The Natural Logarithm

Similar documents
The Natural Logarithm

Section 5: Chain Rule

( ) ( ) ( ) ( u) ( u) = are shown in Figure =, it is reasonable to speculate that. = cos u ) and the inside function ( ( t) du

Chapter 2 The Derivative Applied Calculus 107. We ll need a rule for finding the derivative of a product so we don t have to multiply everything out.

Advanced Integration Techniques: Integration by Parts We may differentiate the product of two functions by using the product rule:

The Fundamental Theorems of Calculus

Challenge Problems. DIS 203 and 210. March 6, (e 2) k. k(k + 2). k=1. f(x) = k(k + 2) = 1 x k

Sections 2.2 & 2.3 Limit of a Function and Limit Laws

Math 2142 Exam 1 Review Problems. x 2 + f (0) 3! for the 3rd Taylor polynomial at x = 0. To calculate the various quantities:

Homework 2 Solutions

1. (16 points) Answer the following derivative-related questions. dx tan sec x. dx tan u = du d. dx du tan u. du tan u d v.

HOMEWORK # 2: MATH 211, SPRING Note: This is the last solution set where I will describe the MATLAB I used to make my pictures.

MATH 31B: MIDTERM 2 REVIEW. x 2 e x2 2x dx = 1. ue u du 2. x 2 e x2 e x2] + C 2. dx = x ln(x) 2 2. ln x dx = x ln x x + C. 2, or dx = 2u du.

5.1 - Logarithms and Their Properties

SMT 2014 Calculus Test Solutions February 15, 2014 = 3 5 = 15.

Note: For all questions, answer (E) NOTA means none of the above answers is correct.

( ) 2. Review Exercise 2. cos θ 2 3 = = 2 tan. cos. 2 x = = x a. Since π π, = 2. sin = = 2+ = = cotx. 2 sin θ 2+

THE 2-BODY PROBLEM. FIGURE 1. A pair of ellipses sharing a common focus. (c,b) c+a ROBERT J. VANDERBEI

Reading from Young & Freedman: For this topic, read sections 25.4 & 25.5, the introduction to chapter 26 and sections 26.1 to 26.2 & 26.4.

Echocardiography Project and Finite Fourier Series

4.1 - Logarithms and Their Properties

1 1 + x 2 dx. tan 1 (2) = ] ] x 3. Solution: Recall that the given integral is improper because. x 3. 1 x 3. dx = lim dx.

Chapter 7: Solving Trig Equations

Section 2.6 Derivatives of products and quotients

CALCULUS EXPLORATION OF THE SECOND FUNDAMENTAL THEOREM OF CALCULUS. Second Fundamental Theorem of Calculus (Chain Rule Version):

! ln 2xdx = (x ln 2x - x) 3 1 = (3 ln 6-3) - (ln 2-1)

Two Coupled Oscillators / Normal Modes

Math 333 Problem Set #2 Solution 14 February 2003

23.2. Representing Periodic Functions by Fourier Series. Introduction. Prerequisites. Learning Outcomes

SOLUTIONS TO ECE 3084

Web Appendix N - Derivations of the Properties of the LaplaceTransform

Predator - Prey Model Trajectories and the nonlinear conservation law

Homework sheet Exercises done during the lecture of March 12, 2014

On Customized Goods, Standard Goods, and Competition

FITTING EQUATIONS TO DATA

Math 334 Fall 2011 Homework 11 Solutions

Chapter 4. Truncation Errors

Section 4.4 Logarithmic Properties

Section 3.5 Nonhomogeneous Equations; Method of Undetermined Coefficients

Math 106: Review for Final Exam, Part II. (x x 0 ) 2 = !

EECE 301 Signals & Systems Prof. Mark Fowler

Chapter 2. First Order Scalar Equations

Answers to 1 Homework

CHAPTER 12 DIRECT CURRENT CIRCUITS

Solutions to Assignment 1

Chapter 2: The Derivative

Ground Rules. PC1221 Fundamentals of Physics I. Kinematics. Position. Lectures 3 and 4 Motion in One Dimension. A/Prof Tay Seng Chuan

72 Calculus and Structures

Exponential and Logarithmic Functions -- ANSWERS -- Logarithms Practice Diploma ANSWERS 1

Week #13 - Integration by Parts & Numerical Integration Section 7.2

Topics covered in tutorial 01: 1. Review of definite integrals 2. Physical Application 3. Area between curves. 1. Review of definite integrals

PHYSICS Solving Equations

VS203B Lecture Notes Spring, 2011 Topic: Diffraction

Unit Root Time Series. Univariate random walk

TEACHER NOTES MATH NSPIRED

Week 1 Lecture 2 Problems 2, 5. What if something oscillates with no obvious spring? What is ω? (problem set problem)

MATH 128A, SUMMER 2009, FINAL EXAM SOLUTION

t + t sin t t cos t sin t. t cos t sin t dt t 2 = exp 2 log t log(t cos t sin t) = Multiplying by this factor and then integrating, we conclude that

THE BERNOULLI NUMBERS. t k. = lim. = lim = 1, d t B 1 = lim. 1+e t te t = lim t 0 (e t 1) 2. = lim = 1 2.

Solutions from Chapter 9.1 and 9.2

MATH 4330/5330, Fourier Analysis Section 6, Proof of Fourier s Theorem for Pointwise Convergence

on the interval (x + 1) 0! x < ", where x represents feet from the first fence post. How many square feet of fence had to be painted?

Topics in Combinatorial Optimization May 11, Lecture 22

Linear Response Theory: The connection between QFT and experiments

Review - Quiz # 1. 1 g(y) dy = f(x) dx. y x. = u, so that y = xu and dy. dx (Sometimes you may want to use the substitution x y

Radical Expressions. Terminology: A radical will have the following; a radical sign, a radicand, and an index.

MA 214 Calculus IV (Spring 2016) Section 2. Homework Assignment 1 Solutions

Chapter 4 The Fourier Series and Fourier Transform

Problem Set 7-7. dv V ln V = kt + C. 20. Assume that df/dt still equals = F RF. df dr = =

Instructor: Barry McQuarrie Page 1 of 5

Welcome Back to Physics 215!

From Complex Fourier Series to Fourier Transforms

Chapter 11. Parametric, Vector, and Polar Functions. aπ for any integer n. Section 11.1 Parametric Functions (pp ) cot

Linear Motion I Physics

EECE 301 Signals & Systems Prof. Mark Fowler

Lecture 2-1 Kinematics in One Dimension Displacement, Velocity and Acceleration Everything in the world is moving. Nothing stays still.

Section 4.4 Logarithmic Properties

Some Ramsey results for the n-cube

Inventory Analysis and Management. Multi-Period Stochastic Models: Optimality of (s, S) Policy for K-Convex Objective Functions

3.6 Derivatives as Rates of Change

ACCUMULATION. Section 7.5 Calculus AP/Dual, Revised /26/2018 7:27 PM 7.5A: Accumulation 1

Module 2 F c i k c s la l w a s o s f dif di fusi s o i n

Kinematics Vocabulary. Kinematics and One Dimensional Motion. Position. Coordinate System in One Dimension. Kinema means movement 8.

Physics 180A Fall 2008 Test points. Provide the best answer to the following questions and problems. Watch your sig figs.

EECE 301 Signals & Systems Prof. Mark Fowler

a 10.0 (m/s 2 ) 5.0 Name: Date: 1. The graph below describes the motion of a fly that starts out going right V(m/s)

Analyze patterns and relationships. 3. Generate two numerical patterns using AC

d dx But have you ever seen a derivation of these results? We ll prove the first result below. cos h 1

Circuit Variables. AP 1.1 Use a product of ratios to convert two-thirds the speed of light from meters per second to miles per second: 1 ft 12 in

Chapter Three Systems of Linear Differential Equations

( ) a system of differential equations with continuous parametrization ( T = R + These look like, respectively:

KEY. Math 334 Midterm I Fall 2008 sections 001 and 003 Instructor: Scott Glasgow

Calculus for AP Physics-C

An random variable is a quantity that assumes different values with certain probabilities.

5.2. The Natural Logarithm. Solution

!!"#"$%&#'()!"#&'(*%)+,&',-)./0)1-*23)

Physics 218 Exam 1. with Solutions Fall 2010, Sections Part 1 (15) Part 2 (20) Part 3 (20) Part 4 (20) Bonus (5)

15. Vector Valued Functions

= ( ) ) or a system of differential equations with continuous parametrization (T = R

Transcription:

The Naural Logarihm 5-4-007 The Power Rule says n = n + n+ + C provie ha n. The formula oes no apply o. An anierivaive F( of woul have o saisfy F( =. Bu he Funamenal Theorem implies ha if > 0, hen Thus, plays he role of F(. Define he naural log funcion ln by By consrucion, if > 0, ln = =. for > 0. ln =, an = ln + C. represens he area uner f( = y from o : f( = / ln Bu why is his calle a logarihm? You ve probably seen logarihms use like his: log 8 = 3 because 3 = 8.

So I s no clear wha ln = has o o wih raising numbers o powers. Well, for one hing, ln has many properies you epec a logarihm o have. For eample, ln = = 0. You epec he log of a prouc o equal he sum of he logs. If a an b are posiive numbers, hen ln(ab = ab = a + In he secon inegral, le u =, so u =, an = a u. When = a, u = ; when = ab, u = b. a a a ab + a a = b + u = lna + lnb. u In oher wors, ln(ab = lna + lnb. In similar fashion, you can verify ha ab a. ln a b = lna lnb an lnr = r ln. Thus, here is some jusificaion in calling ln a logarihm, because i has he same properies you epec logs o have. I urns ou ha he whole sory is backwars! When you iscuss logs as he opposie of powers, you are acually being a lile sloppy. To efine he familiar logs (an eponenials wih mahemaical precision, wha you acually o is o efine ln an is inverse e firs, as I ve one above. Then you efine he oher logs an powers using ln an e. For eample, if a an b are posiive numbers, efine log a b = lnb lna. I s possible o check ha logs base a, as I ve jus efine hem, behave he way you epec logs o behave. Here are some aiional properies of ln. Firs, ln = > 0 for > 0. Therefore, he graph of ln is increasing for > 0. Moreover, ln = < 0 for > 0. Therefore, he graph of ln is concave own for > 0. Ne, consier he following picure: f( = / 3 4

The area uner he curve from o 4 is ln4. I is greaer han he sum of he areas of he hree recangles, so ln 4 > + 3 + 4 = 3 >. If n is a posiive ineger, hen So if > 4 n, hen n ln 4 > n, or ln 4 n > n. ln > ln4 n > n. Since n is an arbirary posiive ineger, I can make ln arbirarily large by making sufficienly large. This proves ha lim ln = +. + Here s he graph of ln: y y = ln Eample. The iffereniaion formula for ln works ogeher wih he oher iffereniaion rules in he usual ways. ln( + + 7 = + + + 7. ln(sin + cos + 3 3 = sin +. 3 [ (ln 7 + ln( 7 ] = [ (ln 7 + 7 ln ] ( = 7(ln 6 (ln( ( ln = (ln ( ln (ln (ln + = ( ln(ln + ln = (ln. ( ( ln +. + 7. If I say ha f( = g( he erivaive of f( is g( hen g( shoul be efine wherever f( is efine. Therefore, i is no really correc o say wihou he qualificaion > 0 ha ln =. For is efine for 0, whereas ln is only efine for > 0. 3

I urns ou ha he correc saemen is: ln = for 0. For > 0, his is he same as he ol formula. For < 0, =, so ln = ln( = =. So he upae anierivaive formula is = ln + C. You can omi he absolue value signs if he quaniy insie is never negaive. For eample, i urns ou ha + = ln( + + C. (Deriving his formula requires an inegraion echnique calle subsiuion. However, you can check ha i s correc by iffereniaing ln( + o ge. I can omi he absolue values aroun he + +, because + is always posiive. Eample. 5 = ln 5 + C. an = ln cos + C = ln sec + C. 3 + = 3 ln 3 + + C. Eample. You can use logarihmic iffereniaion o compue erivaives which are ifficul o compue in oher ways. For eample, suppose you wan o iffereniae y =. You can use he Power Rule, because he eponen is, no a number. You can use he rule a = a lna (which you ll see laer, because he base is, no a number. Insea, ake logs of boh sies: lny = ln = ln. Differeniae implicily: Hence, y y = + ln. y = y ( + ln = ( + ln. By a similar proceure, you can iffereniae complicae proucs an quoiens. For eample, o ( + 3 0 iffereniae y =, ake logs of boh sies: 5 + 3(sin 4 lny = ln ( + 3 0 5 + 3(sin 4 = ln( + 3 0 ln ( 5 + 3(sin 4 = ln( + 3 0 ln 5 + 3 ln(sin 4 = 4

0 ln( + 3 ln(5 + 3 4 ln sin. To simplify, I use (in orer he following properies of logs: Differeniae implicily: Hence, ( 0 y = y + 3 ln a b = lna lnb, lnab = lna + lnb, lnar = r lna. y y = 0 + 3 5 5 + 3 4 cos = sin 5 5 + 3 4 cos sin. ( + 3 0 ( 0 5 + 3(sin 4 + 3 5 (5 + 3 4 cos. sin Eample. Compue ( + 4. Le y = ( + 4. Taking logs an bringing he power own, I ge lny = ln( + 4 = 4 ln( +. Differeniae boh sies, using he Chain Rule on he lef an he Prouc Rule (an Chain Rule on he righ: y ( y = (4 + [ln( + ](4 3. + Muliply boh sies by y o clear he fracion on he lef, hen subsiue y = ( + 4 : ( ( ( ( y = y ( 4 + [ln( + ](4 3 = ( + 4 ( 4 + [ln( + ](4 3. + + c 006 by Bruce Ikenaga 5