The Perceptron. Nuno Vasconcelos ECE Department, UCSD
|
|
- Jonathan Fletcher
- 5 years ago
- Views:
Transcription
1 he Perceprn Nun Vscncels ECE Deprmen, UCSD
2 Clssfcn clssfcn prblem hs pes f vrbles e.g. X - vecr f bservns feures n he rld Y - se clss f he rld X R = fever, bld pressure Y = {dsese, n dsese} X, Y reled b unknn funcn f. = f gl: desgn clssfer h: X Y such hh h h = f
3 Lner dscrmnn he clssfer mplemens he lner decsn rule f g > 0 h = = sgn[ g ] f g < 0 hs he prperes dvdes X n hlf-spces bundr s he plne h: nrml dsnce he rgn b/ g/ s he dsnce frm pn he bundr g = 0 fr pns n he plne h g g > 0 n he sde pns psve sde g < 0 n he negve sde = b b g 3
4 Lner dscrmnn he clssfer mplemens he lner decsn rule f g > 0 h = = sgn[ g ] f g < 0 h gven lnerl seprble rnng se D = {,,..., n, n } n errrs f nd nl f, = nd g > 0 r = - nd g < 0.e..g > 0 g =- = b = b g hs lls ver cncse epressn fr he sun f n rnng errr r zer emprcl rsk 4
5 Lernng s pmzn necessr nd suffcen cndn fr zer emprcl rsk b > 0, hs s neresng becuse lls he frmuln f he lernng prblem s ne f funcn pmzn srng frm rndm guess fr he prmeers nd b e mmze he rerd funcn n = b r, equvlenl, mnmze he cs funcn J n, b b = = 5
6 he grden e hve seen h he grden f funcn f z s f z f = 0 f z, L, n z herem: he grden pns n he drecn f mmum grh grden s he drecn f grees ncrese f f z, nrml he s-cnurs f f. f f, 0 f f, 0 f, 6
7 Crcl pn cndns le f be cnnuusl dfferenble s lcl mnmum f f f nd nl f s lcl mnmum f f f nd nl f f hs zer grden 0 = f nd he Hessn f f s psve defne 0 = f n d d f d R 0 here d d f d R 0, f f = 0 0 f f f n M L 7 0 f f n n L
8 Grden descen hs sugges smple mnmzn echnque pck nl esme 0 f fll he negve grden = η f n n n η f hs s grden descen n η s he lernng re nd needs be crefull chsen f η lrge, descen m dverge f mn eensns re pssble mn pn: nce frmed s pmzn, e cn n generl slve n η f n n 8
9 he perceprn hs s he mn nsgh f Rsenbl, hch led he Perceprn he bsc de s d grden descen n ur cs J e kn h: n, b b = = f he rnng se s lnerl seprble here s les pr,b such h J,b < 0 n mnmum h s equl r beer hn hs ll d Q: cn e fnd ne such mnmum? 9
10 Perceprn lernng he grden s srghfrrd cmpue f = nd grden descen s rvl f = b here s, hever, ne prblem: J,b s n bunded bel f J,b < 0, cn mke J b mulplng nd b b λ > 0 he mnmum s ls hch s que bd, numercll hs s rell jus he nrmlzn prblem h e lred lked bu 0
11 Rsenbl s de resrc enn he pns ncrrecl clssfed ech ern defne se f errrs { < 0} E = b nd mke he cs J p ne h, b = b E J p cnn be negve snce, n E, ll b re negve f e ge zer, e kn e hve he bes pssble slun E emp
12 Perceprn lernng s rvl, jus d grden descen n J p,b b n n = = b n n η η E E hs urns u n be ver effecve f he D s lrge lp ver he enre rnng se ke smll sep he end ne lernve h frequenl s beer s schsc grden descen ke he sep mmedel fer ech pn n gurnee hs s descen sep bu, n verge, u fll he sme drecn fer prcessng enre D ver ppulr n lernng, here D s usull lrge
13 Perceprn lernng he lgrhm s s flls: se k = 0, k = 0, b k = 0 se R = m d { fr = :n { f b k < 0 hen { k = k η b k = b k η R k=k k e ll lk bu R shrl! } } } unl b k 0, n errrs 3
14 Perceprn lernng des hs mke sense? cnsder he emple bel se k = 0, k = 0, b k = 0 se R = m d { = fr = :n { f b k < 0 hen { k = k η b k = b k η R k=k k } } } unl b k 0, n errrs b k k =- 4
15 Perceprn lernng des hs mke sense? cnsder he emple bel se k = 0, k = 0, b k = 0 se R = m d { = fr = :n { f b k < 0 hen { k = k η b k = b k η R k=k k } } } unl b k 0, n errrs b k k =- 5
16 Perceprn lernng des hs mke sense? cnsder he emple bel se k = 0, k = 0, b k = 0 se R = m d { = fr = :n { f b k < 0 hen { k = k η b k = b k η R k=k k } } } unl b k 0, n errrs k η b k k =- 6
17 Perceprn lernng des hs mke sense? cnsder he emple bel se k = 0, k = 0, b k = 0 se R = m d { = fr = :n { f b k < 0 hen { k = k η b k = b k η R k=k k } } } unl b k 0, n errrs k b k =- 7
18 Perceprn lernng des hs mke sense? cnsder he emple bel se k = 0, k = 0, b k = 0 se R = m d { = fr = :n { f b k < 0 hen { k = k η b k = b k η R k=k k } } } unl b k 0, n errrs k b k b k η R =- 8
19 Perceprn lernng des hs mke sense? cnsder he emple bel se k = 0, k = 0, b k = 0 se R = m d { = fr = :n { f b k < 0 hen { k = k η b k = b k η R k=k k } } } unl b k 0, n errrs k b k =- 9
20 Perceprn lernng OK, mkes nuve sense h d e kn ll n ge suck n lcl mnmum? hs s Rsenbl s semnl cnrbun herem: Le D = {,,..., n, n } nd R = m If here s,b such h = nd b >, γ hen he Perceprn ll fnd n errr free hper-plne n ms R erns γ 0
21 Prf n h hrd dene ern b, ssume pn prcessed ern - s, fr smplc, use hmgeneus crdnes. Defnng z = R lls he cmpc nn = b / R b = z = snce nl msclssfed pns re prcessed, e hve z < 0
22 Prf h des evlve? d h l l b b/r z R R b R b η η = = = / / denng he pml slun b =, b/r b z = = η η nd, frm, b = η > ηγ slvng he recursn > ηγ ηγ ηγ ηγ > > > >...
23 Prf hs mens cnvergence f e cn bund he mgnude f. Wh s hs mgnude? Snce g g e hve z η = nd, z z η η = = R z η η = < frm frm def f z slvng he recursn R η < frm 3 R η <
24 Prf cmbnng he R nd R η ηγ. < < < b = < b R R b R R γ γ frm def f snce b/ = b s he dsnce he = R b R γ frm = snce b / = b s he dsnce he rgn, e hve b < R = m nd b g R 4 < γ R
25 Ne hs s n he sndrd prf e.g. Dud, Hr, Srk sndrd prf: regulr lgrhm n R n upde equns gher bund < R/γ hs ppers beer, bu requres chsng η = R /γ hch requres knledge f γ, h e dn hve unl e fnd.e. he prf s nn-cnsrucve, cnn desgn lgrhm h he lgrhm bve jus rks! hence, I lke hs prf beer despe lser bund. 5
26 Perceprn lernng herem: Le D = {,,..., n, n } nd R = m If here s,b such h = nd b > γ, hen he Perceprn ll fnd n errr free hper-plne n ms R erns γ hs resul s he sr f lernng her fr he frs me here s prf h lernng mchne culd cull lern smehng! 6
27 he mrgn ne h b γ, ll hld f nd nl f = b γ = mn = mn b =- hch s h e defned he mrgn hs ss h he bund n me cnvergence s nversel prprnl he mrgn even n hs erl resul, he mrgn ppers s mesure f he dffcul f he lernng prblem R γ 7
28 he rle f R sclng he spce shuld n mke dfference s heher he prblem s slvble R ccuns fr hs f he re re-scled bh R nd γ re re-scled nd he bund R γ remns he sme nce gn, jus quesn f nrmlzn γ R = =- llusres he fc h he nrmlzn = s usull n suffcen 8
29 Sme hsr Rsenbl s resul genered lf ecemen bu lernng n he 50s ler, Mnsk nd Pper denfed serus prblems h he Perceprn here re ver smpl lgc prblems h cnn slve mre n hs n he hmerk hs klled ff he enhussm unl n ld resul b Klmgrv sved he d herem: n cnnuus funcn g defned n [0,] n cn be represened n he frm n d g = Γ j Ψ j j = = 9
30 Sme hsr nng h he Perceprn cn be ren s n h l k lk h P l [ ] = = = n b h 0 sgn sgn hs lks lke hvng Perceprn lers ler : J hper-plnes j n ler : hperplne v J j, h n j j j,..., sgn 0 = = = = = J J J j j j j v h v u 0 sgn 30 = = = J j j J j j j j v v 0 0 sgn sgn
31 Sme hsr hch cn be ren s [ ] J sgn u = sgn g h g = v j nd resembles suggesed he de h g = J j = j = hle ne Perceprn s n gd enugh nn d Γ j Ψ j j = = mbe mul-lered Perceprn MLP ll rk j j 0 v lf rk n MLPs ensued under he nme f neurl nerks evenull, s shn h ms funcns cn be pprmed b MLPs j 0 3
32 Grphcl represenn he Perceprn s usull represened s npu uns: crdnes f eghs: crdnes f hmgeneus crdnes: =, bs erm h = sgn 0 = sgn 3
33 Sgmds he sgn[.] funcn s prblemc n s: n dervve 0 f nn-smh pprmns cn be pprmed n vrus s fr emple b he hperblc ngen f f = nh σ = e e e e σ σ σ σ σ cnrls he pprmn errr, bu hs dervve everhere smh f neurl nerks re mplemened h hese funcns 33
34 Neurl nerk he MLP s funcn pprmn 34
35 35
Chapter 2 Linear Mo on
Chper Lner M n .1 Aerge Velcy The erge elcy prcle s dened s The erge elcy depends nly n he nl nd he nl psns he prcle. Ths mens h prcle srs rm pn nd reurn bck he sme pn, s dsplcemen, nd s s erge elcy s
More informationUse 10 m/s 2 for the acceleration due to gravity.
ANSWERS Prjecle mn s he ecrl sum w ndependen elces, hrznl cmpnen nd ercl cmpnen. The hrznl cmpnen elcy s cnsn hrughu he mn whle he ercl cmpnen elcy s dencl ree ll. The cul r nsnneus elcy ny pn lng he prblc
More informationComparing Possibly Misspeci ed Forecasts
Supplemenl Appendx : Cmprng Pssbly Msspec ed Frecss Andrew J. Pn Duke Unversy 4 Augus Ts supplemenl ppendx cnns w prs. Appendx SA. cnns dervns used n e nlycl resuls presened n e pper. Appendx SA. cnns
More informationMachine Learning Support Vector Machines SVM
Mchne Lernng Support Vector Mchnes SVM Lesson 6 Dt Clssfcton problem rnng set:, D,,, : nput dt smple {,, K}: clss or lbel of nput rget: Construct functon f : X Y f, D Predcton of clss for n unknon nput
More informationADORO TE DEVOTE (Godhead Here in Hiding) te, stus bat mas, la te. in so non mor Je nunc. la in. tis. ne, su a. tum. tas: tur: tas: or: ni, ne, o:
R TE EVTE (dhd H Hdg) L / Mld Kbrd gú s v l m sl c m qu gs v nns V n P P rs l mul m d lud 7 súb Fí cón ví f f dó, cru gs,, j l f c r s m l qum t pr qud ct, us: ns,,,, cs, cut r l sns m / m fí hó sn sí
More informationR th is the Thevenin equivalent at the capacitor terminals.
Chaper 7, Slun. Applyng KV Fg. 7.. d 0 C - Takng he derae f each erm, d 0 C d d d r C Inegrang, () ln I 0 - () I 0 e - C C () () r - I 0 e - () V 0 e C C Chaper 7, Slun. h C where h s he Theenn equalen
More informationENGI 4421 Probability and Statistics Faculty of Engineering and Applied Science Problem Set 10 Solutions Chi-Square Tests; Simple Linear Regression
ENGI 441 Prbbly nd Sscs Fculy f Engneerng nd Appled Scence Prblem Se 10 Sluns Ch-Squre Tess; Smple Lner Regressn 1. Is he fllwng se f bservns f bjecs n egh dfferen drecns cnssen wh unfrm dsrbun? Drecn
More informationCS434a/541a: Pattern Recognition Prof. Olga Veksler. Lecture 9
CS434/541: Pttern Recognton Prof. Olg Veksler Lecture 9 Announcements Fnl project proposl due Nov. 1 1-2 prgrph descrpton Lte Penlt: s 1 pont off for ech d lte Assgnment 3 due November 10 Dt for fnl project
More informationPhysics 201 Lecture 2
Physcs 1 Lecure Lecure Chper.1-. Dene Poson, Dsplcemen & Dsnce Dsngush Tme nd Tme Inerl Dene Velocy (Aerge nd Insnneous), Speed Dene Acceleron Undersnd lgebrclly, hrough ecors, nd grphclly he relonshps
More informationMinimum Squared Error
Minimum Squred Error LDF: Minimum Squred-Error Procedures Ide: conver o esier nd eer undersood prolem Percepron y i > 0 for ll smples y i solve sysem of liner inequliies MSE procedure y i i for ll smples
More informationGo over vector and vector algebra Displacement and position in 2-D Average and instantaneous velocity in 2-D Average and instantaneous acceleration
Mh Csquee Go oe eco nd eco lgeb Dsplcemen nd poson n -D Aege nd nsnneous eloc n -D Aege nd nsnneous cceleon n -D Poecle moon Unfom ccle moon Rele eloc* The componens e he legs of he gh ngle whose hpoenuse
More informationCHAPTER 10: LINEAR DISCRIMINATION
CHAPER : LINEAR DISCRIMINAION Dscrmnan-based Classfcaon 3 In classfcaon h K classes (C,C,, C k ) We defned dscrmnan funcon g j (), j=,,,k hen gven an es eample, e chose (predced) s class label as C f g
More informationMinimum Squared Error
Minimum Squred Error LDF: Minimum Squred-Error Procedures Ide: conver o esier nd eer undersood prolem Percepron y i > for ll smples y i solve sysem of liner inequliies MSE procedure y i = i for ll smples
More informationModeling and Predicting Sequences: HMM and (may be) CRF. Amr Ahmed Feb 25
Modelg d redcg Sequeces: HMM d m be CRF Amr Ahmed 070 Feb 25 Bg cure redcg Sgle Lbel Ipu : A se of feures: - Bg of words docume - Oupu : Clss lbel - Topc of he docume - redcg Sequece of Lbels Noo Noe:
More informationIntroduction to Neural Networks Computing. CMSC491N/691N, Spring 2001
Iroduco o Neurl Neorks Compug CMSC49N/69N, Sprg 00 us: cvo/oupu: f eghs: X, Y j X Noos, j s pu u, for oher us, j pu sgl here f. s he cvo fuco for j from u o u j oher books use Y f _ j j j Y j X j Y j bs:
More informationPrinciple Component Analysis
Prncple Component Anlyss Jng Go SUNY Bufflo Why Dmensonlty Reducton? We hve too mny dmensons o reson bout or obtn nsghts from o vsulze oo much nose n the dt Need to reduce them to smller set of fctors
More information8 factors of x. For our second example, let s raise a power to a power:
CH 5 THE FIVE LAWS OF EXPONENTS EXPONENTS WITH VARIABLES It s no time for chnge in tctics, in order to give us deeper understnding of eponents. For ech of the folloing five emples, e ill stretch nd squish,
More informationSVMs for regression Multilayer neural networks
Lecture SVMs for regresson Muter neur netors Mos Husrecht mos@cs.ptt.edu 539 Sennott Squre Support vector mchne SVM SVM mmze the mrgn round the seprtng hperpne. he decson functon s fu specfed suset of
More informationSVMs for regression Non-parametric/instance based classification method
S 75 Mchne ernng ecture Mos Huskrecht mos@cs.ptt.edu 539 Sennott Squre SVMs for regresson Non-prmetrc/nstnce sed cssfcton method S 75 Mchne ernng Soft-mrgn SVM Aos some fet on crossng the seprtng hperpne
More informationSeptember 20 Homework Solutions
College of Engineering nd Compuer Science Mechnicl Engineering Deprmen Mechnicl Engineering A Seminr in Engineering Anlysis Fll 7 Number 66 Insrucor: Lrry Creo Sepember Homework Soluions Find he specrum
More informationObjective of curve fitting is to represent a set of discrete data by a function (curve). Consider a set of discrete data as given in table.
CURVE FITTING Obectve curve ttg s t represet set dscrete dt b uct curve. Csder set dscrete dt s gve tble. 3 3 = T use the dt eectvel, curve epress s tted t the gve dt set, s = + = + + = e b ler uct plml
More informationARITHMETIC OPERATIONS. The real numbers have the following properties: a b c ab ac
REVIEW OF ALGEBRA Here we review the bsic rules nd procedures of lgebr tht you need to know in order to be successful in clculus. ARITHMETIC OPERATIONS The rel numbers hve the following properties: b b
More informationA Kalman filtering simulation
A Klmn filering simulion The performnce of Klmn filering hs been esed on he bsis of wo differen dynmicl models, ssuming eiher moion wih consn elociy or wih consn ccelerion. The former is epeced o beer
More informationThe soft-margin support vector machine. Nuno Vasconcelos ECE Department, UCSD
he sft-margn supprt vectr machne Nun Vascncels EE Department USD lassfcatn a classfcatn prlem has t tpes f varales e.g. X - vectr f servatns features n the rld Y - state class f the rld X R fever ld pressure
More informationK The slowest step in a mechanism has this
CM 6 Generl Chemisry II Nme SLUTINS Exm, Spring 009 Dr. Seel. (0 pins) Selec he nswer frm he clumn n he righ h bes mches ech descripin frm he clumn n he lef. Ech nswer cn be used, ms, nly nce. E G This
More informatione t dt e t dt = lim e t dt T (1 e T ) = 1
Improper Inegrls There re wo ypes of improper inegrls - hose wih infinie limis of inegrion, nd hose wih inegrnds h pproch some poin wihin he limis of inegrion. Firs we will consider inegrls wih infinie
More informationOptimal Control. Lecture. Prof. Daniela Iacoviello
Opmal Cnrl Lecure Pr. Danela Iacvell Gradng Prjec + ral eam Eample prjec: Read a paper n an pmal cnrl prblem 1 Sudy: backgrund mvans mdel pmal cnrl slun resuls 2 Smulans Yu mus gve me al leas en days bere
More informationReview of linear algebra. Nuno Vasconcelos UCSD
Revew of lner lgebr Nuno Vsconcelos UCSD Vector spces Defnton: vector spce s set H where ddton nd sclr multplcton re defned nd stsf: ) +( + ) (+ )+ 5) λ H 2) + + H 6) 3) H, + 7) λ(λ ) (λλ ) 4) H, - + 8)
More informationAnnouncements. 30 o. The pumpkin is on the left and the watermelon is on the right. The picture on page 138 is better.
Annuncements Em 1 is ne eek. Ems frm revius semesters hve been sted n the ebsite. HITT quiz slutins re sted n ebsite. Td e ill finish Chter 4 nd begin Chter 5. ill st Em 1 brekdn nd revie mteril. Lk fr
More informationSUMMER KNOWHOW STUDY AND LEARNING CENTRE
SUMMER KNOWHOW STUDY AND LEARNING CENTRE Indices & Logrithms 2 Contents Indices.2 Frctionl Indices.4 Logrithms 6 Exponentil equtions. Simplifying Surds 13 Opertions on Surds..16 Scientific Nottion..18
More informationDishonest casino as an HMM
Dshnes casn as an HMM N = 2, ={F,L} M=2, O = {h,} A = F B= [. F L F L 0.95 0.0 0] h 0.5 0. L 0.05 0.90 0.5 0.9 c Deva ubramanan, 2009 63 A generave mdel fr CpG slands There are w hdden saes: CpG and nn-cpg.
More information4.8 Improper Integrals
4.8 Improper Inegrls Well you ve mde i hrough ll he inegrion echniques. Congrs! Unforunely for us, we sill need o cover one more inegrl. They re clled Improper Inegrls. A his poin, we ve only del wih inegrls
More informationAs we have already discussed, all the objects have the same absolute value of
Lecture 3 Prjectile Mtin Lst time we were tlkin but tw-dimensinl mtin nd intrduced ll imprtnt chrcteristics f this mtin, such s psitin, displcement, elcit nd ccelertin Nw let us see hw ll these thins re
More information1. Consider a PSA initially at rest in the beginning of the left-hand end of a long ISS corridor. Assume xo = 0 on the left end of the ISS corridor.
In Eercise 1, use sndrd recngulr Cresin coordine sysem. Le ime be represened long he horizonl is. Assume ll ccelerions nd decelerions re consn. 1. Consider PSA iniilly res in he beginning of he lef-hnd
More informationISSN 075-7 : (7) 0 007 C ( ), E-l: ssolos@glco FPGA LUT FPGA EM : FPGA, LUT, EM,,, () FPGA (feldprogrble ge rrs) [, ] () [], () [] () [5] [6] FPGA LUT (Look-Up-Tbles) EM (Ebedded Meor locks) [7, 8] LUT
More informationTopic 1 Notes Jeremy Orloff
Topic 1 Notes Jerem Orloff 1 Introduction to differentil equtions 1.1 Gols 1. Know the definition of differentil eqution. 2. Know our first nd second most importnt equtions nd their solutions. 3. Be ble
More informationrank Additionally system of equation only independent atfect Gawp (A) possible ( Alb ) easily process form rang A. Proposition with Definition
Defiion nexivnol numer ler dependen rows mrix sid row Gwp elimion mehod does no fec h numer end process i possile esily red rng fc for mrix form der zz rn rnk wih m dcussion i holds rr o Proposiion ler
More informationChapter 14: Optical Parametric Oscillators
Qunum Oc f Phnc n Olcnc hn n, Cnll Un Ch : Ocl Pmc Ocll. Inucn In h Ch w wll cu n cl mc cll. A mc cll lm lk l. Th ffnc h h cl n n h c cm n fm uln n mum u fm nnln cl mum whch h h cn h h 3 cl nnln. Cn n
More informationApproach: (Equilibrium) TD analysis, i.e., conservation eqns., state equations Issues: how to deal with
Schl f Aerspace Chemcal D: Mtvatn Prevus D Analyss cnsdered systems where cmpstn f flud was frzen fxed chemcal cmpstn Chemcally eactng Flw but there are numerus stuatns n prpulsn systems where chemcal
More informationMASSACHUSETTS INSTITUTE OF TECHNOLOGY 6.265/15.070J Fall 2013 Lecture 15 10/30/2013. Ito integral for simple processes
MASSACHUSETTS INSTITUTE OF TECHNOLOGY 6.65/15.7J Fall 13 Lecure 15 1/3/13 I inegral fr simple prcesses Cnen. 1. Simple prcesses. I ismery. Firs 3 seps in cnsrucing I inegral fr general prcesses 1 I inegral
More information2010 Sectional Physics Solution Set
. Crrec nwer: D WYSE CDEMIC CHLLENGE Secnl hyc E 00 Slun Se y 0 y 4.0 / 9.8 /.45 y. Crrec nwer: y 8 0 / 8 /. Crrec nwer: E y y 0 ( 4 / ) ( 4.9 / ) 5.6 y y 4. Crrec nwer: E 5. Crrec nwer: The e rce c n
More information1.B Appendix to Chapter 1
Secon.B.B Append o Chper.B. The Ordnr Clcl Here re led ome mporn concep rom he ordnr clcl. The Dervve Conder ncon o one ndependen vrble. The dervve o dened b d d lm lm.b. where he ncremen n de o n ncremen
More informationAnswers for Lesson 3-1, pp Exercises
Answers for Lesson -, pp. Eercises * ) PQ * ) PS * ) PS * ) PS * ) SR * ) QR * ) QR * ) QR. nd with trnsversl ; lt. int. '. nd with trnsversl ; lt. int. '. nd with trnsversl ; sme-side int. '. nd with
More informationUNIVERSITY OF IOANNINA DEPARTMENT OF ECONOMICS. M.Sc. in Economics MICROECONOMIC THEORY I. Problem Set II
Mcroeconomc Theory I UNIVERSITY OF IOANNINA DEPARTMENT OF ECONOMICS MSc n Economcs MICROECONOMIC THEORY I Techng: A Lptns (Note: The number of ndctes exercse s dffculty level) ()True or flse? If V( y )
More informationApplied Statistics Qualifier Examination
Appled Sttstcs Qulfer Exmnton Qul_june_8 Fll 8 Instructons: () The exmnton contns 4 Questons. You re to nswer 3 out of 4 of them. () You my use ny books nd clss notes tht you mght fnd helpful n solvng
More information5.1 Properties of Inverse Trigonometric Functions.
Inverse Trignmetricl Functins The inverse f functin f( ) f ( ) f : A B eists if f is ne-ne nt ie, ijectin nd is given Cnsider the e functin with dmin R nd rnge [, ] Clerl this functin is nt ijectin nd
More informationThree fundamental questions with Hidden Markov Models
hree fundmenl uesns wh Hdden Mrv Mdels One defned he sruurl elemens f HMM we my wn s hree dfferen nds f uesns:. Gven he mdel λ A B π wh s he prly f servng O.. h s O λ? 2 2. Gven he seuene f servns O..
More information8.6 The Hyperbola. and F 2. is a constant. P F 2. P =k The two fixed points, F 1. , are called the foci of the hyperbola. The line segments F 1
8. The Hperol Some ships nvigte using rdio nvigtion sstem clled LORAN, which is n cronm for LOng RAnge Nvigtion. A ship receives rdio signls from pirs of trnsmitting sttions tht send signls t the sme time.
More informationSupport vector machines for regression
S 75 Mchne ernng ecture 5 Support vector mchnes for regresson Mos Huskrecht mos@cs.ptt.edu 539 Sennott Squre S 75 Mchne ernng he decson oundr: ˆ he decson: Support vector mchnes ˆ α SV ˆ sgn αˆ SV!!: Decson
More informationMTH 146 Class 11 Notes
8.- Are of Surfce of Revoluion MTH 6 Clss Noes Suppose we wish o revolve curve C round n is nd find he surfce re of he resuling solid. Suppose f( ) is nonnegive funcion wih coninuous firs derivive on he
More informationPRINCE SULTAN UNIVERSITY Department of Mathematical Sciences Final Examination First Semester ( ) STAT 271.
PRINCE SULTAN UNIVERSITY Deprtment f Mthemticl Sciences Finl Exmintin First Semester (007 008) STAT 71 Student Nme: Mrk Student Number: Sectin Number: Techer Nme: Time llwed is ½ hurs. Attendnce Number:
More informationChapter Newton-Raphson Method of Solving a Nonlinear Equation
Chpter 0.04 Newton-Rphson Method o Solvng Nonlner Equton Ater redng ths chpter, you should be ble to:. derve the Newton-Rphson method ormul,. develop the lgorthm o the Newton-Rphson method,. use the Newton-Rphson
More informationAverage & instantaneous velocity and acceleration Motion with constant acceleration
Physics 7: Lecure Reminders Discussion nd Lb secions sr meeing ne week Fill ou Pink dd/drop form if you need o swich o differen secion h is FULL. Do i TODAY. Homework Ch. : 5, 7,, 3,, nd 6 Ch.: 6,, 3 Submission
More informationProblems for HW X. C. Gwinn. November 30, 2009
Problems for HW X C. Gwinn November 30, 2009 These problems will not be grded. 1 HWX Problem 1 Suppose thn n object is composed of liner dielectric mteril, with constnt reltive permittivity ɛ r. The object
More informationRequired: Solution: 1.4)
CHAER.) ) 000 000 000 00 k b) 000, 000W 000,000 W 000,000 W 00, 000 W W ) d) k 8.... s 4... s.) 8 r nn n s 9 s p p p.) Gen:.4) n 9 r re r re / dy / r re n n / s dy ) 8 b) nn dy 4 0 0se nds dy 8400s r re
More informationInterval Estimation. Consider a random variable X with a mean of X. Let X be distributed as X X
ECON 37: Ecoomercs Hypohess Tesg Iervl Esmo Wh we hve doe so fr s o udersd how we c ob esmors of ecoomcs reloshp we wsh o sudy. The queso s how comforble re we wh our esmors? We frs exme how o produce
More informationSolutions to Problems from Chapter 2
Soluions o Problems rom Chper Problem. The signls u() :5sgn(), u () :5sgn(), nd u h () :5sgn() re ploed respecively in Figures.,b,c. Noe h u h () :5sgn() :5; 8 including, bu u () :5sgn() is undeined..5
More information0 for t < 0 1 for t > 0
8.0 Sep nd del funcions Auhor: Jeremy Orloff The uni Sep Funcion We define he uni sep funcion by u() = 0 for < 0 for > 0 I is clled he uni sep funcion becuse i kes uni sep = 0. I is someimes clled he Heviside
More informationSupporting information How to concatenate the local attractors of subnetworks in the HPFP
n Effcen lgorh for Idenfyng Prry Phenoype rcors of Lrge-Scle Boolen Newor Sng-Mo Choo nd Kwng-Hyun Cho Depren of Mhecs Unversy of Ulsn Ulsn 446 Republc of Kore Depren of Bo nd Brn Engneerng Kore dvnced
More informationProperties of Logarithms. Solving Exponential and Logarithmic Equations. Properties of Logarithms. Properties of Logarithms. ( x)
Properies of Logrihms Solving Eponenil nd Logrihmic Equions Properies of Logrihms Produc Rule ( ) log mn = log m + log n ( ) log = log + log Properies of Logrihms Quoien Rule log m = logm logn n log7 =
More informationFundamentals of Linear Algebra
-7/8-797 Mchine Lerning for Signl rocessing Fundmentls of Liner Alger Administrivi Registrtion: Anone on witlist still? Homework : Will e hnded out with clss Liner lger Clss - Sep Instructor: Bhiksh Rj
More informationLinear and Nonlinear Optimization
Lner nd Nonlner Optmzton Ynyu Ye Deprtment of Mngement Scence nd Engneerng Stnford Unversty Stnford, CA 9430, U.S.A. http://www.stnford.edu/~yyye http://www.stnford.edu/clss/msnde/ Ynyu Ye, Stnford, MS&E
More informationVECTORS VECTORS VECTORS VECTORS. 2. Vector Representation. 1. Definition. 3. Types of Vectors. 5. Vector Operations I. 4. Equal and Opposite Vectors
1. Defnton A vetor s n entt tht m represent phsl quntt tht hs mgntude nd dreton s opposed to slr tht ls dreton.. Vetor Representton A vetor n e represented grphll n rrow. The length of the rrow s the mgntude
More informationPH2200 Practice Exam I Summer 2003
PH00 Prctice Exm I Summer 003 INSTRUCTIONS. Write yur nme nd student identifictin number n the nswer sheet.. Plese cver yur nswer sheet t ll times. 3. This is clsed bk exm. Yu my use the PH00 frmul sheet
More informationChapter 2. Motion along a straight line. 9/9/2015 Physics 218
Chper Moion long srigh line 9/9/05 Physics 8 Gols for Chper How o describe srigh line moion in erms of displcemen nd erge elociy. The mening of insnneous elociy nd speed. Aerge elociy/insnneous elociy
More informationHow to Prove the Riemann Hypothesis Author: Fayez Fok Al Adeh.
How o Prove he Riemnn Hohesis Auhor: Fez Fok Al Adeh. Presiden of he Srin Cosmologicl Socie P.O.Bo,387,Dmscus,Sri Tels:963--77679,735 Emil:hf@scs-ne.org Commens: 3 ges Subj-Clss: Funcionl nlsis, comle
More information10.7 Power and the Poynting Vector Electromagnetic Wave Propagation Power and the Poynting Vector
L 333 lecmgnec II Chpe 0 lecmgnec W Ppgn Pf. l J. l Khnd Islmc Unves f G leccl ngneeng Depmen 06 0.7 Pwe nd he Pnng Vec neg cn be sped fm ne pn (whee nsme s lced) nhe pn (wh eceve) b mens f M ws. The e
More informationSome Inequalities variations on a common theme Lecture I, UL 2007
Some Inequliies vriions on common heme Lecure I, UL 2007 Finbrr Hollnd, Deprmen of Mhemics, Universiy College Cork, fhollnd@uccie; July 2, 2007 Three Problems Problem Assume i, b i, c i, i =, 2, 3 re rel
More informationChapter Newton-Raphson Method of Solving a Nonlinear Equation
Chpter.4 Newton-Rphson Method of Solvng Nonlner Equton After redng ths chpter, you should be ble to:. derve the Newton-Rphson method formul,. develop the lgorthm of the Newton-Rphson method,. use the Newton-Rphson
More informationAdvanced Electromechanical Systems (ELE 847)
(ELE 847) Dr. Smr ouro-rener Topc 1.4: DC moor speed conrol Torono, 2009 Moor Speed Conrol (open loop conrol) Consder he followng crcu dgrm n V n V bn T1 T 5 T3 V dc r L AA e r f L FF f o V f V cn T 4
More informationPhysics 101 Lecture 4 Motion in 2D and 3D
Phsics 11 Lecure 4 Moion in D nd 3D Dr. Ali ÖVGÜN EMU Phsics Deprmen www.ogun.com Vecor nd is componens The componens re he legs of he righ ringle whose hpoenuse is A A A A A n ( θ ) A Acos( θ) A A A nd
More informationDecompression diagram sampler_src (source files and makefiles) bin (binary files) --- sh (sample shells) --- input (sample input files)
. Iroduco Probblsc oe-moh forecs gudce s mde b 50 esemble members mproved b Model Oupu scs (MO). scl equo s mde b usg hdcs d d observo d. We selec some prmeers for modfg forecs o use mulple regresso formul.
More informationThe solution is often represented as a vector: 2xI + 4X2 + 2X3 + 4X4 + 2X5 = 4 2xI + 4X2 + 3X3 + 3X4 + 3X5 = 4. 3xI + 6X2 + 6X3 + 3X4 + 6X5 = 6.
[~ o o :- o o ill] i 1. Mrices, Vecors, nd Guss-Jordn Eliminion 1 x y = = - z= The soluion is ofen represened s vecor: n his exmple, he process of eliminion works very smoohly. We cn elimine ll enries
More informationF Fou n even has domain o. Domain. TE t. Fire Co I. integer. Logarithmic Ty. Exponential Functions. Things. range. Trigonometric Functions.
Cve Functins Midterm 1 Review Plnmils Rtinl Functins Pwer Functins rignmetric Functins nverse rignmetric Functins Expnentil Functins Functins Dmin Lgrithmic Review Definitins nd bsic prperties Dmin f f
More informationMachine Learning. Support Vector Machines. Le Song. CSE6740/CS7641/ISYE6740, Fall Lecture 8, Sept. 13, 2012 Based on slides from Eric Xing, CMU
Mchne Lernng CSE6740/CS764/ISYE6740 Fll 0 Support Vector Mchnes Le Song Lecture 8 Sept. 3 0 Bsed on sldes fro Erc Xng CMU Redng: Chp. 6&7 C.B ook Outlne Mu rgn clssfcton Constrned optzton Lgrngn dult Kernel
More informationDCDM BUSINESS SCHOOL NUMERICAL METHODS (COS 233-8) Solutions to Assignment 3. x f(x)
DCDM BUSINESS SCHOOL NUMEICAL METHODS (COS -8) Solutons to Assgnment Queston Consder the followng dt: 5 f() 8 7 5 () Set up dfference tble through fourth dfferences. (b) Wht s the mnmum degree tht n nterpoltng
More informationDesign of Diaphragm Micro-Devices. (Due date: Nov 2, 2017)
Design f Diphrgm Micr-Devices (Due de: Nv, 017) 1. Bckgrund A number f micr-devices depend upn diphrgm fr heir perin. Such devices re vlve, pump, nd pressure sensr. The generlized mdel f hese devices is
More informationg i fφdx dx = x i i=1 is a Hilbert space. We shall, henceforth, abuse notation and write g i f(x) = f
1. Appliction of functionl nlysis to PEs 1.1. Introduction. In this section we give little introduction to prtil differentil equtions. In prticulr we consider the problem u(x) = f(x) x, u(x) = x (1) where
More information3 x x 3x x. 3x x x 6 x 3. PAKTURK 8 th National Interschool Maths Olympiad, h h
PAKTURK 8 th Ntionl Interschool Mths Olmpid,.9. Q: Evlute 6.9. 6 6 6... 8 8...... Q: Evlute bc bc. b. c bc.9.9b.9.9bc Q: Find the vlue of h in the eqution h 7 9 7.. bc. bc bc. b. c bc bc bc bc......9 h
More informationPHY2053 Summer C 2013 Exam 1 Solutions
PHY053 Sue C 03 E Soluon. The foce G on o G G The onl cobnon h e '/ = doubln.. The peed of lh le 8fulon c 86,8 le 60 n 60n h 4h d 4d fonh.80 fulon/ fonh 3. The dnce eled fo he ene p,, 36 (75n h 45 The
More information(b) 10 yr. (b) 13 m. 1.6 m s, m s m s (c) 13.1 s. 32. (a) 20.0 s (b) No, the minimum distance to stop = 1.00 km. 1.
Answers o Een Numbered Problems Chper. () 7 m s, 6 m s (b) 8 5 yr 4.. m ih 6. () 5. m s (b).5 m s (c).5 m s (d) 3.33 m s (e) 8. ().3 min (b) 64 mi..3 h. ().3 s (b) 3 m 4..8 mi wes of he flgpole 6. (b)
More informationRank One Update And the Google Matrix by Al Bernstein Signal Science, LLC
Introducton Rnk One Updte And the Google Mtrx y Al Bernsten Sgnl Scence, LLC www.sgnlscence.net here re two dfferent wys to perform mtrx multplctons. he frst uses dot product formulton nd the second uses
More informationAn Introduction to Support Vector Machines
An Introducton to Support Vector Mchnes Wht s good Decson Boundry? Consder two-clss, lnerly seprble clssfcton problem Clss How to fnd the lne (or hyperplne n n-dmensons, n>)? Any de? Clss Per Lug Mrtell
More informationIntroduction to Algebra - Part 2
Alger Module A Introduction to Alger - Prt Copright This puliction The Northern Alert Institute of Technolog 00. All Rights Reserved. LAST REVISED Oct., 008 Introduction to Alger - Prt Sttement of Prerequisite
More informationREAL ANALYSIS I HOMEWORK 3. Chapter 1
REAL ANALYSIS I HOMEWORK 3 CİHAN BAHRAN The quesions re from Sein nd Shkrchi s e. Chper 1 18. Prove he following sserion: Every mesurble funcion is he limi.e. of sequence of coninuous funcions. We firs
More informationFor convenience, we rewrite m2 s m2 = m m m ; where m is repeted m times. Since xyz = m m m nd jxyj»m, we hve tht the string y is substring of the fir
CSCI 2400 Models of Computtion, Section 3 Solutions to Homework 4 Problem 1. ll the solutions below refer to the Pumping Lemm of Theorem 4.8, pge 119. () L = f n b l k : k n + lg Let's ssume for contrdiction
More informationOVERVIEW Using Similarity and Proving Triangle Theorems G.SRT.4
OVRVIW Using Similrity nd Prving Tringle Therems G.SRT.4 G.SRT.4 Prve therems ut tringles. Therems include: line prllel t ne side f tringle divides the ther tw prprtinlly, nd cnversely; the Pythgren Therem
More informationA little harder example. A block sits at rest on a flat surface. The block is held down by its weight. What is the interaction pair for the weight?
Neton s Ls of Motion (ges 9-99) 1. An object s velocit vector v remins constnt if nd onl if the net force cting on the object is zero.. hen nonzero net force cts on n object, the object s velocit chnges.
More information3. Renewal Limit Theorems
Virul Lborories > 14. Renewl Processes > 1 2 3 3. Renewl Limi Theorems In he inroducion o renewl processes, we noed h he rrivl ime process nd he couning process re inverses, in sens The rrivl ime process
More informationPart A. Ch Ch (a) b = = 25. from normal distribution table. Thus, order quantity is 58-12=46. (b) Now b=5
Hmerk Sln Par A. Ch a b = 65 4 = 5 frm nrmal dsrbn able Ths, rder qany s 58-=46 b b=5 frm nrmal dsrbn able Ths, rder qany s 5-=39 Ch 3 a Pssn Prcess, snce hs demand prcess accmlaes he randm demand generaed
More information( ) ( ) ( ) ( ) ( ) ( y )
8. Lengh of Plne Curve The mos fmous heorem in ll of mhemics is he Pyhgoren Theorem. I s formulion s he disnce formul is used o find he lenghs of line segmens in he coordine plne. In his secion you ll
More informationVariable Forgetting Factor Recursive Total Least Squares Algorithm for FIR Adaptive filtering
01 Inernanal Cnference n Elecrncs Engneerng and Infrmacs (ICEEI 01) IPCSI vl 49 (01) (01) IACSI Press Sngapre DOI: 107763/IPCSI01V4931 Varable Frgeng Facr Recursve al Leas Squares Algrhm fr FIR Adapve
More informationThe support vector machine. Nuno Vasconcelos ECE Department, UCSD
he supprt vectr machne Nun Vascncels ECE Department UCSD Outlne e have talked abut classfcatn and lnear dscrmnants then e dd a detur t talk abut kernels h d e mplement a nn-lnear bundar n the lnear dscrmnant
More informationSection 3.2: Negative Exponents
Section 3.2: Negtive Exponents Objective: Simplify expressions with negtive exponents using the properties of exponents. There re few specil exponent properties tht del with exponents tht re not positive.
More informationSummary: Method of Separation of Variables
Physics 246 Electricity nd Mgnetism I, Fll 26, Lecture 22 1 Summry: Method of Seprtion of Vribles 1. Seprtion of Vribles in Crtesin Coordintes 2. Fourier Series Suggested Reding: Griffiths: Chpter 3, Section
More informationLecture 21: Order statistics
Lecture : Order sttistics Suppose we hve N mesurements of sclr, x i =, N Tke ll mesurements nd sort them into scending order x x x 3 x N Define the mesured running integrl S N (x) = 0 for x < x = i/n for
More informationf(x) dx with An integral having either an infinite limit of integration or an unbounded integrand is called improper. Here are two examples dx x x 2
Impope Inegls To his poin we hve only consideed inegls f() wih he is of inegion nd b finie nd he inegnd f() bounded (nd in fc coninuous ecep possibly fo finiely mny jump disconinuiies) An inegl hving eihe
More informationLogarithms. Logarithm is another word for an index or power. POWER. 2 is the power to which the base 10 must be raised to give 100.
Logrithms. Logrithm is nother word for n inde or power. THIS IS A POWER STATEMENT BASE POWER FOR EXAMPLE : We lred know tht; = NUMBER 10² = 100 This is the POWER Sttement OR 2 is the power to which the
More informationDuality # Second iteration for HW problem. Recall our LP example problem we have been working on, in equality form, is given below.
Dulity #. Second itertion for HW problem Recll our LP emple problem we hve been working on, in equlity form, is given below.,,,, 8 m F which, when written in slightly different form, is 8 F Recll tht we
More information