MA1200 Exercise for Chapter 7 Techniques of Differentiation Solutions. First Principle 1. a) To simplify the calculation, note. Then. lim h.
|
|
- Godfrey Matthews
- 5 years ago
- Views:
Transcription
1 MA00 Eercise for Chapter 7 Techiques of Differetiatio Solutios First Priciple a) To simplify the calculatio, ote The b) ( h) lim h 0 h lim h 0 h[ ( h) ( h) h ( h) ] f '( ) Product/Quotiet/Chai Rules a) 8 6 b), c), so 6 d) Quotiet rule: ( ) 5 8 e) Quotiet rule: 5 ( ) 999 f) Chai rule: 000( 5 ) ( 0) g) Product rule + Chai rule: ( )()( ) ( ) ( ) (6 ) h) Chai rule: 6si ()cos() 5 si( ) i) Chai rule: l l(l ) j) Product rule: l k) Chai rule: cot l) Chai rule: ta m) ) Chai rule: e si( e ) o) Product rule: = e ( cos cos si )
2 p) Quotiet rule + Product rule: ( 5) e ( si cos ) e ( 5) ( ) q) Quotiet rule: r) Quotiet rule: ( ) ( s) Product rule + Chai rule: cos()cos() si()si() si(l() ) t) Chai rule: cos(l[cos(l() )] ) cos(l() ) a) Differetiate both sides: So y ' / y (We eed to use Chai rule o the term, ad the we get a ) b) Differetiate both sides: y So y' y I the middle term we used Product rule c) Differetiate both sides: a) d ( y ) y ( y ) y d dt dt 9 t 9 t t 9 whe t = Whe Collectig like terms, we get ( si )(6 5) 6 ) So 9 Taget: y 5 ( ), y Normal: y 5 ( ), y 9 9 (Recall: eq of lie is ; slope of taget = ; slope of ormal slope of taget = ) 7 b) We ca do as i part a) to get t whe t = d t Or we ca write ad we sub i Taget: 5 y 5 y 7 5 a) Take log The use Chai rule o LHS ad Product rule o RHS: So cot cot ( ) csc l( ) b) So ( e e ( ) ) e l( )
3 c) The The (si5 ) [ l(si 5) 5( )cot 5] 6 a) So i geeral, y ( ) e b) Use the formula: If, the So y 5(6 )si(6 7 ) c) Write The So y ( )!( ) (There is o, eed to write )!! 6 ( )! ( )!! d) y whe ( )! 0 (First we kow if differetiate more tha times, all terms become 0; we also kow if differetiate more tha times, oly the term remais We differetiate a few times to guess the geeral form above) Leibiz s rule 7 If, the Leibiz rule states that a) The puttig ito the formula above ad otig that ad for, we get b) y e ( ) e ( 6 ) e [ ( ), () ( ) ( )( )] whe So for
4 y ( ) cos( ) ( 7)cos cos ( 7) ( ) cos ( ) ( ) ( ) cos whe whe c) So for, y ( ) ( ) ( )!( ) whe whe Miscellaeous 8 If y si, prove that y'' y' y 0 Proof: y si y' si cos y" cos cos si cos si So y' ' y' ( ) y cos si si cos si si si cos cos si si 0 d a b c 9 If u a b c, prove that ( u) d u Proof: du d d a b a b a b c ( a b c) ( a b c) (a b) d d d u ( a b c) The d du a b a b u a b a b c a b c ( u) u u d d u u u u
5 if c *0 Fid the values of a ad b (i terms of c) such that f '( c) eists, where f ( ) a b if c Solutio: if c if c f ( ) f ( ) a b if c c a b if c if c I order that f '( c) eists, f () must be cotiuous at c That is, lim f ( ) lim lim f ( ) ac b f ( c) c c c c c c ( ac b) f( ) f( c) lim lim lim c lim c lim c lim c c c c c c c c c c c c c f( ) f( c) ab( acb) a( c) lim lim lim a c c c c c c f ( ) f ( c) f ( ) f ( c) That f '( c) eists meas lim lim a c c c c c Therefore, a ad put i ac b c c c a We have b c b c So b c Fially, we have c c c c c b c A fuctio y of is defied by the equatio si y msi y Epress y eplicitly i terms of mcos Hece, or otherwise, show that d m cos m Proof: si y msi y si cos y cos si y msi y si cos y m cos si y si y si si si ta y y ta cos y mcos mcos mcos The si si d m cos cos si d ta m cos m cos d m cos mcos d d si mcos si mcos m m cos m cos 5
6 Let cot y, fid d Solutio: cot l cot l d y y csc l cot y cot cot ycsc l csc l d cot Solutio: d f ta ta f ' 0 d Show that f ' 0, where f ta ta t Cosider the parametric curve where t y t Fid, d y ad the equatio of the taget lie to the curve at the poit, d d Solutio: Method : t y y y t The d y y, d d Method : d d y d d y d d d dt dt dt dt dt dt dt dt dt d d d d d d d dt d d d d dt dt dt dt d d t t dt dt t So t y t t d y d t 6t dt dt d d y d d y dt dt dt dt dt t6t t dt dt d d d t d d dt 6 cot
7 dt dt d d t As a result, t d, d y dt d d t The slope of the taget lie to the curve at the poit, d Therefore, the equatio of the taget lie to the curve at the poit y I additio, t t also t t t 0 7, is: 5 By Leibitz s theorem o repeated differetiatio, fid the th derivatives of the fuctio y = e cos Solutio: By Leibiz s theorem o repeated differetiatio, we have i i i i 0 0 d d f dgd f d g dgd f d f d g fg g Ci f i i C i, where f, g, i i 0 0 d d i d d d i0 d d d d! Cr, r 0,,,,, ad 0! r! r! i d f i Now, f e e, i 0,,, d i i d g i i d g i i gcos cos, si, i 0,,, i i d d So i i d d cos d e e cos Ci i i d d d i0 If m where m is a oegative iteger, the i i d d cos d e e cos Ci i i d i0 d d m i m i cos d e d d cos d e Ci i i C i i i d d d d i i i0 i0 m m i i i i Ci cose Ci sie i0 i0 If m, where m is a oegative iteger, the i i d d cos d e e cos Ci i i d i0 d d m i m i d cos d e d cos d e Ci i i C i i i d d d d i i i0 i0 m m i i i i Ci cos e Ci sie i0 i0
8 -Ed- 8
MATHEMATICS. 61. The differential equation representing the family of curves where c is a positive parameter, is of
MATHEMATICS 6 The differetial equatio represetig the family of curves where c is a positive parameter, is of Order Order Degree (d) Degree (a,c) Give curve is y c ( c) Differetiate wrt, y c c y Hece differetial
More informationQuiz. Use either the RATIO or ROOT TEST to determine whether the series is convergent or not.
Quiz. Use either the RATIO or ROOT TEST to determie whether the series is coverget or ot. e .6 POWER SERIES Defiitio. A power series i about is a series of the form c 0 c a c a... c a... a 0 c a where
More informationDifferent 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 informationTECHNIQUES OF INTEGRATION
7 TECHNIQUES OF INTEGRATION Simpso s Rule estimates itegrals b approimatig graphs with parabolas. Because of the Fudametal Theorem of Calculus, we ca itegrate a fuctio if we kow a atiderivative, that is,
More informationAP Calculus BC Review Applications of Derivatives (Chapter 4) and f,
AP alculus B Review Applicatios of Derivatives (hapter ) Thigs to Kow ad Be Able to Do Defiitios of the followig i terms of derivatives, ad how to fid them: critical poit, global miima/maima, local (relative)
More informationMAT136H1F - Calculus I (B) Long Quiz 1. T0101 (M3) Time: 20 minutes. The quiz consists of four questions. Each question is worth 2 points. Good Luck!
MAT36HF - Calculus I (B) Log Quiz. T (M3) Time: 2 miutes Last Name: Studet ID: First Name: Please mark your tutorial sectio: T (M3) T2 (R4) T3 (T4) T5 (T5) T52 (R5) The quiz cosists of four questios. Each
More informationMATH 10550, EXAM 3 SOLUTIONS
MATH 155, EXAM 3 SOLUTIONS 1. I fidig a approximate solutio to the equatio x 3 +x 4 = usig Newto s method with iitial approximatio x 1 = 1, what is x? Solutio. Recall that x +1 = x f(x ) f (x ). Hece,
More informationy = f x x 1. If f x = e 2x tan -1 x, then f 1 = e 2 2 e 2 p C e 2 D e 2 p+1 4
. If f = e ta -, the f = e e p e e p e p+ 4 f = e ta -, so f = e ta - + e, so + f = e p + e = e p + e or f = e p + 4. The slope of the lie taget to the curve - + = at the poit, - is - 5 Differetiate -
More informationn 3 ln n n ln n is convergent by p-series for p = 2 > 1. n2 Therefore we can apply Limit Comparison Test to determine lutely convergent.
06 微甲 0-04 06-0 班期中考解答和評分標準. ( poits) Determie whether the series is absolutely coverget, coditioally coverget, or diverget. Please state the tests which you use. (a) ( poits) (b) ( poits) (c) ( poits)
More informationMath 21B-B - Homework Set 2
Math B-B - Homework Set Sectio 5.:. a) lim P k= c k c k ) x k, where P is a partitio of [, 5. x x ) dx b) lim P k= 4 ck x k, where P is a partitio of [,. 4 x dx c) lim P k= ta c k ) x k, where P is a partitio
More information1. (25 points) Use the limit definition of the definite integral and the sum formulas 1 to compute
Math, Calculus II Fial Eam Solutios. 5 poits) Use the limit defiitio of the defiite itegral ad the sum formulas to compute 4 d. The check your aswer usig the Evaluatio Theorem. ) ) Solutio: I this itegral,
More informationTaylor Series Mixed Exercise 6
Taylor Series Mied Eercise Let f() cot ad a f(a) f () ( cos ec ) + cot f (a) f''() cotcosec + ( cosec ) f''(a) f'''() ( cosec cot cosec ) + cotcosec Substitutig ito the Taylor series epasio gives f() +
More informationMath 113, Calculus II Winter 2007 Final Exam Solutions
Math, Calculus II Witer 7 Fial Exam Solutios (5 poits) Use the limit defiitio of the defiite itegral ad the sum formulas to compute x x + dx The check your aswer usig the Evaluatio Theorem Solutio: I this
More informationChapter 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 informationFundamental Concepts: Surfaces and Curves
UNDAMENTAL CONCEPTS: SURACES AND CURVES CHAPTER udametal Cocepts: Surfaces ad Curves. INTRODUCTION This chapter describes two geometrical objects, vi., surfaces ad curves because the pla a ver importat
More informationj=1 dz Res(f, z j ) = 1 d k 1 dz k 1 (z c)k f(z) Res(f, c) = lim z c (k 1)! Res g, c = f(c) g (c)
Problem. Compute the itegrals C r d for Z, where C r = ad r >. Recall that C r has the couter-clockwise orietatio. Solutio: We will use the idue Theorem to solve this oe. We could istead use other (perhaps
More informationx x x Using a second Taylor polynomial with remainder, find the best constant C so that for x 0,
Math Activity 9( Due with Fial Eam) Usig first ad secod Taylor polyomials with remaider, show that for, 8 Usig a secod Taylor polyomial with remaider, fid the best costat C so that for, C 9 The th Derivative
More informationMock Exam 4. 1 Hong Kong Educational Publishing Company. Since y = 6 is a horizontal. tangent, dy dx = 0 when y = 6. Section A
Mock Eam Mock Eam Sectio A. Referece: HKDSE Math M Q d [( + h) + ( + h)] ( + ) ( + ) lim h h + h + h + h + + h lim h h h+ h + h lim h h + h lim ( + h + h + ) h + (). Whe, e + l y + 7()(y) l y y e + l y
More informationTopic 1 2: Sequences and Series. A sequence is an ordered list of numbers, e.g. 1, 2, 4, 8, 16, or
Topic : Sequeces ad Series A sequece is a ordered list of umbers, e.g.,,, 8, 6, or,,,.... A series is a sum of the terms of a sequece, e.g. + + + 8 + 6 + or... Sigma Notatio b The otatio f ( k) is shorthad
More informationChapter 2 The Solution of Numerical Algebraic and Transcendental Equations
Chapter The Solutio of Numerical Algebraic ad Trascedetal Equatios Itroductio I this chapter we shall discuss some umerical methods for solvig algebraic ad trascedetal equatios. The equatio f( is said
More informationHKDSE Exam Questions Distribution
HKDSE Eam Questios Distributio Sample Paper Practice Paper DSE 0 Topics A B A B A B. Biomial Theorem. Mathematical Iductio 0 3 3 3. More about Trigoometric Fuctios, 0, 3 0 3. Limits 6. Differetiatio 7
More informationLIMITS AND DERIVATIVES
Capter LIMITS AND DERIVATIVES. Overview.. Limits of a fuctio Let f be a fuctio defied i a domai wic we take to be a iterval, say, I. We sall study te cocept of it of f at a poit a i I. We say f ( ) is
More informationf(w) w z =R z a 0 a n a nz n Liouville s theorem, we see that Q is constant, which implies that P is constant, which is a contradiction.
Theorem 3.6.4. [Liouville s Theorem] Every bouded etire fuctio is costat. Proof. Let f be a etire fuctio. Suppose that there is M R such that M for ay z C. The for ay z C ad R > 0 f (z) f(w) 2πi (w z)
More informationDefinition 2.1 (The Derivative) (page 54) is a function. The derivative of a function f with respect to x, represented by. f ', is defined by
Chapter DACS Lok 004/05 CHAPTER DIFFERENTIATION. THE GEOMETRICAL MEANING OF DIFFERENTIATION (page 54) Defiitio. (The Derivative) (page 54) Let f () is a fctio. The erivative of a fctio f with respect to,
More informationAP CALCULUS AB 2003 SCORING GUIDELINES (Form B)
SCORING GUIDELINES (Form B) Questio 5 Let f be a fuctio defied o the closed iterval [,7]. The graph of f, cosistig of four lie segmets, is show above. Let g be the fuctio give by g ftdt. (a) Fid g (, )
More informationLIMITS AND DERIVATIVES NCERT
. Overview.. Limits of a fuctio Let f be a fuctio defied i a domai wic we take to be a iterval, say, I. We sall study te cocept of it of f at a poit a i I. We say f ( ) is te epected value of f at a give
More informationMock Exam 1. 1 Hong Kong Educational Publishing Company. Section A 1. Reference: HKDSE Math M Q1 (4 - x) 3
Moc Exam Moc Exam Sectio A. Referece: HKDSE Math M 06 Q ( - x) + C () (-x) + C ()(-x) + (-x) 6 - x + x - x 6 ( x) + x (6 x + x x ) + C 6 6 6 6 C C x + x + x + x 6 6 96 (6 x + x x ) + + + + x x x x \ Costat
More informationCHAPTER 7 SUCCESSIVE DIFFERENTIATION
www.sakshieducatio.com CHAPTER 7 SUCCESSIVE DIFFERENTIATION TOPICS:. Successive differetiatio-th derivative of a fuctio theorems.. Fidig the th derivative of the give fuctio.. Leibitz s theorem ad its
More informationCOMPUTING SUMS AND THE AVERAGE VALUE OF THE DIVISOR FUNCTION (x 1) + x = n = n.
COMPUTING SUMS AND THE AVERAGE VALUE OF THE DIVISOR FUNCTION Abstract. We itroduce a method for computig sums of the form f( where f( is ice. We apply this method to study the average value of d(, where
More informationSeptember 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 informationName: Math 10550, Final Exam: December 15, 2007
Math 55, Fial Exam: December 5, 7 Name: Be sure that you have all pages of the test. No calculators are to be used. The exam lasts for two hours. Whe told to begi, remove this aswer sheet ad keep it uder
More informationn n 2 + 4i = lim 2 n lim 1 + 4x 2 dx = 1 2 tan ( 2i 2 x x dx = 1 2 tan 1 2 = 2 n, x i = a + i x = 2i
. ( poits) Fid the limits. (a) (6 poits) lim ( + + + 3 (6 poits) lim h h h 6 微甲 - 班期末考解答和評分標準 +h + + + t3 dt. + 3 +... + 5 ) = lim + i= + i. Solutio: (a) lim i= + i = lim i= + ( i ) = lim x i= + x i =
More information1988 AP Calculus BC: Section I
988 AP Calculus BC: Sectio I 9 Miutes No Calculator Notes: () I this eamiatio, l deotes the atural logarithm of (that is, logarithm to the base e). () Uless otherwise specified, the domai of a fuctio f
More informationSolutions to quizzes Math Spring 2007
to quizzes Math 4- Sprig 7 Name: Sectio:. Quiz a) x + x dx b) l x dx a) x + dx x x / + x / dx (/3)x 3/ + x / + c. b) Set u l x, dv dx. The du /x ad v x. By Itegratio by Parts, x(/x)dx x l x x + c. l x
More informationLinear Differential Equations of Higher Order Basic Theory: Initial-Value Problems d y d y dy
Liear Differetial Equatios of Higher Order Basic Theory: Iitial-Value Problems d y d y dy Solve: a( ) + a ( )... a ( ) a0( ) y g( ) + + + = d d d ( ) Subject to: y( 0) = y0, y ( 0) = y,..., y ( 0) = y
More information1 Generating functions for balls in boxes
Math 566 Fall 05 Some otes o geeratig fuctios Give a sequece a 0, a, a,..., a,..., a geeratig fuctio some way of represetig the sequece as a fuctio. There are may ways to do this, with the most commo ways
More informationCalculus I Practice Test Problems for Chapter 5 Page 1 of 9
Calculus I Practice Test Problems for Chapter 5 Page of 9 This is a set of practice test problems for Chapter 5. This is i o way a iclusive set of problems there ca be other types of problems o the actual
More informationHOMEWORK #10 SOLUTIONS
Math 33 - Aalysis I Sprig 29 HOMEWORK # SOLUTIONS () Prove that the fuctio f(x) = x 3 is (Riema) itegrable o [, ] ad show that x 3 dx = 4. (Without usig formulae for itegratio that you leart i previous
More informationMAS111 Convergence and Continuity
MAS Covergece ad Cotiuity Key Objectives At the ed of the course, studets should kow the followig topics ad be able to apply the basic priciples ad theorems therei to solvig various problems cocerig covergece
More informationf(x) dx as we do. 2x dx x also diverges. Solution: We compute 2x dx lim
Math 3, Sectio 2. (25 poits) Why we defie f(x) dx as we do. (a) Show that the improper itegral diverges. Hece the improper itegral x 2 + x 2 + b also diverges. Solutio: We compute x 2 + = lim b x 2 + =
More informationClasses of Uniformly Convex and Uniformly Starlike Functions as Dual Sets
JOURNAL OF MATHEMATICAL ANALYSIS AND APPLICATIONS 16, 4047 1997 ARTICLE NO. AY975640 Classes of Uiformly Covex ad Uiformly Starlike Fuctios as Dual Sets I. R. Nezhmetdiov Faculty of Mechaics ad Mathematics,
More informationSubject: Differential Equations & Mathematical Modeling-III
Power Series Solutios of Differetial Equatios about Sigular poits Subject: Differetial Equatios & Mathematical Modelig-III Lesso: Power series solutios of differetial equatios about Sigular poits Lesso
More information1.3 Convergence Theorems of Fourier Series. k k k k. N N k 1. With this in mind, we state (without proof) the convergence of Fourier series.
.3 Covergece Theorems of Fourier Series I this sectio, we preset the covergece of Fourier series. A ifiite sum is, by defiitio, a limit of partial sums, that is, a cos( kx) b si( kx) lim a cos( kx) b si(
More informationMa 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 informationMath 132, Fall 2009 Exam 2: Solutions
Math 3, Fall 009 Exam : Solutios () a) ( poits) Determie for which positive real umbers p, is the followig improper itegral coverget, ad for which it is diverget. Evaluate the itegral for each value of
More informationThe picture in figure 1.1 helps us to see that the area represents the distance traveled. Figure 1: Area represents distance travelled
1 Lecture : Area Area ad distace traveled Approximatig area by rectagles Summatio The area uder a parabola 1.1 Area ad distace Suppose we have the followig iformatio about the velocity of a particle, how
More informationMost text will write ordinary derivatives using either Leibniz notation 2 3. y + 5y= e and y y. xx tt t
Itroductio to Differetial Equatios Defiitios ad Termiolog Differetial Equatio: A equatio cotaiig the derivatives of oe or more depedet variables, with respect to oe or more idepedet variables, is said
More informationLIMIT. f(a h). f(a + h). Lim x a. h 0. x 1. x 0. x 0. x 1. x 1. x 2. Lim f(x) 0 and. x 0
J-Mathematics LIMIT. INTRODUCTION : The cocept of it of a fuctio is oe of the fudametal ideas that distiguishes calculus from algebra ad trigoometr. We use its to describe the wa a fuctio f varies. Some
More informationCHAPTER 5. Theory and Solution Using Matrix Techniques
A SERIES OF CLASS NOTES FOR 2005-2006 TO INTRODUCE LINEAR AND NONLINEAR PROBLEMS TO ENGINEERS, SCIENTISTS, AND APPLIED MATHEMATICIANS DE CLASS NOTES 3 A COLLECTION OF HANDOUTS ON SYSTEMS OF ORDINARY DIFFERENTIAL
More informationSAMPLE. Solutions Manual Exercise Set 1 (page 10) 1.3 Exercise Set 2 (page 19) 1. 2, 1, 2, 1 4.
Solutios Maual Eercise Set page 0,,, 4 + si + Use the Bo method z +siz cos z 4 cos + ct 5 fπ/ sicosπ/ si 0 0 6 + h You get this by dividig by h sice h 0 7 si t + cos h +cost + si h h h 8 a π, b 4π 9 a
More informationMath 210A Homework 1
Math 0A Homework Edward Burkard Exercise. a) State the defiitio of a aalytic fuctio. b) What are the relatioships betwee aalytic fuctios ad the Cauchy-Riema equatios? Solutio. a) A fuctio f : G C is called
More informationDiploma Programme. Mathematics HL guide. First examinations 2014
Diploma Programme First eamiatios 014 33 Topic 6 Core: Calculus The aim of this topic is to itroduce studets to the basic cocepts ad techiques of differetial ad itegral calculus ad their applicatio. 6.1
More information(ii) Q.2 Reduce the following matrix to normal form and find its rank.
B. E. Semester- (All Brach ecept Aero. Questio Bak Each questio has marks. Q. Fid the rak o 9 Q. Reduce the ollowig matri to ormal orm ad id its rak. Q. Fid iverse o a matri b Gauss Jorda Reductio method
More informationSection 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 informationPRACTICE FINAL/STUDY GUIDE SOLUTIONS
Last edited December 9, 03 at 4:33pm) Feel free to sed me ay feedback, icludig commets, typos, ad mathematical errors Problem Give the precise meaig of the followig statemets i) a f) L ii) a + f) L iii)
More informationMath 312 Lecture Notes One Dimensional Maps
Math 312 Lecture Notes Oe Dimesioal Maps Warre Weckesser Departmet of Mathematics Colgate Uiversity 21-23 February 25 A Example We begi with the simplest model of populatio growth. Suppose, for example,
More informationMIDTERM 3 CALCULUS 2. Monday, December 3, :15 PM to 6:45 PM. Name PRACTICE EXAM SOLUTIONS
MIDTERM 3 CALCULUS MATH 300 FALL 08 Moday, December 3, 08 5:5 PM to 6:45 PM Name PRACTICE EXAM S Please aswer all of the questios, ad show your work. You must explai your aswers to get credit. You will
More informationHonors Calculus Homework 13 Solutions, due 12/8/5
Hoors Calculus Homework Solutios, due /8/5 Questio Let a regio R i the plae be bouded by the curves y = 5 ad = 5y y. Sketch the regio R. The two curves meet where both equatios hold at oce, so where: y
More informationExample 2. Find the upper bound for the remainder for the approximation from Example 1.
Lesso 8- Error Approimatios 0 Alteratig Series Remaider: For a coverget alteratig series whe approimatig the sum of a series by usig oly the first terms, the error will be less tha or equal to the absolute
More informationRiemann Sums y = f (x)
Riema Sums Recall that we have previously discussed the area problem I its simplest form we ca state it this way: The Area Problem Let f be a cotiuous, o-egative fuctio o the closed iterval [a, b] Fid
More informationMath 113 Exam 3 Practice
Math Exam Practice Exam 4 will cover.-., 0. ad 0.. Note that eve though. was tested i exam, questios from that sectios may also be o this exam. For practice problems o., refer to the last review. This
More informationCHAPTER 1 SEQUENCES AND INFINITE SERIES
CHAPTER SEQUENCES AND INFINITE SERIES SEQUENCES AND INFINITE SERIES (0 meetigs) Sequeces ad limit of a sequece Mootoic ad bouded sequece Ifiite series of costat terms Ifiite series of positive terms Alteratig
More information3. Z Transform. Recall that the Fourier transform (FT) of a DT signal xn [ ] is ( ) [ ] = In order for the FT to exist in the finite magnitude sense,
3. Z Trasform Referece: Etire Chapter 3 of text. Recall that the Fourier trasform (FT) of a DT sigal x [ ] is ω ( ) [ ] X e = j jω k = xe I order for the FT to exist i the fiite magitude sese, S = x [
More informationPAPER : IIT-JAM 2010
MATHEMATICS-MA (CODE A) Q.-Q.5: Oly oe optio is correct for each questio. Each questio carries (+6) marks for correct aswer ad ( ) marks for icorrect aswer.. Which of the followig coditios does NOT esure
More informationMathematics Extension 1
016 Bored of Studies Trial Eamiatios Mathematics Etesio 1 3 rd ctober 016 Geeral Istructios Total Marks 70 Readig time 5 miutes Workig time hours Write usig black or blue pe Black pe is preferred Board-approved
More informationf(x)g(x) dx is an inner product on D.
Ark9: Exercises for MAT2400 Fourier series The exercises o this sheet cover the sectios 4.9 to 4.13. They are iteded for the groups o Thursday, April 12 ad Friday, March 30 ad April 13. NB: No group o
More informationChapter 2: Numerical Methods
Chapter : Numerical Methods. Some Numerical Methods for st Order ODEs I this sectio, a summar of essetial features of umerical methods related to solutios of ordiar differetial equatios is give. I geeral,
More informationComplete Solutions to Supplementary Exercises on Infinite Series
Coplete Solutios to Suppleetary Eercises o Ifiite Series. (a) We eed to fid the su ito partial fractios gives By the cover up rule we have Therefore Let S S A / ad A B B. Covertig the suad / the by usig
More informationBESSEL EQUATION and BESSEL FUNCTIONS
BESSEL EQUATION ad BESSEL FUNCTIONS Bessel s Equatio Summary of Bessel Fuctios d y dy y d + d + =. If is a iteger, the two idepedet solutios of Bessel s Equatio are J J, Bessel fuctio of the first kid,
More informationDe Moivre s Theorem - ALL
De Moivre s Theorem - ALL. Let x ad y be real umbers, ad be oe of the complex solutios of the equatio =. Evaluate: (a) + + ; (b) ( x + y)( x + y). [6]. (a) Sice is a complex umber which satisfies = 0,.
More informationContinuous Functions
Cotiuous Fuctios Q What does it mea for a fuctio to be cotiuous at a poit? Aswer- I mathematics, we have a defiitio that cosists of three cocepts that are liked i a special way Cosider the followig defiitio
More informationBITSAT MATHEMATICS PAPER III. For the followig liear programmig problem : miimize z = + y subject to the costraits + y, + y 8, y, 0, the solutio is (0, ) ad (, ) (0, ) ad ( /, ) (0, ) ad (, ) (d) (0, )
More informationMa 4121: Introduction to Lebesgue Integration Solutions to Homework Assignment 5
Ma 42: Itroductio to Lebesgue Itegratio Solutios to Homework Assigmet 5 Prof. Wickerhauser Due Thursday, April th, 23 Please retur your solutios to the istructor by the ed of class o the due date. You
More informationPower Series: A power series about the center, x = 0, is a function of x of the form
You are familiar with polyomial fuctios, polyomial that has ifiitely may terms. 2 p ( ) a0 a a 2 a. A power series is just a Power Series: A power series about the ceter, = 0, is a fuctio of of the form
More informationAnalytic 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 informationSection 11.6 Absolute and Conditional Convergence, Root and Ratio Tests
Sectio.6 Absolute ad Coditioal Covergece, Root ad Ratio Tests I this chapter we have see several examples of covergece tests that oly apply to series whose terms are oegative. I this sectio, we will lear
More informationLesson 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 informationIIT JAM Mathematical Statistics (MS) 2006 SECTION A
IIT JAM Mathematical Statistics (MS) 6 SECTION A. If a > for ad lim a / L >, the which of the followig series is ot coverget? (a) (b) (c) (d) (d) = = a = a = a a + / a lim a a / + = lim a / a / + = lim
More informationAdditional Notes on Power Series
Additioal Notes o Power Series Mauela Girotti MATH 37-0 Advaced Calculus of oe variable Cotets Quick recall 2 Abel s Theorem 2 3 Differetiatio ad Itegratio of Power series 4 Quick recall We recall here
More informationTHE SOLUTION OF NONLINEAR EQUATIONS f( x ) = 0.
THE SOLUTION OF NONLINEAR EQUATIONS f( ) = 0. Noliear Equatio Solvers Bracketig. Graphical. Aalytical Ope Methods Bisectio False Positio (Regula-Falsi) Fied poit iteratio Newto Raphso Secat The root of
More informationFall 2013 MTH431/531 Real analysis Section Notes
Fall 013 MTH431/531 Real aalysis Sectio 8.1-8. Notes Yi Su 013.11.1 1. Defiitio of uiform covergece. We look at a sequece of fuctios f (x) ad study the coverget property. Notice we have two parameters
More informationThe type of limit that is used to find TANGENTS and VELOCITIES gives rise to the central idea in DIFFERENTIAL CALCULUS, the DERIVATIVE.
NOTES : LIMITS AND DERIVATIVES Name: Date: Period: Iitial: LESSON.1 THE TANGENT AND VELOCITY PROBLEMS Pre-Calculus Mathematics Limit Process Calculus The type of it that is used to fid TANGENTS ad VELOCITIES
More informationEDEXCEL STUDENT CONFERENCE 2006 A2 MATHEMATICS STUDENT NOTES
EDEXCEL STUDENT CONFERENCE 006 A MATHEMATICS STUDENT NOTES South: Thursday 3rd March 006, Lodo EXAMINATION HINTS Before the eamiatio Obtai a copy of the formulae book ad use it! Write a list of ad LEARN
More information(Figure 2.9), we observe x. and we write. (b) as x, x 1. and we write. We say that the line y 0 is a horizontal asymptote of the graph of f.
The symbol for ifiity ( ) does ot represet a real umber. We use to describe the behavior of a fuctio whe the values i its domai or rage outgrow all fiite bouds. For eample, whe we say the limit of f as
More informationMTH 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 informationReview Questions, Chapters 8, 9. f(y) = 0, elsewhere. F (y) = f Y(1) = n ( e y/θ) n 1 1 θ e y/θ = n θ e yn
Stat 366 Lab 2 Solutios (September 2, 2006) page TA: Yury Petracheko, CAB 484, yuryp@ualberta.ca, http://www.ualberta.ca/ yuryp/ Review Questios, Chapters 8, 9 8.5 Suppose that Y, Y 2,..., Y deote a radom
More informationTopic 5 [434 marks] (i) Find the range of values of n for which. (ii) Write down the value of x dx in terms of n, when it does exist.
Topic 5 [44 marks] 1a (i) Fid the rage of values of for which eists 1 Write dow the value of i terms of 1, whe it does eist Fid the solutio to the differetial equatio 1b give that y = 1 whe = π (cos si
More informationMaximum and Minimum Values
Sec 4.1 Maimum ad Miimum Values A. Absolute Maimum or Miimum / Etreme Values A fuctio Similarly, f has a Absolute Maimum at c if c f f has a Absolute Miimum at c if c f f for every poit i the domai. f
More informationMATH CALCULUS II Objectives and Notes for Test 4
MATH 44 - CALCULUS II Objectives ad Notes for Test 4 To do well o this test, ou should be able to work the followig tpes of problems. Fid a power series represetatio for a fuctio ad determie the radius
More informationProblems from 9th edition of Probability and Statistical Inference by Hogg, Tanis and Zimmerman:
Math 224 Fall 2017 Homework 4 Drew Armstrog Problems from 9th editio of Probability ad Statistical Iferece by Hogg, Tais ad Zimmerma: Sectio 2.3, Exercises 16(a,d),18. Sectio 2.4, Exercises 13, 14. Sectio
More informationComplex 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 informationSolutions to Final Exam Review Problems
. Let f(x) 4+x. Solutios to Fial Exam Review Problems Math 5C, Witer 2007 (a) Fid the Maclauri series for f(x), ad compute its radius of covergece. Solutio. f(x) 4( ( x/4)) ( x/4) ( ) 4 4 + x. Sice the
More informationWe are mainly going to be concerned with power series in x, such as. (x)} converges - that is, lims N n
Review of Power Series, Power Series Solutios A power series i x - a is a ifiite series of the form c (x a) =c +c (x a)+(x a) +... We also call this a power series cetered at a. Ex. (x+) is cetered at
More information+ {JEE Advace 03} Sept 0 Name: Batch (Day) Phoe No. IT IS NOT ENOUGH TO HAVE A GOOD MIND, THE MAIN THING IS TO USE IT WELL Marks: 00. If A (α, β) = (a) A( α, β) = A( α, β) (c) Adj (A ( α, β)) = Sol : We
More informationSubject: Differential Equations & Mathematical Modeling -III. Lesson: Power series solutions of Differential Equations. about ordinary points
Power series solutio of Differetial equatios about ordiary poits Subject: Differetial Equatios & Mathematical Modelig -III Lesso: Power series solutios of Differetial Equatios about ordiary poits Lesso
More informationCurve Sketching Handout #5 Topic Interpretation Rational Functions
Curve Sketchig Hadout #5 Topic Iterpretatio Ratioal Fuctios A ratioal fuctio is a fuctio f that is a quotiet of two polyomials. I other words, p ( ) ( ) f is a ratioal fuctio if p ( ) ad q ( ) are polyomials
More informationAPPENDIX F Complex Numbers
APPENDIX F Complex Numbers Operatios with Complex Numbers Complex Solutios of Quadratic Equatios Polar Form of a Complex Number Powers ad Roots of Complex Numbers Operatios with Complex Numbers Some equatios
More informationWe 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 informationA 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