UNIVERSITY OF CALIFORNIA - SANTA CRUZ DEPARTMENT OF PHYSICS PHYS 116C. Problem Set 2. Benjamin Stahl. October 22, d dx J 0(x) = J 1 (x) (1.
|
|
- Reginald O’Neal’
- 5 years ago
- Views:
Transcription
1 UNIVERSITY OF CALIFORNIA - SANTA CRUZ DEPARTMENT OF PHYSICS PHYS 6C Problem Set Benjamin Stahl October, 4 BOAS, P. 59, PROBLEM.-4 Using Equation.9 from Chapter of Boas, it will be shown that: Starting from Equation.9 an then ifferentiating both sies gives: x J (x) J (x) (.) x J p(x) ( ) n ) n+p x n Γ(n + )Γ(n + + p) ( ) n ( ) (n + p) n+p x n+p n Γ(n + )Γ(n + + p) (.) ( ) n ( ) (n) n+p x n+p ( ) n ( ) (p) n+p + x n+p Γ(n + )Γ(n + + p) Γ(n + )Γ(n + + p) n n Now multiplying both sies of the equation by x gives: x x J ( ) n ( ) (n) n+p p(x) x x n+p ( ) n ) n+p + p n Γ(n + )Γ(n + + p) n Γ(n + )Γ(n + + p) ( ) n ( ) (n) n+p x x n+p + p J p (x) Γ(n + )Γ(n + + p) n (.) Using properties of the Gamma function, this result is further simplifie: x x J ( ) n ( ) (n) n+p p(x) x x n+p + p J p (x) n nγ(n)γ(n + + p) ( ) n ) n+p x Γ(n)Γ(n + + p) n (.4)
2 Now re-inexing this result using n n + gives: x x J ( ) n ) (n+)+p p(x) x n Γ(n + )Γ(n + + p + ) ( ) n+ ) n+p+ x n Γ(n + )Γ(n + + p + ) ( ) n ) n+p+ x Γ(n + )Γ(n + + p + ) n x J p+ (x) + p J p (x) (.5) Plugging in p an simplifying proves the esire result: x x J (x) x J (x) + J (x) x J (x) J (x) (.6) BOAS, P. 59, PROBLEM.-6 Using Equation. from Chapter of Boas, it will shown that: N n+ (x) ( ) n+ J n+ (x) (.) Equation. states: Make the substitution p n+ an simplifying yiels : N p (x) cos(πp)j p(x) J p (x) sin(πp) (.) N n+ (x) cos π ( n+ ) J n+ sin π ( n+ (x) J n+ (x) ) (.) Since the argument of the trigonometric functions is always a half integer multiple of π, the cosine terms goes to an the sine term becomes ±. Using these facts yiels the esire result: N n+ (x) J n+ (x) J n+ (x) ( ) n N n+ (x) ( ) n+ J n+ (x) (.4) BOAS, P. 59, PROBLEM.5-8 Using Equation 5.4 from Chapter of Boas, it will be shown that: J (x)x J (x)x J n+ (x)x & J (x)x J (x)x J n (x)x (.) Equation 5.4 states: J p (x) J p+ (x) J p (x) (.)
3 Starting with p an then integrating both sies from to yiels: J (x) J (x) J (x) J (x) J (x) J (x) J (x) J (x) J (x) J (x) J (x) J (x) lim J (x) J () J (x) J (x) J (x) (.) Now starting with p 4 an then integrating both sies from to yiels: J (x) J (x) J (x) J (x) J (x) J 5 (x) J 4 (x) J 5 (x) J 4 (x) J 5 (x) J 4 (x) J 5 (x) lim J 4(x) J 4 () J 5 (x) J (x) J 5 (x) (.4) Thus, this process will continue inefinitely for o values of p. Therefore it has been shown that: J (x)x J (x)x Starting with p an then integrating both sies from to yiels: J n+ (x)x (.5) J (x) J (x) J (x) J (x) J (x) J (x) J (x) J (x) J (x) J (x) J (x) J (x) lim J (x) J () J (x) J (x) J (x) (.6)
4 Now starting with p an then integrating both sies from to yiels: J (x) J (x) J (x) J (x) J (x) J 4 (x) J (x) J 4 (x) J (x) J 4 (x) J (x) J 4 (x) lim J (x) J () J 4 (x) J (x) Thus, this process will continue inefinitely for even values of p. Therefore it has been shown that: J 4 (x) (.7) J (x)x J (x)x It was presumably shown in Problem 7 of this section in Boas that: J (x)x Thus by the proven relations presente in Equations.5 &.8: & J n (x)x (.8) J (x)x (.9) J n (x)x for all integer values of n (.) 4 BOAS, P. 594, PROBLEM.6-9 The solutions to the following ifferential equation will be foun in terms of Bessel functions: x y + y + y (4.) Rewriting this equation so that it matches most closely to Equation 6. of Boas Chapter : y + x y + 4 x y (4.) Matching this equation term by term to Equation 6. yiels the following relations which are then solve to fin the variables that will be use in the expression of the solution: a a a (bcx c ) 4 x b c x c 4x c c c b c 4 b ( ) 4 b 6 b 4 a p c ( ) ( ) p p 4 9 p Thus using Equation 6. from Boas Chapter, the solution can be written as: ( ) y x a Z p (bx c ) x Z 4x Therefore, in terms of Bessel functions the solution can be expresse as: ( ) ( ) y x AJ 4x + BN 4x (4.) (4.4) (4.5) 4
5 5 BOAS, P. 594, PROBLEM.6-: SECOND PART ONLY As instructe, it will be shown that Equation 6. (y x a Z p (bx c )) from Chapter of Boas is inee a solution to Equation 6. (y" + a x y + (bcx c ) + a p c y ). First, the erivatives of y are taken as follows: x y x a Z p (bx c ) y ax a Z p (bx c ) + bcx a+c Z p (bxc ) y" a(a )x a Z p (bx c ) + abcx a+c Z p (bxc ) + (a + c )bcx a+c Z p (bxc ) + b c x a+c Z p "(bx c ) (5.) Now substituting y an its erivatives into the ifferential equation yiels: a(a )x a Z p (bx c ) + abcx a+c Z p (bxc ) + (a + c )bcx a+c Z p (bxc )+ b c x a+c Z p "(bx c ) + ( a)ax a Z p (bx c ) + ( a)bcx a+c Z p (bxc ) + (bcx c ) x a Z p (bx c )+ (a p c )x a Z p (bx c ). (5.) Simplifying this result gives: bc x a+c Z p (bxc ) + (bcx c ) x a Z p "(bx c ) + (bcx c ) x a p c x a Z p (bx c ) (5.) Multiplying this result by by x a yiels: Now multiplying by x yiels: c bc x c Z p (bxc ) + (bcx c ) Z p "(bx c ) + (bcx c ) p c Z p (bx c ) (5.4) bx c Z p (bxc ) + (bx c ) Z p "(bx c ) + (bx c ) p Z p (bx c ) (5.5) Now, if the substitution (bx c ) w is mae, the above simplifies into Bessel s ifferential equation, an thus 6. is in a fact a solution. w Z p (w) w + w Z p(w) w + w p Z p (w) (5.6) x 6 BOAS, P. 65, PROBLEM.- Using the table presente on pg 64 of Boas, the following limit will be evaluate: J 4 (x) lim x J (x) (6.) Substituting the relationships from the table an simplifying gives the esire result: J 4 (x) lim x J (x) lim x lim x ( x Γ(4+) +O(x 4+ ) Γ(+) ( x (!) ( (!) ( ) lim +O(x + x ) +! (!)! + Ax 6 ) Bx 4 + B x lim 8! 4 6 x + Ax 6 ) + Bx 4 x 4 ( ) 4 + Ax x 4 ( (!) +! ) B + B x 4 (6.) 5
6 7 BOAS, P. 66, PROBLEM.- For a Bessel function of the first kin, J p (x), the small x approximation is given in Boas to be: ) p ( +O x p+ ) (7.) Γ(p + ) The the same Bessel function, asymptotic approximation is given in Boas to be: ( πx cos x p + ) ( ) π +O x 4 (7.) Using these formulae, the function J (x) is plotte along with its small x approximation an asymptotic expansion in Figure 7.:.6.4. J (x) & Approximations J (x) Exact Small x Approximation Asymptotic Approximation Figure 7.: J (x) an its approximations over a large interval in x As can be seen in Figure 7., the asymptotic approximation agrees very well with the actual function for suitably larger values of x. It woul also seem that the small x approximation hols well for sufficiently low values of x; to see this better the plot is regenerate over a smaller interval in x:.6.4. J (x) & Approximations (Zoome) J (x) Exact Small x Approximation Asymptotic Approximation Figure 7.: J (x) an its approximations over a small interval in x From Figure 7., it is clear that the small x approximation hols well for low values of x. 6
7 8 BESSEL PROBLEM Originally, this problem aske that the following equation be ifferentiate an substitute into Bessel s ifferential equation to prove that it was a solution: y π cos(x cosh(t)) t (8.) However, it was announce in class that instea it woul be sufficient to show that oing this resulte in the following: x sin(x cosh(t))sinh(t)t (8.) t First, the above result is re-expresse by evaluating the erivative within the integral: x sin(x cosh(t))sinh(t)t t x cos(x cosh(t))x sinh(t)sinh(t) + x sin(x cosh(t))cosh(t)t x sinh (t)cos(x cosh(t)) + x cosh(t)sin(x cosh(t)) t (8.) Next, the first equation will be ifferentiate to fin y an y : y y x π y y x π sin(x cosh(t))cosh(t)t π cosh(t)cos(x cosh(t))cosh(t)t π cosh(t) sin(x cosh(t)) t cosh (t)cos(x cosh(t))t (8.4) Now, y an its erivatives are substitute into Bessel s ifferential equation. It is assume that p because the problem hints that the ultimate result is an integral representation of N. x π x y + x y + (x p )y x y + x y + x y cosh (t)cos(x cosh(t))t + x cosh(t)sin(x cosh(t))t + x cos(x cosh(t))t π π x cosh (t)cos(x cosh(t))t + x cosh(t)sin(x cosh(t))t x cos(x cosh(t))t x cosh (t)cos(x cosh(t))t cos(x cosh(t)) t + x cosh(t)sin(x cosh(t))t x cosh (t) cos(x cosh(t))t + x cosh(t)sin(x cosh(t))t x sinh (t)cos(x cosh(t))t + x cosh(t)sin(x cosh(t))t x sinh (t)cos(x cosh(t)) + x cosh(t)sin(x cosh(t)) t (8.5) Thus as require it has been shown that the original function oes result in the given integral when it is substitute into Bessel s ifferential equation. 7
Strauss PDEs 2e: Section Exercise 6 Page 1 of 5
Strauss PDEs 2e: Section 4.3 - Exercise 6 Page 1 of 5 Exercise 6 If a 0 = a l = a in the Robin problem, show that: (a) There are no negative eigenvalues if a 0, there is one if 2/l < a < 0, an there are
More informationLinear First-Order Equations
5 Linear First-Orer Equations Linear first-orer ifferential equations make up another important class of ifferential equations that commonly arise in applications an are relatively easy to solve (in theory)
More informationReview of Differentiation and Integration for Ordinary Differential Equations
Schreyer Fall 208 Review of Differentiation an Integration for Orinary Differential Equations In this course you will be expecte to be able to ifferentiate an integrate quickly an accurately. Many stuents
More informationImplicit Differentiation
Implicit Differentiation Thus far, the functions we have been concerne with have been efine explicitly. A function is efine explicitly if the output is given irectly in terms of the input. For instance,
More informationcosh x sinh x So writing t = tan(x/2) we have 6.4 Integration using tan(x/2) = 2 2t 1 + t 2 cos x = 1 t2 We will revisit the double angle identities:
6.4 Integration using tanx/) We will revisit the ouble angle ientities: sin x = sinx/) cosx/) = tanx/) sec x/) = tanx/) + tan x/) cos x = cos x/) sin x/) tan x = = tan x/) sec x/) tanx/) tan x/). = tan
More informationUsing the definition of the derivative of a function is quite tedious. f (x + h) f (x)
Derivative Rules Using te efinition of te erivative of a function is quite teious. Let s prove some sortcuts tat we can use. Recall tat te efinition of erivative is: Given any number x for wic te limit
More informationSection 7.2. The Calculus of Complex Functions
Section 7.2 The Calculus of Complex Functions In this section we will iscuss limits, continuity, ifferentiation, Taylor series in the context of functions which take on complex values. Moreover, we will
More informationHyperbolic Functions. Notice: this material must not be used as a substitute for attending. the lectures
Hyperbolic Functions Notice: this material must not be use as a substitute for attening the lectures 0. Hyperbolic functions sinh an cosh The hyperbolic functions sinh (pronounce shine ) an cosh are efine
More informationFebruary 21 Math 1190 sec. 63 Spring 2017
February 21 Math 1190 sec. 63 Spring 2017 Chapter 2: Derivatives Let s recall the efinitions an erivative rules we have so far: Let s assume that y = f (x) is a function with c in it s omain. The erivative
More informationQF101: Quantitative Finance September 5, Week 3: Derivatives. Facilitator: Christopher Ting AY 2017/2018. f ( x + ) f(x) f(x) = lim
QF101: Quantitative Finance September 5, 2017 Week 3: Derivatives Facilitator: Christopher Ting AY 2017/2018 I recoil with ismay an horror at this lamentable plague of functions which o not have erivatives.
More informationMath 300 Winter 2011 Advanced Boundary Value Problems I. Bessel s Equation and Bessel Functions
Math 3 Winter 2 Avance Bounary Value Problems I Bessel s Equation an Bessel Functions Department of Mathematical an Statistical Sciences University of Alberta Bessel s Equation an Bessel Functions We use
More informationJUST THE MATHS UNIT NUMBER DIFFERENTIATION APPLICATIONS 5 (Maclaurin s and Taylor s series) A.J.Hobson
JUST THE MATHS UNIT NUMBER.5 DIFFERENTIATION APPLICATIONS 5 (Maclaurin s and Taylor s series) by A.J.Hobson.5. Maclaurin s series.5. Standard series.5.3 Taylor s series.5.4 Exercises.5.5 Answers to exercises
More informationDifferentiation ( , 9.5)
Chapter 2 Differentiation (8.1 8.3, 9.5) 2.1 Rate of Change (8.2.1 5) Recall that the equation of a straight line can be written as y = mx + c, where m is the slope or graient of the line, an c is the
More informationExamples of the Fourier Theorem (Sect. 10.3). The Fourier Theorem: Continuous case.
s of the Fourier Theorem (Sect. 1.3. The Fourier Theorem: Continuous case. : Using the Fourier Theorem. The Fourier Theorem: Piecewise continuous case. : Using the Fourier Theorem. The Fourier Theorem:
More informationMath 180, Exam 2, Fall 2012 Problem 1 Solution. (a) The derivative is computed using the Chain Rule twice. 1 2 x x
. Fin erivatives of the following functions: (a) f() = tan ( 2 + ) ( ) 2 (b) f() = ln 2 + (c) f() = sin() Solution: Math 80, Eam 2, Fall 202 Problem Solution (a) The erivative is compute using the Chain
More information4.2 First Differentiation Rules; Leibniz Notation
.. FIRST DIFFERENTIATION RULES; LEIBNIZ NOTATION 307. First Differentiation Rules; Leibniz Notation In this section we erive rules which let us quickly compute the erivative function f (x) for any polynomial
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 3 Review. Lesson 19: Concavity, Inflection Points, and the Second Derivative Test. Lesson 20: Absolute Extrema on an Interval
Exam 3 Review Lessons 17-18: Relative Extrema, Critical Numbers, an First Derivative Test (from exam 2 review neee for curve sketching) Critical Numbers: where the erivative of a function is zero or unefine.
More informationMTH 133 Solutions to Exam 1 October 11, Without fully opening the exam, check that you have pages 1 through 11.
MTH 33 Solutions to Exam October, 7 Name: Section: Recitation Instructor: INSTRUCTIONS Fill in your name, etc. on this first page. Without fully opening the exam, check that you have pages through. Show
More informationd dx But have you ever seen a derivation of these results? We ll prove the first result below. cos h 1
Lecture 5 Some ifferentiation rules Trigonometric functions (Relevant section from Stewart, Seventh Eition: Section 3.3) You all know that sin = cos cos = sin. () But have you ever seen a erivation of
More information1 Heisenberg Representation
1 Heisenberg Representation What we have been ealing with so far is calle the Schröinger representation. In this representation, operators are constants an all the time epenence is carrie by the states.
More informationARAB ACADEMY FOR SCIENCE TECHNOLOGY AND MARITIME TRANSPORT
ARAB ACADEMY FOR SCIENCE TECHNOLOGY AND MARITIME TRANSPORT Course: Math For Engineering Winter 8 Lecture Notes By Dr. Mostafa Elogail Page Lecture [ Functions / Graphs of Rational Functions] Functions
More informationChapter 5. Factorization of Integers
Chapter 5 Factorization of Integers 51 Definition: For a, b Z we say that a ivies b (or that a is a factor of b, or that b is a multiple of a, an we write a b, when b = ak for some k Z 52 Theorem: (Basic
More informationFirst Order Linear Differential Equations
LECTURE 6 First Orer Linear Differential Equations A linear first orer orinary ifferential equation is a ifferential equation of the form ( a(xy + b(xy = c(x. Here y represents the unknown function, y
More informationChapter 2. Exponential and Log functions. Contents
Chapter. Exponential an Log functions This material is in Chapter 6 of Anton Calculus. The basic iea here is mainly to a to the list of functions we know about (for calculus) an the ones we will stu all
More informationUNDERSTANDING INTEGRATION
UNDERSTANDING INTEGRATION Dear Reaer The concept of Integration, mathematically speaking, is the "Inverse" of the concept of result, the integration of, woul give us back the function f(). This, in a way,
More informationThe Chain Rule. Composition Review. Intuition. = 2(1.5) = 3 times faster than (X)avier.
The Chain Rule In the previous section we ha to use a trig ientity to etermine the erivative of. h(x) = sin(2x). We can view h(x) as the composition of two functions. Let g(x) = 2x an f (x) = sin x. Then
More informationMath 19 Sample Final Exam Solutions
SSEA Summer 2017 Math 19 Sample Final Exam Solutions 1. [10 points] For what value of the constant k will the function: ) 1 x 2 sin, x < 0 fx) = x, x + k coskx), x 0 be continuous at x = 0? Explain your
More informationCK- 12 Algebra II with Trigonometry Concepts 1
14.1 Graphing Sine and Cosine 1. A.,1 B. (, 1) C. 3,0 D. 11 1, 6 E. (, 1) F. G. H. 11, 4 7, 1 11, 3. 3. 5 9,,,,,,, 4 4 4 4 3 5 3, and, 3 3 CK- 1 Algebra II with Trigonometry Concepts 1 4.ans-1401-01 5.
More informationMathematics 116 HWK 25a Solutions 8.6 p610
Mathematics 6 HWK 5a Solutions 8.6 p6 Problem, 8.6, p6 Fin a power series representation for the function f() = etermine the interval of convergence. an Solution. Begin with the geometric series = + +
More information11.10a Taylor and Maclaurin Series
11.10a 1 11.10a Taylor and Maclaurin Series Let y = f(x) be a differentiable function at x = a. In first semester calculus we saw that (1) f(x) f(a)+f (a)(x a), for all x near a The right-hand side of
More information2.5 The Chain Rule Brian E. Veitch
2.5 The Chain Rule This is our last ifferentiation rule for this course. It s also one of the most use. The best way to memorize this (along with the other rules) is just by practicing until you can o
More informationHigher. Further Calculus 149
hsn.uk.net Higher Mathematics UNIT 3 OUTCOME 2 Further Calculus Contents Further Calculus 49 Differentiating sinx an cosx 49 2 Integrating sinx an cosx 50 3 The Chain Rule 5 4 Special Cases of the Chain
More informationSET 1. (1) Solve for x: (a) e 2x = 5 3x
() Solve for x: (a) e x = 5 3x SET We take natural log on both sides: ln(e x ) = ln(5 3x ) x = 3 x ln(5) Now we take log base on both sides: log ( x ) = log (3 x ln 5) x = log (3 x ) + log (ln(5)) x x
More informationswapneel/207
Partial differential equations Swapneel Mahajan www.math.iitb.ac.in/ swapneel/207 1 1 Power series For a real number x 0 and a sequence (a n ) of real numbers, consider the expression a n (x x 0 ) n =
More informationA. Incorrect! The letter t does not appear in the expression of the given integral
AP Physics C - Problem Drill 1: The Funamental Theorem of Calculus Question No. 1 of 1 Instruction: (1) Rea the problem statement an answer choices carefully () Work the problems on paper as neee (3) Question
More informationMathcad Lecture #5 In-class Worksheet Plotting and Calculus
Mathca Lecture #5 In-class Worksheet Plotting an Calculus At the en of this lecture, you shoul be able to: graph expressions, functions, an matrices of ata format graphs with titles, legens, log scales,
More informationUnit #6 - Families of Functions, Taylor Polynomials, l Hopital s Rule
Unit # - Families of Functions, Taylor Polynomials, l Hopital s Rule Some problems an solutions selecte or aapte from Hughes-Hallett Calculus. Critical Points. Consier the function f) = 54 +. b) a) Fin
More informationAntiderivatives and Initial Value Problems
Antierivatives an Initial Value Problems Warm up If f (x) =2x, whatisf (x)? Can you think of another function that f (x) coul be? If f (x) =3x 2 +, what is f (x)? Can you think of another function that
More informationChapter 5 Logarithmic, Exponential, and Other Transcendental Functions
Chapter 5 Logarithmic, Exponential, an Other Transcenental Functions 5.1 The Natural Logarithmic Function: Differentiation 5.2 The Natural Logarithmic Function: Integration 5.3 Inverse Functions 5.4 Exponential
More information1 Solutions in cylindrical coordinates: Bessel functions
1 Solutions in cylindrical coordinates: Bessel functions 1.1 Bessel functions Bessel functions arise as solutions of potential problems in cylindrical coordinates. Laplace s equation in cylindrical coordinates
More information7.3 Singular points and the method of Frobenius
284 CHAPTER 7. POWER SERIES METHODS 7.3 Singular points and the method of Frobenius Note: or.5 lectures, 8.4 and 8.5 in [EP], 5.4 5.7 in [BD] While behaviour of ODEs at singular points is more complicated,
More informationSolutions to Practice Problems Tuesday, October 28, 2008
Solutions to Practice Problems Tuesay, October 28, 2008 1. The graph of the function f is shown below. Figure 1: The graph of f(x) What is x 1 + f(x)? What is x 1 f(x)? An oes x 1 f(x) exist? If so, what
More informationOutline. MS121: IT Mathematics. Differentiation Rules for Differentiation: Part 1. Outline. Dublin City University 4 The Quotient Rule
MS2: IT Mathematics Differentiation Rules for Differentiation: Part John Carroll School of Mathematical Sciences Dublin City University Pattern Observe You may have notice the following pattern when we
More informationComputing Derivatives
Chapter 2 Computing Derivatives 2.1 Elementary erivative rules Motivating Questions In this section, we strive to unerstan the ieas generate by the following important questions: What are alternate notations
More informationQuantum Mechanics in Three Dimensions
Physics 342 Lecture 20 Quantum Mechanics in Three Dimensions Lecture 20 Physics 342 Quantum Mechanics I Monay, March 24th, 2008 We begin our spherical solutions with the simplest possible case zero potential.
More informationDifferentiability, Computing Derivatives, Trig Review. Goals:
Secants vs. Derivatives - Unit #3 : Goals: Differentiability, Computing Derivatives, Trig Review Determine when a function is ifferentiable at a point Relate the erivative graph to the the graph of an
More informationFormulas From Calculus
Formulas You Shoul Memorize (an I o mean Memorize!) S 997 Pat Rossi Formulas From Calculus. [sin ()] = cos () 2. [cos ()] = sin () 3. [tan ()] = sec2 () 4. [cot ()] = csc2 () 5. [sec ()] = sec () tan ()
More information4.4 AREAS, INTEGRALS AND ANTIDERIVATIVES
1 4.4 AREAS, INTEGRALS AND ANTIDERIVATIVES This section explores properties of functions defined as areas and examines some of the connections among areas, integrals and antiderivatives. In order to focus
More information2 ODEs Integrating Factors and Homogeneous Equations
2 ODEs Integrating Factors an Homogeneous Equations We begin with a slightly ifferent type of equation: 2.1 Exact Equations These are ODEs whose general solution can be obtaine by simply integrating both
More informationIntroduction to ODE's (0P) Young Won Lim 12/27/14
Introuction to ODE's (0P) Copyright (c) 2011-2014 Young W. Lim. Permission is grante to copy, istribute an/or moify this ocument uner the terms of the GNU Free Documentation License, Version 1.2 or any
More informationBy writing (1) as y (x 5 1). (x 5 1), we can find the derivative using the Product Rule: y (x 5 1) 2. we know this from (2)
3.5 Chain Rule 149 3.5 Chain Rule Introuction As iscusse in Section 3.2, the Power Rule is vali for all real number exponents n. In this section we see that a similar rule hols for the erivative of a power
More informationMAT137 Calculus! Lecture 5
MAT137 Calculus! Lecture 5 Today: 2.5 The Pinching Theorem; 2.5 Trigonometric Limits. 2.6 Two Basic Theorems. 3.1 The Derivative Next: 3.2-3.6 DIfferentiation Rules Deadline to notify us if you have a
More informationAdditional Exercises for Chapter 10
Aitional Eercises for Chapter 0 About the Eponential an Logarithm Functions 6. Compute the area uner the graphs of i. f() =e over the interval [ 3, ]. ii. f() =e over the interval [, 4]. iii. f() = over
More informationMa 530 Power Series II
Ma 530 Power Series II Please note that there is material on power series at Visual Calculus. Some of this material was used as part of the presentation of the topics that follow. Operations on Power Series
More informationLecture XII. where Φ is called the potential function. Let us introduce spherical coordinates defined through the relations
Lecture XII Abstract We introuce the Laplace equation in spherical coorinates an apply the metho of separation of variables to solve it. This will generate three linear orinary secon orer ifferential equations:
More informationChapter 6: Integration: partial fractions and improper integrals
Chapter 6: Integration: partial fractions an improper integrals Course S3, 006 07 April 5, 007 These are just summaries of the lecture notes, an few etails are inclue. Most of what we inclue here is to
More informationDerivatives of Trigonometric Functions
Derivatives of Trigonometric Functions 9-8-28 In this section, I ll iscuss its an erivatives of trig functions. I ll look at an important it rule first, because I ll use it in computing the erivative of
More informationDERIVATIVES: LAWS OF DIFFERENTIATION MR. VELAZQUEZ AP CALCULUS
DERIVATIVES: LAWS OF DIFFERENTIATION MR. VELAZQUEZ AP CALCULUS THE DERIVATIVE AS A FUNCTION f x = lim h 0 f x + h f(x) h Last class we examine the limit of the ifference quotient at a specific x as h 0,
More informationMATH 1300 Lecture Notes Wednesday, September 25, 2013
MATH 300 Lecture Notes Wenesay, September 25, 203. Section 3. of HH - Powers an Polynomials In this section 3., you are given several ifferentiation rules that, taken altogether, allow you to quickly an
More informationCONTINUITY AND DIFFERENTIABILITY
CONTINUITY AND DIFFERENTIABILITY Revision Assignment Class 1 Chapter 5 QUESTION1: Check the continuity of the function f given by f () = 7 + 5at = 1. The function is efine at the given point = 1 an its
More informationChapter 13: General Solutions to Homogeneous Linear Differential Equations
Worked Solutions 1 Chapter 13: General Solutions to Homogeneous Linear Differential Equations 13.2 a. Verifying that {y 1, y 2 } is a fundamental solution set: We have y 1 (x) = cos(2x) y 1 (x) = 2 sin(2x)
More informationL Hôpital s Rule was discovered by Bernoulli but written for the first time in a text by L Hôpital.
7.5. Ineterminate Forms an L Hôpital s Rule L Hôpital s Rule was iscovere by Bernoulli but written for the first time in a text by L Hôpital. Ineterminate Forms 0/0 an / f(x) If f(x 0 ) = g(x 0 ) = 0,
More informationMATH 150 TOPIC 16 TRIGONOMETRIC EQUATIONS
Math 150 T16-Trigonometric Equations Page 1 MATH 150 TOPIC 16 TRIGONOMETRIC EQUATIONS In calculus, you will often have to find the zeros or x-intercepts of a function. That is, you will have to determine
More informationDerivatives and the Product Rule
Derivatives an the Prouct Rule James K. Peterson Department of Biological Sciences an Department of Mathematical Sciences Clemson University January 28, 2014 Outline Differentiability Simple Derivatives
More informationSection 1.3 Evaluating Limits Analytically
Section 1.3 Evaluating Limits Analytically Welcome to BC Calculus Wed August 7 th Use Graph and Table to find limits: BC: 1.1-1.4 Due Monday night (extension) CalcChat WolframAlpha Bin not black hole Epsilon
More informationMath 1B, lecture 8: Integration by parts
Math B, lecture 8: Integration by parts Nathan Pflueger 23 September 2 Introuction Integration by parts, similarly to integration by substitution, reverses a well-known technique of ifferentiation an explores
More informationStrauss PDEs 2e: Section Exercise 4 Page 1 of 6
Strauss PDEs 2e: Section 5.3 - Exercise 4 Page of 6 Exercise 4 Consider the problem u t = ku xx for < x < l, with the boundary conditions u(, t) = U, u x (l, t) =, and the initial condition u(x, ) =, where
More informationAs we know, the three basic trigonometric functions are as follows: Figure 1
Trigonometry Basic Functions As we know, the three basic trigonometric functions are as follows: sin θ = cos θ = opposite hypotenuse adjacent hypotenuse tan θ = opposite adjacent Where θ represents an
More informationMath221: HW# 7 solutions
Math22: HW# 7 solutions Andy Royston November 7, 25.3.3 let x = e u. Then ln x = u, x2 = e 2u, and dx = e 2u du. Furthermore, when x =, u, and when x =, u =. Hence x 2 ln x) 3 dx = e 2u u 3 e u du) = e
More informationComputing Derivatives J. Douglas Child, Ph.D. Rollins College Winter Park, FL
Computing Derivatives by J. Douglas Chil, Ph.D. Rollins College Winter Park, FL ii Computing Inefinite Integrals Important notice regaring book materials Texas Instruments makes no warranty, either express
More informationDifferentiability, Computing Derivatives, Trig Review
Unit #3 : Differentiability, Computing Derivatives, Trig Review Goals: Determine when a function is ifferentiable at a point Relate the erivative graph to the the graph of an original function Compute
More information(c) Find the equation of the degree 3 polynomial that has the same y-value, slope, curvature, and third derivative as ln(x + 1) at x = 0.
Chapter 7 Challenge problems Example. (a) Find the equation of the tangent line for ln(x + ) at x = 0. (b) Find the equation of the parabola that is tangent to ln(x + ) at x = 0 (i.e. the parabola has
More informationTable of Contents Derivatives of Logarithms
Derivatives of Logarithms- Table of Contents Derivatives of Logarithms Arithmetic Properties of Logarithms Derivatives of Logarithms Example Example 2 Example 3 Example 4 Logarithmic Differentiation Example
More informationMath 342 Partial Differential Equations «Viktor Grigoryan
Math 342 Partial Differential Equations «Viktor Grigoryan 6 Wave equation: solution In this lecture we will solve the wave equation on the entire real line x R. This correspons to a string of infinite
More informationIntroductions to ExpIntegralEi
Introductions to ExpIntegralEi Introduction to the exponential integrals General The exponential-type integrals have a long history. After the early developments of differential calculus, mathematicians
More informationCalculus I Announcements
Slide 1 Calculus I Announcements Read sections 3.9-3.10 Do all the homework for section 3.9 and problems 1,3,5,7 from section 3.10. The exam is in Thursday, October 22nd. The exam will cover sections 3.2-3.10,
More informationThe derivative of a function f(x) is another function, defined in terms of a limiting expression: f(x + δx) f(x)
Y. D. Chong (2016) MH2801: Complex Methos for the Sciences 1. Derivatives The erivative of a function f(x) is another function, efine in terms of a limiting expression: f (x) f (x) lim x δx 0 f(x + δx)
More informationTaylor Series and Numerical Approximations
Taylor Series and Numerical Approximations Hilary Weller h.weller@reading.ac.uk August 7, 05 An introduction to the concept of a Taylor series and how these are used in numerical analysis to find numerical
More informationMA4001 Engineering Mathematics 1 Lecture 14 Derivatives of Trigonometric Functions Critical Points
MA4001 Engineering Mathematics 1 Lecture 14 Derivatives of Trigonometric Functions Critical Points Dr. Sarah Mitchell Autumn 2014 An important limit To calculate the limits of basic trigonometric functions
More informationMathematics 104 Fall Term 2006 Solutions to Final Exam. sin(ln t) dt = e x sin(x) dx.
Mathematics 14 Fall Term 26 Solutions to Final Exam 1. Evaluate sin(ln t) dt. Solution. We first make the substitution t = e x, for which dt = e x. This gives sin(ln t) dt = e x sin(x). To evaluate the
More informationNotes on Calculus. Dinakar Ramakrishnan Caltech Pasadena, CA Fall 2001
Notes on Calculus by Dinakar Ramakrishnan 53-37 Caltech Pasaena, CA 9115 Fall 001 1 Contents 0 Logical Backgroun 0.1 Sets... 0. Functions... 3 0.3 Carinality... 3 0.4 EquivalenceRelations... 4 1 Real an
More informationMath Chapter 2 Essentials of Calculus by James Stewart Prepared by Jason Gaddis
Math 231 - Chapter 2 Essentials of Calculus by James Stewart Prepare by Jason Gais Chapter 2 - Derivatives 21 - Derivatives an Rates of Change Definition A tangent to a curve is a line that intersects
More informationMath 210 Midterm #1 Review
Math 20 Miterm # Review This ocument is intene to be a rough outline of what you are expecte to have learne an retaine from this course to be prepare for the first miterm. : Functions Definition: A function
More information3. Use absolute value notation to write an inequality that represents the statement: x is within 3 units of 2 on the real line.
PreCalculus Review Review Questions 1 The following transformations are applied in the given order) to the graph of y = x I Vertical Stretch by a factor of II Horizontal shift to the right by units III
More informationDifferential and Integral Calculus
School of science an engineering El Akhawayn University Monay, March 31 st, 2008 Outline 1 Definition of hyperbolic functions: The hyperbolic cosine an the hyperbolic sine of the real number x are enote
More informationTuesday, Feb 12. These slides will cover the following. [cos(x)] = sin(x) 1 d. 2 higher-order derivatives. 3 tangent line problems
Tuesday, Feb 12 These slides will cover the following. 1 d dx [cos(x)] = sin(x) 2 higher-order derivatives 3 tangent line problems 4 basic differential equations Proof First we will go over the following
More informationLinear and quadratic approximation
Linear an quaratic approximation November 11, 2013 Definition: Suppose f is a function that is ifferentiable on an interval I containing the point a. The linear approximation to f at a is the linear function
More informationFinal Exam Study Guide and Practice Problems Solutions
Final Exam Stuy Guie an Practice Problems Solutions Note: These problems are just some of the types of problems that might appear on the exam. However, to fully prepare for the exam, in aition to making
More informationMA 2232 Lecture 08 - Review of Log and Exponential Functions and Exponential Growth
MA 2232 Lecture 08 - Review of Log an Exponential Functions an Exponential Growth Friay, February 2, 2018. Objectives: Review log an exponential functions, their erivative an integration formulas. Exponential
More informationDifferentiation Rules Derivatives of Polynomials and Exponential Functions
Derivatives of Polynomials an Exponential Functions Differentiation Rules Derivatives of Polynomials an Exponential Functions Let s start with the simplest of all functions, the constant function f(x)
More informationSingle Variable Calculus Warnings
Single Variable Calculus Warnings These notes highlight number of common, but serious, first year calculus errors. Warning. The formula g(x) = g(x) is vali only uner the hypothesis g(x). Discussion. In
More informationInverse Functions. Review from Last Time: The Derivative of y = ln x. [ln. Last time we saw that
Inverse Functions Review from Last Time: The Derivative of y = ln Last time we saw that THEOREM 22.0.. The natural log function is ifferentiable an More generally, the chain rule version is ln ) =. ln
More informationSome functions and their derivatives
Chapter Some functions an their erivatives. Derivative of x n for integer n Recall, from eqn (.6), for y = f (x), Also recall that, for integer n, Hence, if y = x n then y x = lim δx 0 (a + b) n = a n
More information1 Lecture 18: The chain rule
1 Lecture 18: The chain rule 1.1 Outline Comparing the graphs of sin(x) an sin(2x). The chain rule. The erivative of a x. Some examples. 1.2 Comparing the graphs of sin(x) an sin(2x) We graph f(x) = sin(x)
More informationLecture 5: Function Approximation: Taylor Series
1 / 10 Lecture 5: Function Approximation: Taylor Series MAR514 Geoffrey Cowles Department of Fisheries Oceanography School for Marine Science and Technology University of Massachusetts-Dartmouth Better
More informationMath 308 Week 8 Solutions
Math 38 Week 8 Solutions There is a solution manual to Chapter 4 online: www.pearsoncustom.com/tamu math/. This online solutions manual contains solutions to some of the suggested problems. Here are solutions
More informationMath 1A Midterm 2 Fall 2015 Riverside City College (Use this as a Review)
Name Date Miterm Score Overall Grae Math A Miterm 2 Fall 205 Riversie City College (Use this as a Review) Instructions: All work is to be shown, legible, simplifie an answers are to be boxe in the space
More information10.7. DIFFERENTIATION 7 (Inverse hyperbolic functions) A.J.Hobson
JUST THE MATHS SLIDES NUMBER 0.7 DIFFERENTIATION 7 (Inverse hyperbolic functions) by A.J.Hobson 0.7. Summary of results 0.7.2 The erivative of an inverse hyperbolic sine 0.7.3 The erivative of an inverse
More informationSection 7.2 Addition and Subtraction Identities. In this section, we begin expanding our repertoire of trigonometric identities.
Section 7. Addition and Subtraction Identities 47 Section 7. Addition and Subtraction Identities In this section, we begin expanding our repertoire of trigonometric identities. Identities The sum and difference
More information