the voltage/current flowing in the circuit
i. For a transition happening at
a. Capacitor acts as an open circuit under dc conditions
=
−
0 )
b. Inductor acts as a short circuit under dc conditions
−
0 )
i. For a transition happening at
t = ... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/4b6a5fe626d65de20ceb97d8d23f38a4_transient1_rl_rc.pdf |
-
C
+
vc
-
Figure 21
The RC circuit shown on has two time constants.
For 0<t<t1 the time constant is 1 ( 1
.
R
+
2R C
For t>t1 the time constant is 2
.
R C
2)
τ =
τ =
The solution now is
v t
( )
c
=
Vs
(1
−
e
v t
( )
c
=
Vs
(1
−
e
−
t
1
τ
)
−
t
1
1
τ
−
t
2
τ
)
e
For
t
≤
t
1
For
t
>
t
1
(0.35)
The plot of vc(t) is... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/4b6a5fe626d65de20ceb97d8d23f38a4_transient1_rl_rc.pdf |
6.087 Lecture 9 – January 22, 2010
Review
Using External Libraries
Symbols and Linkage
Static vs. Dynamic Linkage
Linking External Libraries
Symbol Resolution Issues
Creating Libraries
Data Structures
B-trees
Priority Queues
1
Review: Void pointers
• Void pointer – points to any data type:
int x; void ∗ px... | https://ocw.mit.edu/courses/6-087-practical-programming-in-c-january-iap-2010/4b9152a12a01274abfaf5a3c2686564b_MIT6_087IAP10_lec09.pdf |
(or hash map): array of linked lists for storing
and accessing data efficiently
• Each element associated with a key (can be an integer,
string, or other type)
• Hash function computes hash value from key (and table
size); hash value represents index into array
• Multiple elements can have same hash value – result... | https://ocw.mit.edu/courses/6-087-practical-programming-in-c-january-iap-2010/4b9152a12a01274abfaf5a3c2686564b_MIT6_087IAP10_lec09.pdf |
-Wall -c hello.c -o hello.o
1
• -c: compile, but do not link hello.c; result will compile the
code into machine instructions but not make the program
executable
• addresses for lines of code and static and global variables
not yet assigned
• need to perform link step on hello.o (using gcc or ld) to
•
assign mem... | https://ocw.mit.edu/courses/6-087-practical-programming-in-c-january-iap-2010/4b9152a12a01274abfaf5a3c2686564b_MIT6_087IAP10_lec09.pdf |
R msg
U puts@@GLIBC_2.2.5
• Addresses for static (allocated at compile time) symbols
• Symbol puts located in shared library GLIBC_2.2.5 (GNU
C standard library)
• Shared symbol puts not assigned memory until run time
1
Athena is MIT's UNIX-based computing environment. OCW does not provide access to it.
10
Stat... | https://ocw.mit.edu/courses/6-087-practical-programming-in-c-january-iap-2010/4b9152a12a01274abfaf5a3c2686564b_MIT6_087IAP10_lec09.pdf |
define symbol exactly as
expected from header file declaration
• changing function in shared library can break your program
• version information used to minimize this problem
• reason why common libraries like libc rarely modify or
remove functions, even broken ones like gets()
14
Linking external libraries
• Pr... | https://ocw.mit.edu/courses/6-087-practical-programming-in-c-january-iap-2010/4b9152a12a01274abfaf5a3c2686564b_MIT6_087IAP10_lec09.pdf |
not part of C standard library; need to
link against library libdl: -ldl compiler flag
18
• Our puts() gets used since ours is static, and puts() in
C standard library not resolved until run-time
• If statically linked against C standard library, linker finds
two puts() definitions and aborts (multiple definitions not
... | https://ocw.mit.edu/courses/6-087-practical-programming-in-c-january-iap-2010/4b9152a12a01274abfaf5a3c2686564b_MIT6_087IAP10_lec09.pdf |
.087 Lecture 9 – January 22, 2010
Review
Using External Libraries
Symbols and Linkage
Static vs. Dynamic Linkage
Linking External Libraries
Symbol Resolution Issues
Creating Libraries
Data Structures
B-trees
Priority Queues
21
Creating libraries
• Libraries contain C code like any other program
• Static or... | https://ocw.mit.edu/courses/6-087-practical-programming-in-c-january-iap-2010/4b9152a12a01274abfaf5a3c2686564b_MIT6_087IAP10_lec09.pdf |
Data structures
• Many data structures designed to support certain
algorithms
• B-tree - generalized binary search tree, used for databases
and file systems
• Priority queue - ordering data by “priority,” used for sorting,
event simulation, and many other algorithms
23
B-tree structure
• Binary search tree with ... | https://ocw.mit.edu/courses/6-087-practical-programming-in-c-january-iap-2010/4b9152a12a01274abfaf5a3c2686564b_MIT6_087IAP10_lec09.pdf |
not in tree
3. search list of keys for element (using linear or binary
search)
4. if element in list, return element
5. otherwise, element between keys, and repeat search on
child node for that range
• Tree is balanced – search takes O(log n) time
29
Deletion
• Deletion complicated by minimum children restrict... | https://ocw.mit.edu/courses/6-087-practical-programming-in-c-january-iap-2010/4b9152a12a01274abfaf5a3c2686564b_MIT6_087IAP10_lec09.pdf |
ibonacci heap
• We’ll focus on simple binary heaps
• Usually implemented as an array with top element at
beginning
• Can sort data using a heap – O(n log n) worst case
in-place sort!
35
Extracting data
•
Heap-ordering property maximum priority element at
top of heap
⇒
• Can peek by looking at top element
• ... | https://ocw.mit.edu/courses/6-087-practical-programming-in-c-january-iap-2010/4b9152a12a01274abfaf5a3c2686564b_MIT6_087IAP10_lec09.pdf |
18.417 Introduction to Computational Molecular Biology
Lecture 12: October 19, 2004
Lecturer: Ross Lippert
Scribe: Tushara C. Karunaratna
Editor: Peter Lee
Suffix Arrays and BWTs
Notation
We use � to denote the alphabet.
We use S to denote the text string, and n to denote the length of the text.
We use P to denote... | https://ocw.mit.edu/courses/18-417-introduction-to-computational-molecular-biology-fall-2004/4bcb2459df6be70d5b0da5e8d4742aa0_lecture_12.pdf |
n + 2n = 7n words, which is 28n bytes assuming a machine word size
of four bytes.
12-1
12-2
Lecture 12: October 19, 2004
It is possible to reduce the space requirement to 20n bytes using a technique due to
Kurtz.
Suffix Arrays
Suffix arrays are a more space efficient alternative to suffix trees. They were first
develo... | https://ocw.mit.edu/courses/18-417-introduction-to-computational-molecular-biology-fall-2004/4bcb2459df6be70d5b0da5e8d4742aa0_lecture_12.pdf |
tree. Thus, a naive method of computing the suffix array is by first computing the
suffix tree. This naive method takes only O(n) time, but could take a large amount of
space during the intermediate step of constructing the suffix tree. Manber and Myers,
Lecture 12: October 19, 2004
12-3
in their seminal paper, show how... | https://ocw.mit.edu/courses/18-417-introduction-to-computational-molecular-biology-fall-2004/4bcb2459df6be70d5b0da5e8d4742aa0_lecture_12.pdf |
��x arrays.
Figure 12.2 shows the graph of the suffix array permutation A for the string
acataggagacatacga$. This graph does not show us any obvious structure.
Figure 12.3 shows the graph of the auxilliary permutation A−1[A[i] + 1]. This graph
does indeed show us that suffix arrays have structure: the graph consists of... | https://ocw.mit.edu/courses/18-417-introduction-to-computational-molecular-biology-fall-2004/4bcb2459df6be70d5b0da5e8d4742aa0_lecture_12.pdf |
S[A[i] − 1] A−1[A[i] + 1] A[i]
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
3
0
9
10
11
13
15
16
17
7
8
12
1
2
5
14
4
6
a
g
g
$
t
g
t
c
c
a
a
a
c
a
g
a
a
a
17 $
16 a$
9 acatacga$
0 acataggagacatacga$
13 acga$
7 agacatacga$
4 aggagacatacga$
11 atacga$
2 ataggagacatacga$
10 catacga$
... | https://ocw.mit.edu/courses/18-417-introduction-to-computational-molecular-biology-fall-2004/4bcb2459df6be70d5b0da5e8d4742aa0_lecture_12.pdf |
used is to incrementally determine the interval [lo, hi) in the suffix
lookup(P):
lo = 0, hi = length(B)
i = length(P)
while i>0:
i = i-1
lo = block(P[i]) + occ(P[i],lo)
hi = block(P[i]) + occ(P[i],hi)
return hi-lo
Figure 12.4: Pattern lookup using the BWT.
array in which the i-suffix of P is a prefix. Thus, patte... | https://ocw.mit.edu/courses/18-417-introduction-to-computational-molecular-biology-fall-2004/4bcb2459df6be70d5b0da5e8d4742aa0_lecture_12.pdf |
0
1
1
0
1
0
0
0
0
1
0
i #$ #a #c #g #t B[i]
a
0
0
g
1
g
2
$
3
t
4
g
5
t
6
c
7
c
8
9
a
a
10
a
11
c
12
a
13
g
14
a
15
a
16
a
17
1
1
2
3
2
2
2
2
1
1
4
5
3
3
4
Table 12.5: Counts with W = 3.
pattern is to choose W large enough so that the data structure fits into memor... | https://ocw.mit.edu/courses/18-417-introduction-to-computational-molecular-biology-fall-2004/4bcb2459df6be70d5b0da5e8d4742aa0_lecture_12.pdf |
18.034, Honors Differential Equations
Prof. Jason Starr
Lecture 7
2/18/04
1. One more example w/ slope fields. From p. 72 Example 2.4.4. y’=y-y2-0.2sin(t).
Observed that on line
y =
1.2
, y’ is always negative. On line y=0.7, y’ is always positive.
Thus solution curves that enter the region 0.7
0.7
≤
y
≤
1.2
ge... | https://ocw.mit.edu/courses/18-034-honors-differential-equations-spring-2004/4bf2d11f4c44e6399f03782fd83824fb_lec7.pdf |
determine equilibrium sol’ns, state line, stability of equilibrium sol’ns:
(1) Find all zeros of f(y).
(3) Draw the state line.
(2) Determine the sign of f(y) b/w these values.
(4) Sketch sol’n curve and determine stability
unstability/ semi stability of equilibria.
18.034, Honors Differential Equations
Prof. Ja... | https://ocw.mit.edu/courses/18-034-honors-differential-equations-spring-2004/4bf2d11f4c44e6399f03782fd83824fb_lec7.pdf |
Massachusetts Institute of Technology
Department of Electrical Engineering and Computer Science
6.245: MULTIVARIABLE CONTROL SYSTEMS
by A. Megretski �
Solving the H2 optimization problem1
There are several ways to derive a solution to the H2 optimization problem. The path de
veloped below relies on reduction of t... | https://ocw.mit.edu/courses/6-245-multivariable-control-systems-spring-2004/4c031c3f765729d3087b5558bdb4bb17_lec3_6245_2004.pdf |
0,
t��
(3.5)
and there exist a real matrix-valued function V = V (t), each entry of which is a linear
combination of terms of the form tk e−st with k → {0, 1, . . . }, Re(s) > 0, such that
where
U (t) = �(t)B1 + V (t)D21, lim �(t) = 0,
t��
˙�(t) = �(t)A + V (t)C2, �(0) = 0.
(3.6)
(3.7)
Equality (3.5) represe... | https://ocw.mit.edu/courses/6-245-multivariable-control-systems-spring-2004/4c031c3f765729d3087b5558bdb4bb17_lec3_6245_2004.pdf |
� = Ax + B2u + w1,
y = C2x + w2
(3.8)
(3.9)
with w = [w1; w2]. Then (3.5) follows from (3.8). One way to see this is by applying the
one-sided Laplace transform (denoted by tildes) to (3.8) to get
s˜
x = A˜
x + B2 ˜ w1.
u + ˜
3
Since
and ˜w is arbitrary, (3.5) follows.
˜ w, ˜
x = X ˜ u = U ˜
˜
˜
˜ w,... | https://ocw.mit.edu/courses/6-245-multivariable-control-systems-spring-2004/4c031c3f765729d3087b5558bdb4bb17_lec3_6245_2004.pdf |
matrix function G = G(t) can be achieved as a closed loop impulse
response from w to Kx − u if and only if there exists a real matrix-valued function V =
V (t), each entry of which is a linear combination of terms of the form tk e
with k →
{0, 1, . . . }, Re(s) > 0, such that
−st
where
G(t) = �(t)B1 + V (t)D21, li... | https://ocw.mit.edu/courses/6-245-multivariable-control-systems-spring-2004/4c031c3f765729d3087b5558bdb4bb17_lec3_6245_2004.pdf |
to Theorem 3.1, every zero-state response of the
system (i.e. when X(0) = B1 is replaced by X(0) = 0) must coincide with a zero-state
response of (3.10),(3.11).
3.2 Abstract H2 Optimization
We will use the term “abstract H2 optimization” to refer to an auxiliary optimization
problem: an optimal program control in ... | https://ocw.mit.edu/courses/6-245-multivariable-control-systems-spring-2004/4c031c3f765729d3087b5558bdb4bb17_lec3_6245_2004.pdf |
1, the closed
loop H2 norm can be expressed as the integral
�
�f i =
�
0
∞C1X(t) + D12U (t)∞F dt,
2
5
where
∞M ∞2
F = trace(M →M )
denotes the square of the Frobenius norm of matrix M , and X, U are the closed loop
impulse responses from w to x and u respectively, constrained only by
X˙ = AX + B2U, X(... | https://ocw.mit.edu/courses/6-245-multivariable-control-systems-spring-2004/4c031c3f765729d3087b5558bdb4bb17_lec3_6245_2004.pdf |
(t)D21|2dt,
�
0
minimizing which leads to solving a family of independent abstract H2 optimization prob
lems with
a = A→ , b = C2
→ , c = B1
→ , d = D→
→ .
21, p0 = Ki
3.2.2 The “easy” version of the KYP Lemma
The so-called “Kalman-Yakubovich-Popov Lemma” (also frequently referred to as the
“Positive Real Lemma... | https://ocw.mit.edu/courses/6-245-multivariable-control-systems-spring-2004/4c031c3f765729d3087b5558bdb4bb17_lec3_6245_2004.pdf |
0 and there exist (unique) matrices β = β→ � 0 and k such that a + bk is a
Hurwitz matrix, and
|cp + dq|2 + 2p β(ap + bq) = (q − kp)→d→d(q − kp)
→
(3.18)
for all vectors p, q.
If conditions (a)-(d) are satisfied then the optimal q is defined by the relation q = kp, i.e.
q�(t) = ke(a+bk)t p0,
and the minimal cost e... | https://ocw.mit.edu/courses/6-245-multivariable-control-systems-spring-2004/4c031c3f765729d3087b5558bdb4bb17_lec3_6245_2004.pdf |
β = βαβ,
A solution β for which
H =
α
�
� −�→
.
�
�
a + bk = � − αβ
is a Hurwitz matrix is called a stabilizing solution of the ARE. A stabilizing solution, if
it exists, is unique. It defines the “optimal controller” gain k in q = kp, though, formally,
abstract H2 optimization is about program control optimiza... | https://ocw.mit.edu/courses/6-245-multivariable-control-systems-spring-2004/4c031c3f765729d3087b5558bdb4bb17_lec3_6245_2004.pdf |
= Kf i ˆx(t),
dˆx(t)
dt
= Aˆx(t) + B2u(t) + K →
se(C2 ˆx(t) − y(t)),
and provides the minimal H2 norm (squared) of
8
(3.21)
(3.22)
J� = trace(B→
1Pf iB1 + trace(D12Kf iPseK →
f eD→
12).
Proof The overall closed loop H2 norm (squared) is given by
�
J =
�
0
∞C1X(t) + D12U (t)∞2
F dt.
According to the ... | https://ocw.mit.edu/courses/6-245-multivariable-control-systems-spring-2004/4c031c3f765729d3087b5558bdb4bb17_lec3_6245_2004.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
6.080 / 6.089 Great Ideas in Theoretical Computer Science
Spring 2008
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
Scribe Notes: Introduction to Computational Complexity
Jason Furtado
February 28, 2008
1
Motivating Com... | https://ocw.mit.edu/courses/6-080-great-ideas-in-theoretical-computer-science-spring-2008/4c40dede95d3b2fe07e6aa435b46b471_lec7.pdf |
this lookup table is that it would have to be incredibly large. The amount
of storage necessary would be many times larger than the size of the observable universe (which
has maybe 1080 atoms). Therefore, such a system would be infeasible to create. The argument has
now changed from theoretical possibility to a ques... | https://ocw.mit.edu/courses/6-080-great-ideas-in-theoretical-computer-science-spring-2008/4c40dede95d3b2fe07e6aa435b46b471_lec7.pdf |
said, ”but I don’t understand. Given any graph, there’s at most a
finite number of non-intersecting paths. Every finite set has a minimal element, so just enumerate
them all and you’re done.” The colleague’s argument does, indeed, show that the shortest path
problem is computable. But from a modern perspective, showin... | https://ocw.mit.edu/courses/6-080-great-ideas-in-theoretical-computer-science-spring-2008/4c40dede95d3b2fe07e6aa435b46b471_lec7.pdf |
. . . , xn.
Then how many gates to do we need? (For simplicity, let’s suppose that a two-input XOR gate is
available.) Right, n − 1 gates are certainly sufficient: XOR x1 and x2, then XOR the result with
x3, then XOR the result with x4, etc. Can we show that n − 1 gates are necessary?
Claim: You need at least n − 1 g... | https://ocw.mit.edu/courses/6-080-great-ideas-in-theoretical-computer-science-spring-2008/4c40dede95d3b2fe07e6aa435b46b471_lec7.pdf |
a remarkably simple argument for why
such a Boolean function must exist.
2
First, how many Boolean functions are there on n inputs? Well, you can think of a Boolean
function with n inputs as a truth table with 2n entries, and each entry can be either 0 or 1. This
leads to 22n possible Boolean functions: not merel... | https://ocw.mit.edu/courses/6-080-great-ideas-in-theoretical-computer-science-spring-2008/4c40dede95d3b2fe07e6aa435b46b471_lec7.pdf |
Shannon’s counting argument is that it proves that such complex
functions must exist, yet without giving us a single specific example of such a function. Arguments
of this kind are called non-constructive.
3 Hartmanis-Stearns
So the question remains: can we find any concrete problem, one people actually care about, t... | https://ocw.mit.edu/courses/6-080-great-ideas-in-theoretical-computer-science-spring-2008/4c40dede95d3b2fe07e6aa435b46b471_lec7.pdf |
if it halts at all – in any
case, in fewer than n3 steps.
Now run P �(P �, n). There is a contradiction! If P � halts, then it will run forever by case 1. If
P � runs forever then it will halt by case 2.
3
The conclusion is that P could not have existed in the first place. In other words, not only is
the halting ... | https://ocw.mit.edu/courses/6-080-great-ideas-in-theoretical-computer-science-spring-2008/4c40dede95d3b2fe07e6aa435b46b471_lec7.pdf |
b for all n.
Intuitively, the function f (n) grows at the same rate or more quickly than g(n) as n goes to
infinity.
Big-Theta
We say f (n) = Θ(g(n)) if f (n) = O(g(n)) AND f (n) = Ω(g(n)).
Intuitively, the function f (n) grows at the same rate as g(n) as n goes to infinity.
Little-o
We say f (n) = o(g(n)) if for ... | https://ocw.mit.edu/courses/6-080-great-ideas-in-theoretical-computer-science-spring-2008/4c40dede95d3b2fe07e6aa435b46b471_lec7.pdf |
Electricity and Magnetism
• Recap:
– Fundamental Forces
– E.S. Induction
– Coulombs law (qualitative)
• Coulombs Law (quantitative)
• Induction (Demos)
Feb 11 2002
What we learned last time (II)
• Strength of Electrostatic Force (qualitatively):
– If distance gets larger, force gets weaker
– If charge gets bigger, for... | https://ocw.mit.edu/courses/8-02x-physics-ii-electricity-magnetism-with-an-experimental-focus-spring-2005/4c60171da92653f26eb3fffe1105f9fb_2_11_2002_edited.pdf |
1 Principal Component Analysis in High Dimensions and the Spike
Model
1.1 Dimension Reduction and PCA
When faced with a high dimensional dataset, a natural approach is to try to reduce its dimension,
either by projecting it to a lower dimension space or by finding a better representation for the data.
During this course... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/4c9fa7ce658a63174f562fdf44b55626_MIT18_S096F15_Ses2_4.pdf |
will start with the first interpretation of PCA and then show that it is equivalent to the second.
1.1.1 PCA as best d-dimensional affine fit
We are trying to approximate each xk by
xk ≈ µ +
(βk)i vi,
d
(cid:88)
i=1
10
(6)
where v1, . . . , vd is an orthonormal basis for the d-dimensional subspace, µ ∈ Rp represents the t... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/4c9fa7ce658a63174f562fdf44b55626_MIT18_S096F15_Ses2_4.pdf |
(9) by solving
n
(cid:88)
k=1
min
V, βk
V T
V =I
(cid:107)x −
k
µn
− βk(cid:107)2
2 .
V
(9)
Let us proceed by optimizing for βk. Since the problem decouples for each k, we can focus on, for
each k,
min (cid:107)xk − µn − V βk(cid:107)2 = min
β
β
k
k
2
(cid:13)
(cid:13)
(cid:13)
(cid:13)
(cid:13)
xk − µn
−
d
(cid:88)
i=... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/4c9fa7ce658a63174f562fdf44b55626_MIT18_S096F15_Ses2_4.pdf |
n) does not depend on V , minimizing (9) is equivalent to
− (xk −
Tµn) V V T (xk − µn) .
max
V T V =I
n
(cid:88)
k=1
(xk − µn) V V T (xk − µn) .
T
(12)
A few more simple algebraic manipulations using properties of the trace:
n
(cid:88)
k=1
(xk − µn) V V T (xk − µn) =
T
=
n
(cid:88)
k=1
n
(cid:88)
(cid:104)
(x
T
k − µn)... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/4c9fa7ce658a63174f562fdf44b55626_MIT18_S096F15_Ses2_4.pdf |
, xn projected on this subspace has the most
variance. Equivalently we can ask for the points
T
n
k=1
,
vT
1 xk
.
..
vT
d xk
12
to have as much variance as possible. Hence, we are interested in solving
max
V T V =I
n (cid:13)
(cid:13)
(cid:88)
(cid:13)V T
(cid:13)
(cid:13)
k=1
xk −
1
n
n
(c... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/4c9fa7ce658a63174f562fdf44b55626_MIT18_S096F15_Ses2_4.pdf |
�n (which takes O(np2) work) and then finding its
spectral decomposition (which takes O(p3) work). This means that the computational complexity of
(see [HJ85] and/or [Gol96]).
this procedure is
An alternative is to use the Singular Value Decomposition (1). Let X = [x1 · · · xn] recall that,
(cid:8)
max np2, p3
O (cid:0)... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/4c9fa7ce658a63174f562fdf44b55626_MIT18_S096F15_Ses2_4.pdf |
oftentimess possible to reduce the noise while keeping the signal. (2) One may be interested
in running an algorithm that would be too computationally expensive to run in high dimensions,
1If there is time, we might discuss some of these methods later in the course.
13
dimension reduction may help there, etc. In these... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/4c9fa7ce658a63174f562fdf44b55626_MIT18_S096F15_Ses2_4.pdf |
).
14
The conjecture in [MZ11] is that the optimal basis is the eigendecomposition of Σ. It is known
that this is the case for d = 1 (see [MZ11]) but the question remains open for d > 1. It is not very
difficult to see that one can assume, without loss of generality, that Σ is diagonal.
A particularly intuitive way of s... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/4c9fa7ce658a63174f562fdf44b55626_MIT18_S096F15_Ses2_4.pdf |
p
V T V =I
S⊂[p]
|
=d
S|
∈S
i
i g(cid:1)2
(cid:35)
.
1.2 PCA in high dimensions and Marcenko-Pastur
1, . . . , xn ∈ R are independent draws of a gaussian random
Let us assume that the data points x
variable g ∼ N (0, Σ) for some covariance Σ ∈ Rp×p. In this case when we use PCA we are hoping
to find low dimens... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/4c9fa7ce658a63174f562fdf44b55626_MIT18_S096F15_Ses2_4.pdf |
histogram (left) and
a scree plot of the eigenvalues of a sample of Sn (when Σ = I) for p = 500 and n = 1000. The red
line is the eigenvalue distribution predicted by the Marchenko-Pastur distribution (15), that we will
discuss below.
As one can see in the image, there are many eigenvalues considerably larger than 1 (a... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/4c9fa7ce658a63174f562fdf44b55626_MIT18_S096F15_Ses2_4.pdf |
:34)(cid:18)
1
p
E Tr
1
n
XX T
(cid:19) (cid:35)
k
(cid:16) (cid:17)
= E Tr Sk
n = E
1
p
p
1 (cid:88)
p
i=1
λk
i (Sn) =
(cid:90) γ+
γ
−
k
λ dFγ(λ),
and that the quantities 1 E Tr
natorics). The distribution dFγ(λ) can then be computed from its moments.
n
p
can be estimated (these estimates rely essentially in combi-
(c... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/4c9fa7ce658a63174f562fdf44b55626_MIT18_S096F15_Ses2_4.pdf |
Sn).
4The i-th diagonal element of Σ in the SVD √1 X = U ΣV .
n
17
This means that we can compute αR in the limit (since we know the limiting distribution of λi (Sn))
and get (since p = n we have γ = 1, γ = 0, and γ
+ = 2)
−
lim αR(n) =
n
→∞
(cid:90)
2
0
1
λ 2 dF1(λ) =
1
2π
(cid:90)
2
0
1
λ 2
(cid:112) −
(2
λ
λ) λ
=
8... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/4c9fa7ce658a63174f562fdf44b55626_MIT18_S096F15_Ses2_4.pdf |
(0, I + βvvT ).
A natural question is whether this rank 1 perturbation can be seen in Sn. Let us build some
intuition with an example, the following is the histogram of the eigenvalues of a sample of Sn for
p = 500, n = 1000, v is the first element of the canonical basis v = e1, and β = 1.5:
18
The images suggests that... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/4c9fa7ce658a63174f562fdf44b55626_MIT18_S096F15_Ses2_4.pdf |
a rigorous proof, but it is not one at the present form!
First of all, it is not difficult to see that we can assume that v = e1 (since everything else is rotation
invariant). We want to understand the behavior of the leading eigenvalue of
Sn =
n1
(cid:88)
n
i=1
x T
ix
1
i = XX T ,
n
5Notice that the Marchenko-Pastur the... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/4c9fa7ce658a63174f562fdf44b55626_MIT18_S096F15_Ses2_4.pdf |
�= λ
(cid:21)
(cid:20) v1
v2
which can be rewritten as
1
n
(1 + β)ZT
1
1 Z1v1 + (cid:112)1 + βZT
n
(cid:112)
2 Z1v1 + ZT
1 + βZT
1 Z2v2 = λv1
ˆ
1
n 2 Z2v2 = λv2.
ˆ
1
n
,
(16)
(17)
(17) is equivalent to
1
n
(cid:112)
1 + βZT
2 Z1v1 =
(cid:18)
1
ˆλ I − ZT
n 2 Z2
(cid:19)
v2.
ˆIf λ I − 1 ZT
n
2 Z2 is invertible (this w
on... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/4c9fa7ce658a63174f562fdf44b55626_MIT18_S096F15_Ses2_4.pdf |
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