Math Spring 2013 Solutions to Assignment # 3 Completion Date: Wednesday May 15, (1/z) 2 (1/z 1) 2 = lim
|
|
- Shauna Hoover
- 6 years ago
- Views:
Transcription
1 Mat Spring 013 Solutions to Assignment # 3 Completion Date: Wednesday May 15, 013 Question 1. [p 56, #10 (a)] 4z Use te teorem of Sec. 17 to sow tat z (z 1) = 4. We ave z 4z (z 1) = z 0 4 (1/z) (1/z 1) = z 0 4 (1 z) = 4. Question. [p 56, #13] Sow tat a set is unbounded (Sec. 11) if only if every neigborood of te point at infinity contains at least one point in S. If S C is unbounded, ten for eac n 1, tere is a point z n S wit z n n. Now, given ɛ > 0, coose n 0 1 wit 0 < 1 n 0 < ɛ, ten z n0 n 0 > 1 ɛ, z n0 is in te ɛ-neigborood {z : z > 1/ɛ} of te point of infinity. Tus, every neigborood of te point at infinity contains at least one point in S. Conversely, if every neigborood of te point at infinity contains at least one point in S, ten for eac n 1, we can coose a point z n S wit z n n (tat is, z n is in te 1/n-neigborood of te point at infinity). Ten S cannot be bounded, since z n S for all n 1, z n = +, tere is no n M > 0 suc tat z M for all z S. Question 3. [p 6, #1] Use te results in Sec. 0 to find f (z) wen (a) f(z) = 3z z + 4; (b) f(z) = ( 1 4z ) 3 ; (c) f(z) = z 1 ( ) 1 + z 4 (z 1/); (d) f(z) = z + 1 z (z 0). (a) f (z) = 6z.
2 (b) f (z) = 3(1 4z ) ( 8z) = 4z(1 4z ). (c) f (z) = 1 (z + 1) (z 1) 3 (z + 1) =, for z 1/. (z + 1) (d) f (z) = 4(1 + z ) 3 z z (1 + z ) 4 z z 4 = (1 + z ) 3 z 3 (3z 1), for z 0. Question 4. [p 6, #3] Apply definition (3), Sec. 19, of derivative to give a direct proof tat f (z) = 1 z wen f(z) = 1 z (z 0). If f(z) = 1 z for z 0, ten f f(z + ) f(z) (z) = = 1 z + 1 z = z (z + ) z(z + ), tat is, for z 0. f (z) = z(z + ) = 1 z(z + ) = 1 z Question 5. [p 63, #8 (b)] Use te metod in Example, Sec. 19, to sow tat f (z) does not exist at any point z wen f(z) = Im z. Let f(z) = Im(z), ten for z, C, wit 0, we ave Im(z + ) Im(z) Now, if 0 troug real values, Im() = 0, = Im(z) + Im() Im(z) = Im(). Im(z + ) Im(z) real = real However, if 0 troug imaginary values, say = it were t R t 0, ten 0 = 0. (1) Im() Im(z + ) Im(z) imag = t it = i, = ( i) = i. () imag Terefore, (1) () imply tat Im(z + ) Im(z) doesn t exist, tat is, f (z) does not exist for any z C.
3 Question 6. [p 71, #1] Use te teorem in Sec. 1 to sow tat f (z) does not exist at any point if (a) f(z) = z; (b) f(z) = z z; (c) f(z) = x + i xy ; (d) f(z) = e x e i y. (a) If f(z) = z = x iy, ten u(x, y) = x v(x, y) = y, so tat = 1 1 =, te Caucy-Riemann equations do not old at any point z C. Terefore f (z) does not exist for any z C. (b) If f(z) = z z = iy, ten u(x, y) = 0 v(x, y) = y, so tat = 0 =, again te Caucy-Riemann equations do not old at any point z C. Terefore f (z) does not exist for any z C. (c) If f(z) = x + ixy, ten u(x, y) = x v(x, y) = xy, so tat =, = 0, = xy = y. Now te Caucy-Riemann equations are = xy 0 = y tese equations ave no solutions (x, y). Terefore, f (z) does not exist at any point z C. (d) If f(z) = e x e i y = e x cos y ie x sin y, ten = ex cos y, = ex sin y, = ex cos y = ex sin y. Now te Caucy-Riemann equations are since e x 0 for all x R, tese equations are e x cos y = 0 e x sin y = 0 cos y = 0 sin y = 0, but tis is impossible since cos y+sin y = 1. Terefore, tere are no solutions to te Caucy-Riemann equations, f (z) does not exist for any z C.
4 Question 7. [p 71, #3] From te results obtained in Secs. 1, determine were f (z) exists find its value wen (a) f(z) = 1 z ; (b) f(z) = x + i y ; (c) f(z) = z Im z; Ans : (a) f (z) = 1 z (z 0); (b) f (x + i x) = x; (c) f (0) = 0. (a) If f(z) = 1 z, ten so tat for z 0. Now, u(x, y) = f(z) = z z = x x + y i y x + y, x x + y v(x, y) = y x + y = y x (x + y ) = = xy (x + y ) =. Since te partial derivatives are all continuous at eac z C, z 0, te Caucy-Riemann equations old at eac z C, z 0, ten f (z) exists for all z 0, f (z) = + i = y x (x + y ) + ixy (x + y ), tat is, for z 0. f (z) = x y ixy (x + y ) = (z) z 4 = 1 z (b) If f(z) = x + i y, ten u(x, y) = x v(x, y) = y, = x, = 0, = y = 0, so te Caucy-Riemann equations old only for tose points z = x + i y wit x = y. Since all partial derivatives are continuous everywere, f (z) exists only for te points z = x + i x = x(1 + i), x R, f (x + ix) = (x, x) + i (x, x) = x. (c) If f(z) = zim(z) = (x + i y) y = xy + i y, ten u(x, y) = xy v(x, y) = y, so tat = y, = x, = y = 0, te Caucy-Riemann equations old if only if y = y x = 0,
5 tat is, if only if x = y = 0. Since all partial derivatives are continuous, ten f (z) exists only for z = 0, f (0) = (0, 0) + i (0, 0) = 0. Question 8. [p 71, #4 (b)] Use te teorem in Sec. 3 to sow tat te function f(z) = re i θ/ (r > 0, α < θ < α + π) is differentiable in te indicated domain of definition, ten use expression (7) in tat section to find f (z). Ans : f (z) = 1 f(z). If f(z) = re iθ/, r > 0, α < θ < α + π, ten so tat u(r, θ) = r cos θ/ v(r, θ) = r sin θ/. f(z) = r (cos θ/ + i sin θ/), Now, so tat for r > 0, α < θ < α + π. Also, so tat for r > 0, α < θ < α + π. r = 1 r cos θ/ r θ = cos θ/, r = 1 r cos θ/ = 1 r θ r θ = sin θ/ r = 1 sin θ/, r 1 r θ = 1 sin θ/ = r r Te partial derivatives are all continuous for eac (r, θ) wit r > 0, α < θ < α+π, te Caucy-Riemann equations old for eac suc point, so tat f (z) exists for all suc points, ( f (z) = e iθ r + i ) = 1 r r (cos θ/ + i sin θ/) e iθ = 1 r e iθ/ = 1 f(z), for r > 0, α < θ < α + π. Question 9. [p 7, #5] Sow tat wen f(z) = x 3 + i (1 y) 3, it is legitimate to write f (z) = u x + i v x = 3x only wen z = i.
6 If f(z) = x 3 + i(1 y) 3, ten u(x, y) = x 3 v(x, y) = (1 y) 3, so tat Te Caucy-Riemann equations become = 3x, = 3(1 y) = 0, = 0. 3x + 3(1 y) = 0 tese old only at te point z = (0, 1) = i. Since all partial derivatives are continuous everywere, ten f (i) exists, f (i) = (0, 1) + i (0, 1) = 0. Question 10. [p 7, #10] (a) Recall (Sec. 5) tat if z = x + i y ten x = z + z y = z z. i By formally applying te cain rule in calculus to a function F (x, y) of two real variables, derive te expression F z = F z + F z = 1 ( F + i F ). (b) Define te operator z = 1 ( + i ), suggested by part (a), to sow tat if te first-order partial derivatives of te real imaginary parts of a function f(z) = u(x, y) + i v(x, y) satisfy te Caucy-Riemann equations, ten z = 1 [(u x v y ) + i (v x + u y )] = 0. Tus derive te complex form z = 0 of te Caucy-Riemann equations. (a) Since z = x + i y z = x i y, ten x = z + z y = z z, i F z = F z + F z = 1 ( F + 1 ) F i = 1 ( F + i F ). (b) Now, if f(z) = u(x, y) + i v(x, y) te real-valued functions u v satisfy te Caucy-Riemann equations, ten from part (a), we ave tat is, since = z = 1 (u + i v) + i (u + i v) = 1 + i + i 1, z = 1 ( ) + i ( + ) = 0, =.
Analytic Functions. Differentiable Functions of a Complex Variable
Analytic Functions Differentiable Functions of a Complex Variable In tis capter, we sall generalize te ideas for polynomials power series of a complex variable we developed in te previous capter to general
More informationMath 242: Principles of Analysis Fall 2016 Homework 7 Part B Solutions
Mat 22: Principles of Analysis Fall 206 Homework 7 Part B Solutions. Sow tat f(x) = x 2 is not uniformly continuous on R. Solution. Te equation is equivalent to f(x) = 0 were f(x) = x 2 sin(x) 3. Since
More informationExam 1 Solutions. x(x 2) (x + 1)(x 2) = x
Eam Solutions Question (0%) Consider f() = 2 2 2 2. (a) By calculating relevant its, determine te equations of all vertical asymptotes of te grap of f(). If tere are none, say so. f() = ( 2) ( + )( 2)
More informationContinuity and Differentiability of the Trigonometric Functions
[Te basis for te following work will be te definition of te trigonometric functions as ratios of te sides of a triangle inscribed in a circle; in particular, te sine of an angle will be defined to be te
More informationMath 212-Lecture 9. For a single-variable function z = f(x), the derivative is f (x) = lim h 0
3.4: Partial Derivatives Definition Mat 22-Lecture 9 For a single-variable function z = f(x), te derivative is f (x) = lim 0 f(x+) f(x). For a function z = f(x, y) of two variables, to define te derivatives,
More information3.1. COMPLEX DERIVATIVES 89
3.1. COMPLEX DERIVATIVES 89 Teorem 3.1.4 (Cain rule) Let f : D D C and g : D C be olomorpic functions on te domains D and D respectively. Suppose tat bot f and g are C 1 -smoot. 10 Ten (g f) (z) g (f(z))
More information4.2 - Richardson Extrapolation
. - Ricardson Extrapolation. Small-O Notation: Recall tat te big-o notation used to define te rate of convergence in Section.: Definition Let x n n converge to a number x. Suppose tat n n is a sequence
More informationPoisson Equation in Sobolev Spaces
Poisson Equation in Sobolev Spaces OcMountain Dayligt Time. 6, 011 Today we discuss te Poisson equation in Sobolev spaces. It s existence, uniqueness, and regularity. Weak Solution. u = f in, u = g on
More information(4.2) -Richardson Extrapolation
(.) -Ricardson Extrapolation. Small-O Notation: Recall tat te big-o notation used to define te rate of convergence in Section.: Suppose tat lim G 0 and lim F L. Te function F is said to converge to L as
More informationDepartment of Mathematics, K.T.H.M. College, Nashik F.Y.B.Sc. Calculus Practical (Academic Year )
F.Y.B.Sc. Calculus Practical (Academic Year 06-7) Practical : Graps of Elementary Functions. a) Grap of y = f(x) mirror image of Grap of y = f(x) about X axis b) Grap of y = f( x) mirror image of Grap
More information3.4 Worksheet: Proof of the Chain Rule NAME
Mat 1170 3.4 Workseet: Proof of te Cain Rule NAME Te Cain Rule So far we are able to differentiate all types of functions. For example: polynomials, rational, root, and trigonometric functions. We are
More informationMATH1131/1141 Calculus Test S1 v8a
MATH/ Calculus Test 8 S v8a October, 7 Tese solutions were written by Joann Blanco, typed by Brendan Trin and edited by Mattew Yan and Henderson Ko Please be etical wit tis resource It is for te use of
More informationMath 161 (33) - Final exam
Name: Id #: Mat 161 (33) - Final exam Fall Quarter 2015 Wednesday December 9, 2015-10:30am to 12:30am Instructions: Prob. Points Score possible 1 25 2 25 3 25 4 25 TOTAL 75 (BEST 3) Read eac problem carefully.
More informationMinimal surfaces of revolution
5 April 013 Minimal surfaces of revolution Maggie Miller 1 Introduction In tis paper, we will prove tat all non-planar minimal surfaces of revolution can be generated by functions of te form f = 1 C cos(cx),
More informationSection 15.6 Directional Derivatives and the Gradient Vector
Section 15.6 Directional Derivatives and te Gradient Vector Finding rates of cange in different directions Recall tat wen we first started considering derivatives of functions of more tan one variable,
More informationContinuity. Example 1
Continuity MATH 1003 Calculus and Linear Algebra (Lecture 13.5) Maoseng Xiong Department of Matematics, HKUST A function f : (a, b) R is continuous at a point c (a, b) if 1. x c f (x) exists, 2. f (c)
More informationHow to Find the Derivative of a Function: Calculus 1
Introduction How to Find te Derivative of a Function: Calculus 1 Calculus is not an easy matematics course Te fact tat you ave enrolled in suc a difficult subject indicates tat you are interested in te
More informationThe complex exponential function
Capter Te complex exponential function Tis is a very important function!. Te series For any z C, we define: exp(z) := n! zn = + z + z2 2 + z3 6 + z4 24 + On te closed disk D(0,R) := {z C z R}, one as n!
More informationContinuity and Differentiability Worksheet
Continuity and Differentiability Workseet (Be sure tat you can also do te grapical eercises from te tet- Tese were not included below! Typical problems are like problems -3, p. 6; -3, p. 7; 33-34, p. 7;
More informationMath 31A Discussion Notes Week 4 October 20 and October 22, 2015
Mat 3A Discussion Notes Week 4 October 20 and October 22, 205 To prepare for te first midterm, we ll spend tis week working eamples resembling te various problems you ve seen so far tis term. In tese notes
More informationUniversity Mathematics 2
University Matematics 2 1 Differentiability In tis section, we discuss te differentiability of functions. Definition 1.1 Differentiable function). Let f) be a function. We say tat f is differentiable at
More information1 The concept of limits (p.217 p.229, p.242 p.249, p.255 p.256) 1.1 Limits Consider the function determined by the formula 3. x since at this point
MA00 Capter 6 Calculus and Basic Linear Algebra I Limits, Continuity and Differentiability Te concept of its (p.7 p.9, p.4 p.49, p.55 p.56). Limits Consider te function determined by te formula f Note
More informationSolutions to the Multivariable Calculus and Linear Algebra problems on the Comprehensive Examination of January 31, 2014
Solutions to te Multivariable Calculus and Linear Algebra problems on te Compreensive Examination of January 3, 24 Tere are 9 problems ( points eac, totaling 9 points) on tis portion of te examination.
More informationFunctions of the Complex Variable z
Capter 2 Functions of te Complex Variable z Introduction We wis to examine te notion of a function of z were z is a complex variable. To be sure, a complex variable can be viewed as noting but a pair of
More informationDifferentiation in higher dimensions
Capter 2 Differentiation in iger dimensions 2.1 Te Total Derivative Recall tat if f : R R is a 1-variable function, and a R, we say tat f is differentiable at x = a if and only if te ratio f(a+) f(a) tends
More informationMA119-A Applied Calculus for Business Fall Homework 4 Solutions Due 9/29/ :30AM
MA9-A Applied Calculus for Business 006 Fall Homework Solutions Due 9/9/006 0:0AM. #0 Find te it 5 0 + +.. #8 Find te it. #6 Find te it 5 0 + + = (0) 5 0 (0) + (0) + =.!! r + +. r s r + + = () + 0 () +
More informationFunction Composition and Chain Rules
Function Composition and s James K. Peterson Department of Biological Sciences and Department of Matematical Sciences Clemson University Marc 8, 2017 Outline 1 Function Composition and Continuity 2 Function
More informationConvexity and Smoothness
Capter 4 Convexity and Smootness 4.1 Strict Convexity, Smootness, and Gateaux Differentiablity Definition 4.1.1. Let X be a Banac space wit a norm denoted by. A map f : X \{0} X \{0}, f f x is called a
More informationMATH 155A FALL 13 PRACTICE MIDTERM 1 SOLUTIONS. needs to be non-zero, thus x 1. Also 1 +
MATH 55A FALL 3 PRACTICE MIDTERM SOLUTIONS Question Find te domain of te following functions (a) f(x) = x3 5 x +x 6 (b) g(x) = x+ + x+ (c) f(x) = 5 x + x 0 (a) We need x + x 6 = (x + 3)(x ) 0 Hence Dom(f)
More informationNumerical Differentiation
Numerical Differentiation Finite Difference Formulas for te first derivative (Using Taylor Expansion tecnique) (section 8.3.) Suppose tat f() = g() is a function of te variable, and tat as 0 te function
More informationMATH1151 Calculus Test S1 v2a
MATH5 Calculus Test 8 S va January 8, 5 Tese solutions were written and typed up by Brendan Trin Please be etical wit tis resource It is for te use of MatSOC members, so do not repost it on oter forums
More information1. Consider the trigonometric function f(t) whose graph is shown below. Write down a possible formula for f(t).
. Consider te trigonometric function f(t) wose grap is sown below. Write down a possible formula for f(t). Tis function appears to be an odd, periodic function tat as been sifted upwards, so we will use
More information1 Calculus. 1.1 Gradients and the Derivative. Q f(x+h) f(x)
Calculus. Gradients and te Derivative Q f(x+) δy P T δx R f(x) 0 x x+ Let P (x, f(x)) and Q(x+, f(x+)) denote two points on te curve of te function y = f(x) and let R denote te point of intersection of
More information1. Questions (a) through (e) refer to the graph of the function f given below. (A) 0 (B) 1 (C) 2 (D) 4 (E) does not exist
Mat 1120 Calculus Test 2. October 18, 2001 Your name Te multiple coice problems count 4 points eac. In te multiple coice section, circle te correct coice (or coices). You must sow your work on te oter
More informationPolynomial Interpolation
Capter 4 Polynomial Interpolation In tis capter, we consider te important problem of approximatinga function fx, wose values at a set of distinct points x, x, x,, x n are known, by a polynomial P x suc
More informationMA455 Manifolds Solutions 1 May 2008
MA455 Manifolds Solutions 1 May 2008 1. (i) Given real numbers a < b, find a diffeomorpism (a, b) R. Solution: For example first map (a, b) to (0, π/2) and ten map (0, π/2) diffeomorpically to R using
More informationBob Brown Math 251 Calculus 1 Chapter 3, Section 1 Completed 1 CCBC Dundalk
Bob Brown Mat 251 Calculus 1 Capter 3, Section 1 Completed 1 Te Tangent Line Problem Te idea of a tangent line first arises in geometry in te context of a circle. But before we jump into a discussion of
More information4. The slope of the line 2x 7y = 8 is (a) 2/7 (b) 7/2 (c) 2 (d) 2/7 (e) None of these.
Mat 11. Test Form N Fall 016 Name. Instructions. Te first eleven problems are wort points eac. Te last six problems are wort 5 points eac. For te last six problems, you must use relevant metods of algebra
More informationLecture XVII. Abstract We introduce the concept of directional derivative of a scalar function and discuss its relation with the gradient operator.
Lecture XVII Abstract We introduce te concept of directional derivative of a scalar function and discuss its relation wit te gradient operator. Directional derivative and gradient Te directional derivative
More informationSome Review Problems for First Midterm Mathematics 1300, Calculus 1
Some Review Problems for First Midterm Matematics 00, Calculus. Consider te trigonometric function f(t) wose grap is sown below. Write down a possible formula for f(t). Tis function appears to be an odd,
More informationGradient Descent etc.
1 Gradient Descent etc EE 13: Networked estimation and control Prof Kan) I DERIVATIVE Consider f : R R x fx) Te derivative is defined as d fx) = lim dx fx + ) fx) Te cain rule states tat if d d f gx) )
More informationSFU UBC UNBC Uvic Calculus Challenge Examination June 5, 2008, 12:00 15:00
SFU UBC UNBC Uvic Calculus Callenge Eamination June 5, 008, :00 5:00 Host: SIMON FRASER UNIVERSITY First Name: Last Name: Scool: Student signature INSTRUCTIONS Sow all your work Full marks are given only
More informationClick here to see an animation of the derivative
Differentiation Massoud Malek Derivative Te concept of derivative is at te core of Calculus; It is a very powerful tool for understanding te beavior of matematical functions. It allows us to optimize functions,
More informationSolution. Solution. f (x) = (cos x)2 cos(2x) 2 sin(2x) 2 cos x ( sin x) (cos x) 4. f (π/4) = ( 2/2) ( 2/2) ( 2/2) ( 2/2) 4.
December 09, 20 Calculus PracticeTest s Name: (4 points) Find te absolute extrema of f(x) = x 3 0 on te interval [0, 4] Te derivative of f(x) is f (x) = 3x 2, wic is zero only at x = 0 Tus we only need
More informationTest 2 Review. 1. Find the determinant of the matrix below using (a) cofactor expansion and (b) row reduction. A = 3 2 =
Test Review Find te determinant of te matrix below using (a cofactor expansion and (b row reduction Answer: (a det + = (b Observe R R R R R R R R R Ten det B = (((det Hence det Use Cramer s rule to solve:
More informationCalculus I Homework: The Derivative as a Function Page 1
Calculus I Homework: Te Derivative as a Function Page 1 Example (2.9.16) Make a careful sketc of te grap of f(x) = sin x and below it sketc te grap of f (x). Try to guess te formula of f (x) from its grap.
More informationChapter 5 Differentiation
Capter 5 Differentiation Course Title: Real Analsis 1 Course Code: MTH31 Course instrutor: Dr Atiq ur Reman Class: MS-II Course URL: wwwmatitorg/atiq/fa15-mt31 Derivative of a funtion: Let f be defined
More informationName: Answer Key No calculators. Show your work! 1. (21 points) All answers should either be,, a (finite) real number, or DNE ( does not exist ).
Mat - Final Exam August 3 rd, Name: Answer Key No calculators. Sow your work!. points) All answers sould eiter be,, a finite) real number, or DNE does not exist ). a) Use te grap of te function to evaluate
More informationMathematics 123.3: Solutions to Lab Assignment #5
Matematics 3.3: Solutions to Lab Assignment #5 Find te derivative of te given function using te definition of derivative. State te domain of te function and te domain of its derivative..: f(x) 6 x Solution:
More informationCalculus I - Spring 2014
NAME: Calculus I - Spring 04 Midterm Exam I, Marc 5, 04 In all non-multiple coice problems you are required to sow all your work and provide te necessary explanations everywere to get full credit. In all
More informationMath 185 Homework Exercises II
Math 185 Homework Exercises II Instructor: Andrés E. Caicedo Due: July 10, 2002 1. Verify that if f H(Ω) C 2 (Ω) is never zero, then ln f is harmonic in Ω. 2. Let f = u+iv H(Ω) C 2 (Ω). Let p 2 be an integer.
More informationMVT and Rolle s Theorem
AP Calculus CHAPTER 4 WORKSHEET APPLICATIONS OF DIFFERENTIATION MVT and Rolle s Teorem Name Seat # Date UNLESS INDICATED, DO NOT USE YOUR CALCULATOR FOR ANY OF THESE QUESTIONS In problems 1 and, state
More informationREVIEW LAB ANSWER KEY
REVIEW LAB ANSWER KEY. Witout using SN, find te derivative of eac of te following (you do not need to simplify your answers): a. f x 3x 3 5x x 6 f x 3 3x 5 x 0 b. g x 4 x x x notice te trick ere! x x g
More information2.8 The Derivative as a Function
.8 Te Derivative as a Function Typically, we can find te derivative of a function f at many points of its domain: Definition. Suppose tat f is a function wic is differentiable at every point of an open
More informationExam 1 Review Solutions
Exam Review Solutions Please also review te old quizzes, and be sure tat you understand te omework problems. General notes: () Always give an algebraic reason for your answer (graps are not sufficient),
More informationLECTURE 14 NUMERICAL INTEGRATION. Find
LECTURE 14 NUMERCAL NTEGRATON Find b a fxdx or b a vx ux fx ydy dx Often integration is required. However te form of fx may be suc tat analytical integration would be very difficult or impossible. Use
More informationNumerical Analysis MTH603. dy dt = = (0) , y n+1. We obtain yn. Therefore. and. Copyright Virtual University of Pakistan 1
Numerical Analysis MTH60 PREDICTOR CORRECTOR METHOD Te metods presented so far are called single-step metods, were we ave seen tat te computation of y at t n+ tat is y n+ requires te knowledge of y n only.
More informationPolynomial Interpolation
Capter 4 Polynomial Interpolation In tis capter, we consider te important problem of approximating a function f(x, wose values at a set of distinct points x, x, x 2,,x n are known, by a polynomial P (x
More informationChapter 2. Limits and Continuity 16( ) 16( 9) = = 001. Section 2.1 Rates of Change and Limits (pp ) Quick Review 2.1
Capter Limits and Continuity Section. Rates of Cange and Limits (pp. 969) Quick Review..... f ( ) ( ) ( ) 0 ( ) f ( ) f ( ) sin π sin π 0 f ( ). < < < 6. < c c < < c 7. < < < < < 8. 9. 0. c < d d < c
More informationKey Concepts. Important Techniques. 1. Average rate of change slope of a secant line. You will need two points ( a, the formula: to find value
AB Calculus Unit Review Key Concepts Average and Instantaneous Speed Definition of Limit Properties of Limits One-sided and Two-sided Limits Sandwic Teorem Limits as x ± End Beaviour Models Continuity
More information1 2 x Solution. The function f x is only defined when x 0, so we will assume that x 0 for the remainder of the solution. f x. f x h f x.
Problem. Let f x x. Using te definition of te derivative prove tat f x x Solution. Te function f x is only defined wen x 0, so we will assume tat x 0 for te remainder of te solution. By te definition of
More information1 Lecture 13: The derivative as a function.
1 Lecture 13: Te erivative as a function. 1.1 Outline Definition of te erivative as a function. efinitions of ifferentiability. Power rule, erivative te exponential function Derivative of a sum an a multiple
More information232 Calculus and Structures
3 Calculus and Structures CHAPTER 17 JUSTIFICATION OF THE AREA AND SLOPE METHODS FOR EVALUATING BEAMS Calculus and Structures 33 Copyrigt Capter 17 JUSTIFICATION OF THE AREA AND SLOPE METHODS 17.1 THE
More informationChapter 2 Limits and Continuity
4 Section. Capter Limits and Continuity Section. Rates of Cange and Limits (pp. 6) Quick Review.. f () ( ) () 4 0. f () 4( ) 4. f () sin sin 0 4. f (). 4 4 4 6. c c c 7. 8. c d d c d d c d c 9. 8 ( )(
More informationTHE IMPLICIT FUNCTION THEOREM
THE IMPLICIT FUNCTION THEOREM ALEXANDRU ALEMAN 1. Motivation and statement We want to understand a general situation wic occurs in almost any area wic uses matematics. Suppose we are given number of equations
More informationFunction Composition and Chain Rules
Function Composition an Cain Rules James K. Peterson Department of Biological Sciences an Department of Matematical Sciences Clemson University November 2, 2018 Outline Function Composition an Continuity
More informationMATH 1A Midterm Practice September 29, 2014
MATH A Midterm Practice September 9, 04 Name: Problem. (True/False) If a function f : R R is injective, ten f as an inverse. Solution: True. If f is injective, ten it as an inverse since tere does not
More informationWeierstrass and Hadamard products
August 9, 3 Weierstrass and Hadamard products Paul Garrett garrett@mat.umn.edu ttp://www.mat.umn.edu/ garrett/ [Tis document is ttp://www.mat.umn.edu/ garrett/m/mfms/notes 3-4/b Hadamard products.pdf].
More information(a) At what number x = a does f have a removable discontinuity? What value f(a) should be assigned to f at x = a in order to make f continuous at a?
Solutions to Test 1 Fall 016 1pt 1. Te grap of a function f(x) is sown at rigt below. Part I. State te value of eac limit. If a limit is infinite, state weter it is or. If a limit does not exist (but is
More informationSection 3.1: Derivatives of Polynomials and Exponential Functions
Section 3.1: Derivatives of Polynomials and Exponential Functions In previous sections we developed te concept of te derivative and derivative function. Te only issue wit our definition owever is tat it
More informationChapter 1D - Rational Expressions
- Capter 1D Capter 1D - Rational Expressions Definition of a Rational Expression A rational expression is te quotient of two polynomials. (Recall: A function px is a polynomial in x of degree n, if tere
More information5.1 We will begin this section with the definition of a rational expression. We
Basic Properties and Reducing to Lowest Terms 5.1 We will begin tis section wit te definition of a rational epression. We will ten state te two basic properties associated wit rational epressions and go
More informationExample: f(x) = x 3. 1, x > 0 0, x 0. Example: g(x) =
2.1 Instantaneous rate of cange, or, an informal introduction to derivatives Let a, b be two different values in te domain of f. Te average rate of cange of f between a and b is f(b) f(a) b a. Geometrically,
More information158 Calculus and Structures
58 Calculus and Structures CHAPTER PROPERTIES OF DERIVATIVES AND DIFFERENTIATION BY THE EASY WAY. Calculus and Structures 59 Copyrigt Capter PROPERTIES OF DERIVATIVES. INTRODUCTION In te last capter you
More informationMAT 145. Type of Calculator Used TI-89 Titanium 100 points Score 100 possible points
MAT 15 Test #2 Name Solution Guide Type of Calculator Used TI-89 Titanium 100 points Score 100 possible points Use te grap of a function sown ere as you respond to questions 1 to 8. 1. lim f (x) 0 2. lim
More informationConvexity and Smoothness
Capter 4 Convexity and Smootness 4. Strict Convexity, Smootness, and Gateaux Di erentiablity Definition 4... Let X be a Banac space wit a norm denoted by k k. A map f : X \{0}!X \{0}, f 7! f x is called
More informationOrder of Accuracy. ũ h u Ch p, (1)
Order of Accuracy 1 Terminology We consider a numerical approximation of an exact value u. Te approximation depends on a small parameter, wic can be for instance te grid size or time step in a numerical
More informationPre-Calculus Review Preemptive Strike
Pre-Calculus Review Preemptive Strike Attaced are some notes and one assignment wit tree parts. Tese are due on te day tat we start te pre-calculus review. I strongly suggest reading troug te notes torougly
More informationJANE PROFESSOR WW Prob Lib1 Summer 2000
JANE PROFESSOR WW Prob Lib Summer 2000 Sample WeBWorK problems. WeBWorK assignment Derivatives due 2//06 at 2:00 AM..( pt) If f x 9, find f 0. 2.( pt) If f x 7x 27, find f 5. 3.( pt) If f x 7 4x 5x 2,
More informationThe Derivative as a Function
Section 2.2 Te Derivative as a Function 200 Kiryl Tsiscanka Te Derivative as a Function DEFINITION: Te derivative of a function f at a number a, denoted by f (a), is if tis limit exists. f (a) f(a + )
More informationA = h w (1) Error Analysis Physics 141
Introduction In all brances of pysical science and engineering one deals constantly wit numbers wic results more or less directly from experimental observations. Experimental observations always ave inaccuracies.
More information10 Derivatives ( )
Instructor: Micael Medvinsky 0 Derivatives (.6-.8) Te tangent line to te curve yf() at te point (a,f(a)) is te line l m + b troug tis point wit slope Alternatively one can epress te slope as f f a m lim
More informationLesson 6: The Derivative
Lesson 6: Te Derivative Def. A difference quotient for a function as te form f(x + ) f(x) (x + ) x f(x + x) f(x) (x + x) x f(a + ) f(a) (a + ) a Notice tat a difference quotient always as te form of cange
More informationHOMEWORK HELP 2 FOR MATH 151
HOMEWORK HELP 2 FOR MATH 151 Here we go; te second round of omework elp. If tere are oters you would like to see, let me know! 2.4, 43 and 44 At wat points are te functions f(x) and g(x) = xf(x)continuous,
More informationNUMERICAL DIFFERENTIATION. James T. Smith San Francisco State University. In calculus classes, you compute derivatives algebraically: for example,
NUMERICAL DIFFERENTIATION James T Smit San Francisco State University In calculus classes, you compute derivatives algebraically: for example, f( x) = x + x f ( x) = x x Tis tecnique requires your knowing
More informationLogarithmic functions
Roberto s Notes on Differential Calculus Capter 5: Derivatives of transcendental functions Section Derivatives of Logaritmic functions Wat ou need to know alread: Definition of derivative and all basic
More informationApplications of the van Trees inequality to non-parametric estimation.
Brno-06, Lecture 2, 16.05.06 D/Stat/Brno-06/2.tex www.mast.queensu.ca/ blevit/ Applications of te van Trees inequality to non-parametric estimation. Regular non-parametric problems. As an example of suc
More informationDerivatives. if such a limit exists. In this case when such a limit exists, we say that the function f is differentiable.
Derivatives 3. Derivatives Definition 3. Let f be a function an a < b be numbers. Te average rate of cange of f from a to b is f(b) f(a). b a Remark 3. Te average rate of cange of a function f from a to
More information3.1 Extreme Values of a Function
.1 Etreme Values of a Function Section.1 Notes Page 1 One application of te derivative is finding minimum and maimum values off a grap. In precalculus we were only able to do tis wit quadratics by find
More informationMATH 173: Problem Set 5 Solutions
MATH 173: Problem Set 5 Solutions Problem 1. Let f L 1 and a. Te wole problem is a matter of cange of variables wit integrals. i Ff a ξ = e ix ξ f a xdx = e ix ξ fx adx = e ia+y ξ fydy = e ia ξ = e ia
More informationWeek #15 - Word Problems & Differential Equations Section 8.2
Week #1 - Word Problems & Differential Equations Section 8. From Calculus, Single Variable by Huges-Hallett, Gleason, McCallum et. al. Copyrigt 00 by Jon Wiley & Sons, Inc. Tis material is used by permission
More informationLecture 10: Carnot theorem
ecture 0: Carnot teorem Feb 7, 005 Equivalence of Kelvin and Clausius formulations ast time we learned tat te Second aw can be formulated in two ways. e Kelvin formulation: No process is possible wose
More informationUNIVERSITY OF MANITOBA DEPARTMENT OF MATHEMATICS MATH 1510 Applied Calculus I FIRST TERM EXAMINATION - Version A October 12, :30 am
DEPARTMENT OF MATHEMATICS MATH 1510 Applied Calculus I October 12, 2016 8:30 am LAST NAME: FIRST NAME: STUDENT NUMBER: SIGNATURE: (I understand tat ceating is a serious offense DO NOT WRITE IN THIS TABLE
More information2.4 Exponential Functions and Derivatives (Sct of text)
2.4 Exponential Functions an Derivatives (Sct. 2.4 2.6 of text) 2.4. Exponential Functions Definition 2.4.. Let a>0 be a real number ifferent tan. Anexponential function as te form f(x) =a x. Teorem 2.4.2
More informationChapter Primer on Differentiation
Capter 0.01 Primer on Differentiation After reaing tis capter, you soul be able to: 1. unerstan te basics of ifferentiation,. relate te slopes of te secant line an tangent line to te erivative of a function,.
More informationThe total error in numerical differentiation
AMS 147 Computational Metods and Applications Lecture 08 Copyrigt by Hongyun Wang, UCSC Recap: Loss of accuracy due to numerical cancellation A B 3, 3 ~10 16 In calculating te difference between A and
More informationMath 1241 Calculus Test 1
February 4, 2004 Name Te first nine problems count 6 points eac and te final seven count as marked. Tere are 120 points available on tis test. Multiple coice section. Circle te correct coice(s). You do
More informationWYSE Academic Challenge 2004 Sectional Mathematics Solution Set
WYSE Academic Callenge 00 Sectional Matematics Solution Set. Answer: B. Since te equation can be written in te form x + y, we ave a major 5 semi-axis of lengt 5 and minor semi-axis of lengt. Tis means
More information2.3 Product and Quotient Rules
.3. PRODUCT AND QUOTIENT RULES 75.3 Product and Quotient Rules.3.1 Product rule Suppose tat f and g are two di erentiable functions. Ten ( g (x)) 0 = f 0 (x) g (x) + g 0 (x) See.3.5 on page 77 for a proof.
More information1 + t5 dt with respect to x. du = 2. dg du = f(u). du dx. dg dx = dg. du du. dg du. dx = 4x3. - page 1 -
Eercise. Find te derivative of g( 3 + t5 dt wit respect to. Solution: Te integrand is f(t + t 5. By FTC, f( + 5. Eercise. Find te derivative of e t2 dt wit respect to. Solution: Te integrand is f(t e t2.
More information