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Pj ψ(X) = j ψ 1 M j ≤ ≤ (cid:2) (cid:3) 1 ≥ − 1 M 2 M j,k=1 KL (Pj, Pk) + log 2 P log(M 1) − 6 6 6 5.4. Lower bounds based on many hypotheses 111 where the infimum is taken over all tests with values in 1, . . . , M . } { 1, . . . , M be a random variable such that IP(Z = i) = 1/M PZ . Note that PZ is a mixture distrib...
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1 x) log(1 x) − − qj = X) IP(Z = j IP(Z = ψ(X) | | X) j=ψ(X) qj = 1. It implies by Jensen’s inequality that is such that qj ≥ 0 and qj log(qj ) = j=ψ(X) X P − j=ψ(X) X qj log 1 qj (cid:16) (cid:17) log ≥ − qj qj j=ψ(X) (cid:16) X (cid:17) log(M = − 1) . − By the same convexity argument, we get h(x) log 2. It yields ≥ −...
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Z n X j=1 M = = ≤ = M j=1 X Z n 1 M X Z j=1 M1 M 2 j,k=1 X Z M 1 M 2 KL(Pj, Pk) log M , − j,k=1 X Together with (5.8), it yields 1 M 2 M 1 j,k= X Since KL(Pj , Pk) log M log 2 − ≥ − − 1 IP(Z = ψ(X)) log(M ) − IP(Z = ψ(X)) = M1 M j=1 X this implies the desired result. Pj(ψ(X) = j) max Pj (ψ(X) = j) , 1 j ≤ ≤ M ≤ Fano’s ...
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(IPj, IP θk n θ k) = | − | j 2σ2 2 2 ≤ α log(M ) . Moreover, since M 5, ≥ 1 M 2 M j,k=1 KL (IPj, IPk) + log 2 P log(M 1) − α log(M ) + log 2 ≤ lo g(M 1) − ≤ 2α + . 1 2 The proof then follows from Fano’s inequality. 2 ≤ Theorem 5.11 indicates that we must take φ ασ log(M ). Therefore, the 2n larger the M , the larger th...
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2), there exist binary ∈ ii) M = e d ( γ 2 ⌊ 2 γ d e 2 (cid:1) . (cid:0) ⌋ ≥ Proof. Let ωj,i, 1 ≤ with parameter 1/2 and observe that d, 1 ≤ ≤ ≤ j i M to be i.i.d Bernoulli random variables ρ(ωj, ωk) = X d − ∼ Bin(d, 1/2) . Therefore it follows from a union bound that IP j = k , ρ(ωj, ωk) < ∃ (cid:2) 1 2 − γ d M (M 1) ...
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� − 1 2 ω , . . . ω 1 M 0, 1 } ∈ { d that 5.5 APPLICATION TO THE GAUSSIAN SEQUENCE MODEL We are now in a position to apply Theorem 5.11 by choosing θ1, . . . , θM based on ω1, . . . , ωM from the Varshamov-Gilbert Lemma. Lower bounds for estimation ⌋ ≥ ed/16 Take γ = 1/4 and apply the Varshamov-Gilbert Lemma to obtain ...
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imator θls = Y. ˆ Note that this rate is minimax over sets Θ that are strictly smaller than IRd (see Problem 5.4). Indeed, it is minimax over any subset of IRd that contains θ1, . . . , θM . 6 6 5.5. Application to the Gaussian sequence model 115 Lower bounds for sparse estimation It appears from Table 5.1 that when e...
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∈ be k random variables such that U1 is drawn uniformly at (cid:0)ra(cid:1)ndom 1, . . . , d } { and for any i = 2, . . . , k, conditionally on U1, . . . , Ui 1, the ran- from 1, . . . , d } dom variable Ui is drawn uniformly at random from . 1} Then define U1, . . . , Ui 1, . . . , d }\{ { { − − k 1 if i 0 otherwise . ...
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equally likely. } Note that ρ(ω, x0) k − ≥ k Zi , i=1 X supp(x0)). Indeed the left hand side is the number of where Zi = 1I(Ui ∈ coordinates on which the vectors ω, x0 disagree and the right hand side is the number of coordinates in supp(x0) on which the two vectors disagree. In Ber(k/d) and for any i = 2, . . . , d, c...
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s (cid:1) Zi h i=1 X (cid:0) (cid:1)i(cid:0) 2k d (es − 1) + 1 (cid:1) ≤ .. . 2k d ≤ = 2k (cid:0) k 1) + 1 (es − (cid:1) 6 5.5. Application to the Gaussian sequence model 117 For s = log(1 + d ). Putting everything together, we get 2k ωj = ωk : ρ(ωj, ωk) < k ≤ exp log M + k log 2 IP ∃ (cid:0) sk − 2 k 2 k 2 − − (cid:1...
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(ωj, ωk) log(1 + ) (ii) θj− | for β = 2 d ρ(ωj, ωk) log(1+ ) 2k θk|2 = α . Applying now Theorem 5.11 yields 8 ≤ d 2k ≥ 4 2 2 β σ 8n 2kβ2σ2 n 2 2 β σ n β2σ2 n k log(1 + ) d 2k log(1+ ) d 2k ≤ 2ασ2 n log(M ) , p inf sup IPθ | ˆθ θ IRd ∈ (cid:0) k θ |0≤ | ˆ θ θ 2 |2 ≥ − α2σ2 64n It implies the following corollary. k log(1...
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that for the lower bound. |1 = n . We can essentially do the same log d θ′ θ σ | | q (0, 1) be a parameter to be chosen later Assume that d √ n and let β and define k to be the smallest integer such that ≥ ∈ k R ≥ βσ r n log(ed/√ n) . Let ω1, . . . , ωM be obtained from the sparse Varshamov-Gilbert Lemma 5.14 with this ...
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. Recall that that ε > 0, the minimax rate of estimation over model is IRd such n1/2+ε, B1(R) in the Gaussian sequence |1 ≤ ∈ ≥ (R) ⊂ B 1 | 0(k)) = min(R , Rσ 2 φ( B log d n ) . Moreover, it is attained by the constrained least squares estimator θls R and by the trivial estimator θ = 0 otherwise. σ ˆ ˆ B1(R) if log d n...
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j=0,1 ≥ 1 4 e− KL(IP0,IP1) . Problem 5.3. For any R > 0, θ B2(θ, R) the (Euclidean) ball of radius R and centered at θ. For any ε > 0 let N = N (ε) be the largest integer such that there exist θ1, . . . , θN ∈ B2(0, 1) for which IRd, denote by ∈ for all i = j. We call the set an ε-packing of { (a) Show that there exist...
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2β+1 . fj = C √ n N ωjiϕi i=1 X N for some appropriately chosen where C is a constant, ωj ∈ { N and 1 is the trigonometric basis.] 0, 1 } ϕj}j { ≥ 6 6 Bibliography [AS08] [Ber09] [Bil95] [Bir83] [BLM13] [BRT09] [BT09] [Cav11] [CT06] Noga Alon and Joel H. Spencer. The probabilistic method. Wiley- Interscience Series in...
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. Inverse problems in statistics. In Inverse prob- lems and high-dimensional estimation, volume 203 of Lect. Notes Stat. Proc., pages 3–96. Springer, Heidelberg, 2011. Thomas M. Cover and Joy A. Thomas. Elements of information theory. Wiley-Interscience [John Wiley & Sons], Hoboken, NJ, second edition, 2006. 121 Bibli...
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pringer, New York, 2013. Branko Grunbaum. Convex polytopes, volume 221 of Graduate Texts in Mathematics. Springer-Verlag, New York, second edi- tion, 2003. Prepared and with a preface by Volker Kaibel, Victor Klee and Gu¨nter M. Ziegler. Gene H. Golub and Charles F. Van Loan. Matrix computa- tions. Johns Hopkins Studie...
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Gabriel Peyr´e. Harry Markowitz. Portfolio selection. The journal of finance, 7(1):77–91, 1952. Arkadi Nemirovski. Topics in non-parametric statistics. In Lec- tures on probability theory and statistics (Saint-Flour, 1998), volume 1738 of Lecture Notes in Math., pages 85–277. Springer, Berlin, 2000. G. Pisier. Remarques...
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Turbulent Flow and Transport 9 Dispersion in Pipe and Channel flow 9.1 Dispersion in laminar pipe flow. Purely diffusive dispersion, purely convective dispersion, and Taylor (or Taylor−Aris) dispersion. Scaling laws that define the conditions under which the various types of dispersion occur. Radial concentration ...
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6.088 Intro to C/C++ Day 4: Object-oriented programming in C++ Eunsuk Kang and Jean Yang Today’s topics Why objects? Object-oriented programming (OOP) in C++ �classes �fields & methods �objects �representation invariant 2 Why objects? At the end of the day... computers just manipulate 0’s and 1’s Figure by MIT...
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Characteristics? Responsibilities? 11 Write a program that simulates the growth of virus population in humans over time. Each virus cell reproduces itself at some time interval. Patients may undergo drug treatment to inhibit the reproduction process, and clear the virus cells from their body. However, some of t...
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, in % float resistance; static const float defaultReproductionRate = 0.1; // resistance against drugs, in % public: constructors Virus(float newResistance); Virus(float newReproductionRate, float newResistance); Virus* reproduce(float immunity); bool survive(float immunity); }; method don’t forget the semi-c...
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.g. virus must be able to reproduce) implementation: parts that may change frequently (e.g. representation of resistance inside virus) 26 Protect your private parts! Client Interface Implementation Why is this bad? X 27 Access control: public vs. private class Virus { float reproductionRate; float resistance;...
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this virus cell survives, given the patient's immunity bool Virus::survive(float immunity) { // If the patient's immunity is too strong, then this cell cannot survive if (immunity > resistance) return false; return true; } const float Virus::defaultReproductionRate; 31 Header inclusion #include <stdlib.h> #...
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void takeDrug(); bool simulateStep(); }; 37 Patient class declaration #include “Virus.h” #define MAX_VIRUS_POP 1000 Array of pointers to objects class Patient { Virus* virusPop[MAX_VIRUS_POP]; int numVirusCells; float immunity; // degree of immunity, in % Constructor public: Patient(float initImmunity, int in...
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rand()/RAND_MAX; virusPop[i] = new Virus(resistance); } numVirusCells = initNumVirusCells; } 43 Using dynamically allocated objects bool Patient::simulateStep() { Virus* virus; bool survived = false; ... for (int i = 0; i < numVirusCells; i++){ virus = virusPop[i]; survived = virus->survive(immunity); if (...
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{ ... bool checkRep(); public: ... }; void Patient::takeDrug() { assert(checkRep()); ... assert(checkRep()); } Patient::Patient(float initImmunity, int initNumViruses) { ... assert(checkRep()); } 51 Preserving rep. invariant Will calling checkRep() slow down my program? Yes, but you can take them out...
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MEASURE AND INTEGRATION: LECTURE 1 Preliminaries. We need to know how to measure the “size” or “vol­ ume” of subsets of a space X before we can integrate functions f : X → R or f : X C.→ We’re familiar with volume in Rn . What about more general spaces X? We need a measure function µ : {subsets of X} → [0, ∞]. For t...
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Y is→ continuous if f −1(U ) ∈ τX for all U ∈ τY . “Inverse images of open sets are open.” i=1 i=1 Let (X, M) be a measure space (i.e., M is a σ­algebra for the space X). Then f : X → Y is measurable if f −1(U ) ∈ M for all U ∈ τY . “Inverse images of open sets are measurable.” Basic properties of measurable funct...
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�. We just need to show (NTS) that f is measurable. Let R ⊂ R2 be a rectangle of the form I1 × I2 where each Ii ⊂ R(i = 1, 2) is an open interval. Then f −1(R) = u−1(I1) ∩ v−1(I2). Let x ∈ f −1(R) so that f (x) ∈ R. Then u(x) ∈ I1 and v(x) ∈ I2. Since u is measurable, u−1(I1) ∈ M, and since v is measurable, v−1(I2)...
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(Also holds for complex measurable functions.) (d) If E ⊂ X is measurable (i.e., E ∈ M), then the characteristic function of E, χE (x) = � 1 0 if x ∈ E; otherwise. Proposition 0.3. Let F be any collection of subsets of X. Then there exists a smallest σ­algebra M∗ such that F ⊂ M∗. We call M ∗ the σ­algebra g...
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then there exists a smallest σ­algebra B containing the open sets. Ele­ ments of B are called Borel sets. If f : (X, B) → (Y, τ ) and f −1(U ) ∈ B for all U ∈ τ , then f is called Borel measurable. In particular, continuous functions are Borel measurable. Terminology: • Fσ (“F­sigma”) = countable union of closed se...
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1 Instruction Set Evolution in the Sixties: GPR, Stack, and Load-Store Architectures Arvind Computer Science and Artificial Intelligence Laboratory M.I.T. Based on the material prepared by Arvind and Krste Asanovic 6.823 L3- 2 Arvind The Sixties • Hardware costs started dropping - memories beyond 32K words see...
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model, had many more innovative features – tagged data – virtual memory – multiple processors and memories September 14, 2005 6.823 L3- 6 Arvind A Stack Machine Processor stack : A Stack machine has a stack as a part of the processor state Main Store typical operations: push, pop, +, *, ... Instruction...
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Size and Memory References 6.823 L3- 10 Arvind a b c * + a d c * + e - / program push a push b push c * + push a push d push c * + push e - / stack (size = 2) R0 R0 R1 R0 R1 R2 R0 R1 R0 R0 R1 R0 R1 R2 R0 R1 R2 R3 R0 R1 R2 R0 R1 R0 R1 R2 R0 R1 R0 memory refs a b c, ss(a) sf(a) a d, ss(a+...
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1 R2 R1 R0 R0 R1 R2 R1 R0 R0 a and c are “loaded” twice ⇒ not the best use of registers! September 14, 2005      Register Usage in a GPR Machine (a + b * c) / (a + d * c - e) 6.823 L3- 13 Arvind a c b R1 R0 d R1 R0 e R0 R3 More control over register usage since registers can be na...
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any element in the data area – jump to any instruction in the code area – move any element in the stack frame to the top machinery to carry out +, -, etc. ⇔ SP DP PC stack a b c . . . data push a push b push c * + push e / code September 14, 2005 Stack versus GPR Organization Amdahl, Blaa...
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language – Fixed-height stack design simplified implementation – Stack trashed on context swap (fast context switches) – Inmos T800 was world’s fastest microprocessor in late 80’s • Forth machines – Direct support for Forth execution in small embedded real- time environments – Several manufacturers (Rockwell, Pat...
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Control Store Read only 1µsec 256K - 512 KB 64-bit 5 nsec/level Transistor Registers Conventional circuits IBM 360 instruction set architecture completely hid the underlying technological differences between various models. With minor modifications it survives till today September 14, 2005 IBM S/390 z900 Mi...
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2 SS format: store to store instructions M[(B1) + D1] ← M[(B1) + D1] op M[(B2) + D2] iterate “length” times Most operations on decimal and character strings use this format MVC move characters MP multiply two packed decimal strings CLC compare two character strings ... Multiple memory operations per instruct...
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addr result addr September 14, 2005 CDC 6600: A Load/Store Architecture 6.823 L3- 29 Arvind • Separate instructions to manipulate three types of reg. 8 60-bit data registers (X) 8 18-bit address registers (A) 8 18-bit index registers (B) • All arithmetic and logic instructions are reg-to-reg 3 6 opcode i ...
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” instruction set changes continually – Technology allows larger CPUs over time – Technology constraints change (e.g., now it is power) – Compiler technology improves (e.g., register allocation) – Programming styles change (assembly, HLL, object-oriented, …) – Applications change (e.g., multimedia, ....) – Bad ne...
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Simple Probabilistic Reasoning 6.873/HST951 Harvard-MIT Division of Health Sciences and Technology HST.951J: Medical Decision Support Change over 30 years • 1970’s: human knowledge, not much data • 2000’s: vast amounts of data, traditional human knowledge (somewhat) in doubt • Could we “re-discover” all of me...
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(1-specificity) 1 What makes a better test? TPR (sensitivity) 1 superb OK worthless 0 0 FPR (1-specificity) 1 How certain are we after a test? T+ TP=p(T+|D+) D+ p(D+) FN=p(T-|D+) D? T­ T+ p(D-)=1-p(D+) FP=p(T+|D-) D- Bayes’ Rule: TN=p(T-|D-) T- Pi(D j)P(S| D j) Pi+1( D j) = n � Pi(Dk)P(S|D k) k=1 ...
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• value may depend on how we got there (see below) • therefore, value of a treatment can be determined by expectation • Test: lead to few results, revise probability distribution of diseases, and impose disutility • Questions: lead to few results, revise probability distribution Treatment Outcome (not as in ARF...
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.01 0.01 0.8 0.4 0.1 0.1 0.001 0.01 0.1 0.001 0.001 0.8 0.2 0.3 0.2 0.8 0.2 0.6 0.8 0.7 0.1 0.2 0.4 0.2 0.4 0.1 0.001 0.001 0.8 0.2 0.001 0.001 0.1 0.2 0.9 0.8 0.5 0.8 0.6 Questions casts in urine sediment • Blood pressure at onset • proteinuria • • hematuria • history of pr...
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Pi(Dk)P(S|Dk) k=1 Bayes’ rule Value of treatment • Three results: improved, unchanged, worsened – each has an innate value, modified by “tolls” paid on the way • Probabilities depend on underlying disease probability distribution I V(I) Ip Tx Up U V(U) W V(W) Modeling treatment Steroids improved unchan...
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vasc scl cgae mh How large is the tree? • Infinite, or at least (27+3+8)^(27+3+8), ~10^60 • What can we do? – Assume any action is done only once – Order: • questions • tests • treatments • 27! x 4 x 3 x 2 x 8, ~10^30 • Search, with a myopic evaluation function – like game-tree search; what’s the static ...
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the patient's sex? 1 Male 2 Pregnant Female 3 Non-pregnant Female Reply: 1 . . . Local Sensitivity Analysis Case-specific Likelihood Ratios Therapy Planning Based on Utilities Global Sensitivity Analysis • When asking questions, “how bad could it get for the leading hypothesis?” – Assume all future answers are w...
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3.052 Nanomechanics of Materials and Biomaterials Tuesday 02/27/07 I Prof. C. Ortiz, MIT-DMSE LECTURE 6: AFM IMAGING II : ARTIFACTS AND APPLICATIONS Outline : LAST TIME : BASIC PRINCIPLES OF ATOMIC FORCE MICROSCOPY ................................................ 2 FACTORS AFFECTING RESOLUTION ....................
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imaged in fluid environments (near-physiological conditions), 3) Unlike STM samples do not need to be conductive, 4) Sub-nm resolutions have been achieved on biological samples (detailed the molecular conformation, spatial arrangement, structural dimensions, rate dependent processes, etc.) information on -Basic ...
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m δt(max) m PROBE TIP SHARPNESS Sheng, et al. J. Microscopy 1999, 196, 1. Image removed due to copyright restrictions. Image removed due to copyright restrictions. 3-D model of sharp probe tip on a protein, from Lieber et al, 2000 (http://cnst.rice.edu) 3 3.052 Nanomechanics of Materials and Biomaterial...
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750 nm scan courtesy C. Tolksdorf, Digital Instruments/Veeco, Santa Barbara, USA, and R. Schneider and G. Muskhelishvili, Istitut für Genetik und Mikrobiologie, Germany. Courtesy of Veeco Instruments and G. Muskhelishvili. Used with permission. Courtesy of Zhifeng Shao. Used with permission. http://people.virginia...
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Biophys. J. 2006 91, 2532. DOPC Courtesy of the Biophysical Society. Used with permission. 7 3.052 Nanomechanics of Materials and Biomaterials Tuesday 02/27/07 NANOMECHANICS OF SUPPORTED LIPID BILAYERS Prof. C. Ortiz, MIT-DMSE Courtesy of the Biophysical Society...
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4. Cyclic groups Lemma 4.1. Let G be a group and let Hi, i ∈ I be a collection of subgroups of G. Then the intersection is a subgroup of G H = Hi, i∈I Proof. First note that H is non-empty, as the identity belongs to every Hi. We have to check that H is closed under products and inverses. Suppose that g and h ...
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is closed under taking products and inverses, it is clear that H must contain K. On the other hand, as K is a subgroup of G, K must contain H. But then H = K. D Definition 4.4. Let G be a group. We say that a subset S of G gen­ erates G, if the smallest subgroup of G that contains S is G itself. Definition 4.5. Let...
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(g) and G is cyclic. D It is interesting to go back to the problem of classifying groups of finite order and see how these results change our picture of what is going on. Now we know that every group of order 1, 2, 3 and 5 must be cyclic. Suppose that G has order 4. There are two cases. If G has an element a of o...
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we must put c somewhere in the row that contains a and we cannot put it in the last column, as this already contains c. Continuing in this way, it turns out there is only one way to fill in the whole table ∗ e e a a b b c c e a b c a b c e c b c e a b a e So now we have a complete classification of all fini...
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subgroup (g) generated by g, we might as well assume that G is cyclic, generated by g. Suppose that gl = e. I claim that in this case G = { e, g, g , g , g , . . . , g l−1 }. 3 Indeed it suffices to show that the set is closed under multiplication 4 2 and taking inverses. Suppose that gi and gj are in the set. Then...
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a = b between 0 and k − 1. Suppose that a < b. Then gb−a = e. But this contradicts the fact that k is the smallest D integer such that gk = e. Lemma 4.10. Let G be a finite group of order n and let g be an element of G. Then gn = e. Proof. We know that gk = e where k is the order of g. But k divides n. So n = km...
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n. The group of rotations of an n-gon forms a cyclic group of order n. Indeed any rotation may be expressed as a power of a rotation R through 2π/n. On the other hand, Rn = 1. However there is another way to write down a cyclic group of order n. Suppose that one takes the integers Z. Look at the subgroup nZ. Then...
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a + pn b' = b + qn, and where p and q are integers. Then 'a + b' = (a + pn) + (b + qn) = (a + b) + (p + q)n. So we are okay [a + b] = [a + b'], ' 5 MIT OCW: 18.703 Modern AlgebraProf. James McKernan and addition is well-defined. The set of left cosets with this law of addition is denote Z/nZ, the integers mod...
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0]. So if you throw away [0] then you have to throw away [2]. In fact given n, you should throw away all those integers that are not coprime to n, at the very least. In fact this is enough. Definition-Lemma 4.12. Let n be a positive integer. The group of units, Un, for the integers modulo n is the subset of Z/nZ o...
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inverse of [a]. We want an integer b such that This means that [ab] = 1. ab + mn = 1, for some integer m. But a and n are coprime. So by Euclid’s algorithm, D such integers exist. Definition 4.13. The Euler φ function is the function ϕ(n) which assigns the order of Un to n. Lemma 4.14. Let a be any integer...
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if a is coprime to p. Proof. Follows from (4.14). D How about ϕ(pk)? Let us do an easy example. Suppose we take p = 3, k = 2. Then of the eight numbers between 1 and 8, two are multiples of 3, 3 and 6 = 2 · 3. More generally, if a number between 1 and pk − 1 is not coprime to p, then it is a multiple of p. But t...
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. 52 = 25 = 1 mod 6. How about U8? Well ϕ(8) = 4. So either U8 is either cyclic of order 4, or every element has order 2. 1, 3, 5 and 7 are the numbers coprime to 2. Now 32 = 9 = 1 mod 8, 8 MIT OCW: 18.703 Modern AlgebraProf. James McKernan and 52 = 25 = 1 mod 8, 72 = 49 = 1 mod 8. So [3]2 = [5]2 = [7]2 = ...
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18.404/6.840 Lecture 22 Last time: - Finished NL = coNL - Time and Space Hierarchy Theorems Today: (Sipser §9.2) - A “natural” intractable problem - Oracles and P versus NP 1 Review: Hierarchy Theorems Theorems: SPACE ! " # ⊆, SPACE " # for space constructible ". TIME ! " #...
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EXPTIME-complete if 1) & ∈ EXPTIME 2) Same for EXPSPACE-complete For all ( ∈ EXPTIME, ( ≤* & Theorem: If B is EXPTIME-complete then & ∉ P Theorem: If B is EXPSPACE-complete then & ∉ PSPACE (and & ∉ P) intractable Next will exhibit an EXPSPACE-complete problem 3 ...
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-. Give a polynomial-time reduction / mapping ! to $%&'(↑. . / 0 = 23, 25 0 ∈ ! iff 6 23 = 6 25 all strings except a rejecting computation history for + on 0. Construct 23 so that 6 23 Construct 25 = Δ∗ ( Δ is the alphabet for computation histories, i.e., Δ = Γ ∪ % ∪ # ) • = 23 construction: 23 = 2<=>?@A=BA ∪ 2<=...
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H2 = Hstart 2 AL ⋯ H? abababa # ⋯ # 2 AL ⋯ = reject ⋯ Hreject *+,-./0,10 generates all strings that do not start with Hstart = =>424? ⋯ 4A ˽ … ˽ *+,-./0,10 = M> ∪ M2 ∪ M? ∪ ⋯ ∪ MA ∪ Mblanks ∪ M# Remember: Δ is the alphabet for computation histories, i.e., Δ = Γ ∪ % ∪ # ) Notation: Δd = Δ ∪ {f} Δ.+ = Δ wit...
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:. *8 = *+,-.<7,47 ∪ *+,-./012 ∪ *+,-.425267 Rejecting computation history for 9 on :: 2 BM >?:8:@ ⋯ :B ˽ … ˽ # 2 BM ababa ⋯ abababa # ⋯ # I8 = Istart I@ 2 BM ⋯ >reject ⋯ Ireject 267 generates all strings that do not contain >re ject *+,-.425 *+,-.425267 = Δ.O ∗ reject *+,-./012 generates all strings th...
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to solve the coNP problem: 7 and 9 are equivalent 3. Accept if 7 and 9 are equivalent. Reject if not.” 8 ...
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ed !"#$%↑ is EXPSPACE-complete and thus !"#$%↑ ∉ PSPACE 4. Defined oracle TMs 5. Showed P( = NP( for some oracle * 6. Discussed relevance to the P vs NP question 10 !"REX ∈ PSPACE Theorem: !"REX ∈ PSPACE Proof: Show !"RE' ∈ NPSPACE “On input (), (...
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Transition to the Systems Age • Beginning ~ 1940 (according to Blanchard & Fabrycky) • Rescuing Prometheus • Thomas P. Hughes, Prof. of History and Sociology of Technology, U. of Penn. • Tells the story of four major projects – SAGE – Atlas – CA/T – ARPANET Figure removed for copyright reasons. Schematic of SAGE ...
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Complete Key Aspects of the CA/T • Greater “messy complexity” than either SAGE or Atlas (T. Hughes) • Bechtel / Parsons Brinkerhoff coordinates • ~1/3 of budget spent on remediation • Highly publicized mistakes – Voids in concrete of Zakim Bridge – Planning maps missing the Fleet Center – "Based on anecdotal evi...
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Support Vector Machines Stephan Dreiseitl University of Applied Sciences Upper Austria at Hagenberg Harvard-MIT Division of Health Sciences and Technology HST.951J: Medical Decision Support Overview • Motivation • Statistical learning theory • VC dimension • Optimal separating hyperplanes • Kernel functions •...
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test set (generalization error) from performance on training set? Statistical learning theory Average error on a data set D for model with parameter α: n Remp(α ) = 2 1 n ∑| y(α, xi ) − ti | i =1 Expected error of same model given unseen data distributed like D: R(α ) = 2 ∫ | y(α, x) − t | dP( x, t) 1 Stat...
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• Best classifier minimizes right-hand side Structural risk minimization R(α ) ≤ Remp(α ) + h(log(2n / h) + 1) − log(η / 4) n Model Remp VC conf. Upper bound f1(α) f2(α) f3(α) f4(α) f5(α) best Model selection • Cross-validation: use test sets to estimate error • Penalize model complexity: – Akaike info...
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(w • xi + w0) –1 ≥ 0 Optimal hyperplanes • Optimal hyperplane has largest margin (“large margin classifiers”) • Parameter estimation problem turned into constrained optimization problem • Unique solution w = Σαixi over all inputs xi on the margin (“support vectors”) • Decision function g(x) = sign(Σαixi ·x + w0)...
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 1 y2  y 2  2 y2  2 y1 = x1 = ( x1 y1 + x2 y2)2 2  y1   x1        ⋅   =        x2   y2  Nonlinear SVM • Recall: Input data xi enters calculation only via dot products xi ·xj or Φ(xi)·Φ(xj) • Kernel trick: K(xi ,xj) = Φ(xi)·Φ(xj) • Advantage: no need to calculate Φ • Advantage:...
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VM examples Cubic polynomial Degree 4 poly. SVM examples Gaussian, σ = 1 Gaussian, σ = 3 Performance comparison • Log. regression ⇔ ANN ⇔ SVM • Real-world data set • 1619 lesion images • 107 morphometric features: – Global (size, shape) • size • shape – Local (color distributions) • Use ROC analysis C...
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Soft Lithography and Materials Properties in MEMS Carol Livermore Massachusetts Institute of Technology * With thanks to Steve Senturia and Joel Voldman, from whose lecture notes some of these materials are adapted. Cite as: Carol Livermore, course materials for 6.777J / 2.372J Design and Fabrication of Microelectro...
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2007, Lecture 5 - 3 Sample process: multilayer SU-8 microfluidics > Spin coat, prebake, expose, and postbake first layer > Spin coat, prebake, expose, and postbake second layer > Develop both layers > Cap with SU-8 coated transparent plate > Expose to crosslink SU-8 “glue”, final Described in Jackman, J. Micromech....
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Ware (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. CL: 6.777J/2.372J Spring 2007, Lecture 5 - 6 SU-8 Removal > When using SU-8 as a resist and not a structural material, it must be removed! • Enduring challenge – best option is not to strip the SU-8 > Postbaked material ...
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CH3 CH3 CH3 > Upon treatment in oxygen CH3 Si O Si O Si CH3 CH3 CH3 CH3 n plasma, PDMS seals to itself, glass, silicon, silicon nitride, and some plastic materials. Plasma oxidation Air (~ 10 min) contact PDMS surfaces irreversible seal: formation of covalent bonds Courtesy of Hang Lu and Rebecca Jackman. Used with ...
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Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. CL: 6.777J/2.372J Spring 2007, Lecture 5 - 11 Microcontact printing Stamp Stamp Stamp > Apply ink to an elastomer stamp > Bring stamp i...
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deformable Master Master > Imprint in a thermoplastic material by heating and applying pressure > Typical material: PMMA (polymethyl-methacrylate) > Or imprint in UV-curable fluid, like polyurethane > Process usually leaves trace material in “clear” areas, which may be removed by dry etch > Can replicate nanoscale...
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materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. CL: 6.777J/2.372J Spring 2007, Lecture 5 - 16 Outline > Soft Lithography • Materials and processes • Patt...
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. "Protein Patterning.“ Biomaterials 19, nos. 7-9 (April 1998): 595-609. Cite as: Carol Livermore, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. ...
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Ware (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. CL: 6.777J/2.372J Spring 2007, Lecture 5 - 21 Biomaterials processing by stencils > Stencils • Use PDMS stamps as dry resists • Physically pattern biomaterials • Can use with most any substrate • Potential damage to cells...
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many MEMS devices is in coupling one domain to another, and this coupling is typically described by material properties • Mechanical to electrical • Electrical to thermal • Thermal to fluids > The failure modes of many MEMS devices are in coupling one domain to another • For example, package stress interacting with ...
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CL: 6.777J/2.372J Spring 2007, Lecture 5 - 26 Planning around material properties > Well-controlled material properties can pose a design challenge • Many factors, may point design in opposite directions > Poorly-controlled material properties are worse • Every device has specifications, which must be met by either g...
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Inverse of viscosity Cite as: Carol Livermore, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. CL: 6.777J/2.372J Spring 2007, Lecture 5 - 29 Scala...
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J/2.372J Spring 2007, Lecture 5 - 31 Outline > Soft Lithography • Materials and processes • Patterning biomaterials > Material Properties in MEMS • Role of material properties in MEMS • Some examples • Determining material properties Cite as: Carol Livermore, course materials for 6.777J / 2.372J Design and Fabrication...
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> Characterize by four-point measurement of test structures with known geometry. Cite as: Carol Livermore, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Mont...
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Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. CL: 6.777J/2.372J Spring 2007, Lecture 5 - 36 Residual film stress > Stress in a film deposited on a Si wafer, in the absence of external loading. > Two flavor...
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TE mismatch plus high T processing or operation creates stress (and deformation and/or destruction) > Examples: • Bonding glass (quartz or Pyrex) to Si • Thermal stress in a film that is deposited at high T > CTE is tabulated, and one of the less variable material properties. Cite as: Carol Livermore, course materi...
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is described by a fourth-rank tensor ] J ⋅σ⋅Π+ρ=E [ e Cite as: Carol Livermore, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. CL: 6.777J/2.372J ...
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.777J/2.372J Spring 2007, Lecture 5 - 42 Piezoresistivity in Silicon > Coefficients depend on doping, and decrease rapidly above about 1019 cm-3 > Coefficients are functions of temperature > Typical values Type Units n-type p-type Resistivity Ω-cm 11.7 7.8 π11 10-11 Pa-1 -102.2 6.6 π12 10-11 Pa-1 53.4 -1.1 π44 10-11 ...
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perpendicular to the applied electric field (actuation) • Straining a piezoelectric creates an electric field both parallel and perpendicular to the imposed strain (sensing) > Interaction between stored mechanical energy and stored electrostatic energy • Permits both sensing and actuation • Unlike piezoresistivity, ...
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