Chapter 6. Phillip Hall - Room 537, Huxley

Similar documents
MATH 2710: NOTES FOR ANALYSIS

MTH 3102 Complex Variables Practice Exam 1 Feb. 10, 2017

0.1 Practical Guide - Surface Integrals. C (0,0,c) A (0,b,0) A (a,0,0)

CMSC 425: Lecture 7 Geometric Programming: Sample Solutions

CHAPTER 5 TANGENT VECTORS

Introduction to Group Theory Note 1

Solutions to Test #2 (Kawai) MATH 2421

Elementary Analysis in Q p

Topic 7: Using identity types

Sets of Real Numbers

CALCULUS I. Practice Problems Integrals. Paul Dawkins

Planar Transformations and Displacements

1 Properties of Spherical Harmonics

Short Solutions to Practice Material for Test #2 MATH 2421

The Second Law: The Machinery

Trigonometric Identities

Elements of Asymptotic Theory. James L. Powell Department of Economics University of California, Berkeley

CHAPTER 2: SMOOTH MAPS. 1. Introduction In this chapter we introduce smooth maps between manifolds, and some important

AP Calculus Testbank (Chapter 10) (Mr. Surowski)

Applications to stochastic PDE

4. Score normalization technical details We now discuss the technical details of the score normalization method.

HENSEL S LEMMA KEITH CONRAD

CHAPTER 3: TANGENT SPACE

ε i (E j )=δj i = 0, if i j, form a basis for V, called the dual basis to (E i ). Therefore, dim V =dim V.

Lecture 1.2 Pose in 2D and 3D. Thomas Opsahl

CMSC 425: Lecture 4 Geometry and Geometric Programming

16.2. Infinite Series. Introduction. Prerequisites. Learning Outcomes

2 Asymptotic density and Dirichlet density

2 Asymptotic density and Dirichlet density

Round-off Errors and Computer Arithmetic - (1.2)

Solutions to Problem Set 5

F(p) y + 3y + 2y = δ(t a) y(0) = 0 and y (0) = 0.

Section 0.10: Complex Numbers from Precalculus Prerequisites a.k.a. Chapter 0 by Carl Stitz, PhD, and Jeff Zeager, PhD, is available under a Creative

Lecture 25: The Sine and Cosine Functions. tan(x) 1+y

On a Markov Game with Incomplete Information

Lilian Markenzon 1, Nair Maria Maia de Abreu 2* and Luciana Lee 3

Almost 4000 years ago, Babylonians had discovered the following approximation to. x 2 dy 2 =1, (5.0.2)

16.2. Infinite Series. Introduction. Prerequisites. Learning Outcomes

Solutions to Homework #05 MATH ln z 2 + x 2 1 = 2x2 z 2 + x 2 + ln z2 + x 2. = x. ln z 2 + x 2 2z z 2 + x 2 = 2xz

Availability and Maintainability. Piero Baraldi

2. Review of Calculus Notation. C(X) all functions continuous on the set X. C[a, b] all functions continuous on the interval [a, b].

DIFFERENTIAL GEOMETRY. LECTURES 9-10,

= =5 (0:4) 4 10 = = = = = 2:005 32:4 2: :

Factorability in the ring Z[ 5]

General Random Variables

Autonomous Equations / Stability of Equilibrium Solutions. y = f (y).

Chapter 2 Arithmetic Functions and Dirichlet Series.

ATM The thermal wind Fall, 2016 Fovell

19th Bay Area Mathematical Olympiad. Problems and Solutions. February 28, 2017

MAS 4203 Number Theory. M. Yotov

Chapter 5 Notes. These notes correspond to chapter 5 of Mas-Colell, Whinston, and Green.

Further differentiation and integration

SECTION 5: FIBRATIONS AND HOMOTOPY FIBERS

Elementary theory of L p spaces

Worksheet on Derivatives. Dave L. Renfro Drake University November 1, 1999

10.2 The Unit Circle: Cosine and Sine

Math 121: Calculus 1 - Fall 2012/2013 Review of Precalculus Concepts

Simplifications to Conservation Equations

Elliptic Curves Spring 2015 Problem Set #1 Due: 02/13/2015

Math 104B: Number Theory II (Winter 2012)

Chapter 7: Special Distributions

THE CHARACTER GROUP OF Q

MATH 248A. THE CHARACTER GROUP OF Q. 1. Introduction

Quaternionic Projective Space (Lecture 34)

FOCUS ON THEORY. We recall that a function g(x) is differentiable at the point a if the limit

Radian Measure and Angles on the Cartesian Plane

FOURIER SERIES PART III: APPLICATIONS

Various Proofs for the Decrease Monotonicity of the Schatten s Power Norm, Various Families of R n Norms and Some Open Problems

You may not start to read the questions printed on the subsequent pages until instructed to do so by the Invigilator.

An Overview of Witt Vectors

Andrea Mantile. Fractional Integral Equations and Applications to Point Interaction Models in Quantum Mechanics TESI DI DOTTORATO DI RICERCA

PROFIT MAXIMIZATION. π = p y Σ n i=1 w i x i (2)

Real Analysis 1 Fall Homework 3. a n.

Sums of independent random variables

Chater Matrix Norms and Singular Value Decomosition Introduction In this lecture, we introduce the notion of a norm for matrices The singular value de

On the minimax inequality and its application to existence of three solutions for elliptic equations with Dirichlet boundary condition

Outline. EECS150 - Digital Design Lecture 26 Error Correction Codes, Linear Feedback Shift Registers (LFSRs) Simple Error Detection Coding

Micro I. Lesson 5 : Consumer Equilibrium

Participation Factors. However, it does not give the influence of each state on the mode.

ON MINKOWSKI MEASURABILITY

TRACES OF SCHUR AND KRONECKER PRODUCTS FOR BLOCK MATRICES

MATH 829: Introduction to Data Mining and Analysis Consistency of Linear Regression

3 Properties of Dedekind domains

John Weatherwax. Analysis of Parallel Depth First Search Algorithms

arxiv: v1 [physics.data-an] 26 Oct 2012

Generalized Least-Squares Regressions II: Theory and Classication

An Inverse Problem for Two Spectra of Complex Finite Jacobi Matrices

MODULE 2: DIFFUSION LECTURE NO. 2

NONLINEAR OPTIMIZATION WITH CONVEX CONSTRAINTS. The Goldstein-Levitin-Polyak algorithm

Averaging sums of powers of integers and Faulhaber polynomials

A QUATERNIONIC APPROACH to GEOMETRY of CURVES on SPACES of CONSTANT CURVATURE

ECON Answers Homework #2

The Arm Prime Factors Decomposition

p-adic Measures and Bernoulli Numbers

Galois Fields, Linear Feedback Shift Registers and their Applications

(IV.D) PELL S EQUATION AND RELATED PROBLEMS

ON FREIMAN S 2.4-THEOREM

Matemáticas I: C (1 point) Given the function f (x; y) = xy p 1 + x2 + y 2

A CONCRETE EXAMPLE OF PRIME BEHAVIOR IN QUADRATIC FIELDS. 1. Abstract

Lecture 3 January 16

Transcription:

Chater 6 6 Partial Derivatives.................................................... 72 6. Higher order artial derivatives...................................... 73 6.2 Matrix of artial derivatives.........................................74 6.3 Chain rule....................................................... 74 6.4 Change of variables................................................77 Philli Hall - Room 537, Huxle email: hil.hallimerial.ac.uk

Chater 6 Partial Derivatives De nition 5 Let f : R 2 R be a function of two variables, f (x; ) 2 R. The artial derivative of f with resect to x (x-variable) at a oint (x ; ) 2 R 2 is de ned as the limit x (x f (x + h; ) f (x ; ) ; ) := lim h h (6.) rovided this limit exists. Equivalentl, we de ne the artial derivative of f with resect to as the limit (x f (x ; + h) f (x ; ) ; ) := lim (6.2) h h rovided it exists. In other words, we de ne the artial derivatives of f with resect to x (res. ) keeing the other variable xed and aling the standard de nition of derivative as if f was a function of one variable. Alternativel to the notation x, we will also denote the artial derivatives as f x or x f (similarl, = f = f). The artial derivatives x and will be considered as functions x ; : R2 R that, to an oint (x; ) 2 R 2, associate the artial derivatives (6.) and (6.2). Examle 6 Let Then f(x; ) = x 2 3 + sin(x + 2) + : x (x; ) = 2x3 + cos (x + 2) (x; ) = 32 x 2 + 2 cos (x + 2) + : 72

6. Higher order artial derivatives 73 6. Higher order artial derivatives The artial derivatives x f and f of f : R 2 R are themselves functions of both variables x and. Therefore, we can di erentiate them and consider their artial derivatives with resect to x and, x x = 2 f x 2 ; x = 2 f x ; x = 2 f x ; = 2 f : One of the most imortant results in artial di erentiation is that, rovided that higher order artial derivatives exist and are continuous functions, the order in which artial derivatives are taken is not imortant. Theorem 7 Let f : R 2 R be a function such that both x f and x f are continuous. Then 2 f x = 2 f x : This means that if, for examle, we want to di erentiate a function f : R 2 R three times with resect to x and ve times with resect to, we will carr out these artial derivatives consecutivel in an order and we will siml write 8 f 5 x 3 : In general, De nition 5 can be extended in a natural wa to functions of more than 2 variables. That is, if f : R n R is a function of (x ; :::; x n ) 2 R n, then, for examle, f (x ; :::; x i + h; :::; x n ) f (x ; :::; x i ; :::; x n ) = lim x i h h rovided this limit exists. Equall, one can consider higher order artial derivatives di erentiating with resect to an variable x ; :::; x n iterativel, 2 f x j x i ; 3 f x j x i x k ; ::: etc. i; j; k 2 f; :::; ng: If f : R n R m is not real-valued but vector-valued, 7 f : R n R m (x ; :::; x n ) (f (x ; ::; x n ) ; :::; f m (x ; ::; x n )) we can artiall di erentiate an of its m comonents (f ; :::; f m ) and consider j x i ; j 2 f; :::; mg i 2 f; :::; ng:

6.2 Matrix of artial derivatives 74 6.2 Matrix of artial derivatives Partial derivatives of functions are usuall dislaed in a matrix, the so-called matrix of artial derivatives. If f : R n R m is a R m -valued function of n variables, we arrange its artial derivatives as x x 2 x n 2 2 2 x x 2 x n T f := B. (6.3).... C A m x m x 2 We will denote this matrix b T f. For examle, if f : R n R is real valued (onl one comonent, f = f ), then the matrix (6.3) reduces to a vector T f = ; ; :::; : x x 2 x n In this articular case, the vector of artial derivatives is called the gradient of f and it is alternativel denoted b rf. Examle 8 If m x n f(x; ) = x 2 3 + sin(x + 2) + is the function of Examle 6. Then its gradient is rf = 2x 3 + cos (x + 2) ; 3 2 x 2 + 2 cos (x + 2) + and the matrix of second order artial derivatives is 2 3 sin (x + 2) 6x 2 2 sin (x + 2) 6x 2 2 sin (x + 2) 6x 2 : 4 sin (x + 2) Observe that 6.3 Chain rule 2 f x = 6x2 2 sin (x + 2) = 2 f x : In Chater Section 3, we saw that in order to di erentiate the comosition of two functions of one variable we had to al the chain rule. More exlicitl, if f; g : R R, then (f g) (x) = f (g(x)) g (x) : (6.4) The roblem now is the following. Suose that we have two vector-valued functions G : R n R m and F : R m R r. We want to write the artial derivatives of the comosite function F G in terms of the artial derivatives of F and G. F G : R n G R m F R r :

6.3 Chain rule 75 Theorem 9 The matrix of artial derivatives of F G equals the roduct of the matrix of artial derivatives of F comosed with G with the matrix of artial derivatives of G, i.e., T (F G) = (T F G) T G: (6.5) That is, instead of multiling the two derivatives as in the classic version of the chain rule (6.4), in the multidimensional case we multil the matrices of artial derivatives. Examle 2 Let G : R 2 R 2 (x; ) 7 ln ; x 3 : and F : R2 R 2 (x; ) 7 x sin ; x 2 + 2 so that (F G) (x; ) = ln() sin x 3 ; qln 2 + x 6 (6.6) Then, T F = sin x x 2 + 2 x cos x 2 + 2 and T G = 3x 2 : The matrix of artial derivatives of (6.6) is T (F G) = (T F G)T G. Since sin x cos (T F G) (x; ) = = sin x3 ln() cos x 3 ln A x 2 + 2 x x 2 + 2 (x=ln ;=x 3 ) we have T (F G) = sin x3 ln() cos x 3 ln A 3x 2 = 3x2 ln() cos x 3 3x 5 x 3 ln 2 +x 6 ln 2 +x 6 x 3 ln 2 +x 6 ln 2 +x 6 sin x3 ln ln 2 +x 6 ln 2 +x 6 A : (6.7) Exercise 2 Comute T (F G) directl from (6.6) and verif that it coincides with (6.7). It is not alwas straightforward to al the chain rule (6.5). For examle, if G; F : R 2 R are two function of two variables, we could consider the new function H (x; ) := F (G (x; ) ; ) : However, this function is not the comosition of F with G because, strictl seaking G is real valued and F deends on two variables so F G has no meaning. In order to obtain the artial derivative of H in terms of those of F and G, we need to work a little bit (but not much). Observe that H can be written as H = F e G if we de ne e G from G as eg : R 2 R 2 (x; ) 7 (G (x; ) ; ) :

6.3 Chain rule 76 Therefore, b (6.5), T H = T F G e T G: e Since we have Exlicitl, T F e G = T H = F F x (G (x; ) ; ) F G x (G (x; ) ; ) x ; (G (x; ) ; ) and T e G = G x G ; F G x (G (x; ) ; ) + F (G (x; ) ; ) : H x = F G (G (x; ) ; ) x x ; H = F G (G (x; ) ; ) x + F (G (x; ) ; ) : Examle 22 (Total Derivative) Let : R R 3, (t) = (x (t) ; (t) ; z (t)), be the trajector of a article on R 3, where t stands for the time. Let f : R 4 R, f = f (t; x; ; z), be a hsical observable that deends both on time and osition. The function f (t; (t)) = f (t; x (t) ; (t) ; z (t)) is a one-variable function. Its derivative df dt (with resect to time) is called the total derivative of f (with resect to t). Observe that we can write where Consequentl, df dt = (T f e) T e = t (t) f (t; (t)) = (f e) (t) 7 e : R R 4 t (t; x (t) ; (t) ; z (t)) : x ( (t)) ( (t)) z ( (t)) Bx (t) C (t) A z (t) = (t) + t x ( (t)) x (t) + ( (t)) (t) + z ( (t)) z (t): (6.8) Since ~v = (x (t); (t); z (t)) is the velocit of the article, Equation (6.8) is sometimes found in the literature as df dt = 3X t + (t; (t)) v i ; x i where we have rede ned the variables as x = x, x 2 =, and x 3 = z. i=

6.4 Change of variables 77 6.4 Change of variables In this section, we will learn how the artial derivatives of a function change under a change of variables. Roughl seaking, a change of variables of, let us sa, R n, is a di erentiable ma ' : R n R n and, consequentl, artial derivatives are transformed according to the chain rule (6.5). However, in some situations, we ma need to invert the change of variables and comute the matrix of artial derivatives T ' of the inverse ma ' : R n R n. The comutation of T ' ma be highl non-trivial comared with T ', the matrix of artial derivatives of the direct transformation. We will exemlif this with olar coordinates on R 2 and we will learn how to exress di erential oerators given in Cartesian coordinates in olar coordinates. The next result is a generalisation of the inverse function di erentiation rule (Chater Section 3). Proosition 23 Let ' : R n R n be bijective ma such that all its artial derivatives exist and consider ' : R n R n its inverse ma (i.e., ' ' = ' ' = Id). Let x 2 R n and = ' (x) 2 R n. Then T ' () = (T ') ' () = (T ') (x) : That is, the matrix of artial derivatives of ' evaluated at 2 R n equals the inverse of the matrix T ' evaluated at x. Examle 24 (olar coordinates) Let ' : R 2 R 2 be given b 7 ' : R 2 R 2 (r; ) (r cos ; r sin ) : This change of coordinates is usuall exressed as x = r cos ; = r sin : (6.9) Let f (x; ) be a function de ned on R 2. Regarded as a function of (r; ), B the chain rule, ef (r; ) = f (r cos ; r sin ) : (6.) Using (6.9), f e r = x x r + r ; f e = x x + : f e r = cos + x sin e f = r sin + r cos : x

6.4 Change of variables 78 These equations allow us to exress the artial derivatives of f considered as a function of the olar coordinates (r; ) in terms of the artial derivatives of f with resect to x and. In matrix notation, f e r f e A = (T ') x = cos r sin sin r cos x : If we now want to nd ( x f; f) in terms of ( r e f; e f), we have two otions. On the one hand, we can invert ' r = x 2 + 2 = tan x and then al the chain rule. On the other hand, we can invert T ', (T ') = r cos sin r r sin cos (6.) so that x = r r cos r sin sin cos f e r f e For examle, x (x; ) = cos f e r (r; ) r sin f e (r; ) where r and are given b (6.). A : (6.2) Di erential oerators such as the Lalacian involving artial derivatives are ver frequent in hsics. Suose now that we want to exress a di erential oerator, for examle the Lalacian = 2 x 2 + 2 2 in olar coordinates, that is, in terms of the artial di erential oerators r and. The advantage of having the Lalacian exressed in both Cartesian and olar coordinates is twofold. On the one hand, f is eas to calculate if f deends on the variables (x; ) but if, on the contrar, f e is a function of the olar coordinates (r; ) we need, rst of all, to rewrite r and in terms of x and and al the chain rule twice. On the other hand, some simle functions of (r; ) have a ver comlicated functional form when exressed in terms of (x; ). In those situations, aling the Lalacian in olar coordinates is easier. To start with, observe that, at the level of di erential oerators, Equation (6.2) reads = r cos sin : r r sin cos x r

6.4 Change of variables 79 Carring out the matrix roduct, From (6.3) and with the same notation as in (6.), we have Equivalentl, 2 f x 2 = x x = cos r = cos 2 2 f e r 2 + r 2 cos sin f e x = cos r r sin ; (6.3) = sin r + r cos : (6.4) r sin cos f e r r sin f e r cos sin 2 f e r + r sin2 f e r r sin cos 2 f e r + r 2 sin cos f e + r 2 sin2 2 f e 2 : (6.5) 2 f 2 = = sin r + r cos sin f e r + r cos f e = sin 2 2 f e r 2 r 2 sin cos f e + r sin cos 2 f e r + r cos2 f e r + r cos sin 2 e f r Adding (6.5) and (6.6) and using cos 2 + sin 2 =, we have r 2 cos sin f e + r 2 cos2 2 f e 2 : (6.6) f e = 2 f e r 2 + f e r r + 2 f e r 2 2 :