Solutions. tan 2 θ(tan 2 θ + 1) = cot6 θ,

Size: px
Start display at page:

Download "Solutions. tan 2 θ(tan 2 θ + 1) = cot6 θ,"

Transcription

1 Solutios 99. Let A ad B be two poits o a parabola with vertex V such that V A is perpedicular to V B ad θ is the agle betwee the chord V A ad the axis of the parabola. Prove that V A V B cot3 θ. Commet. A lot of studets wored harder o this problem tha was ecessary. It should be oted that all parabolas are similar (as ideed all circles are similar); this meas that you ca establish a geeral result about parabolas by dealig with a coveiet oe. Let us see why this is so. Oe defiitio of a parabola is that it is the locus of poits that are equidistat from a give poit (called the focus) ad a give lie (called the directrix) that does ot cotai the poit. Ay poit-lie pair ca be used, ad each such pair ca be trasformed ito aother by a similarity trasformatio. (Traslate oe poit o to the other, mae a rotatio to mae the two lies parallel ad perform a dilatio about the poit that maes the two lies coicide.) The same trasformatio will tae the prabola defied by oe pair to the parabola defied by the other. You should poit out i your solutio that there is o loss of geerality i taig the particular case of a parabola whose equatio i the plae is y ax. But you do ot have to be eve that geeral; it is eough to assume that the parabola has the equatio y x or x y. (Exercise: Determie the focus ad the directrix for these parabolas.) Some of the solvers did ot appear to be aware that parabolas eed ot have vertical or horizotal axes; the axis of a parabola ca poit i ay directio. Solutio. Wolog, suppose that the parabola is give by y x, so that its vertex is the origi ad its axis is the x axis. Suppose A (u, v) is a poit o the parabola whose radius vector maes a agle θ with the axis; the v/u ta θ. Hece /u v /u ta θ, so that A (cot θ, cot θ). Similarly, it ca be show that B (ta θ, ta θ). Hece ad the result follows. V A V B cot θ(cot θ + ) ta θ(ta θ + ) cot6 θ, 00. Let be a positive iteger exceedig. Determie the umber of permutatios (a, a,, a ) of (,,, ) for which there exists exactly oe idex i with i ad a i > a i+. Commet. Some solvers foud it difficult to appreciate what was goig o i this problem. It is ofte a good begiig strategy to actually write out the appropriate permutatios for low values of. This does two thigs for you. First, it gives you a sese of what goes ito costructig the right permutatios ad so how your argumet ca be framed. Secodly, it gives you some data agaist which you ca chec your fial aswer. Solutio. For, let p be the umber of permutatios of the first atural umbers that satisfy the coditio. Suppose that a i for some i with i. The (a, a,, a i ) ad (a i+,, a ) must both be i icreasig order, so that the appropriate permutatio is determied uiquely oce its first i etries are foud. There are ( i ) ways of choosig these etries. If a, the there are p ways of orderig the first umbers to give a appropriate permutatio. Hece p [ Thus, substitutig for each p i i tur, we have that ( )] + p + p. i p ( ) + ( ) + + ( ) + ( ) + ( ) ( + ).

2 Solutio. [H. Li; M. Zaharia] For, let p be the umber of acceptable permutatios. We have that p. Cosider first the placig of the umbers,,, i some order. If they appear i their atural order, the we ca slip i before ay oe of them to get a acceptable permutatio; there are ways of doig this. If there exists a sigle cosecutive pair (r, s) of umbers for which r < s ad r follows s, the we ca slip betwee s ad r or at the ed to get a acceptable permutatio. There are p possibilities. If there is more tha oe pair (r, s) of cosecutive pairs with r < s ad r followig s, the o placemet of will yield a acceptable permutatio. Hece p p + ( ) so that whece p ( + ). p + + (p + ) (p + ) (p + 3) 4, Solutio 3. [R. Barrigto Leigh] Let ad let (x, y) be a pair of itegers for which y < x ad x y. There are such pairs, (, + ), (, + ),, (, ). For each such pair, we cosider suitable permutatios for which x ad y are adjacet i the order (x, y). The the umbers,,, y must precede ad x +,, must follow the pair. The remaiig umers from x + to x + y ca go either before or after the pair; there are possibilities. Oce it is decided whether each of these goes before or after the pair, there is oly oe possible arragemet. Hece the umber of permutatios of the required type is ( ) [( + ) ( + ) ] [( ) ( ) ( + ) ( + ). Solutio 4. Let i ad cosider the umber of suitable permutatios for which a i > a i+. There are ( i) possible choices of {a, a,, a i } with a < a < < a i, ad except for the sigle case of {,,, i}, the maximum elemet a i of each of them exceeds the miimum elemet a i+ of its complemet {a i+,, a }. Hece the umber of permutatios is [( ) ] i i0 [( ) ] ( + ). i Solutio 5. (Variat of Solutio 4.) We ca form a acceptable permutatio i the followig way. Let. Select ay subset of umbers i oe of ( ) ways ad place them i ascedig order at the begiig of the arragemet ad place the other at the ed, agai i ascedig order. This fails to wor oly whe the set chose is {,,, }. Hece the total umber of ways is [( ) ] [ ( )] ( ). 0. Let (a, a,, a ) be a arithmetic progressio ad (b, b,, b ) be a geometric progressio, each of positive real umbers, for which a b ad a b. Prove that a + a + + a b + b + + b.

3 Solutio. The result is obvious if a a b b, as the all of the a i ad b j are equal. Suppose that the progressios are otrivial ad that the commo ratio of the geometric progressio is r. Observe that (r + ) (r + r ) (r )(r ) > 0. The b + b + + b b ( + r + r + r r ) b (r + r ) < b (r + ) [b r + b ] [b + b ] [a + a ] a + a + + a. Solutio. For r, we have that by the arithmetic-geometric meas iequality. b r b ( r)/() b (r )/() a ( r)/() a (r )/() r a + r a a r, 0. For each positive iteger, let a + (/) + (/3) + + (/). Prove that, for each positive iteger, 3a + 5a + 7a ( + )a ( + ) a ( + ). The Solutio. Observe that, for, ( + ) + ( + 3) + + ( + ) ( ) ( ) ( + ). 3a + 5a + 7a ( + )a ( ) ( ) ( ) + ( ) + + ( + ) ( ) [( + ) ] ( + ) ( + ) a ( + ). Solutio. Observe that for each positive iteger, [( + ) a ( + )] [ a ( )] (a a ) + ( + )a ( + ) (/) + ( + )a ( + )a. 3

4 Hece 3a + 5a + + ( + )a 3a + {[( + ) a ( + )] [ a ( )]} 3a + [( + ) a ( + )] [4a ] ( + ) a ( + ) + a ( + ) a ( + ). Solutio 3. We use a iductio argumet. The result holds for. Suppose it holds for. The 3a + 5a + + ( )a + ( + )a a ( ) + ( + )a ( a ) ( ) + ( + )a ( + ) a [ + ( )] ( + ) a ( + ). Solutio 4. [R. Furmaia] Let a 0 0, The a i a i + (/i) for i, so that (i + )a i [(i + ) i ]a i [(i + ) a i i a i i (/i)] ( + ) a a 0 i ( + ) a ( + ). Solutio 5. [A. Verroe] Let a 0 0. For, ( + ) a ( + )a [ ( + ) a + ( )] ( ( + )a + ( + ) ) ( ) ( + )a + ( ( + )) + ( ) ( + )a + ( + ) + from which the result follows. (To see the secod last equality, write out the sums ad istead of summig alog the +, sum alog the /( + ).] 4

5 Solutio 6. [T. Yi] Recall Abel s Partial Summatio Formula: u v (u + u + + u )v (u + u + + u )(v + v ). (Prove this. Compare with itegratio by parts i calculus.) Applyig this to u + ad v a, we fid that u + + u ( + ) ad v + v /( + ), whereupo ( + )a ( + ) a a ( + ) + + [ ] ( + ) ( + ) a a + [a ] ( + ) a ( + ). 03. Every midpoit of a edge of a tetrahedro is cotaied i a plae that is perpedicular to the opposite edge. Prove that these six plaes itersect i a poit that is symmetric to the cetre of the circumsphere of the tetrahedro with respect to its cetroid. Solutio. Let O be the cetre of the circumsphere of the tetrahedro ABCD ad G be its cetroid. The OG 4 ( OA + OB + OC + OD). Let N be the poit determied by Let P be the midpoit of the edge AB. The ad ON OG ( OA + OB + OC + OD). P N ON OP ON ( OA + OB) ( OC + OD) P N CD ( OD + OC) ( OD + OC) ( OD OC ) 0. Hece P N CD, so that the segmet P N is cotaied i a plae that is orthogoal to CD. A similar result holds for the other five edges. The result follows. Solutio. [O. Bormasheo] Let O be the circumcetre ad let G be the cetroid of the tetrahedro. Let M be the midpoit of the edge AB ad N the midpoit of the edge CD. The cetroid of the triagle ABC lies at a poit E o MC for which CE EM, so that CM 3EM. The cetroid of the tetrahedro is the positio of the cetre of gravity whe uit masses are placed at its vertices, ad so is the positio of the cetre of gravity of a uit mass placed at D ad a triple mass at E. Thus G is o DE ad satisfies DG 3GE. Cosider triagle CDE. We have that CM ME EG GD DN NC ( 3) ( ), 3 so that, by the coverse to Meelaus Theorem, G, M ad N are colliear. Cosider triagle M CN ad trasversal DGE. By Meelaus Theorem, ME EC CD DN NG ( ) GM ( ) NG GM, 5

6 whece NG GM ad G is the midpoit of MN. Suppose that K is the poit o OG produced so that OG GK. Sice OK ad MN itersect i G at their respective midpoits, OMKN is a plaar parallelogram ad ON KM. Sice OC OD, triagle OCD is isosceles, ad so ON CD. Hece KM CD. Therefore, K lies o the plae through the midpoit M of AB ad perpedicular to CD. By symmetry, K lies o the other plaes through the midpoits of a edge ad perpedicular to the opposite edge. 04. Each of people i a certai village has at least oe of eight differet ames. No two people have exactly the same set of ames. For a arbitrary set of ames (with 7), the umber of people cotaiig at least oe of the ames amog his/her set of ames is eve. Determie the value of. Solutio. Let P be a perso with the least umber of ames. The remaiig people have at least oe of the ames ot possessed by P, so by the coditio of the problem applied to the set of ames ot possessed by P, is eve ad so is odd. Let x be oe of the eight ames, ad suppose, if possible, that o perso has x as his/her sole ame. The all people have at least oe of the remaiig ames which yields the cotradictio that must be eve. Hece, for each ame, there is a perso with oly that ame. Suppose there is o perso with oly a pair {x, y} of ames. The there are people who have a ame other tha x ad y, which yields agai a cotradictio, sice is odd. Hece, for each pair of ames, there is exactly oe perso possessig those two ames. We ca cotiue the argumet. Suppose, if possible, there is o perso possessig exactly the three ames x, y ad z. The except for the six people with the ame sets {x}, {y}, {z}, {x, y}, {y, z}, {z, x}, everyoe possesses at least oe of the ames other tha x, y, z, which leads to a cotradictio. Evetually, we ca argue that, for each ovoid set of the eight ames, there is exactly oe perso with that set of ames. Sice there are 55 8 such subsets, there must be 55 people. Solutio. [R. Furmaia] For i 8, let S i be the set of people whose ames iclude the ith ame. By the coditio of the problem for, the cardiality, #S i, of S i must be eve. Suppose, for 7, it has bee show that ay itersectio of fewer tha of the S i has eve cardiality. Cosider a itersectio of of the S i, say S S S. By the coditio of the problem, #(S S S ), the umber of people with at least oe of the first ames, is eve. But, from the Priciple of Iclusio-Exclusio, we have that #(S S S ) #S i i j #(S i S j ) + i,j, #(S i S j S ) + ( ) #(S S S ). By the iductio hypothesis, each term i the series o the right but the last is eve, ad so the last is eve as well. Cosider the largest set of ames, say {i,, i r } possessed by ay oe perso. This set ca appear oly oce, so that r j S i j is a sigleto. By the above paragraph, the itersectio must have eight members (o fewer) ad so some perso possesses all eight ames. If a set of ames does ot belog to ay perso, let T be a maximal such set with 7 ames, say the first ames. By maximality, each superset of T be be a set of ames for someoe. The supersets cosist of the ames alog with all of the 8 possible subsets of the remaiig ames. But the superset of ames are possessed by all the people i S S S, ad this set has eve cardiality ad so caot have cardiality 8. This is a cotradictio. Thus every possible ovoid set of ames must occur ad Let f(x) be a covex realvalued fuctio defied o the reals, ad x < x < < x. Prove that x f(x ) + x f(x 3 ) + + x f(x ) x f(x ) + x 3 f(x ) + + x f(x ). 6

7 Solutio. The case is obvious. For 3, we have that x f(x )+x f(x 3 ) + x 3 f(x ) x f(x ) x 3 f(x ) x f(x 3 ) (x 3 x )f(x ) + (x x )f(x 3 ) (x 3 x )f(x ) [ (x3 x ) (x 3 x ) (x 3 x ) f(x ) + (x ] x ) (x 3 x ) f(x 3) f(x ) 0. Suppose, as a iductio hypothesis, that the result holds for all values of up to 3. The x f(x ) + x f(x 3 ) + + x f(x + ) + x + f(x ) by the result for ad 3. [x f(x ) + + x f(x )] + [x f(x + ) + x + f(x ) x f(x )] [x f(x ) + + x f(x )] + [x + f(x ) + x f(x + ) x f(x )] x f(x ) + + x + f(x ) + x f(x + ), Solutio. [J. Kramar] For i, let λ i (x i x )/(x x ), so that 0 λ λ λ λ ad x i λ i x + ( λ i )x. The f(x )λ + ( λ )f(x ) (f(x ) f(x ))λ + f(x )(λ + λ λ ) (f(x ) f(x ))(λ λ λ λ ) + f(x )(λ + λ λ λ ) (f(x ) f(x )) (λ i λ i+ λ i λ i ) + f(x ) (λ i+ λ i ) i [λ i+ λ i ][λ i f(x ) + ( λ i )f(x )] i [λ i+ λ i ]f(x i ). i Multiplyig by x x ad rearragig terms yields that x f(x ) + x f(x ) from which the desired result follows. [ i i ] f(x i )(x i+ x i ) + x f(x ) + x f(x ) 7

THE UNIVERSITY OF TORONTO UNDERGRADUATE MATHEMATICS COMPETITION. In Memory of Robert Barrington Leigh. March 9, 2014

THE UNIVERSITY OF TORONTO UNDERGRADUATE MATHEMATICS COMPETITION. In Memory of Robert Barrington Leigh. March 9, 2014 THE UNIVERSITY OF TORONTO UNDERGRADUATE MATHEMATICS COMPETITION I Memory of Robert Barrigto Leigh March 9, 4 Time: 3 hours No aids or calculators permitted. The gradig is desiged to ecourage oly the stroger

More information

Solutions for May. 3 x + 7 = 4 x x +

Solutions for May. 3 x + 7 = 4 x x + Solutios for May 493. Prove that there is a atural umber with the followig characteristics: a) it is a multiple of 007; b) the first four digits i its decimal represetatio are 009; c) the last four digits

More information

Complex Numbers Solutions

Complex Numbers Solutions Complex Numbers Solutios Joseph Zoller February 7, 06 Solutios. (009 AIME I Problem ) There is a complex umber with imagiary part 64 ad a positive iteger such that Fid. [Solutio: 697] 4i + + 4i. 4i 4i

More information

CALCULUS BASIC SUMMER REVIEW

CALCULUS BASIC SUMMER REVIEW CALCULUS BASIC SUMMER REVIEW NAME rise y y y Slope of a o vertical lie: m ru Poit Slope Equatio: y y m( ) The slope is m ad a poit o your lie is, ). ( y Slope-Itercept Equatio: y m b slope= m y-itercept=

More information

Substitute these values into the first equation to get ( z + 6) + ( z + 3) + z = 27. Then solve to get

Substitute these values into the first equation to get ( z + 6) + ( z + 3) + z = 27. Then solve to get Problem ) The sum of three umbers is 7. The largest mius the smallest is 6. The secod largest mius the smallest is. What are the three umbers? [Problem submitted by Vi Lee, LCC Professor of Mathematics.

More information

UNIVERSITY OF NORTHERN COLORADO MATHEMATICS CONTEST. First Round For all Colorado Students Grades 7-12 November 3, 2007

UNIVERSITY OF NORTHERN COLORADO MATHEMATICS CONTEST. First Round For all Colorado Students Grades 7-12 November 3, 2007 UNIVERSITY OF NORTHERN COLORADO MATHEMATICS CONTEST First Roud For all Colorado Studets Grades 7- November, 7 The positive itegers are,,, 4, 5, 6, 7, 8, 9,,,,. The Pythagorea Theorem says that a + b =

More information

Math 155 (Lecture 3)

Math 155 (Lecture 3) Math 55 (Lecture 3) September 8, I this lecture, we ll cosider the aswer to oe of the most basic coutig problems i combiatorics Questio How may ways are there to choose a -elemet subset of the set {,,,

More information

Math 451: Euclidean and Non-Euclidean Geometry MWF 3pm, Gasson 204 Homework 3 Solutions

Math 451: Euclidean and Non-Euclidean Geometry MWF 3pm, Gasson 204 Homework 3 Solutions Math 451: Euclidea ad No-Euclidea Geometry MWF 3pm, Gasso 204 Homework 3 Solutios Exercises from 1.4 ad 1.5 of the otes: 4.3, 4.10, 4.12, 4.14, 4.15, 5.3, 5.4, 5.5 Exercise 4.3. Explai why Hp, q) = {x

More information

(A sequence also can be thought of as the list of function values attained for a function f :ℵ X, where f (n) = x n for n 1.) x 1 x N +k x N +4 x 3

(A sequence also can be thought of as the list of function values attained for a function f :ℵ X, where f (n) = x n for n 1.) x 1 x N +k x N +4 x 3 MATH 337 Sequeces Dr. Neal, WKU Let X be a metric space with distace fuctio d. We shall defie the geeral cocept of sequece ad limit i a metric space, the apply the results i particular to some special

More information

Chapter 6 Infinite Series

Chapter 6 Infinite Series Chapter 6 Ifiite Series I the previous chapter we cosidered itegrals which were improper i the sese that the iterval of itegratio was ubouded. I this chapter we are goig to discuss a topic which is somewhat

More information

6.3 Testing Series With Positive Terms

6.3 Testing Series With Positive Terms 6.3. TESTING SERIES WITH POSITIVE TERMS 307 6.3 Testig Series With Positive Terms 6.3. Review of what is kow up to ow I theory, testig a series a i for covergece amouts to fidig the i= sequece of partial

More information

Problem Cosider the curve give parametrically as x = si t ad y = + cos t for» t» ß: (a) Describe the path this traverses: Where does it start (whe t =

Problem Cosider the curve give parametrically as x = si t ad y = + cos t for» t» ß: (a) Describe the path this traverses: Where does it start (whe t = Mathematics Summer Wilso Fial Exam August 8, ANSWERS Problem 1 (a) Fid the solutio to y +x y = e x x that satisfies y() = 5 : This is already i the form we used for a first order liear differetial equatio,

More information

Injections, Surjections, and the Pigeonhole Principle

Injections, Surjections, and the Pigeonhole Principle Ijectios, Surjectios, ad the Pigeohole Priciple 1 (10 poits Here we will come up with a sloppy boud o the umber of parethesisestigs (a (5 poits Describe a ijectio from the set of possible ways to est pairs

More information

USA Mathematical Talent Search Round 3 Solutions Year 27 Academic Year

USA Mathematical Talent Search Round 3 Solutions Year 27 Academic Year /3/27. Fill i each space of the grid with either a or a so that all sixtee strigs of four cosecutive umbers across ad dow are distict. You do ot eed to prove that your aswer is the oly oe possible; you

More information

(ii) Two-permutations of {a, b, c}. Answer. (B) P (3, 3) = 3! (C) 3! = 6, and there are 6 items in (A). ... Answer.

(ii) Two-permutations of {a, b, c}. Answer. (B) P (3, 3) = 3! (C) 3! = 6, and there are 6 items in (A). ... Answer. SOLUTIONS Homewor 5 Due /6/19 Exercise. (a Cosider the set {a, b, c}. For each of the followig, (A list the objects described, (B give a formula that tells you how may you should have listed, ad (C verify

More information

Bertrand s Postulate

Bertrand s Postulate Bertrad s Postulate Lola Thompso Ross Program July 3, 2009 Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 1 / 33 Bertrad s Postulate I ve said it oce ad I ll say it agai: There s always a

More information

Ma 530 Introduction to Power Series

Ma 530 Introduction to Power Series Ma 530 Itroductio to Power Series Please ote that there is material o power series at Visual Calculus. Some of this material was used as part of the presetatio of the topics that follow. What is a Power

More information

Simple Polygons of Maximum Perimeter Contained in a Unit Disk

Simple Polygons of Maximum Perimeter Contained in a Unit Disk Discrete Comput Geom (009) 1: 08 15 DOI 10.1007/s005-008-9093-7 Simple Polygos of Maximum Perimeter Cotaied i a Uit Disk Charles Audet Pierre Hase Frédéric Messie Received: 18 September 007 / Revised:

More information

Mathematical Foundations -1- Sets and Sequences. Sets and Sequences

Mathematical Foundations -1- Sets and Sequences. Sets and Sequences Mathematical Foudatios -1- Sets ad Sequeces Sets ad Sequeces Methods of proof 2 Sets ad vectors 13 Plaes ad hyperplaes 18 Liearly idepedet vectors, vector spaces 2 Covex combiatios of vectors 21 eighborhoods,

More information

18th Bay Area Mathematical Olympiad. Problems and Solutions. February 23, 2016

18th Bay Area Mathematical Olympiad. Problems and Solutions. February 23, 2016 18th Bay Area Mathematical Olympiad February 3, 016 Problems ad Solutios BAMO-8 ad BAMO-1 are each 5-questio essay-proof exams, for middle- ad high-school studets, respectively. The problems i each exam

More information

Sequences, Mathematical Induction, and Recursion. CSE 2353 Discrete Computational Structures Spring 2018

Sequences, Mathematical Induction, and Recursion. CSE 2353 Discrete Computational Structures Spring 2018 CSE 353 Discrete Computatioal Structures Sprig 08 Sequeces, Mathematical Iductio, ad Recursio (Chapter 5, Epp) Note: some course slides adopted from publisher-provided material Overview May mathematical

More information

How to Maximize a Function without Really Trying

How to Maximize a Function without Really Trying How to Maximize a Fuctio without Really Tryig MARK FLANAGAN School of Electrical, Electroic ad Commuicatios Egieerig Uiversity College Dubli We will prove a famous elemetary iequality called The Rearragemet

More information

Section 1.1. Calculus: Areas And Tangents. Difference Equations to Differential Equations

Section 1.1. Calculus: Areas And Tangents. Difference Equations to Differential Equations Differece Equatios to Differetial Equatios Sectio. Calculus: Areas Ad Tagets The study of calculus begis with questios about chage. What happes to the velocity of a swigig pedulum as its positio chages?

More information

Homework 3. = k 1. Let S be a set of n elements, and let a, b, c be distinct elements of S. The number of k-subsets of S is

Homework 3. = k 1. Let S be a set of n elements, and let a, b, c be distinct elements of S. The number of k-subsets of S is Homewor 3 Chapter 5 pp53: 3 40 45 Chapter 6 p85: 4 6 4 30 Use combiatorial reasoig to prove the idetity 3 3 Proof Let S be a set of elemets ad let a b c be distict elemets of S The umber of -subsets of

More information

First selection test, May 1 st, 2008

First selection test, May 1 st, 2008 First selectio test, May st, 2008 Problem. Let p be a prime umber, p 3, ad let a, b be iteger umbers so that p a + b ad p 2 a 3 + b 3. Show that p 2 a + b or p 3 a 3 + b 3. Problem 2. Prove that for ay

More information

Problem 4: Evaluate ( k ) by negating (actually un-negating) its upper index. Binomial coefficient

Problem 4: Evaluate ( k ) by negating (actually un-negating) its upper index. Binomial coefficient Problem 4: Evaluate by egatig actually u-egatig its upper idex We ow that Biomial coefficiet r { where r is a real umber, is a iteger The above defiitio ca be recast i terms of factorials i the commo case

More information

Week 5-6: The Binomial Coefficients

Week 5-6: The Binomial Coefficients Wee 5-6: The Biomial Coefficiets March 6, 2018 1 Pascal Formula Theorem 11 (Pascal s Formula For itegers ad such that 1, ( ( ( 1 1 + 1 The umbers ( 2 ( 1 2 ( 2 are triagle umbers, that is, The petago umbers

More information

Math 61CM - Solutions to homework 3

Math 61CM - Solutions to homework 3 Math 6CM - Solutios to homework 3 Cédric De Groote October 2 th, 208 Problem : Let F be a field, m 0 a fixed oegative iteger ad let V = {a 0 + a x + + a m x m a 0,, a m F} be the vector space cosistig

More information

Review Problems 1. ICME and MS&E Refresher Course September 19, 2011 B = C = AB = A = A 2 = A 3... C 2 = C 3 = =

Review Problems 1. ICME and MS&E Refresher Course September 19, 2011 B = C = AB = A = A 2 = A 3... C 2 = C 3 = = Review Problems ICME ad MS&E Refresher Course September 9, 0 Warm-up problems. For the followig matrices A = 0 B = C = AB = 0 fid all powers A,A 3,(which is A times A),... ad B,B 3,... ad C,C 3,... Solutio:

More information

SEQUENCE AND SERIES NCERT

SEQUENCE AND SERIES NCERT 9. Overview By a sequece, we mea a arragemet of umbers i a defiite order accordig to some rule. We deote the terms of a sequece by a, a,..., etc., the subscript deotes the positio of the term. I view of

More information

SOLUTIONS TO PRISM PROBLEMS Junior Level 2014

SOLUTIONS TO PRISM PROBLEMS Junior Level 2014 SOLUTIONS TO PRISM PROBLEMS Juior Level 04. (B) Sice 50% of 50 is 50 5 ad 50% of 40 is the secod by 5 0 5. 40 0, the first exceeds. (A) Oe way of comparig the magitudes of the umbers,,, 5 ad 0.7 is 4 5

More information

Infinite Sequences and Series

Infinite Sequences and Series Chapter 6 Ifiite Sequeces ad Series 6.1 Ifiite Sequeces 6.1.1 Elemetary Cocepts Simply speakig, a sequece is a ordered list of umbers writte: {a 1, a 2, a 3,...a, a +1,...} where the elemets a i represet

More information

SAFE HANDS & IIT-ian's PACE EDT-10 (JEE) SOLUTIONS

SAFE HANDS & IIT-ian's PACE EDT-10 (JEE) SOLUTIONS . If their mea positios coicide with each other, maimum separatio will be A. Now from phasor diagram, we ca clearly see the phase differece. SAFE HANDS & IIT-ia's PACE ad Aswer : Optio (4) 5. Aswer : Optio

More information

Find a formula for the exponential function whose graph is given , 1 2,16 1, 6

Find a formula for the exponential function whose graph is given , 1 2,16 1, 6 Math 4 Activity (Due by EOC Apr. ) Graph the followig epoetial fuctios by modifyig the graph of f. Fid the rage of each fuctio.. g. g. g 4. g. g 6. g Fid a formula for the epoetial fuctio whose graph is

More information

International Contest-Game MATH KANGAROO Canada, Grade 11 and 12

International Contest-Game MATH KANGAROO Canada, Grade 11 and 12 Part A: Each correct aswer is worth 3 poits. Iteratioal Cotest-Game MATH KANGAROO Caada, 007 Grade ad. Mike is buildig a race track. He wats the cars to start the race i the order preseted o the left,

More information

Different kinds of Mathematical Induction

Different kinds of Mathematical Induction Differet ids of Mathematical Iductio () Mathematical Iductio Give A N, [ A (a A a A)] A N () (First) Priciple of Mathematical Iductio Let P() be a propositio (ope setece), if we put A { : N p() is true}

More information

INEQUALITIES BJORN POONEN

INEQUALITIES BJORN POONEN INEQUALITIES BJORN POONEN 1 The AM-GM iequality The most basic arithmetic mea-geometric mea (AM-GM) iequality states simply that if x ad y are oegative real umbers, the (x + y)/2 xy, with equality if ad

More information

Randomized Algorithms I, Spring 2018, Department of Computer Science, University of Helsinki Homework 1: Solutions (Discussed January 25, 2018)

Randomized Algorithms I, Spring 2018, Department of Computer Science, University of Helsinki Homework 1: Solutions (Discussed January 25, 2018) Radomized Algorithms I, Sprig 08, Departmet of Computer Sciece, Uiversity of Helsiki Homework : Solutios Discussed Jauary 5, 08). Exercise.: Cosider the followig balls-ad-bi game. We start with oe black

More information

Math 220A Fall 2007 Homework #2. Will Garner A

Math 220A Fall 2007 Homework #2. Will Garner A Math 0A Fall 007 Homewor # Will Garer Pg 3 #: Show that {cis : a o-egative iteger} is dese i T = {z œ : z = }. For which values of q is {cis(q): a o-egative iteger} dese i T? To show that {cis : a o-egative

More information

Chapter 0. Review of set theory. 0.1 Sets

Chapter 0. Review of set theory. 0.1 Sets Chapter 0 Review of set theory Set theory plays a cetral role i the theory of probability. Thus, we will ope this course with a quick review of those otios of set theory which will be used repeatedly.

More information

sin(n) + 2 cos(2n) n 3/2 3 sin(n) 2cos(2n) n 3/2 a n =

sin(n) + 2 cos(2n) n 3/2 3 sin(n) 2cos(2n) n 3/2 a n = 60. Ratio ad root tests 60.1. Absolutely coverget series. Defiitio 13. (Absolute covergece) A series a is called absolutely coverget if the series of absolute values a is coverget. The absolute covergece

More information

Baltic Way 2002 mathematical team contest

Baltic Way 2002 mathematical team contest Baltic Way 00 mathematical team cotest Tartu, November, 00 Problems ad solutios 1. Solve the system of equatios a 3 + 3ab + 3ac 6abc = 1 b 3 + 3ba + 3bc 6abc = 1 c 3 + 3ca + 3cb 6abc = 1 i real umbers.

More information

Section 5.1 The Basics of Counting

Section 5.1 The Basics of Counting 1 Sectio 5.1 The Basics of Coutig Combiatorics, the study of arragemets of objects, is a importat part of discrete mathematics. I this chapter, we will lear basic techiques of coutig which has a lot of

More information

Lesson 10: Limits and Continuity

Lesson 10: Limits and Continuity www.scimsacademy.com Lesso 10: Limits ad Cotiuity SCIMS Academy 1 Limit of a fuctio The cocept of limit of a fuctio is cetral to all other cocepts i calculus (like cotiuity, derivative, defiite itegrals

More information

PROPERTIES OF AN EULER SQUARE

PROPERTIES OF AN EULER SQUARE PROPERTIES OF N EULER SQURE bout 0 the mathematicia Leoard Euler discussed the properties a x array of letters or itegers ow kow as a Euler or Graeco-Lati Square Such squares have the property that every

More information

Analytic Continuation

Analytic Continuation Aalytic Cotiuatio The stadard example of this is give by Example Let h (z) = 1 + z + z 2 + z 3 +... kow to coverge oly for z < 1. I fact h (z) = 1/ (1 z) for such z. Yet H (z) = 1/ (1 z) is defied for

More information

We will conclude the chapter with the study a few methods and techniques which are useful

We will conclude the chapter with the study a few methods and techniques which are useful Chapter : Coordiate geometry: I this chapter we will lear about the mai priciples of graphig i a dimesioal (D) Cartesia system of coordiates. We will focus o drawig lies ad the characteristics of the graphs

More information

Davenport-Schinzel Sequences and their Geometric Applications

Davenport-Schinzel Sequences and their Geometric Applications Advaced Computatioal Geometry Sprig 2004 Daveport-Schizel Sequeces ad their Geometric Applicatios Prof. Joseph Mitchell Scribe: Mohit Gupta 1 Overview I this lecture, we itroduce the cocept of Daveport-Schizel

More information

Sequences and Series of Functions

Sequences and Series of Functions Chapter 6 Sequeces ad Series of Fuctios 6.1. Covergece of a Sequece of Fuctios Poitwise Covergece. Defiitio 6.1. Let, for each N, fuctio f : A R be defied. If, for each x A, the sequece (f (x)) coverges

More information

MTH Assignment 1 : Real Numbers, Sequences

MTH Assignment 1 : Real Numbers, Sequences MTH -26 Assigmet : Real Numbers, Sequeces. Fid the supremum of the set { m m+ : N, m Z}. 2. Let A be a o-empty subset of R ad α R. Show that α = supa if ad oly if α is ot a upper boud of A but α + is a

More information

JEE ADVANCED 2013 PAPER 1 MATHEMATICS

JEE ADVANCED 2013 PAPER 1 MATHEMATICS Oly Oe Optio Correct Type JEE ADVANCED 0 PAPER MATHEMATICS This sectio cotais TEN questios. Each has FOUR optios (A), (B), (C) ad (D) out of which ONLY ONE is correct.. The value of (A) 5 (C) 4 cot cot

More information

U8L1: Sec Equations of Lines in R 2

U8L1: Sec Equations of Lines in R 2 MCVU U8L: Sec. 8.9. Equatios of Lies i R Review of Equatios of a Straight Lie (-D) Cosider the lie passig through A (-,) with slope, as show i the diagram below. I poit slope form, the equatio of the lie

More information

Math F215: Induction April 7, 2013

Math F215: Induction April 7, 2013 Math F25: Iductio April 7, 203 Iductio is used to prove that a collectio of statemets P(k) depedig o k N are all true. A statemet is simply a mathematical phrase that must be either true or false. Here

More information

Sequences I. Chapter Introduction

Sequences I. Chapter Introduction Chapter 2 Sequeces I 2. Itroductio A sequece is a list of umbers i a defiite order so that we kow which umber is i the first place, which umber is i the secod place ad, for ay atural umber, we kow which

More information

Assignment 1 : Real Numbers, Sequences. for n 1. Show that (x n ) converges. Further, by observing that x n+2 + x n+1

Assignment 1 : Real Numbers, Sequences. for n 1. Show that (x n ) converges. Further, by observing that x n+2 + x n+1 Assigmet : Real Numbers, Sequeces. Let A be a o-empty subset of R ad α R. Show that α = supa if ad oly if α is ot a upper boud of A but α + is a upper boud of A for every N. 2. Let y (, ) ad x (, ). Evaluate

More information

MA131 - Analysis 1. Workbook 2 Sequences I

MA131 - Analysis 1. Workbook 2 Sequences I MA3 - Aalysis Workbook 2 Sequeces I Autum 203 Cotets 2 Sequeces I 2. Itroductio.............................. 2.2 Icreasig ad Decreasig Sequeces................ 2 2.3 Bouded Sequeces..........................

More information

Chapter 4. Fourier Series

Chapter 4. Fourier Series Chapter 4. Fourier Series At this poit we are ready to ow cosider the caoical equatios. Cosider, for eample the heat equatio u t = u, < (4.) subject to u(, ) = si, u(, t) = u(, t) =. (4.) Here,

More information

CSE 191, Class Note 05: Counting Methods Computer Sci & Eng Dept SUNY Buffalo

CSE 191, Class Note 05: Counting Methods Computer Sci & Eng Dept SUNY Buffalo Coutig Methods CSE 191, Class Note 05: Coutig Methods Computer Sci & Eg Dept SUNY Buffalo c Xi He (Uiversity at Buffalo CSE 191 Discrete Structures 1 / 48 Need for Coutig The problem of coutig the umber

More information

End-of-Year Contest. ERHS Math Club. May 5, 2009

End-of-Year Contest. ERHS Math Club. May 5, 2009 Ed-of-Year Cotest ERHS Math Club May 5, 009 Problem 1: There are 9 cois. Oe is fake ad weighs a little less tha the others. Fid the fake coi by weighigs. Solutio: Separate the 9 cois ito 3 groups (A, B,

More information

PUTNAM TRAINING INEQUALITIES

PUTNAM TRAINING INEQUALITIES PUTNAM TRAINING INEQUALITIES (Last updated: December, 207) Remark This is a list of exercises o iequalities Miguel A Lerma Exercises If a, b, c > 0, prove that (a 2 b + b 2 c + c 2 a)(ab 2 + bc 2 + ca

More information

CIS Spring 2018 (instructor Val Tannen)

CIS Spring 2018 (instructor Val Tannen) CIS 160 - Sprig 2018 (istructor Val Tae) Lecture 5 Thursday, Jauary 25 COUNTING We cotiue studyig how to use combiatios ad what are their properties. Example 5.1 How may 8-letter strigs ca be costructed

More information

62. Power series Definition 16. (Power series) Given a sequence {c n }, the series. c n x n = c 0 + c 1 x + c 2 x 2 + c 3 x 3 +

62. Power series Definition 16. (Power series) Given a sequence {c n }, the series. c n x n = c 0 + c 1 x + c 2 x 2 + c 3 x 3 + 62. Power series Defiitio 16. (Power series) Give a sequece {c }, the series c x = c 0 + c 1 x + c 2 x 2 + c 3 x 3 + is called a power series i the variable x. The umbers c are called the coefficiets of

More information

MATH spring 2008 lecture 3 Answers to selected problems. 0 sin14 xdx = x dx. ; (iv) x +

MATH spring 2008 lecture 3 Answers to selected problems. 0 sin14 xdx = x dx. ; (iv) x + MATH - sprig 008 lecture Aswers to selected problems INTEGRALS. f =? For atiderivatives i geeral see the itegrals website at http://itegrals.wolfram.com. (5-vi (0 i ( ( i ( π ; (v π a. This is example

More information

University of Colorado Denver Dept. Math. & Stat. Sciences Applied Analysis Preliminary Exam 13 January 2012, 10:00 am 2:00 pm. Good luck!

University of Colorado Denver Dept. Math. & Stat. Sciences Applied Analysis Preliminary Exam 13 January 2012, 10:00 am 2:00 pm. Good luck! Uiversity of Colorado Dever Dept. Math. & Stat. Scieces Applied Aalysis Prelimiary Exam 13 Jauary 01, 10:00 am :00 pm Name: The proctor will let you read the followig coditios before the exam begis, ad

More information

6.003 Homework #3 Solutions

6.003 Homework #3 Solutions 6.00 Homework # Solutios Problems. Complex umbers a. Evaluate the real ad imagiary parts of j j. π/ Real part = Imagiary part = 0 e Euler s formula says that j = e jπ/, so jπ/ j π/ j j = e = e. Thus the

More information

Lecture Overview. 2 Permutations and Combinations. n(n 1) (n (k 1)) = n(n 1) (n k + 1) =

Lecture Overview. 2 Permutations and Combinations. n(n 1) (n (k 1)) = n(n 1) (n k + 1) = COMPSCI 230: Discrete Mathematics for Computer Sciece April 8, 2019 Lecturer: Debmalya Paigrahi Lecture 22 Scribe: Kevi Su 1 Overview I this lecture, we begi studyig the fudametals of coutig discrete objects.

More information

The Growth of Functions. Theoretical Supplement

The Growth of Functions. Theoretical Supplement The Growth of Fuctios Theoretical Supplemet The Triagle Iequality The triagle iequality is a algebraic tool that is ofte useful i maipulatig absolute values of fuctios. The triagle iequality says that

More information

A sequence of numbers is a function whose domain is the positive integers. We can see that the sequence

A sequence of numbers is a function whose domain is the positive integers. We can see that the sequence Sequeces A sequece of umbers is a fuctio whose domai is the positive itegers. We ca see that the sequece,, 2, 2, 3, 3,... is a fuctio from the positive itegers whe we write the first sequece elemet as

More information

Name of the Student:

Name of the Student: SUBJECT NAME : Discrete Mathematics SUBJECT CODE : MA 65 MATERIAL NAME : Problem Material MATERIAL CODE : JM08ADM010 (Sca the above QR code for the direct dowload of this material) Name of the Studet:

More information

Joe Holbrook Memorial Math Competition

Joe Holbrook Memorial Math Competition Joe Holbrook Memorial Math Competitio 8th Grade Solutios October 5, 07. Sice additio ad subtractio come before divisio ad mutiplicatio, 5 5 ( 5 ( 5. Now, sice operatios are performed right to left, ( 5

More information

If the escalator stayed stationary, Billy would be able to ascend or descend in = 30 seconds. Thus, Billy can climb = 8 steps in one second.

If the escalator stayed stationary, Billy would be able to ascend or descend in = 30 seconds. Thus, Billy can climb = 8 steps in one second. BMT 01 INDIVIDUAL SOLUTIONS March 01 1. Billy the kid likes to play o escalators! Movig at a costat speed, he maages to climb up oe escalator i 4 secods ad climb back dow the same escalator i 40 secods.

More information

Intermediate Math Circles November 4, 2009 Counting II

Intermediate Math Circles November 4, 2009 Counting II Uiversity of Waterloo Faculty of Mathematics Cetre for Educatio i Mathematics ad Computig Itermediate Math Circles November 4, 009 Coutig II Last time, after lookig at the product rule ad sum rule, we

More information

MODEL TEST PAPER II Time : hours Maximum Marks : 00 Geeral Istructios : (i) (iii) (iv) All questios are compulsory. The questio paper cosists of 9 questios divided ito three Sectios A, B ad C. Sectio A

More information

Math 475, Problem Set #12: Answers

Math 475, Problem Set #12: Answers Math 475, Problem Set #12: Aswers A. Chapter 8, problem 12, parts (b) ad (d). (b) S # (, 2) = 2 2, sice, from amog the 2 ways of puttig elemets ito 2 distiguishable boxes, exactly 2 of them result i oe

More information

MA131 - Analysis 1. Workbook 3 Sequences II

MA131 - Analysis 1. Workbook 3 Sequences II MA3 - Aalysis Workbook 3 Sequeces II Autum 2004 Cotets 2.8 Coverget Sequeces........................ 2.9 Algebra of Limits......................... 2 2.0 Further Useful Results........................

More information

1. By using truth tables prove that, for all statements P and Q, the statement

1. By using truth tables prove that, for all statements P and Q, the statement Author: Satiago Salazar Problems I: Mathematical Statemets ad Proofs. By usig truth tables prove that, for all statemets P ad Q, the statemet P Q ad its cotrapositive ot Q (ot P) are equivalet. I example.2.3

More information

Homework 1 Solutions. The exercises are from Foundations of Mathematical Analysis by Richard Johnsonbaugh and W.E. Pfaffenberger.

Homework 1 Solutions. The exercises are from Foundations of Mathematical Analysis by Richard Johnsonbaugh and W.E. Pfaffenberger. Homewor 1 Solutios Math 171, Sprig 2010 Hery Adams The exercises are from Foudatios of Mathematical Aalysis by Richard Johsobaugh ad W.E. Pfaffeberger. 2.2. Let h : X Y, g : Y Z, ad f : Z W. Prove that

More information

ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER / Statistics

ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER / Statistics ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER 1 018/019 DR. ANTHONY BROWN 8. Statistics 8.1. Measures of Cetre: Mea, Media ad Mode. If we have a series of umbers the

More information

PRELIM PROBLEM SOLUTIONS

PRELIM PROBLEM SOLUTIONS PRELIM PROBLEM SOLUTIONS THE GRAD STUDENTS + KEN Cotets. Complex Aalysis Practice Problems 2. 2. Real Aalysis Practice Problems 2. 4 3. Algebra Practice Problems 2. 8. Complex Aalysis Practice Problems

More information

2 Geometric interpretation of complex numbers

2 Geometric interpretation of complex numbers 2 Geometric iterpretatio of complex umbers 2.1 Defiitio I will start fially with a precise defiitio, assumig that such mathematical object as vector space R 2 is well familiar to the studets. Recall that

More information

page Suppose that S 0, 1 1, 2.

page Suppose that S 0, 1 1, 2. page 10 1. Suppose that S 0, 1 1,. a. What is the set of iterior poits of S? The set of iterior poits of S is 0, 1 1,. b. Give that U is the set of iterior poits of S, evaluate U. 0, 1 1, 0, 1 1, S. The

More information

Objective Mathematics

Objective Mathematics . If sum of '' terms of a sequece is give by S Tr ( )( ), the 4 5 67 r (d) 4 9 r is equal to : T. Let a, b, c be distict o-zero real umbers such that a, b, c are i harmoic progressio ad a, b, c are i arithmetic

More information

CALCULATION OF FIBONACCI VECTORS

CALCULATION OF FIBONACCI VECTORS CALCULATION OF FIBONACCI VECTORS Stuart D. Aderso Departmet of Physics, Ithaca College 953 Daby Road, Ithaca NY 14850, USA email: saderso@ithaca.edu ad Dai Novak Departmet of Mathematics, Ithaca College

More information

September 2012 C1 Note. C1 Notes (Edexcel) Copyright - For AS, A2 notes and IGCSE / GCSE worksheets 1

September 2012 C1 Note. C1 Notes (Edexcel) Copyright   - For AS, A2 notes and IGCSE / GCSE worksheets 1 September 0 s (Edecel) Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets September 0 Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets September 0 Copyright

More information

Sequences A sequence of numbers is a function whose domain is the positive integers. We can see that the sequence

Sequences A sequence of numbers is a function whose domain is the positive integers. We can see that the sequence Sequeces A sequece of umbers is a fuctio whose domai is the positive itegers. We ca see that the sequece 1, 1, 2, 2, 3, 3,... is a fuctio from the positive itegers whe we write the first sequece elemet

More information

WBJEE MATHEMATICS

WBJEE MATHEMATICS WBJEE - 06 MATHEMATICS Q.No. 0 A C B B 0 B B A B 0 C A C C 0 A B C C 05 A A B C 06 B C B C 07 B C A D 08 C C C A 09 D D C C 0 A C A B B C B A A C A B D A A A B B D C 5 B C C C 6 C A B B 7 C A A B 8 C B

More information

Lecture Notes for Analysis Class

Lecture Notes for Analysis Class Lecture Notes for Aalysis Class Topological Spaces A topology for a set X is a collectio T of subsets of X such that: (a) X ad the empty set are i T (b) Uios of elemets of T are i T (c) Fiite itersectios

More information

R is a scalar defined as follows:

R is a scalar defined as follows: Math 8. Notes o Dot Product, Cross Product, Plaes, Area, ad Volumes This lecture focuses primarily o the dot product ad its may applicatios, especially i the measuremet of agles ad scalar projectio ad

More information

Lecture 23 Rearrangement Inequality

Lecture 23 Rearrangement Inequality Lecture 23 Rearragemet Iequality Holde Lee 6/4/ The Iequalities We start with a example Suppose there are four boxes cotaiig $0, $20, $50 ad $00 bills, respectively You may take 2 bills from oe box, 3

More information

MAT 271 Project: Partial Fractions for certain rational functions

MAT 271 Project: Partial Fractions for certain rational functions MAT 7 Project: Partial Fractios for certai ratioal fuctios Prerequisite kowledge: partial fractios from MAT 7, a very good commad of factorig ad complex umbers from Precalculus. To complete this project,

More information

Ray-triangle intersection

Ray-triangle intersection Ray-triagle itersectio ria urless October 2006 I this hadout, we explore the steps eeded to compute the itersectio of a ray with a triagle, ad the to compute the barycetric coordiates of that itersectio.

More information

Math 61CM - Solutions to homework 1

Math 61CM - Solutions to homework 1 Math 61CM - Solutios to homework 1 Cédric De Groote October 1 st, 2018 Problem 1: Use mathematical iductio to check the validity of the formula j 3 = 2 ( + 1) 2 for = 1, 2,.... Note: The priciple of mathematical

More information

Mathematics Extension 2

Mathematics Extension 2 004 HIGHER SCHOOL CERTIFICATE EXAMINATION Mathematics Etesio Geeral Istructios Readig time 5 miutes Workig time hours Write usig black or blue pe Board-approved calculators may be used A table of stadard

More information

NICK DUFRESNE. 1 1 p(x). To determine some formulas for the generating function of the Schröder numbers, r(x) = a(x) =

NICK DUFRESNE. 1 1 p(x). To determine some formulas for the generating function of the Schröder numbers, r(x) = a(x) = AN INTRODUCTION TO SCHRÖDER AND UNKNOWN NUMBERS NICK DUFRESNE Abstract. I this article we will itroduce two types of lattice paths, Schröder paths ad Ukow paths. We will examie differet properties of each,

More information

Math 4707 Spring 2018 (Darij Grinberg): homework set 4 page 1

Math 4707 Spring 2018 (Darij Grinberg): homework set 4 page 1 Math 4707 Sprig 2018 Darij Griberg): homewor set 4 page 1 Math 4707 Sprig 2018 Darij Griberg): homewor set 4 due date: Wedesday 11 April 2018 at the begiig of class, or before that by email or moodle Please

More information

Z ß cos x + si x R du We start with the substitutio u = si(x), so du = cos(x). The itegral becomes but +u we should chage the limits to go with the ew

Z ß cos x + si x R du We start with the substitutio u = si(x), so du = cos(x). The itegral becomes but +u we should chage the limits to go with the ew Problem ( poits) Evaluate the itegrals Z p x 9 x We ca draw a right triagle labeled this way x p x 9 From this we ca read off x = sec, so = sec ta, ad p x 9 = R ta. Puttig those pieces ito the itegralrwe

More information

2.1. The Algebraic and Order Properties of R Definition. A binary operation on a set F is a function B : F F! F.

2.1. The Algebraic and Order Properties of R Definition. A binary operation on a set F is a function B : F F! F. CHAPTER 2 The Real Numbers 2.. The Algebraic ad Order Properties of R Defiitio. A biary operatio o a set F is a fuctio B : F F! F. For the biary operatios of + ad, we replace B(a, b) by a + b ad a b, respectively.

More information

Seunghee Ye Ma 8: Week 5 Oct 28

Seunghee Ye Ma 8: Week 5 Oct 28 Week 5 Summary I Sectio, we go over the Mea Value Theorem ad its applicatios. I Sectio 2, we will recap what we have covered so far this term. Topics Page Mea Value Theorem. Applicatios of the Mea Value

More information

Chimica Inorganica 3

Chimica Inorganica 3 himica Iorgaica Irreducible Represetatios ad haracter Tables Rather tha usig geometrical operatios, it is ofte much more coveiet to employ a ew set of group elemets which are matrices ad to make the rule

More information

The z-transform. 7.1 Introduction. 7.2 The z-transform Derivation of the z-transform: x[n] = z n LTI system, h[n] z = re j

The z-transform. 7.1 Introduction. 7.2 The z-transform Derivation of the z-transform: x[n] = z n LTI system, h[n] z = re j The -Trasform 7. Itroductio Geeralie the complex siusoidal represetatio offered by DTFT to a represetatio of complex expoetial sigals. Obtai more geeral characteristics for discrete-time LTI systems. 7.

More information