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MIT OpenCourseWare http://ocw.mit.edu 6.641 Electromagnetic Fields, Forces, and Motion, Spring 2005 Please use the following citation format: Markus Zahn, 6.641 Electromagnetic Fields, Forces, and Motion, Spring 2005. (Massachusetts Institute of Technology: MIT OpenCourseWare). http://ocw.mit.edu (accessed MM DD, ...
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, z dxdz ) + ∫ Az (x, y, z + ∆ ) z dxdy − ∫ Az (x, y, z dxdy ) 3' 1 3 Φ ≈ ∆ ∆ ∆ ⎧ ⎪ x y z ⎨ ⎪ ⎩ ( ⎡A x, y, z ⎣ x ) ( - ∆x, y, z − A x x ∆x )⎤ ⎦ + ⎣ y ( ⎡A x, y + ∆y, z y ( − A x, y, z )⎦ ⎤ ) ∆y + ⎡ A x, y, z + z ⎣ z ( z ∆ −) A x, y, z ( ⎤ ⎫ ⎦ ⎪ ⎬ ∆z ⎪⎭ ) ≈ ∆V ⎢ ⎡ ∂Ax ⎣ ∂x + ∂Ay ∂y + ∂Az...
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dS = ∑ (cid:118)∫ i A dS i i S i=1 N→∞ dS i N = lim ∑ (∇ i A) ∆Vi N→∞ V 0 ∆ → n i=1 = ∫ ∇ i A V dV ∫V ∇ i A dV = (cid:118)∫ S A i da 3. Gauss’ Law in Differential Form (cid:118)∫ ε0 E i da = ∫ ∇ i (ε0E dV = ) ∫ ρ dV S V V ∇ i ε0E = ρ ) ( µ H i da = ∇ i µ H dV = 0 ( 0 ) ∫ V 0 S (cid:118)∫ ∇...
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⎣ ∂x - ∂A ⎤ x ⎥ ∂y ⎦ = det ⎢ − ⎡ − − ⎤ i z ⎥ i y ⎢ i x ∂ ⎥ ⎢ ∂ ∂ ⎥ ∂z ⎥ ⎢ ∂x ∂y ⎥ ⎢ ⎢Ax Ay Az ⎥ ⎦ ⎣ = ∇ × A 6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 2 Page 6 of 10 2. Stokes’ Integral Theorem Courtesy of Krieger Publishing. Used with permission. lim ∑ ...
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ations 1. Vector Identity lim A i ds = 0 = ∇ × A i da = ∇ i ∇ × A dV C 0→ ∫ (cid:118)∫ ( ) ) C V (cid:118)∫ ( S ∇ i (∇ × A ) = 0 2. Charge Conservation ⎧ ⎪ ∇ × H = J + ε0 ∇ i ⎨ ⎪ ⎩ ⎫ ∂E ⎪ ⎬ ∂t ⎪⎭ 0 = ∇ i ⎡ ⎢J + ε0 ⎢ ⎣ ⎤ ∂E ⎥ ∂t ⎥ ⎦ 0 = ∇ i J + ∂ρ ∂t 3. Magnetic Field ∇ i ⎨∇ × E = - µ0 ⎧ ⎪ ⎪ ...
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∇ Φ = 2 ( ) (Poisson’s Eq.) ∇ × H = J ρ ε 0 Φ x, y, z = ) ( ∫∫∫ x ',y ',z ' 4πε0 ⎣ ρ (x ', y ', z ') dx ' dy ' dz ' + (y − y ')2 ⎡(x − x ')2 1 + (z − z ')2 ⎤ ⎦ 2 ∇ µ0 i ( H) = 0 ⇒ µ H = ∇ × A 0 ∇ 2 A = − µ 0 J, ∇ i A = 0 A x, y, z ( ) = ∫∫∫ x ',y ',z ' 4π ⎡ ( ) µ0 J x ', y ', z ' dx dy dz + (y −...
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3.46 PHOTONIC MATERIALS AND DEVICES Lecture 1: Optical Materials Design Part 1 Lecture Notes Goal: To develop principles for optical materials design. Approach: Physical basis of properties; use properties in design. Electromagnetic Field Apply voltage: E = ( ,r t ) Apply current: H = ( ,r t ) K K K K Maxwe...
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ε = ε0 (1+ χ) electric permittivity of medium ε ε0 = dielectric constant (static) ε ε 0 11.7 16 43 3600 Si Ge LiNbO3 BaTiO3 Static: ν = 0 3.46 Photonic Materials and Devices Prof. Lionel C. Kimerling Lecture 1: Optical Materials Design Part 1 Page 2 of 6 ...
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AsP Laser InGaAsP p-i-n Si p-i-n Ge APD InGaAs 1.55 Single-mode Systems Design Laser Ps σλ mW nm Power Spectral Bandwidth Fiber dB/km α σ τ /L ns/km L km Attenuation Response Time Length Detector photons/bit Sensitivity Data Rate bits/s n0 B0 3.46 Photonic Materials and Devices Prof. Lion...
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les in material respond as harmonic oscillators Dynamic Relation (time dependent) between P(t) and E(t) K K K K 2 dP d P + a2 + a P ( ) = a E t 1 dt 2 dt 3 accel. vel. x (position) Linear differential equation Resonances Driven simple harmonic oscillators K 2d P dt 2 K K 2P + ω ε χ E 2 0 K dP dt ...
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Ten Lectures and Forty-Two Open Problems in the Mathematics of Data Science Afonso S. Bandeira December, 2015 Preface These are notes from a course I gave at MIT on the Fall of 2015 entitled: “18.S096: Topics in Mathematics of Data Science”. These notes are not in final form and will be continuously edited and/or correc...
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. . . 0.2.1 Koml´os Conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.2.2 Matrix AM-GM inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.3 Brief Review of some linear algebra tools . . . . . . . . . . . . . . . . . . . . . . . . . . Singular Value Decomposition . . . ...
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.4 Which d should we pick? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1.5 A related open problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.1 A related open problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Spi...
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.2.2 Diffusion Maps of point clouds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.3 A simple example . . . . . . . . . . . . . . . 2.2.4 2.3 Semi-supervised learning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.1...
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and the Small Set Expansion Hypothesis . . . . . . . . . . . . . . . . . 3.5 Computing Eigenvectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6 Multiple Clusters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Concentration Inequalities, Scalar and Matrix Ve...
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. . . . . . . . . . . . 4.3.1 Additive Chernoff Bound . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3.2 Multiplicative Chernoff Bound . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3.3 Deviation bounds on χ2 variables . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 Matrix Concentration . ...
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. 4.7.2 4.8 Another open problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6 A matrix concentration inequality for Rademacher Series 5 Johnson-Lindenstrauss Lemma and Gordons Theorem 5.1 The Johnson-Lindenstrauss Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.1 Optim...
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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Finding a dual certificate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 A different approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 Partial Fourier matrices satisfying the Restricted Isometr...
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. . . . . 7.1 Group Testing 7.2 Some Coding Theory and the proof of Theorem 7.3 98 98 . . . . . . . . . . . . . . . . . . . 102 7.2.1 Boolean Classification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 7.2.2 The proof of Theorem 7.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104 I...
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endieck Constant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 8.5 The Paley Graph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 8.6 An interesting conjecture regarding cuts and bisections . . . . . . . . . . . . . . . . . . 115 9 Community detection and the S...
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. . . . . . . . . . . . . 122 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122 9.7 Convex Duality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122 9.8 Building the dual certificate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124 9.9 Matrix Concentration...
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ynchronization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131 10.2.1 Orientation estimation in Cryo-EM . . . . . . . . . . . . . . . . . . . . . . . . . 134 10.2.2 Synchronization over Z2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 10.3 Signal Alignment . . . . . . . . ....
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idefiniteness • 3.3: Multy-way Cheeger’s inequality • 4.1: Non-commutative Khintchine improvement • 4.2: Latala-Riemer-Schutt Problem • 4.3: Matrix Six deviations Suffice • 4.4: OSNAP problem • 4.5: Random k-lifts of graphs • 4.6: Feige’s Conjecture • 5.1: Deterministic Restricted Isometry Property matrices • 5.2: Certify...
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Tightness of the Multireference Alignment SDP • 10.4: Consistency and sample complexity of Multireference Alignment 0.2 A couple of Open Problems We start with a couple of open problems: 0.2.1 Koml´os Conjecture We start with a fascinating problem in Discrepancy Theory. Open Problem 0.1 (Koml´os Conjecture) Given n, le...
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3) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) 1 n! (cid:88) n (cid:89) Aσ(j) σ∈Sym(n) j=1 1 n! (cid:88) σ∈Sym(n) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) n (cid:89) Aσ(j) j=1 (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) ≤ (cid:13) (cid:13) (c...
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] for conjecture (b). Recently these conjectures have been solved for the particular case of n = 3, in [Zha14] for (a) and in [IKW14] for (b). 0.3 Brief Review of some linear algebra tools In this Section we’ll briefly review a few linear algebra tools that will be important during the course. If you need a refresh on a...
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3) 0. In that case we can write M = (cid:16) V Λ1/2(cid:17) (cid:16) V Λ1/2(cid:17) T . A decomposition of M of the form M = U U T (such as the one above) is called a Cholesky decompo- sition. The spectral norm of M is defined as | (cid:107)M (cid:107) = max |λk(M ) . k 0.3.3 Trace and norm Given a matrix M ∈ Rn×n, its ...
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v∈Rn (cid:107)v(cid:107)2=1 (2) (3) It is easy to see (for example, using the spectral decomposition of M ) that (3) is maximized by the leading eigenvector of M and max vT M v = λmax(M ). ∈Rn v (cid:107) (cid:107)2=1 v It is also not very difficult to see (it follows for example from a Theorem of Fan (see, for example, ...
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Dimension reduction through Johnson-Lindenstrauss Lemma and Gordon’s Escape Through a Mesh Theorem. 6. Compressed Sensing/Sparse Recovery, Matrix Completion, etc. If time permits, I will present Number Theory inspired constructions of measurement matrices. 7. Group Testing. Here we will use combinatorial tools to estab...
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particularly useful if, say, one wants to visualize the data in two or three dimensions. There are a couple of different ways we can try to choose this projection: 1. Finding the d-dimensional affine subspace for which the projections of x1, . . . , xn on it best approximate the original points x1, . . . , xn. 2. Finding ...
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×d as the vectors vi are orthonormal. We will measure goodness of fit in terms of least squares and attempt to solve xk ≈ µ + V βk, n (cid:88) k=1 min µ, V, βk V V =I T (cid:107) xk − (µ + V βk) 2 (cid:107)2 (7) (8) We start by optimizing for µ. It is easy to see that the first order conditions for µ correspond to ∇µ n (...
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(cid:13) (cid:13) (cid:13) (cid:13) xk − µn − d (cid:88) i=1 ( βk)i vi . 2 (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) 2 by (βk (10) Since v1, . . . , v which can be succinctly written as β d are orthonormal, it is easy to see that T k = V (xk − µn). Thus, (9) is equivalent to the solution is given ∗) = vT i (xk − µn)...
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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) Tr V V T (xk − µn) (cid:104) V T (xk Tr − µn) (xk − µn) V T k=1 (cid:34) V T (cid:88) n = Tr (xk − µn) (xk − µn) V T (cid:105) (cid:105) (cid:35) k=1 =...
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    vT 1 xk . .. vT d xk  n     k=1 , 13 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 (cid:88) r=1 T V xr (cid:13) 2 (cid:13) (cid:13) (cid:13) (cid:13) . (14) Note that n (cid:88) k=1 (cid:1...
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µ T A naive way of doing this would be to construct Σn (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 ...
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, PCA is useful for many other purposes, for example: (1) often times the data belongs to a lower dimensional space but is corrupted by high dimensional noise. When using PCA it is oftentimess possible to reduce the noise while keeping the signal. (2) One may be interested in running an algorithm that would be too comp...
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generated by d elements of the basis V . The question is: What is the basis V for which E [ΓV ] is maximized? 2Note that Tr (Σn) = (cid:80)p k=1 λk (Σn). 15 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 remain...
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∼ N (0, Σ). The observation regarding the different ordering of the steps amounts to saying that the eigenbasis of Σ is the optimal solution for  (cid:34) argmax max E (cid:88) (cid:0)vT V ∈Rp× 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 ∈ ...
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, Σ) we know that µn → 0 (and, clearly, n essentially the same as Σn.3 n−1 → 1) the spectral properties of Sn will be Let us start by looking into a simple example, Σ = I. In that case, the distribution has no low dimensional structure, as the distribution is rotation invariant. The following is a histogram (left) and ...
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γ , γ+]. This is plotted as the red line in the figure above. − Remark 1.2 We will not show the proof of the Marchenko-Pastur Theorem here (you can see, for example, [Bai99] for several different proofs of it), but an approach to a proof is using the so-called moment method. The core of the idea is to note that one can c...
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the average singular value of √1 X. n The conjecture is that, for every n ≥ 1, αR(n + 1) ≥ αR(n). Moreover, for the analogous quantity αC(n) defined over the complex numbers, meaning simply that each entry of X is an iid complex valued standard gaussian CN (0, 1) the reverse inequality is conjectured for all n ≥ 1: Noti...
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Spike Models and BBP transition What if there actually is some (linear) low dimensional structure on the data? When can we expect to capture it with PCA? A particularly simple, yet relevant, example to analyse is when the covariance matrix Σ is an identity with a rank 1 perturbation, which we refer to as a spike model ...
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astur distribution, and what is the limit value that we expect it to take? As we will see below, there is a critical value of β below which we don’t expect to see a change in the distribution of eivenalues and above which we expect one of the eigenvalues to pop out of the support, this is known as BBP transition (after...
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(p−1), both populated with i.i.d. standard gaussian entries (N (0, 1)). Then, S n = XX = T 1 n 1 n (cid:20) (1 + √ β)ZT 1 + βZT 1 Z 1 2 Z 1 √ T 1 Z2 1 + βZ T Z2 Z2 (cid:21) . ˆ Now, let λ and v = (cid:21) (cid:20) v1 v2 where v2 ∈ Rp−1 and v1 ∈ R, denote, respectively, an eigenvalue and associated eigenvector for Sn. B...
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[Pau]) then we can rewrite it as (cid:18) ˆ v2 = λ I − ZT 2 Z2 1 n 1 − (cid:19) 1 (cid:112)1 + βZT 2 Z1v1, n which we can then plug in (16) to get 1 n (1 + β)Z1 Z1v1 + (cid:112)1 + βZT T 1 Z2 1 n (cid:18) 1 ˆ λ I − ZT n 2 Z2 (cid:19) − 1 1 n (cid:112)1 + βZT 2 Z1v1 = ˆ λv1 If v1 = 0 (again, not properly justified here, ...
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T V = I p × ),×p and Σ is a diagonal matrix. Take D = Σ then × have orthonormal columns 1 n 2 1 n 2 Z2 = V Σ2V T = V DV T , ZT 1 n meaning that the diagonal entries of D correspond to the eigenvalues of 1 ZT 2 Z2 which we expect to be distributed (in the limit) according to the Marchenko-Pastur distribution for − p 1 n...
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D1/2 (cid:0)U T Z1 (cid:1) (cid:21) . U (cid:0) + (cid:20) 1 1 + ZT n 1 1 n 1 (cid:0) n 1 n (cid:20) 1 Since the columns of U are orthonormal, g := U T Z1 ∈ Rp−1 is an isotropic gaussian (g ∼ N (0, 1)), in fact, EggT = EU T Z (cid:0)U T Z (cid:1)T 1 1 = EU T Z1Z1 U = U T E (cid:2)Z1ZT T 1 (cid:3) U = U U = I(p−1)×(p T ...
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+ γ x ˆ λ− γ − x (cid:21) dF ( γ x) , which can be easily solved with the help of a program that computes integrals symbolically (such as Mathematica) to give (you can also see [Pau] for a derivation): ˆλ = (1 + β) (cid:18) 1 + (cid:19) , γ β (20) 22 which is particularly elegant (specially considering the size of some...
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igner matrix (and it will show up later in this course). Given an integer n, a standard gaussian Wigner matrix W ∈ Rn×n is a symmetric matrix with independent (0, 1) entries (except for the fact that Wij = Wji). In the limit, the eigenvalues of √1 n W are distributed according to the so-called semi-circular law N dSC(x...
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1} . q(ξ) = lim EQ 1 n→∞ n (cid:18) ξ n 11T + (cid:19) . 1 √ W n What is the value of ξ , defined as ∗ ξ = inf ∗ {ξ ≥ 0 : q(ξ) > 2}. It is known that, if 0 ≤ ξ ≤ 1, q(ξ) = 2 [MS15]. One can show that 1 Q(B) ≤ λmax(B). In fact, n max {Tr(BX) : X (cid:23) 0, Xii = 1} ≤ max {Tr(BX) : X (cid:23) 0, Tr X = n} . It is also no...
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these things later in the course) is optimal for detection (see [MS15]).7 ∗ Remark 1.5 We remark that Open Problem 1.3 as since been solved [MS15]. 7Later in the course we will discuss clustering under the Stochastic Block Model quite thoroughly, and will see how this same SDP is known to be optimal for exact recovery ...
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clique number of 2. • An independence set of a graph G is a subset S of its nodes such that no two nodes in S share an edge. Equivalently it is a clique of the complement graph Gc := (V, Ec). The independence number of G is simply the clique number of Sc. The Petersen graph has an independence number of 4. 8The Peterso...
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need a simple estimate for what follows (it is a very useful consequence of Stirling’s approximation, e.g.). Proposition 2.2 For every k ≤ n positive integers, (cid:17)k (cid:16) n k ≤ (cid:19) (cid:18)n r ≤ (cid:16) ne k (cid:17) k . We will show a simple lower bound on R(r). But first we introduce a random graph const...
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2 (|S 2 2 ) | . This means that E [X] = By Proposition 2.2 we have, (cid:88) 2 r ) 2(|S| 2 ) ∈(V S = (cid:18)n r (cid:19) 2 2(r 2) = (cid:18)n r (cid:19) 2 r(r−1) . 2 2 E [X] ≤ (cid:17)r (cid:16) ne r 2 r(r−1) 2 2 = 2 (cid:18) n 2 r−1 2 e r (cid:19)r . r−1 2 and r ≥ 3 then E [X] < 1. This means that Prob{X < 1} > 0 and...
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of R(s)? In particular, improve on the base of the exponent on either the lower bound ( 2) or the upper bound (4). √ 27 • Construct a family of graphs G = (V, E) with increasing number of vertices for which there exists ε > 0 such that9 |V | (cid:46) (1 + ε)r. It is known that 43 ≤ R(5) ≤ 49. There is a famous quote in...
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| c 2 | R(G) ≤ 2 log2(n). Proof. Given n, we are interested in upper bounding Prob {R(G) ≥ (cid:100)2 log2 n(cid:101)}. and we proceed by union bounding (and making use of Proposition 2.2): Prob {R(G) ≥ (cid:100)2 log2 n(cid:101)} = Prob (cid:8)   (cid:91)   (cid:88) {S is a clique or independent set} S is a cliqu...
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. 10We say an event happens with high probability if its probability is ≥ 1 − n−Ω(1). 28 The following is one of the most fascinating conjectures in Graph Theory Open Problem 2.2 (Erd˝os-Hajnal Conjecture [EH89]) Prove or disprove the following: For any finite graph H, there exists a constant δH > 0 such that any graph ...
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let us consider a random graph G that (cid:1) consists of taking a graph drawn from G n, 1 , picking ω of its nodes (say at random) and adding an edge between every pair of those ω nodes, thus “planting” a clique of size ω. This will create a clique of size ω in G. If ω > 2 log2 n this clique is larger than any other c...
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prob- √ n log n . (cid:28) n. For example, for ω ≈ ability) for ω √ 11There is an amplification technique that allows one to find the largest clique for ω ≈ c n for arbitrarily small c in polynomial time, where the exponent in the runtime depends on c. The rough idea is to consider all subsets of a certain finite size and...
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{ X(t + 1) = j|X(t) = i} = w ij deg(i) , where deg(i) = (cid:80) j wij. Let M be the matrix of these probabilities, Mij = wij deg(i) . It is easy to see that Mij ≥ 0 and M 1 = 1 (indeed, M is a transition probability matrix). Defining D as the diagonal matrix with diagonal entries Dii = deg(i) we have M = D−1W. If we st...
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= λkvk. Also, × M = D− 1 2 SD 1 2 = D− 1 2 V ΛV T D 1 2 = (cid:16) D− 1 2 V (cid:17) (cid:16) 1 2 V D Λ (cid:17)T . We define Φ = D− 1 Then 2 V with columns Φ = [ϕ1, . . . , ϕn] and Ψ = D 2 V with columns Ψ = [ψ1, . . . , ψn]. 1 M = ΦΛΨT , and Φ, Ψ form a biorthogonal system in the sense that ΦT Ψ = In n or, equivalentl...
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(i) n     ,  (24) Note that M 1 = 1, meaning that one of the right eigenvectors ϕk is simply a multiple of 1 and so it does not distinguish the different nodes of the graph. We will show that this indeed corresponds to the the first eigenvalue. Proposition 2.7 All eigenvalues λk of M satisfy |λk| ≤ 1. 31 Proof. Let ...
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M and its decomposition M = ΦΛΨT as described above. The Diffusion Map is a map φt : V → Rn−1 given by φt (vi) =  λt 2ϕ2(i) λt ϕ3(i)  3  .  ..  λt nϕn(i)      . This map is still a map to n − 1 dimensions. But note now that each coordinate has a factor of λt k which, if λk is small will be rather small for mod...
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Prob {X(t) = j|X(0) = i2}]2 . Proof. as Note that (cid:80)n 1 deg(j ) [Prob X(t) = j|X(0) = i1} − Prob {X(t) = j { |X(0) = i2}] can be rewritten 2 j=1 n (cid:88) j=1 1 deg(j) (cid:34) n (cid:88) k=1 and λt kϕk(i1)ψk(j) − n (cid:88) k=1 (cid:35)2 λt kϕk(i2)ψk(j) = n (cid:88) j=1 1 deg(j) (cid:34) n (cid:88) k=1 λt k (ϕk...
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which forms an orthonormal basis, meaning that (cid:13) n (cid:13) (cid:13)(cid:88) (cid:13) (cid:13) k=1 k (ϕk(i1) − ϕk(i2)) D− λt 1 2 ψk (cid:13) 2 (cid:13) (cid:13) (cid:13) (cid:13) = = n (cid:88) (cid:0) k=1 n (cid:88) (cid:0) k=2 (cid:1) λt k (ϕk(i1) − ϕk(i2)) 2 λt kϕ t k(i1) − λkϕk(i2) (cid:1) , 2 where the last...
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is only designed to find linear structure of the data and the low dimensionality of the dataset may be non-linear. For example, let’s say our dataset is images of the face of someone taken from different angles and lighting conditions, for example, the dimensionality of this dataset is limited by the amount of muscles in...
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3: The Diffusion Map of the complete graph on 4 nodes in 3 dimensions appears to be a regular tetrahedron suggesting that there is no low dimensional structure in this graph. This is not surprising, since every pair of nodes is connected we don’t expect this graph to have a natural representation in low dimensions. 2.2....
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between pairs of nodes vi, vj finding their geodesic distance and then using, for example, Multidimensional 35 © Science. All rights reserved. This content is excluded from our Creative Commons license. For more information, see http://ocw.mit.edu/help/faq-fair-use/. Figure 4: A swiss roll point cloud (see, for example,...
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) where nodes correspond to points. More precisely, let l denote the number of labeled points with labels f1, . . . , fl, and u the number of unlabeled points (with n = l + u), the first l nodes v1, . . . , vl correspond to labeled points and the rest vl+1, . . . , vn are unlabaled. We want to find a function f : V → {−1...
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5) (cid:104) T (ei − ej) f T (cid:105) wijf T (ei − ej) (ei − ej) f T = f T  (cid:88)  i<j T wij (ei − ej) (ei − ej)   f The matrix (cid:80) i<j wij (ei − ej) (ei − ej) will play a central role throughout this course, it is called T the graph Laplacian [Chu97]. LG := (cid:88) i<j wij (ei − ej) (ei − ej) . T Note th...
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∂2f ∂x2 k (x)dx = B. T. − (cid:90) f (x)∆f (x)dx, which helps motivate the use of the term graph Laplacian. Let us consider our problem min f :V →R: f (i)=fi i=1,...,l f T LGf. We can write (cid:20) Dl 0 0 Du D = (cid:21) , W = (cid:20) Wll Wlu Wul Wuu (cid:21) (cid:20) , LG = − Dl Wl l − Wl u −Wul Du − Wuu (cid:21) , ...
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)(cid:3) =i 1 (i) deg n (cid:88) j=1 w ijf (j), which immediately implies that the maximum and minimum value of f needs to be attained at a labeled point. 2.3.1 An interesting experience and the Sobolev Embedding Theorem Let us try a simple experiment. Let’s say we have a grid on [−1, 1]d dimensions (with say md points...
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at the unit (cid:107)∇f (x)(cid:107)2dx. Let us consider the following function on B0(1) (the ball x fε( ) = | (cid:26) 1 − 2 |x −1 ε if |x| ≤ ε otherwise. A quick calculation suggest that (cid:90) B 0(1) 2 (cid:107)∇fε(x)(cid:107) dx = 1 (cid:90) B0(ε) ε2 dx = vol(B0(ε)) 1 dx2 ≈ εd−2, ε meaning that, if d > 2, the per...
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15 shows the outcome of the same experiment with the f T Lf replaced by f T L2f and con- firms our intuition that the discontinuity issue should disappear (see, e.g., [NSZ09] for more on this phenomenon). 2 41 Figure 11: In this example we are given many unlabeled points, the unlabeled points help us learn the geometry ...
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Sl| (cid:88) i∈Sl x i. Lloyd’s algorithm [Llo82] (also known as the k-means algorithm), is an iterative algorithm that alternates between • Given centers µ1, . . . , µk, assign each point xi to the cluster l = argminl=1,...,k (cid:107)xi − µl(cid:107) . • Update the centers µl = 1 |Sl| (cid:80) i∈S x i. l Unfortunately...
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given by color coding. The extra regularization seems to fix the issue of discontinuities. • The formulation is computationally hard, so algorithms may produce suboptimal instances. • The solutions of k-means are always convex clusters. This means that k-means may have diffi- culty in finding cluster such as in Figure 17. ...
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− 1 dimensions (since in k − 1 dimensions we can have k linearly separable sets). If indeed the clusters were linearly separable after embedding then one could attempt to use k-means on the embedding to find the clusters, this is precisely the motivation for Spectral Clustering. Algorithm 3.1 (Spectral Clustering) Given...
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Map set a threshold τ (one can try all different possibilities) and set i → ϕ2(i). S = {i ∈ V : ϕ2(i) ≤ τ }. 46 In what follows we’ll give a different motivation for Algorithm 3.2. 3.3.1 Normalized Cut Given a graph G = (V, E, W ), a natural measure to measure a vertex partition (S, Sc) is cut(S) = (cid:88) (cid:88) i∈S ...
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3.4 (Cheeger’s cut) Given a graph and a vertex partition (S, Sc), the cheeger cut (also known as conductance, and sometimes expansion) of S is given by h(S) = cut(S) min{vol(S), vol(Sc)} , where vol(S) = (cid:80) i∈S deg(i). Also, the Cheeger’s constant of G is given by hG = min h(S). S⊂V A similar object is the Normal...
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M = ψT 1 , where T 1 1 This means that (cid:104) deg(i) (cid:105) vol(G) 1≤i≤n is the stationary distribution of this random walk. Indeed it is easy 1 ψ1 = D 2 v1 = Dϕ1 = [deg(i)]1≤i≤n . t M a certain distribution pt then X(t + 1) has a distribution pt+1 given by to check that, if X(t) has t+1 = pT pT Proposition 3.5 G...
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) deg(i) vol(G) i∈S j∈Sc wij (cid:80) (cid:80) ∈S i (cid:80) i∈S deg(i) cut(S) vol(S) . Analogously, which concludes the proof. Prob {X(t + 1) ∈ S|X(t) ∈ Sc } = cut(S) vol(Sc) , (cid:50) 48 3.3.2 Normalized Cut as a spectral relaxation Below we will show that Ncut can be written in terms of a minimization of a quadrati...
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) vol(Sc) vol(G) (cid:19) 1 2 , a = corresponding to yi =    (cid:16) vol(Sc) vol(S) vol(G) (cid:16) − vol(S) vol(Sc) vol(G) 1 (cid:17) 2 (cid:17) 1 2 if i ∈ S if i ∈ Sc. Proof. The proof involves only doing simple algebraic manipulations together with noticing that (cid:50) vol(S) + vol(Sc) = vol(G). Proposition ...
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(S) vol(G) (cid:20) vol(Sc) vol(S) (cid:20) vol(Sc) vol(S) 1 vol(G) 1 vol(G) wij wij wij (cid:19) 1 2 (cid:18) + vol(S) vol(Sc) vol(G) 1 2 (cid:19) (cid:35) 2 + + v ol(S) vol(Sc) (cid:21) + 2 vol(S) vol(Sc) + vol(S) vol(S) + (cid:21) vol(Sc) vol(Sc) = Ncut(S). This means that finding the minimum Ncut corresponds to solv...
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= 1 (cid:17) (cid:16) T 1 D 2 1 z = 0. (28) 2 W D− 1 Note that LG = I − D− 1 2 . We order eigenvalues in increasing order 0 = λ1 (LG) ≤ λ2 (LG) ≤ · · · ≤ λn (L G). The eigenvector associated to the smallest eigenvector is given by D 2 1 this means that (by the variational interpretation of the eigenvalues) that the min...
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Cheeger’s inequality was first established for manifolds by Jeff Cheeger in 1970 [Che70], the graph version is due to Noga Alon and Vitaly Milman [Alo86, AM85] in the mid 80s. The upper bound in Cheeger’s inequality (corresponding to Lemma 3.8) is more interesting but more difficult to prove, it is often referred to as the...
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such that and x1 ≤ ... ≤ xn, xm = 0, and x2 1 + x2 n = 1, xT LGx xT Dx ≤ δ, where m be the index for which vol({1, . . . , m−1}) ≤ vol({m, . . . , n}) but vol({1, . . . , m}) > vol({m, . . . , n}). We consider a random construction of S with the following distribution. S = {i ∈ V : xi ≤ τ } where τ ∈ [x1, xn] is drawn ...
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1 2 1 2 (cid:88) w ij |xi − xj| (|xi| + |xj|) i,j (cid:115)(cid:88) ij w (x − x )2 i ij j (cid:115)(cid:88) ij wij( 2 |xi| + |xj|) , 52 where the second inequality follows from the Cauchy-Schwarz inequality. From the construction of x we know that (cid:88) w ij(xi − 2 xj) = 2xT LGx ≤ 2δxT Dx. Also, ij wij( |xi| + |x |)...
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j . Which means that Hence, E min{vol S, vol Sc} = n (cid:88) i=1 deg(i)x2 i = xT Dx. Note however that because we do not necessarily have E cut(S) E min{vol S, vol Sc} E cut(S) E min{vol S,vol Sc } √ 2δ. ≤ is not necessarily the same as E cut(S) min{vol S,vol Sc} and so, cut(S) E min{vol S, vol Sc ≤ 2δ. } √ However, s...
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see [BS14] for a nice description of it). This conjecture is known [RS10] to imply the Unique- Games Conjecture [Kho10], that we will discuss in future lectures. Conjecture 3.10 (Small-Set Expansion Hypothesis [RS10]) For every (cid:15) > 0 there exists δ > 0 such that it is N P -hard to distinguish between the cases 1...
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an eigenvalue existed, it would have been extremely likely that the power method would have found it. This issue is addressed in the open problem below. Open Problem 3.2 Given a symmetric matrix M with small condition number, is there a quasi- linear time (on n and the number of non-zero entries of M ) procedure that c...
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to see that The following is known. Theorem 3.11 ([LGT12]) Let G = (V, E, W ) be a graph and k a positive integer Also, ρG(k) ≤ O (cid:0)k2(cid:1) (cid:112) λk, ρG(k) ≤ O (cid:16)(cid:112)λ2k log k (cid:17) . (29) Open Problem 3.3 Let G = (V, E, W ) be a graph and k a positive integer, is the following true? ρG(k) ≤ po...
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control fluctuations from the mean. Theorem 4.2 (Chebyshev’s inequality) Let X be a random variable with E[X 2] < ∞. Then, Prob{|X − EX| > t} ≤ Var(X) t2 . Proof. Apply Markov’s inequality to the random variable (X − E[X])2 to get: (cid:3) (cid:2) − EX)2 (X t2 Prob X EX > t = Prob | (X EX)2 > { − {| − } ≤ t2 E } = Var(X...
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− (cid:19) . t2 2na2 √ The inequality implies that fluctuations larger than O ( 2n log n we get that the probability is at most 2 .n for t = a (cid:80) n Proof. We first get a probability bound for the event i=1 Xi > t. The proof, again, will follow from Markov. Since we want an exponentially small probability, we use a ...
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17) = cosh(λa) Hence, Together with (32), this gives cosh(x) ≤ ex2/2, for all x ∈ R E[eλXi] ≤ E[e(λXi)2/2] ≤ e(λa)2/2. Prob (cid:40) n (cid:88) i =1 (cid:41) Xi > t ≤ e−tλ n (cid:89) i=1 2 e(λa) /2 15This follows immediately from the Taylor expansions: cosh(x) = (cid:80)∞ n=0 (2n)! , ex2/2 = (cid:80)∞ n=0 x2n 2n n! , a...
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) =1 i ≤ 2e− t > t Xi (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:41) 2na 2 2 Prob Remark 4.4 Let’s say that we have random variables r1, . . . , rn i.i.d. distributed as (cid:50) ri = −   1 with probability p/2 0 with probability 1 − p 1 with probability p/2.  Then, E(ri) = 0 and |ri| ≤ 1 so Hoeffding’s inequa...
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= i = σ2 (cid:16) σ 2 a2 58 Theorem 4.5 (Bernstein’s Inequality) Let X1, X2, . . . , Xn be independent centered bounded ran- dom variables, i.e., |Xi| ≤ a and E[X 2 i] = 0, with variance E[Xi ] = σ . Then, (cid:32) (cid:33) 2 Prob (cid:12)(cid:41) (cid:40)(cid:12) n (cid:12) (cid:12)(cid:88) (cid:12) (cid:12) (cid:12) ...
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, to prove the Theorem. Prob (cid:41) Xi > t (cid:40) n (cid:88) i=1 = Prob{eλ (cid:80) Xi > eλt} ≤ E[eλ (cid:80) Xi] eλt n = e−λt (cid:89) E[eλXi] Now comes the source of the improvement over Hoeffding’s, i=1 (cid:34) E[eλXi] = E 1 + λXi + ∞ (cid:88) λmX m i (cid:35) m! m=2 λmam−2σ2 m! ∞ λa)m ( (cid:88) ∞ (cid:88) m=2 ...
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we try to find the value of λ > 0 that minimizes (cid:26) (cid:27) (eλa − − 1 λa) nσ2 2 a min λ − λt + Differentiation gives −t + nσ2 a2 (aeλa − a) = 0 which implies that the optimal choice of λ is given by (cid:18) λ∗ = log 1 + 1 a (cid:19) at nσ2 If we set u = at nσ2 , then λ∗ = 1 log(1 + u). a Now, the value of the mi...
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0 4.2 Gaussian Concentration One of the most important results in concentration of measure is Gaussian concentration, although being a concentration result specific for normally distributed random variables, it will be very useful throughout these lectures. Intuitively it says that if F : Rn → R is a function that is st...
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107)2 ≤ σ. By subtracting a constant to F , we can assume that EF (X) = 0. Also, it is enough to show a one-sided bound Prob {F (X) − EF (X) ≥ t} ≤ exp (cid:18) − (cid:19) , 2 π2 t2 σ2 since obtaining the same bound for −F (X) and taking a union bound would gives the result. We start by using the same idea as in the pr...
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