Register now to access 7 million high quality study materials (What's Course Hero?) Course Hero is the premier provider of high quality online educational resources. With millions of study documents, online tutors, digital flashcards and free courseware, Course Hero is helping students learn more efficiently and effectively. Whether you're interested in exploring new subjects or mastering key topics for your next exam, Course Hero has the tools you need to achieve your goals.

29 Pages

cn5

Course: ECN 470, Fall 2009
School: ASU
Rating:

Word Count: 1875

Document Preview

IV TOPIC CALCULUS [1] Limit Suppose that we have a function y = f(x). What would happen to y as x xo? Definition: Let y = f(x). Then, the limit value of y as x a is denoted by limxaf(x). EX 1: y = 1 + 2x. y = 1/x, x 0. As x 0, y 1. limx0y = 1. EX 2: As x 0, y or - Limit does not exists. As x 0 from the right, y . limxo 1 . x As x 0 from the left, y -. limx0 1 . x IV-1 EX 3: y= 1 . x1...

Register Now

Unformatted Document Excerpt

Coursehero >> Arizona >> ASU >> ECN 470

Course Hero has millions of student submitted documents similar to the one
below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.

Course Hero has millions of student submitted documents similar to the one below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.
IV TOPIC CALCULUS [1] Limit Suppose that we have a function y = f(x). What would happen to y as x xo? Definition: Let y = f(x). Then, the limit value of y as x a is denoted by limxaf(x). EX 1: y = 1 + 2x. y = 1/x, x 0. As x 0, y 1. limx0y = 1. EX 2: As x 0, y or - Limit does not exists. As x 0 from the right, y . limxo 1 . x As x 0 from the left, y -. limx0 1 . x IV-1 EX 3: y= 1 . x1 limx2 y = 1; limx1y ; limx1y . limxy = 0; EX 4: y= limx-y = 0. 1x 2 , x 1. 1x 0 = 0 (?) 0 limx1y = Nope!!!. 1x 2 (1x)(1x) limx1 = limx1 = limx1(1+x) = 2!!! 1x 1x EX 5: x2 y= , x 0. x limx0y = limx0x = 0. limxy = limxx = . Note: Before computing a limit, better to simplify the function f. Although f(x) may not be defined at x = xo, f(x) could have a limit. IV-2 EX 6: y= xa . x a ) = 1. x limxy = limx(1 + Theorem: Suppose that lim f(x) and lim g(x) exists. (Here, lim = limxx .) Then, o (R1) lim [f(x) g(x)] = lim f(x) lim g(x). (R2) lim[f(x)g(x)] = [lim f(x)][lim g(x)] (R3) lim[f(x)/g(x)] = [lim f(x)]/[lim g(x)], if lim g(x) 0. EX: y = ex/x2. limxy = [limxex]/[limxx2] (?) Nope!!!, since limxex and limxx2 do not exist. In fact, limxex/x2 = . limx0y = [limx0ex]/[limx0x2] (?) Nope!!!, since limx0x2 = 0. In fact, limx0y = . EX: y = (x2-1)/(x-1), x 1. limx1y = [lim (x2-1)]/[lim (x-1)] (?) Nope!!!, since limx1(x-1) = 0. In fact, limx1y = limx1(x+1) = 2. IV-3 EX: y = (x2+2)/(x2+2x+2). limxy = [limx(x2+2)]/[limx(x2+2x+2)] (?) Nope!!!, since limx(x2+2) does not exists. limx x 22 x 2 2x 2 = limx 1 2/x 2 1 2/x 2/x 2 = [lim (1+2/x2)]/[lim (1+2/x+2/x2)] = 1. L'Hpital's Theorem Suppose that we have two functions f(x) and g(x). Suppose: lim f(x) = 0 and lim g(x) = 0 , or, lim f(x) = and lim g(x) = . f(x) f (x) Then, lim = lim . g(x) g (x) IV-4 [2] Derivative y = f(x) We want to know changes in y (y) when x changes by x. rate of change = y x f(x 0 x) f(x 0) y x x x 0 x dy f(x x) f(x) . y f (x) = lim dx x x0 dy = d x x x 0 derivative evaluated at x0 = f(x0) Note: If lim lim , then, we say that derivative does not exist. x0 x0 Question: What is derivative? IV-5 y y=f(x) x0 x / : tangent line at x = x0 f(x0) : measures slope of tangent line IV-6 [3] Continuity vs. differentiability Definition: Consider a function y = f(x). Suppose: 1) f is defined at x0; 2) lim f(x) lim f(x) limf(x) ; xx 0 xx 0 xx 0 3) lim f(x) f(x 0) . xx 0 Then, f is called continuous at x = x0. EX 1) y = f(x) = 6x, for all x. Is this function continuous at x = 1? y y = 6x x IV-7 1) f(1) = 6; 2) lim y 6 ; x1 3) lim y f(1) x1 Continuous. 1 x EX 2) y Is this function continuous at x = 0 1) f(0) not defined. y y = 1/x x y is not continuous at x = 0. IV-8 y x+1 2 1 x y=x+2 -2 EX 3) y = x1 if x 1 x2 if x < 1 1) f(1) = 2. 2) lim f(x) 2 ; lim f(x) 3 x1 x1 Not continuous. Definition: Suppose that at x0, f(x0) = lim f(x 0 x) f(x 0) x x0 exists. IV-9 Then, we say thaty f is differentiable at x0. Note: If lim lim , we say that derivative does not exist. x0 x0 Theorem: If f is differentiable at x0, then, f is continuous at x 0. <proof> Note that: f(x) - f(x0) = f(x) f(x 0) x x0 (x x 0) . Then, xx 0 lim [f(x) f(x 0)] = lim f(x) f(x 0) x x0 xx 0 (x x 0) = f(x0) 0 = 0. Thus, lim f(x) f(x 0) 0 lim f(x) = f(xo). xx 0 xx 0 Corollary: If f is not continuous, then f is not differentiable. EX 1) y = x1 x 1 x2 x < 1 Not continuous at x = 1 IV-10 not differentiable. Note: If a function is continuous, it may or may not be differentiable. Definition: |a| = a if a 0 a if a < 0 EX 1) |-4| = 4; |4| = 4 EX2) y = f(x) = |x-2| + 1 y= x1 if x 2 x3 if x < 2 Continuous? 1) f(2) defined and f(2) = 1 2) lim f(x) lim f(x) 1 x2 x2 3) limx2f(x) = f(2). So, f is continuous. Differentiable? x0 lim f(2 x) f(2) (2 x) 1 1 = lim x x x0 = lim x =1 x x0 IV-11 x0 lim f(2 x) f(2) (2 x) 3 1 = lim x x x0 = lim x0 x = -1 x not differentiable IV-12 [4] Rules for Derivatives y = f(x) = k. dy 0. dx (1) (2) y = axn y = anxn-1 EX 1) y = 3x6 y = 3 6 x6-1 = 18x5 1 2 EX 2) y = 3 x 3x 1 x y 3 2 1 1 2 3 2 3 1 3 x = = 2 2 1 2 x x2 1 (3) d d d (f(x) g(x)) = f(x) g(x) f (x) g (x) dx dx dx EX) y = 2x2 + x dy 4x 1 . dx IV-13 (4) d (f(x) g(x)) = dx d d f(x) g(x) f(x) g(x) dx dx d d f(x) g(x) f(x) g(x) dx dx [g(x)] 2 (5) d f(x) dx g(x) = . EX) f(x) = 6x2 + 6x g(x) = 3x3 + 6x2 f(x) = 12x + 6; g(x) = 9x2 + 12x d f(x) g(x) = (12x + 6)(3x3 + 6x2) + (6x2 + 6x)(9x2 + 12x) dx d f dx g = (12x 6)(3x 3 6x 2) (6x 2 6x)(9x 2 12x) (3x 3 6x 2) 2 IV-14 [5] Chain Rule xyz z = f(y) y = g(x) dz dz dy f (y) g (x) dx dy dx xyzw dw dw dz dy . dx dz dy dx EX 1) z = 3y2; y = 2x + 5 xyz dz dz dy = (6y) 2 = 12y = 12(2x + 5) = 24x + 60 dx dy dx dz 24 1 60 84 dx x 1 EX 2) y = 3(2x2+1)3. Set z = 2x2 + 1. Then, y = 3z3; z = 2x2 + 1. dy/dx = (dy/dz)(dz/dx) = (9z2)(4x) = 36xz2 = 36x(2x2+1)2. 3) EX TR = f(Q); Q = g(L) dTR dTR dQ = MR MPL = MRPL dL dQ dL IV-15 [6] Inverse Function (EX 1) Consider the following example. This is a function. Looking at the reversed relation: (EX 2) Consider the following example. IV-16 Now, look at the reversed relation: Definition: If the reversed relation is also a function, we say that there exists the inverse function, and we denote it by f-1. Definition: A function y = f(x) is strictly increasing (decreasing) iff f(x1) > f(x2), whenever x1 > (<) x2. Theorem: If a function y = f(x) is strictly increasing or strictly decresing, then x = f -1(y) (inverse function) exists. Note: IV-17 (1) "Function": If you know what happened yesterday (cause), you can predict what will happen today (result). (2) "Inverse function": If you know what happens today (result), you can figure out what happened yesterday (cause). EX) y = f(x) = 5x + 25, - < x < . x = y/5 - 5 = f--1(y). Theorem: Suppose that a function y = f(x) has an inverse function. Then, dx 1 . dy dy dx EX) For the above example, dy/dx = 5 ; dx/dy = 1/5. Clearly, dx/dy = 1/(dy/dx). IV-18 [7] Partial Differentiation Consider a function: y = f(x1,x2,...,xn), where xi's are independent variables and y is a dependent variables. Suppose we would like to know the rate of change of y when x1 changes. Definition: dy y y . ceteris paribus lim dx1 x1 x1 x1 (Here, we treat all other variables as constant!!!!) EX 1) y = 3x12 + x1x2 + 4x22 y/x1 = 6x1 + x2 . y/x2 = x1 + 8x2 . EX 2) y = (x1 + 4)(3x1 + 4x2) . y/x1 = [(x1+4)/x1](3x1+4x2) + (x1+4)[(3x1+4x2)/x1] = 3x1 + 4x2 + (x1+4) 3 = 6x1 + 4x2 + 12. y/x2 = 4x1 + 16 (Show this at home.) 2 EX 3) y x1x2 x1 x2 . IV-19 (x1x2 ) y x1 2 2 x1 (x1 x2) (x1x2 ) (x1 x2)2 2 (x1 x2) x1 x2 3 2 x2 (x1 x2) x1x2 (1) (x1 x2) 2 2 (x1 x2) . Simliarly, y/x2 = (2x12x2 - x1x22)/(x1-x2)2. IV-20 [8] Jacobian Determinant y1 = f1(x1, x2, ... xn, xn+1, xn+2, ... xm) y2 = f2(x1, x2, .... , xn, xn+1, xn+2, ... , xm) yn = fn(x1, x2, .... , xn, xn+1, xn+2, ... , xm). There are n equations: Let's treat xn+1, ... xm as given. (Treat them as constants) Sometimes, we wish to know whether all the functions are distinctive, that is, whether there are redundant functions or not. For this case, we use Jacobian matrix: y 1 y 1 x 1 x 2 y 2 y 2 J= x 1 x 2 y n y n x 1 x 2 ... y n x n y 1 x n y 2 x n ... ... If |J| = 0, we say that the functions are functionally dependent. If |J| 0, we say that the functions are functionally independent. IV-21 EX 1) y 1 2x 1 x 2 ; y 2 x 1 x 2 y 1 x 1 y 2 x 1 y 1 x 2 y 2 x 2 2 3 2 2 2x 1 0 ; 3x 2 2 2x 1x 2 ; x1 2 4x 3 6x x 3 0 |J| 1 1 2 2x x x 2 1 1 2 EX 2) y1 = x1 + 2x2 y 2 x 1 4x 1x 2 4x 2 = (x1 + 2x2)2 1 2 4x 8x 2(2x 4x ) 0 |J| 1 2 1 2 2x 1 4x 2 4x 1 8x 2 2 2 4x 1 3x 2 2 IV-22 [9] Differentials dy : derivatives dx dy, dx : differentials (small changes) dy f (x) dy f ...

Find millions of documents on Course Hero - Study Guides, Lecture Notes, Reference Materials, Practice Exams and more. Course Hero has millions of course specific materials providing students with the best way to expand their education.

Below is a small sample set of documents:

ASU - ECN - 470
TOPIC V COMPARATIVE STATICS[1] MotivationP SD(Y1) D(Yo) Q Here, D(Yo) denotes the demand curve when Y = Y o. Similarly, D(Y1) denotes the demand curve when Y = Y1. Observe that as Y changes, P and Q change. In economic models, equilibrium
ASU - ECN - 470
TOPIC VII UNCONSTRAINED OPTIMIZATION II[1] Relative Maximum and Minimum(1)Review for y = f(x): f(x) = 0 get x*. f(x*) &gt; 0 x* is relative min. point. f(x*) &lt; 0 x* is relative max. point. FOC: SOC:(2)For y = f(x1,x2,.,xn): FOC:f f f
ASU - ECN - 470
1. Matrix Algebra and Linear Economic ModelsReferences Ch. 1 3 (Turkington); Ch. 4 5.2 (Klein). [1] Motivation One market equilibrium Model Assume perfectly competitive market: Both buyers and sellers are price-takers. Demand: Qd = a + bP , a &gt;
Texas A&M - GEOL - 648
Stable Isotope Geology GEOL 648Fall, 2007 E. GrossmanANALYSIS OF CARBONATES ON DELTAPLUSXP AND GAS BENCH II LAB 2The purpose of this lab is to perform carbon and oxygen isotopic analyses of carbonate minerals using the Gas Bench II system and D
ASU - ECN - 470
2. Further Topics in Matrix Algebra(1) Quadratic Form Consider: a11 a12 . a1n x1 a x a22 . a2 n 21 ; xn1 = 2 , = [aij ] = : : : : an1 an 2 . ann xn Annwhere A is symmetric ( aij = a ji ). The following form is called a qua
ASU - ECN - 470
S. C. AHNASSIGNMENT 1 Due February 16 (Monday)1998, SPRING1. (10 pts.) Let S1 = {3,6,9} and S2 = {a,b}. Is {(3,a),(6a),(9,a)} a function from S1 to S2. Why or why not? Explain. 2. (10 pts.) Suppose that the domain of the function y = x 2 - x +
ASU - ECN - 470
ECN 485ASSIGNMENT 2 Due March 25 (Wednesday)Spring 19981. (30 pts, 6 pts. on each.) Let 2 u 0 1 3 (a) (b) (c) (d) (e) ; v 5 4 7 1 ;w 6 2 0 9Find 7u + 3w Find 2u - (v + w) Find uv Can u, v and w span 4? Why or Why not? Explain it. Check whethe
ASU - ECN - 470
Spring Semester 1998Mathematical EconomicsDr. Seung Chan (Min) Ahn !&quot;#\$% &amp;%' ()*&quot;# +,- ()*&amp;+\$%' !+./ !&quot; #\$ %&amp;# '%\$() *#\$ +,-.#/# 0 \$)1#*1&amp; # # # !# .2&amp;# #&quot;# %)+# #!# 3*)# #&quot;#4+&quot;(4)&amp;
Cincinnati - P - 508
%!PS-Adobe-2.0 %Creator: dvips(k) 5.86 Copyright 1999 Radical Eye Software %Title: W05MPart1.dvi %Pages: 2 %PageOrder: Ascend %BoundingBox: 0 0 612 792 %DocumentFonts: CMBX12 CMR12 CMMI8 CMMI12 CMEX10 CMR8 CMSY8 CMTI12 %EndComments %DVIPSWebPage: (ww
Washington University in St. Louis - MEXMRS - 0513
PDS_VERSION_ID = PDS3 RECORD_TYPE = STREAM OBJECT = TEXT PUBLICATION_DATE = 2006-12-06 NOTE
Cincinnati - P - 508
Mid-term Examination Part 1 Introduction to Quantum Mechanics (Physics 508) February 2005This is the first part of the mid-term exam. It is due at the end of the in-class exam period on Monday, February 7, along with the second part which will be gi
Cincinnati - P - 508
%!PS-Adobe-2.0 %Creator: dvips(k) 5.86 Copyright 1999 Radical Eye Software %Title: MPart1A.dvi %Pages: 2 %PageOrder: Ascend %BoundingBox: 0 0 612 792 %EndComments %DVIPSWebPage: (www.radicaleye.com) %DVIPSCommandLine: dvips MPart1A -o %DVIPSParameter
Cincinnati - P - 508
%!PS-Adobe-2.0 %Creator: dvips(k) 5.86 Copyright 1999 Radical Eye Software %Title: MPart2.dvi %Pages: 2 %PageOrder: Ascend %BoundingBox: 0 0 612 792 %EndComments %DVIPSWebPage: (www.radicaleye.com) %DVIPSCommandLine: dvips MPart2 -o %DVIPSParameters:
Cincinnati - P - 508
%!PS-Adobe-2.0 %Creator: dvips(k) 5.92b Copyright 2002 Radical Eye Software %Title: W07MPart1.dvi %Pages: 2 %PageOrder: Ascend %BoundingBox: 0 0 596 842 %DocumentFonts: CMBX12 CMR12 CMSY10 CMMI12 CMR8 CMMI8 CMSY8 %EndComments %DVIPSWebPage: (www.radi
Cincinnati - P - 508
%!PS-Adobe-2.0 %Creator: dvips(k) 5.86 Copyright 1999 Radical Eye Software %Title: jan05.dvi %Pages: 5 %PageOrder: Ascend %Orientation: Landscape %BoundingBox: 0 0 612 792 %DocumentFonts: CMBX12 CMMIB10 CMBSY10 CMR17 CMEX10 CMBXTI10 %EndComments %DVI
Oregon State - ENGR - 202
Final Specification Presentation ScheduleMonday - Jun-01 8 - 8:30 8:30 - 9 9 - 9:30 9:30 - 10 10 - 10:30 10:30 - 11 11 - 11:30 11:30 - 12 12 - 12:30 12:30 - 1 1 - 1:30 1:30 - 2 2 - 2:30 2:30 - 3 3 - 3:30 3:30 - 4 4 - 4:30 4:30 - 5 5 - 5:30 5:30 - 6
Arizona - SCHEDULE - 081
Crs. #(units) Call # Sec #Course/Section Title Time DaysInstructor Bldg. Room #Crs. #(units) Call # Sec #Course/Section Title Time DaysInstructor Bldg. Room #NAVAL SCIENCE (N S )Thomas C. Abel, Professor of Naval Science, Dept. Head,
Oregon State - ENGR - 202
Preliminary Specification Presentation ScheduleMonday - Apr-27 8 - 8:30 8:30 - 9 9 - 9:30 9:30 - 10 10 - 10:30 10:30 - 11 11 - 11:30 11:30 - 12 12 - 12:30 12:30 - 1 1 - 1:30 1:30 - 2 2 - 2:30 2:30 - 3 3 - 3:30 3:30 - 4 4 - 4:30 4:30 - 5 5 - 5:30 5:3
Oregon State - ENGR - 202
ENGR202 Presentation 1 RubricTopic Understanding Score Criteria 3 Student demonstrates all of the items below. 2 Student demonstrates two of the item below. 1 Student demonstrates only one of the items below. 0 Student demonstrates none of the i
Oregon State - ENGR - 202
ENGR202 Presentation 2 RubricTopic Understanding Score Criteria 3 Student demonstrates all of the items below. 2 Student demonstrates two of the item below. 1 Student demonstrates only one of the items below. 0 Student demonstrates none of the i
Oregon State - ENGR - 202
ENGR202 Prototype RubricTopic Completeness of Prototype Score Criteria 10 Student demonstrates a generator that outputs specified energy level, or has clearly define explanation of why generator does not output desired energy. 6 Mechanical or elec
Oregon State - ENGR - 202
ECE322 Lab Grading ScheduleWeek of: 1/7/2007 1/14/2007 Section Title Section 1 - Tools Section 2 - AC Rectifier To Turn in Nothing Prelab for Section 2 Study Questions from Section 1 Report From Section 1 Nothing Prelab Questions Part 1, 2, 3, and 4
Oregon State - ENGR - 202
ENGR202 Final Report Rubric Description Technical Knowledge6 Supports points through correct use of relevant factual knowledge and theories. Some theories and factual knowledge may be drawn from outside of the class. Technical information provided
Oregon State - ENGR - 202
Appendix A: Project Design Specification DocumentAPPENDIX AProject Design Specification Document 2009 Oregon State UniversityENGR 202 ManualPage 53Appendix A: Project Design Specification DocumentDOCUMENT OVERVIEWModern engineering requ
Oregon State - ENGR - 202
ENGR 202: ELECTRICAL FUNDAMENTALS IIA TekBots CourseCopyright InformationCopyright 2009 Oregon State University School of Electrical Engineering &amp; Computer Science (EECS) This document is the property of Oregon State University and the School of
Mississippi State - ECE - 4522
Continuous Playback FM TransmitterA new concept in advertising.Team MembersTeam leader: Joseph Massey Group members: Melissa Bourg Craig Ivey Charles Ingram Faculty Advisor: Dr. Pat DonohoeOverview Problem: Businesses do not have an effi
Mississippi State - ECE - 4522
Continuous Playback FM TransmitterA new concept in advertising.Overview Problem: Businesses do not have an efficient method of advertising to nearby automobile passengers. Solution: Develop a low-power radio transmitter that will relay a custom
Arizona - SCHEDULE - 094
Crs. #(units) Call # Sec #Course/Section Title Time DaysInstructor Bldg. Room #Crs. #(units) Call # Sec #Course/Section Title Time DaysInstructor Bldg. Room #NAVAL SCIENCE (N S )Department Head: Colonel Thomas C. Abel, Professor of N
Arizona - SECTION - 505
An alternative form of the Mie-Grneisen equation of state: We can write the Grneisen parameter in the general form:V (T , S ) ln T - =- T (V , S ) ln V Sin Jacobian notation. We can expand this Jacobian by introducing a new variable, th
Ohio State - STAT - 763
Functions on a finite domainGeneral Perspective on Penalized Least Squares MethodsTwo illustrative examples Domain: X = {1, . . . , k}. Consider F = {f |f : X R}. Let fj = f (j) for j = 1, . . . , k. Each f F is represented by f = (f1 , . .
Ohio State - STAT - 763
age log.income 21 11.156322 12.813122 13.096022 11.695222 11.532722 12.765722 12.587922 11.982922 13.458823 12.206123 12.043623 11.925023 13.700123 12.793923 13.347123 13.056223 12.0