Project 6: Minigoals Towards Simplifying and Rewriting Expressions

 Tracey Davis
 3 months ago
 Views:
Transcription
1 MAT 51 Wldis Projet 6: Minigols Towrds Simplifying nd Rewriting Expressions The distriutive property nd like terms You hve proly lerned in previous lsses out dding like terms ut one prolem with the wy tht we often think out like terms is tht it isn t lwys ler wht is ment y like. Relly, omining like terms is relly just pplying the distriutive property in reverse: + = ( + ) Or, more generlly (for ny numer of terms, where njust denotes the numer of terms) we ould write: n = ( n ) Any time we hve sequene of terms, eh of whih is produt (two things eing multiplied together), nd the seond ftor in eh produt is the SAME, then we n used the distriutive property to rewrite the expression. So, now let s work on prtiing when we n pply the distriutive property. Exmples: Whih of these hve the form of the left side of one of the distriutive properties ove? For eh one, CIRCLE, nd identify,, or 1,, n, : 5x 3x 5x 3y 8p 2 q 2 12p 2 q The distriutive property n e pplied if we rewrite the sutrtion s dding negtive first: 5x 3x = 5x + 3x So 5 x + 3 x And = 5, = 3, = x Or we ould write 1 = 5, 2 = 3, = x The distriutive property n NOT e pplied here euse these two terms don t hve nything in ommon. I nnot find nything of the form +, even if I rewrite the sutrtion s dding negtive first. The distriutive property n e pplied if we rewrite the sutrtion s dding negtive first: 8p 2 q 2 12p 2 q 2 = 8p 2 q p 2 q 2 So 8 p 2 q p 2 q 2 And = 8, = 12, = p 2 q 2 Or we ould write 1 = 8, 2 = 12, = p 2 q 2 The distriutive property n e pplied if we rewrite the sutrtion s dding negtive first: = So And 1 = 8, 2 = 5, 3 = 2, = 5 Note: here we hve to use the formul with 1,, n in it, euse here we hve more thn two terms. The distriutive property n NOT e pplied here euse these two terms don t hve nything in ommon. I nnot find nything of the form +, even if I use the ommuttive property = to reorder the ftors in eh term first. Note: 2 is NOT the sme s 2, nd 3 is NOT the sme s 2. The distriutive property n e pplied if we think out how to group ll the ftors in the first term, nd then group the 5 together nd tret it s one unit: = (5) + 4 So (5) + 4
2 3x(x 1) + 2(x 1) 5x 3x 2 Here I hve s nd s in my expression nd in the distriutive property identity, so I m going to hnge the letters in the first distriutive property like this: + = ( + ) pp + qq = (p + q)r Then p = 5, q = 4, r = Or we ould write 1 = 5, 2 = 4, = The distriutive property n e pplied: So 3x (x 1) + 2 (x 1) And = 3x, = 2, = (x 1) Or we ould write 1 = 3x, 2 = 2, = (x 1) The distriutive property n e pplied if we 1) rewrite the sutrtion s dding negtive, nd think out how to group ll the ftors in the seond term, fter rememering tht x 2 n e repled with x x: 5x 3x 2 = 5x + 3 x x = 5x + ( 3x) x So 5 x + ( 3x) x And = 5, = 3x, = x Or we ould write 1 = 5, 2 = 3x, = x Do we lwys wnt to use the distriute property in exmples like this? For eh of these exmples, we ould see how we n pply the distriutive property in mny ses to omine terms, even in mny ses tht you proly didn t previously think of s like terms. In more omplex ses like these, we my or my not wnt to pply the distriutive property in reverse it will depend on our gols with the prtiulr expression. But it is importnt to understnd tht the distriutive property CAN e used in ll of these ses. Now you try! Whih of these hve the form of the left side of one of the distriutive properties ove? For eh one: ) Stte whether or not one of the distriutive properties n e pplied; ) If it n, explin whih identities, if ny, need to e pplied first; if it n NOT, explin WHY not; ) CIRCLE ; nd d) identify,, (or 1,, n,, depending on the distriutive property tht you re using) from one of the two given distriutive properties: 1. Two terms: + = ( + ) 2. Any numer of terms: n = ( n ) Use the exmples ove s model
3 4. 4x 3 y 2 + 5x 3 y xx 3y 8. 2x(2x + 1) + 3(2x + 1)
4 Rewriting expressions with speifi gol in mind Typilly when we rewrite expressions, it is with speifi gol in mind. We my wnt to simplify them, or we my wnt to put them into prtiulr form tht llows us to see reltionships (for exmple, when we ome to equtions, we often put them in prtiulr form in order to e le to solve them or grph them). When we hve gol in mind for rewriting n expression (or eqution), we then hve to put together mny of the things tht we hve lredy een doing so tht we pply identities to the expressions one fter nother. This requires severl steps: 1) We hve to identify our end gol (e.g. Why re we rewriting the expression/eqution? Wht does it need to look like when we re done?). 2) We hve to selet prtiulr identities tht ould e pplied to the urrent expression (either in its urrent form, or with little rewriting), nd we need to onsider eh of these possile identities s wy of rewriting the expression. If the rewriting is somewht omplex, this my require us to set some smller gols tht we need to meet long the wy to meet our lrger gol. 3) Eh time we pply n identity to the expression, we hve to sk ourselves if this hs rought us loser to our gol or not.. If not, we my need to put tht work side nd to strt over y pplying different identity to the originl eqution.. If it hs gotten us loser, then we need to think out wht identity we should pply to this new rewritten funtion in order to get even loser to our gol. 4) We keep repeting this proess until we reh our gol. As prolems grow more omplex, we might hve to pply mny, mny identities sequentilly to get to our finl result. Let s just jump in nd strt trying to simplify expressions y using the identities we hve een using so fr. To keep things simple, we will only need to use the identities on the list elow s we work to simplify the next set of prolems. x = 1x = + + = + ( + ) + = ( + ) = + + = () = () = ( + ) = + ( + ) = + x n = x x n mmmm ttttt x = 1 x (wheeeeee x 0) (wheeeeee 0) d (wheeeeee, d 0) + = + (wheeeeee 0) = x 2 = x (wheeeeee x 0)
5
6 Exmples: A) Rewrite the following expression, with the im of simplifying it s muh s possile: (6x 2 + 5) (2x 2 + 6) If our gol is to simplify this, we will likely wnt to rewrite this without the prentheses so tht we n try to omine wht is in eh set of prentheses with eh other. If tht is possile, it will very likely mke the expression simpler. So, our first minigol is: Rewrite the expression without prentheses. If the expression ws of this form: ( + ) +, then we ould use the identity ( + ) + = + + to rewrite this expression without the first set of prentheses. But we need to rewrite ll of the sutrtion in this expression s ddition first. So this leds us to nother minigol, whih must ome BEFORE our gol of rewriting the expression without prentheses: Rewrite ll sutrtion in the expression s ddition. Let s put these two gols in the order tht we need to do them, nd list the identities needed to omplish eh gol: Minigol Identities needed 1) Rewrite ll sutrtion in the expression s ddition. = + 2) Rewrite the expression without prentheses. We will ( + ) + = + + strt with the first set of prentheses nd then fous on the seond set. Now let s try to do these two steps, nd see wht hppens: Step 1: Rewrite the sutrtion in the expression s ddition, using the identity = + = (6x 2 + 5), = (2x 2 + 6) = + ( ) (6x 2 + 5) (2x 2 + 6) = (6x 2 + 5) + (2x 2 + 6) So (6x 2 + 5) (2x 2 + 6) = (6x 2 + 5) + (2x 2 + 6) ( ) = ( ) + ( ) Step 2: Rewrite the expression without (the first set of) prentheses, using the identity ( + ) + = + +. Fousing on rewriting without the first set of prentheses: = 6x 2, = 5, = (2x 2 + 6) ( + ) + = + + ( ) (6x 2 ) = ( ) + (5) + (2x 2 + 6) = (6x 2 ) So (6x 2 + 5) + (2x 2 + 6) = 6x (2x 2 + 6) + (5) + (2x 2 + 6) Now we need to fous on rewriting the expression without the seond set of prentheses. The suexpression (2x 2 + 6) lmost looks like the left side of the identity ( + ) = +, exept tht the vrile n t stnd in for negtive sign lone the negtive sign hs to elong to numer or n expression. So now we form some new minigols, on our wy to the gol of rewriting the expression without prentheses: Minigol: Put 2x into the form ( + ). We n do this if we notie tht we n pply the identity x = 1x to the expression (2x 2 + 6). So, let s do tht: First let s rewrite the identity x = 1x with nother vrile to void onfusion: q = 1q. q = (2x 2 + 6), q = 1q ( ) q q = 1 ( ) So 6x (2x 2 + 6) = 6x (2x 2 + 6) q (2x 2 + 6) q = 1 (2x 2 + 6)
7 Now, k to our first mini gol: Minigol: Rewrite the expression without the seond set of prentheses. We see tht we n now use the identity ( + ) = + to rewrite the seond hlf of the expression without prentheses: = 1, = 2x 2, = 6 ( + ) = + ( ) ( 1) (2x 2 ) + (6) = ( 1) (2x 2 ) + ( 1) (6) ( ) So 6x (2x 2 + 6) = 6x ( 1)(2x 2 ) + ( 1)(6) + ( ) = ( ) ( ) So now we hve simplified the originl expression to this equivlent expression: 6x ( 1)(2x 2 ) + ( 1)(6) How n we simplify this expression further? The most ovious step is for us to perform the multiplition in the lst few terms: 6x ( 1)(2x 2 ) + ( 1)(6) = 6x ( 1)(2x 2 ) + 6 euse ( 1)(6) = 6 6x ( 1)(2x 2 ) + 6 = 6x x euse ( 1)(2x 2 ) = 1 2 x 2 = 2x 2 So now we hve simplified the originl expression to this equivlent expression: 6x x Wht might help to mke this simpler? It would e good for us to look to see if we n omine ny of these terms together. For exmple, if 5 nd 6 were next to eh other, we ould dd them. Similrly, if 6x 2 nd 2x 2 were next to eh other, we ould use the identity ( + ) = + to omine them. So, our next minigol is: ( ) Minigol: Reorder the terms so tht like terms re next to eh other. I n use the identity + = + to do this: = 5, = 2x 2 + = + ( ) = ( ) (5) + ( 2x 2 ) = ( 2x 2 ) + (5) So 6x x = 6x 2 + 2x Now we n simply dd the finl two like terms: 6x 2 + 2x = 6x 2 + 2x euse = 1 And now our finl minigol is: Minigol: Used the identity ( + ) = + to omine the first two terms. We noted tht the first two terms hve the form of the right side of the identity. = 6, = 2, = x 2 ( + ) = + ( ) (6) + ( 2) (x 2 ) = (6) (x 2 ) + ( 2) (x 2 ) ( ) = ( ) So, 6x 2 + 2x = (6 + 2)x euse (6 + 2)x 2 = 6x 2 + 2x 2 ( ) And finlly, we n dd the 6 nd the 2 inside the prentheses, so we go hed nd do tht: (6 + 2) x = 4 x euse = 4 So now we hve repled the originl expression with the equivlent expression 4x This is out s simple s it gets, lthough, we ould do one finl step nd rewrite the ddition of negtive s sutrtion, if we wnt: = 4x 2, = 1 = + ( ) ( ) = ( ) + ( ) (4x 2 ) (1) ( ) = (4x 2 ) + (1)
8 So, 4x = 4x 2 1. This is definitely s simple s we ould mke this expression, so our finl expression is 4x 2 1 nd: (6x 2 + 5) (2x 2 + 6) = 4x 2 1 Now you try! We re going to strt trying to rewrite expressions with speifi gol in mind. However, first you re going to strt y working on prolems where you will e given severl minigols in order. You should try to omplete eh minigol, nd tht will help led you to the overll gol. Be sure to do the minigols in order, nd e sure to work sequentilly, pplying identities to the results of the previous step, NOT to the originl expression. In mny sts, you will first hve to use severl identities to rewrite the expression efore you n omplish the minigol. Use the previous exmple s guide to help you work through these prolems. 1. Rewrite the following expression, with the im of simplifying it s muh s possile: (2x + 3) (3x + 1) Minigol 1: Apply the identity ( + ) = + in order to rewrite the expression without prentheses. Minigol 2: Apply the identity + = ( + ) in order to omine like terms. 2. Rewrite the following expression, with the im of simplifying it s muh s possile: (2x + 3)(3x + 1) Minigol 1: Apply the identity ( + ) = + in order to rewrite the expression without prentheses. Minigol 2: Apply the identity ( + ) = + in order to rewrite the expression without prentheses. Minigol 3: Apply the identity + = ( + ) in order to omine like terms.
9 3. Rewrite the following expression, with the im of simplifying it s muh s possile: 3x 2 2x + 5x 2 5 Minigol 1: Apply the identity + = ( + ) in order to omine like terms. 4. Rewrite the following expression, with the im of simplifying it s muh s possile: 2xy 3 2xx (x, y 0) Minigol 1: Apply the identity = to rewrite the expression so tht the whole expression is under single rdil. Minigol 2: Rewrite ll multiplition under the rdil s produt of squres, insofr s possile. Prt A: Rewrite ll the multiplition under the rdil sign so tht numers re next to numers, nd the sme letters re next to eh other, y using the identity = s mny times s neessry.
10 Prt B: Use the identity = n to rewrite ll exponents s multiplition. n ttttt Prt C: Use the identity = 2 to rewrite ll multiplition s multiplition of squres (inluding numers nd vriles), wherever possile. Minigol 3: Use the identity 2 = to rewrite the expression without rdils, where possile. 5. Rewrite the following expression, with the im of simplifying it s muh s possile: 6x 2 3x 3x (x 0) Minigol 1: Use the identity + = + to rewrite this expression s two frtions, so tht you n fous on one frtion t time. Minigol 2: Use the identity = to rewrite the expression so tht something of the form p is isolted y itself. d d p
11 6. Rewrite the following expression, with the im of simplifying it s muh s possile: 2x 2 y 3 Minigol 1: Use the identity = d to rewrite the expression so tht something of the form p p x 4 is isolted y itself. 7. Put the following expression into x form, where,, re rel numers: 2x(3x 6) + 5 Minigol 1: Apply the identity ( + ) = + in order to rewrite the expression without prentheses. Minigol 2: Apply the identity + = ( + ) in order to omine like terms.
12
NONDETERMINISTIC FSA
Tw o types of nondeterminism: NONDETERMINISTIC FS () Multiple strtsttes; strtsttes S Q. The lnguge L(M) ={x:x tkes M from some strtstte to some finlstte nd ll of x is proessed}. The string x = is
More informationSection 7.1 Area of a Region Between Two Curves
Section 7.1 Are of Region Between Two Curves White Bord Chllenge The circle elow is inscried into squre: Clcultor 0 cm Wht is the shded re? 400 100 85.841cm White Bord Chllenge Find the re of the region
More informationMA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp.
MA123, Chpter 1: Formuls for integrls: integrls, ntiderivtives, nd the Fundmentl Theorem of Clculus (pp. 27233, Gootmn) Chpter Gols: Assignments: Understnd the sttement of the Fundmentl Theorem of Clculus.
More informationexpression simply by forming an OR of the ANDs of all input variables for which the output is
2.4 Logic Minimiztion nd Krnugh Mps As we found ove, given truth tle, it is lwys possile to write down correct logic expression simply y forming n OR of the ANDs of ll input vriles for which the output
More information(a) A partition P of [a, b] is a finite subset of [a, b] containing a and b. If Q is another partition and P Q, then Q is a refinement of P.
Chpter 7: The Riemnn Integrl When the derivtive is introdued, it is not hrd to see tht the it of the differene quotient should be equl to the slope of the tngent line, or when the horizontl xis is time
More information1 Nondeterministic Finite Automata
1 Nondeterministic Finite Automt Suppose in life, whenever you hd choice, you could try oth possiilities nd live your life. At the end, you would go ck nd choose the one tht worked out the est. Then you
More informationPythagoras theorem and surds
HPTER Mesurement nd Geometry Pythgors theorem nd surds In IEEM Mthemtis Yer 8, you lernt out the remrkle reltionship etween the lengths of the sides of rightngled tringle. This result is known s Pythgors
More informationQUADRATIC EQUATION. Contents
QUADRATIC EQUATION Contents Topi Pge No. Theory 004 Exerise  0509 Exerise  093 Exerise  3 45 Exerise  4 6 Answer Key 78 Syllus Qudrti equtions with rel oeffiients, reltions etween roots nd oeffiients,
More information8. Complex Numbers. We can combine the real numbers with this new imaginary number to form the complex numbers.
8. Complex Numers The rel numer system is dequte for solving mny mthemticl prolems. But it is necessry to extend the rel numer system to solve numer of importnt prolems. Complex numers do not chnge the
More informationContinuous Random Variables Class 5, Jeremy Orloff and Jonathan Bloom
Lerning Gols Continuous Rndom Vriles Clss 5, 8.05 Jeremy Orloff nd Jonthn Bloom. Know the definition of continuous rndom vrile. 2. Know the definition of the proility density function (pdf) nd cumultive
More information5: The Definite Integral
5: The Definite Integrl 5.: Estimting with Finite Sums Consider moving oject its velocity (meters per second) t ny time (seconds) is given y v t = t+. Cn we use this informtion to determine the distnce
More informationThis enables us to also express rational numbers other than natural numbers, for example:
Overview Study Mteril Business Mthemtis 0506 Alger The Rel Numers The si numers re,,3,4, these numers re nturl numers nd lso lled positive integers. The positive integers, together with the negtive integers
More informationNondeterministic Finite Automata
Nondeterministi Finite utomt The Power of Guessing Tuesdy, Otoer 4, 2 Reding: Sipser.2 (first prt); Stoughton 3.3 3.5 S235 Lnguges nd utomt eprtment of omputer Siene Wellesley ollege Finite utomton (F)
More informationThe Fundamental Theorem of Algebra
The Fundmentl Theorem of Alger Jeremy J. Fries In prtil fulfillment of the requirements for the Mster of Arts in Teching with Speciliztion in the Teching of Middle Level Mthemtics in the Deprtment of Mthemtics.
More informationWe know that if f is a continuous nonnegative function on the interval [a, b], then b
1 Ares Between Curves c 22 Donld Kreider nd Dwight Lhr We know tht if f is continuous nonnegtive function on the intervl [, b], then f(x) dx is the re under the grph of f nd bove the intervl. We re going
More informationMath 113 Exam 2 Practice
Mth Em Prctice Februry, 8 Em will cover sections 6.5, 7.7.5 nd 7.8. This sheet hs three sections. The first section will remind you bout techniques nd formuls tht you should know. The second gives number
More information12.1 Nondeterminism Nondeterministic Finite Automata. a a b ε. CS125 Lecture 12 Fall 2016
CS125 Lecture 12 Fll 2016 12.1 Nondeterminism The ide of nondeterministic computtions is to llow our lgorithms to mke guesses, nd only require tht they ccept when the guesses re correct. For exmple, simple
More informationGrammar. Languages. Content 5/10/16. Automata and Languages. Regular Languages. Regular Languages
5//6 Grmmr Automt nd Lnguges Regulr Grmmr Contextfree Grmmr Contextsensitive Grmmr Prof. Mohmed Hmd Softwre Engineering L. The University of Aizu Jpn Regulr Lnguges Context Free Lnguges Context Sensitive
More informationNUMERICAL INTEGRATION. The inverse process to differentiation in calculus is integration. Mathematically, integration is represented by.
NUMERICAL INTEGRATION 1 Introduction The inverse process to differentition in clculus is integrtion. Mthemticlly, integrtion is represented by f(x) dx which stnds for the integrl of the function f(x) with
More informationThermodynamics. Question 1. Question 2. Question 3 3/10/2010. Practice Questions PV TR PV T R
/10/010 Question 1 1 mole of idel gs is rought to finl stte F y one of three proesses tht hve different initil sttes s shown in the figure. Wht is true for the temperture hnge etween initil nd finl sttes?
More informationArithmetic & Algebra. NCTM National Conference, 2017
NCTM Ntionl Conference, 2017 Arithmetic & Algebr Hether Dlls, UCLA Mthemtics & The Curtis Center Roger Howe, Yle Mthemtics & Texs A & M School of Eduction Relted Common Core Stndrds First instnce of vrible
More information63. Representation of functions as power series Consider a power series. ( 1) n x 2n for all 1 < x < 1
3 9. SEQUENCES AND SERIES 63. Representtion of functions s power series Consider power series x 2 + x 4 x 6 + x 8 + = ( ) n x 2n It is geometric series with q = x 2 nd therefore it converges for ll q =
More informationUsing integration tables
Using integrtion tbles Integrtion tbles re inclue in most mth tetbooks, n vilble on the Internet. Using them is nother wy to evlute integrls. Sometimes the use is strightforwr; sometimes it tkes severl
More informationPolynomial Approximations for the Natural Logarithm and Arctangent Functions. Math 230
Polynomil Approimtions for the Nturl Logrithm nd Arctngent Functions Mth 23 You recll from first semester clculus how one cn use the derivtive to find n eqution for the tngent line to function t given
More informationVectors. Chapter14. Syllabus reference: 4.1, 4.2, 4.5 Contents:
hpter Vetors Syllus referene:.,.,.5 ontents: D E F G H I J K Vetors nd slrs Geometri opertions with vetors Vetors in the plne The mgnitude of vetor Opertions with plne vetors The vetor etween two points
More informationExam 2 Solutions ECE 221 Electric Circuits
Nme: PSU Student ID Numer: Exm 2 Solutions ECE 221 Electric Circuits Novemer 12, 2008 Dr. Jmes McNmes Keep your exm flt during the entire exm If you hve to leve the exm temporrily, close the exm nd leve
More informationPrefixFree RegularExpression Matching
PrefixFree RegulrExpression Mthing YoSu Hn, Yjun Wng nd Derik Wood Deprtment of Computer Siene HKUST PrefixFree RegulrExpression Mthing p.1/15 Pttern Mthing Given pttern P nd text T, find ll sustrings
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 informationSolutions to Problem Set #1
CSE 233 Spring, 2016 Solutions to Prolem Set #1 1. The movie tse onsists of the following two reltions movie: title, iretor, tor sheule: theter, title The first reltion provies titles, iretors, n tors
More informationPreLie algebras, rooted trees and related algebraic structures
PreLie lgers, rooted trees nd relted lgeri strutures Mrh 23, 2004 Definition 1 A prelie lger is vetor spe W with mp : W W W suh tht (x y) z x (y z) = (x z) y x (z y). (1) Exmple 2 All ssoitive lgers
More informationChapter 4 Regular Grammar and Regular Sets. (Solutions / Hints)
C K Ngpl Forml Lnguges nd utomt Theory Chpter 4 Regulr Grmmr nd Regulr ets (olutions / Hints) ol. (),,,,,,,,,,,,,,,,,,,,,,,,,, (),, (c) c c, c c, c, c, c c, c, c, c, c, c, c, c c,c, c, c, c, c, c, c, c,
More informationSeries. Teacher. Fractions
Series E Teher Frtions Copyright 009 P Lerning. All rights reserved. First edition printed 009 in Austrli. A tlogue reord for this ook is ville from P Lerning Ltd. ISBN 979090 Ownership of ontent The
More informationChapter Gauss Quadrature Rule of Integration
Chpter 7. Guss Qudrture Rule o Integrtion Ater reding this hpter, you should e le to:. derive the Guss qudrture method or integrtion nd e le to use it to solve prolems, nd. use Guss qudrture method to
More informationECON 331 Lecture Notes: Ch 4 and Ch 5
Mtrix Algebr ECON 33 Lecture Notes: Ch 4 nd Ch 5. Gives us shorthnd wy of writing lrge system of equtions.. Allows us to test for the existnce of solutions to simultneous systems. 3. Allows us to solve
More informationQUADRATIC EQUATIONS OBJECTIVE PROBLEMS
QUADRATIC EQUATIONS OBJECTIVE PROBLEMS +. The solution of the eqution will e (), () 0,, 5, 5. The roots of the given eqution ( p q) ( q r) ( r p) 0 + + re p q r p (), r p p q, q r p q (), (d), q r p q.
More informationWaveguide Guide: A and V. Ross L. Spencer
Wveguide Guide: A nd V Ross L. Spencer I relly think tht wveguide fields re esier to understnd using the potentils A nd V thn they re using the electric nd mgnetic fields. Since Griffiths doesn t do it
More information4.1. Probability Density Functions
STT 1 4.14. 4.1. Proility Density Functions Ojectives. Continuous rndom vrile  vers  discrete rndom vrile. Proility density function. Uniform distriution nd its properties. Expected vlue nd vrince of
More informationUniversitaireWiskundeCompetitie. Problem 2005/4A We have k=1. Show that for every q Q satisfying 0 < q < 1, there exists a finite subset K N so that
Problemen/UWC NAW 5/7 nr juni 006 47 Problemen/UWC UniversitireWiskundeCompetitie Edition 005/4 For Session 005/4 we received submissions from Peter Vndendriessche, Vldislv Frnk, Arne Smeets, Jn vn de
More informationSCHOOL OF ENGINEERING & BUILT ENVIRONMENT. Mathematics
SCHOOL OF ENGINEERING & BUIL ENVIRONMEN Mthemtics An Introduction to Mtrices Definition of Mtri Size of Mtri Rows nd Columns of Mtri Mtri Addition Sclr Multipliction of Mtri Mtri Multipliction 7 rnspose
More informationWave Equation on a Two Dimensional Rectangle
Wve Eqution on Two Dimensionl Rectngle In these notes we re concerned with ppliction of the method of seprtion of vriles pplied to the wve eqution in two dimensionl rectngle. Thus we consider u tt = c
More informationLIP. Laboratoire de l Informatique du Parallélisme. Ecole Normale Supérieure de Lyon
LIP Lortoire de l Informtique du Prllélisme Eole Normle Supérieure de Lyon Institut IMAG Unité de reherhe ssoiée u CNRS n 1398 Onewy Cellulr Automt on Cyley Grphs Zsuzsnn Rok Mrs 1993 Reserh Report N
More informationLecture 27: Diffusion of Ions: Part 2: coupled diffusion of cations and
Leture 7: iffusion of Ions: Prt : oupled diffusion of tions nd nions s desried y NernstPlnk Eqution Tody s topis Continue to understnd the fundmentl kinetis prmeters of diffusion of ions within n eletrilly
More informationa a a a a a a a a a a a a a a a a a a a a a a a In this section, we introduce a general formula for computing determinants.
Section 9 The Lplce Expnsion In the lst section, we defined the determinnt of (3 3) mtrix A 12 to be 22 12 21 22 2231 22 12 21. In this section, we introduce generl formul for computing determinnts. Rewriting
More informationFractions arise to express PART of a UNIT 1 What part of an HOUR is thirty minutes? Fifteen minutes? tw elve minutes? (The UNIT here is HOUR.
6 FRACTIONS sics MATH 0 F Frctions rise to express PART of UNIT Wht prt of n HOUR is thirty minutes? Fifteen minutes? tw elve minutes? (The UNIT here is HOUR.) Wht frction of the children re hppy? (The
More informationMapping the delta function and other Radon measures
Mpping the delt function nd other Rdon mesures Notes for Mth583A, Fll 2008 November 25, 2008 Rdon mesures Consider continuous function f on the rel line with sclr vlues. It is sid to hve bounded support
More informationSemantic Analysis. CSCI 3136 Principles of Programming Languages. Faculty of Computer Science Dalhousie University. Winter Reading: Chapter 4
Semnti nlysis SI 16 Priniples of Progrmming Lnguges Fulty of omputer Siene Dlhousie University Winter 2012 Reding: hpter 4 Motivtion Soure progrm (hrter strem) Snner (lexil nlysis) Front end Prse tree
More informationSECOND HARMONIC GENERATION OF Bi 4 Ti 3 O 12 FILMS
SECOND HARMONIC GENERATION OF Bi 4 Ti 3 O 12 FILMS INSITU PROBING OF DOMAIN POLING IN Bi 4 Ti 3 O 12 THIN FILMS BY OPTICAL SECOND HARMONIC GENERATION YANIV BARAD, VENKATRAMAN GOPALAN Mterils Reserh Lortory
More informationAutomata and Languages
Automt nd Lnguges Prof. Mohmed Hmd Softwre Engineering Lb. The University of Aizu Jpn Grmmr Regulr Grmmr Contextfree Grmmr Contextsensitive Grmmr Regulr Lnguges Context Free Lnguges Context Sensitive
More informationSection 11.5 Notes Page Partial Fraction Decomposition. . You will get: +. Therefore we come to the following: x x
Setio Notes Pge Prtil Frtio Deompositio Suppose we were sked to write the followig s sigle frtio: We would eed to get ommo deomitors: You will get: Distributig o top will give you: 8 This simplifies to:
More informationThe RiemannStieltjes Integral
Chpter 6 The RiemnnStieltjes Integrl 6.1. Definition nd Eistene of the Integrl Definition 6.1. Let, b R nd < b. ( A prtition P of intervl [, b] is finite set of points P = { 0, 1,..., n } suh tht = 0
More informationGénération aléatoire uniforme pour les réseaux d automates
Génértion létoire uniforme pour les réseux d utomtes Niols Bsset (Trvil ommun ve Mihèle Sori et Jen Miresse) Université lire de Bruxelles Journées Alé 2017 1/25 Motivtions (1/2) p q Automt re omnipresent
More informationIf only one fertilizer x is used, the dependence of yield z(x) on x first was given by Mitscherlich (1909) in form of the differential equation
Mitsherlih s Lw: Generliztion with severl Fertilizers Hns Shneeerger Institute of Sttistis, University of ErlngenNürnerg, Germny 00, 5 th August Astrt: It is shown, tht the ropyield z in dependene on
More informationUniversity of Washington Department of Chemistry Chemistry 453 Winter Quarter 2010 Homework Assignment 4; Due at 5p.m. on 2/01/10
University of Wshington Deprtment of Chemistry Chemistry 45 Winter Qurter Homework Assignment 4; Due t 5p.m. on // We lerned tht the Hmiltonin for the quntized hrmonic oscilltor is ˆ d κ H. You cn obtin
More informationIntroduction to vectors
Introduction to vectors mctyintrovector20091 Avectorisquntityththsothmgnitude(orsize)nddirection. Bothofthese propertiesmustegiveninordertospecifyvectorcompletely.inthisunitwedescriehowto writedownvectors,howtoddndsutrctthem,ndhowtousethemingeometry.
More informationa,b a 1 a 2 a 3 a,b 1 a,b a,b 2 3 a,b a,b a 2 a,b CS Determinisitic Finite Automata 1
CS4 45 Determinisitic Finite Automt : Genertors vs. Checkers Regulr expressions re one wy to specify forml lnguge String Genertor Genertes strings in the lnguge Deterministic Finite Automt (DFA) re nother
More informationDiophantine Steiner Triples and PythagoreanType Triangles
Forum Geometricorum Volume 10 (2010) 93 97. FORUM GEOM ISSN 15341178 Diophntine Steiner Triples nd PythgorenType Tringles ojn Hvl bstrct. We present connection between Diophntine Steiner triples (integer
More informationGraph Theory. Simple Graph G = (V, E). V={a,b,c,d,e,f,g,h,k} E={(a,b),(a,g),( a,h),(a,k),(b,c),(b,k),...,(h,k)}
Grph Theory Simple Grph G = (V, E). V ={verties}, E={eges}. h k g f e V={,,,,e,f,g,h,k} E={(,),(,g),(,h),(,k),(,),(,k),...,(h,k)} E =16. 1 Grph or MultiGrph We llow loops n multiple eges. G = (V, E.ψ)
More information11.1 Exponential Functions
. Eponentil Functions In this chpter we wnt to look t specific type of function tht hs mny very useful pplictions, the eponentil function. Definition: Eponentil Function An eponentil function is function
More informationDesign and Analysis of Distributed Interacting Systems
Design nd Anlysis of Distriuted Intercting Systems Lecture 6 LTL Model Checking Prof. Dr. Joel Greenyer My 16, 2013 Some Book References (1) C. Bier, J.P. Ktoen: Principles of Model Checking. The MIT
More informationLecture 8. Band theory con.nued
Lecture 8 Bnd theory con.nued Recp: Solved Schrodinger qu.on for free electrons, for electrons bound in poten.l box, nd bound by proton. Discrete energy levels rouse. The Schrodinger qu.on pplied to periodic
More informationRegular Rational Diophantine Sextuples
Regulr Rtionl Diophntine Sextuples Philip E Gis philegis@gmil.om A polynomil eqution in six vriles is given tht generlises the definition of regulr rtionl Diophntine triples, qudruples nd quintuples to
More informationLecture 9: LTL and Büchi Automata
Lecture 9: LTL nd Büchi Automt 1 LTL Property Ptterns Quite often the requirements of system follow some simple ptterns. Sometimes we wnt to specify tht property should only hold in certin context, clled
More informationContinuous Random Variables
STAT/MATH 395 A  PROBABILITY II UW Winter Qurter 217 Néhémy Lim Continuous Rndom Vribles Nottion. The indictor function of set S is relvlued function defined by : { 1 if x S 1 S (x) if x S Suppose tht
More information1.3 The log laws. Inverse functions. We have defined logx as the index needed to write x as a power of 10. For example:
1.3 The log lws Inverse functions. We hve defined log s the inde needed to write s power of. For emple: since 1.8 = 66.07 we hve log66.07 = 1.8 Note tht wht this sys is tht the functions f() = nd g() =
More informationNonLinear & Logistic Regression
NonLiner & Logistic Regression If the sttistics re boring, then you've got the wrong numbers. Edwrd R. Tufte (Sttistics Professor, Yle University) Regression Anlyses When do we use these? PART 1: find
More informationOverview HC9. Parsing: TopDown & LL(1) ContextFree Grammars (1) Introduction. CFGs (3) ContextFree Grammars (2) Vertalerbouw HC 9: Ch.
Overview H9 Vertlerouw H 9: Prsing: opdown & LL(1) do 3 mei 2001 56 heo Ruys h. 8  Prsing 8.1 ontextfree Grmmrs 8.2 opdown Prsing 8.3 LL(1) Grmmrs See lso [ho, Sethi & Ullmn 1986] for more thorough
More informationPDE Notes. Paul Carnig. January ODE s vs PDE s 1
PDE Notes Pul Crnig Jnury 2014 Contents 1 ODE s vs PDE s 1 2 Section 1.2 Het diffusion Eqution 1 2.1 Fourier s w of Het Conduction............................. 2 2.2 Energy Conservtion.....................................
More informationKnowledge Spaces and Learning Spaces 1
Knowledge Spes nd Lerning Spes 1 JenPul Doignon 2 JenClude Flmgne 3 Astrt How to design utomted proedures whih (i) urtely ssess the knowledge of student, nd (ii) effiiently provide dvies for further
More informationClassification: Rules. Prof. Pier Luca Lanzi Laurea in Ingegneria Informatica Politecnico di Milano Polo regionale di Como
Metodologie per Sistemi Intelligenti Clssifiction: Prof. Pier Luc Lnzi Lure in Ingegneri Informtic Politecnico di Milno Polo regionle di Como Rules Lecture outline Why rules? Wht re clssifiction rules?
More informationThe Bernoulli Numbers John C. Baez, December 23, x k. x e x 1 = n 0. B k n = n 2 (n + 1) 2
The Bernoulli Numbers John C. Bez, December 23, 2003 The numbers re defined by the eqution e 1 n 0 k. They re clled the Bernoulli numbers becuse they were first studied by Johnn Fulhber in book published
More informationCubical Structures for HigherDimensional Type Theories
Cubicl Structures for HigherDimensionl Type Theories Ed Morehouse October 30, 2015 1 / 36 A higherdimensionl type theory depends on notion of higherdimensionl bstrct spces. Mny choices: globulr, simplicil,
More informationBasic Angle Rules 5. A Short Hand Geometric Reasons. B Two Reasons. 1 Write in full the meaning of these short hand geometric reasons.
si ngle Rules 5 6 Short Hnd Geometri Resons 1 Write in full the mening of these short hnd geometri resons. Short Hnd Reson Full Mening ) se s isos Δ re =. ) orr s // lines re =. ) sum s t pt = 360. d)
More informationFoundation of Diagnosis and Predictability in Probabilistic Systems
Foundtion of Dignosis nd Preditility in Proilisti Systems Nthlie Bertrnd 1, Serge Hddd 2, Engel Lefuheux 1,2 1 Inri Rennes, Frne 2 LSV, ENS Chn & CNRS & Inri Sly, Frne De. 16th FSTTCS 14 Dignosis of disrete
More informationF is n ntiderivtive èor èindeæniteè integrlè off if F 0 èxè =fèxè. Nottion: F èxè = ; it mens F 0 èxè=fèxè ëthe integrl of f of x dee x" Bsic list: xn
Mth 70 Topics for third exm Chpter 3: Applictions of Derivtives x7: Liner pproximtion nd diæerentils Ide: The tngent line to grph of function mkes good pproximtion to the function, ner the point of tngency.
More informationFrom the Numerical. to the Theoretical in. Calculus
From the Numericl to the Theoreticl in Clculus Teching Contemporry Mthemtics NCSSM Ferury 67, 003 Doug Kuhlmnn Phillips Acdemy Andover, MA 01810 dkuhlmnn@ndover.edu How nd Why Numericl Integrtion Should
More informationTransition systems (motivation)
Trnsition systems (motivtion) Course Modelling of Conurrent Systems ( Modellierung neenläufiger Systeme ) Winter Semester 2009/0 University of DuisurgEssen Brr König Tehing ssistnt: Christoph Blume In
More informationMatrix Eigenvalues and Eigenvectors September 13, 2017
Mtri Eigenvlues nd Eigenvectors September, 7 Mtri Eigenvlues nd Eigenvectors Lrry Cretto Mechnicl Engineering 5A Seminr in Engineering Anlysis September, 7 Outline Review lst lecture Definition of eigenvlues
More informationExponents and Polynomials
C H A P T E R 5 Eponents nd Polynomils ne sttistic tht cn be used to mesure the generl helth of ntion or group within ntion is life epectncy. This dt is considered more ccurte thn mny other sttistics becuse
More informationContinuity. Recall the following properties of limits. Theorem. Suppose that lim. f(x) =L and lim. lim. [f(x)g(x)] = LM, lim
Recll the following properties of limits. Theorem. Suppose tht lim f() =L nd lim g() =M. Then lim [f() ± g()] = L + M, lim [f()g()] = LM, if M = 0, lim f() g() = L M. Furthermore, if f() g() for ll, then
More informationChapter 1, Part 1. Regular Languages. CSC527, Chapter 1, Part 1 c 2012 Mitsunori Ogihara 1
Chpter 1, Prt 1 Regulr Lnguges CSC527, Chpter 1, Prt 1 c 2012 Mitsunori Ogihr 1 Finite Automt A finite utomton is system for processing ny finite sequence of symols, where the symols re chosen from finite
More informationAccelerator Physics. G. A. Krafft Jefferson Lab Old Dominion University Lecture 5
Accelertor Phyic G. A. Krfft Jefferon L Old Dominion Univerity Lecture 5 ODU Accelertor Phyic Spring 15 Inhomogeneou Hill Eqution Fundmentl trnvere eqution of motion in prticle ccelertor for mll devition
More informationI. Equations of a Circle a. At the origin center= r= b. Standard from: center= r=
11.: Circle & Ellipse I cn Write the eqution of circle given specific informtion Grph circle in coordinte plne. Grph n ellipse nd determine ll criticl informtion. Write the eqution of n ellipse from rel
More informationFachgebiet Rechnersysteme1. 1. Boolean Algebra. 1. Boolean Algebra. Verification Technology. Content. 1.1 Boolean algebra basics (recap)
. Boolen Alger Fchgeiet Rechnersysteme. Boolen Alger Veriiction Technology Content. Boolen lger sics (recp).2 Resoning out Boolen expressions . Boolen Alger 2 The prolem o logic veriiction: Show tht two
More information4. Statements Reasons
Chpter 9 Answers PrentieHll In. Alterntive Ativity 9. Chek students work.. Opposite sides re prllel. 3. Opposite sides re ongruent. 4. Opposite ngles re ongruent. 5. Digonls iset eh other. 6. Students
More informationThermal Diffusivity. Paul Hughes. Department of Physics and Astronomy The University of Manchester Manchester M13 9PL. Second Year Laboratory Report
Therml iffusivity Pul Hughes eprtment of Physics nd Astronomy The University of nchester nchester 3 9PL Second Yer Lbortory Report Nov 4 Abstrct We investigted the therml diffusivity of cylindricl block
More informationf (x)dx = f(b) f(a). a b f (x)dx is the limit of sums
Green s Theorem If f is funtion of one vrible x with derivtive f x) or df dx to the Fundmentl Theorem of lulus, nd [, b] is given intervl then, ording This is not trivil result, onsidering tht b b f x)dx
More informationarxiv: v2 [cs.lo] 26 Dec 2016
On Negotition s Concurrency Primitive II: Deterministic Cyclic Negotitions Jvier Esprz 1 nd Jörg Desel 2 1 Fkultät für Informtik, Technische Universität München, Germny 2 Fkultät für Mthemtik und Informtik,
More informationMATH 174A: PROBLEM SET 5. Suggested Solution
MATH 174A: PROBLEM SET 5 Suggested Solution Problem 1. Suppose tht I [, b] is n intervl. Let f 1 b f() d for f C(I; R) (i.e. f is continuous relvlued function on I), nd let L 1 (I) denote the completion
More information6.5 Plate Problems in Rectangular Coordinates
6.5 lte rolems in Rectngulr Coordintes In this section numer of importnt plte prolems ill e emined ug Crte coordintes. 6.5. Uniform ressure producing Bending in One irection Consider first the cse of plte
More informationComplementing Büchi Automata with a Subsettuple Construction
DEPARTEMENT D INFORMATIQUE DEPARTEMENT FÜR INFORMATIK Bd de Pérolles 90 CH1700 Friourg www.unifr.ch/informtics WORKING PAPER Complementing Büchi Automt with Susettuple Construction J. Allred & U. UltesNitsche
More informationComplementing Büchi Automata
Complementing Bühi Automt Guillume Sdegh Tehnil Report n o 090, My 009 revision 07 Model heking is field of forml verifition whih ims to utomtilly hek the ehvior of system with the help of logi formulæ.
More informationPartial Differential Equations
Prtil Differentil Equtions Notes by Robert Piché, Tmpere University of Technology reen s Functions. reen s Function for OneDimensionl Eqution The reen s function provides complete solution to boundry
More informationOrthogonal Polynomials and LeastSquares Approximations to Functions
Chpter Orthogonl Polynomils nd LestSqures Approximtions to Functions **4/5/3 ET. Discrete LestSqures Approximtions Given set of dt points (x,y ), (x,y ),..., (x m,y m ), norml nd useful prctice in mny
More informationDiscrete Leastsquares Approximations
Discrete Lestsqures Approximtions Given set of dt points (x, y ), (x, y ),, (x m, y m ), norml nd useful prctice in mny pplictions in sttistics, engineering nd other pplied sciences is to construct curve
More informationThe final exam will take place on Friday May 11th from 8am 11am in Evans room 60.
Mth 104: finl informtion The finl exm will tke plce on Fridy My 11th from 8m 11m in Evns room 60. The exm will cover ll prts of the course with equl weighting. It will cover Chpters 1 5, 7 15, 17 21, 23
More informationComplexity in Modal Team Logic
ThI Theoretische Informtik Complexity in Modl Tem Logic JulinSteffen Müller Theoretische Informtik 18. Jnur 2012 Theorietg 2012 Theoretische Informtik Inhlt 1 Preliminries 2 Closure properties 3 Model
More informationC1M14. Integrals as Area Accumulators
CM Integrls s Are Accumultors Most tetbooks do good job of developing the integrl nd this is not the plce to provide tht development. We will show how Mple presents Riemnn Sums nd the ccompnying digrms
More informationH 4 H 8 N 2. Example 1 A compound is found to have an accurate relative formula mass of It is thought to be either CH 3.
. Spetrosopy Mss spetrosopy igh resolution mss spetrometry n e used to determine the moleulr formul of ompound from the urte mss of the moleulr ion For exmple, the following moleulr formuls ll hve rough
More informationChapter 19: The Second Law of Thermodynamics
hpter 9: he Seon Lw of hermoynmis Diretions of thermoynmi proesses Irreversile n reversile proesses hermoynmi proesses tht our in nture re ll irreversile proesses whih proee spontneously in one iretion
More informationHypergraph regularity and quasirandomness
Hypergrph regulrity nd qusirndomness Brendn Ngle Annik Poerschke Vojtěch Rödl Mthis Schcht Astrct Thomson nd Chung, Grhm, nd Wilson were the first to systemticlly study qusirndom grphs nd hypergrphs,
More information