Polynomial Approximations for the Natural Logarithm and Arctangent Functions. Math 230


 Allan Arnold
 1 years ago
 Views:
Transcription
1 Polynomil Approimtions for the Nturl Logrithm nd Arctngent Functions Mth 23 You recll from first semester clculus how one cn use the derivtive to find n eqution for the tngent line to function t given point on its grph. For emple, to find n eqution for the tngent line to the grph of y = f() t the point where =, you first find f (). This is the slope of the tngent line, so n eqution for the tngent line in slopeintercept form is y = f () + f(). We cll the function L() = f() + f () the lineriztion of f() t the point where =. Its grph is the tngent line, so this lineriztion, which is polynomil of degree (t most) one, is the best pproimtion of its kind for the function f() if we look t vlues of tht re close to. Of course, one cn t epect line to be very good pproimtion to grph in generl, but one would epect tht grphs of higher degree polynomils (prbols, cubic curves, etc.) could give better pproimtions. Thus, we would like to find higher degree polynomils tht give good pproimtions to functions. We do ectly tht here for few specil functions. We strt with the formul n = n+. This formul (perhps you lredy sw it derived in clss) is vlid for ll nd ll nonnegtive integers n. We cn rewrite this formul s n = n+. Now we see tht the polynomils n (remember n cn be ny nonnegtive integer here) re good pproimtions to the function s long s the bsolute vlue of the difference, tht is, n+, is smll. Clerly this is the cse if itself is reltively smll nd the eponent n + is reltively lrge. For emple, ssuming /2 nd n is t lest, we know tht the polynomil n pproimtes n+ = n+ (/2) /2 with n error no lrger thn = 2.
2 So we hve successfully found polynomils of higher degree tht re good pproimtions to the function, t lest when <. Using the sme formul bove, we cn lso find polynomils tht pproimte ln( + ) nd polynomils tht pproimte rctn(). Replce the in the formul bove with t nd we get t + t 2 t ( ) n t n = + ( )n t n+. + t This formul is vlid for ll t nd ll nonnegtive integers n. Rewrite this s t + t 2 t ( ) n t n + t = ( )n t n+. + t Integrte both sides with respect to t from to (ssuming > ) nd we hve t + t 2 t ( ) n t n + t dt = ( ) n t n+ dt, + t n+ ( ) n t n+ + + ( )n ln( + ) = dt. 4 n + + t Now in order to show tht the polynomils n+ + + ( )n 4 n + re good pproimtions to ln( + ), we need to show tht the bsolute vlue of the differences, ( ) n t n+ dt + t, re smll. Recll from first semester clculus the generl inequlity b). Also recll the inequlity f(t) dt f(t) dt f(t) dt (ssuming g(t) dt if f(t) g(t) for ll t between nd b. We will use both of these inequlities. We will use = nd b =, so we need to ssume for now tht. We will use f(t) = ( )n t n+. Even though is vrible, when integrting with + t respect to t we cn tret it like constnt. Since we re ssuming, we cn sfely ssert tht f(t) n+ + t for ll vlues of t between nd. This mens tht we cn use g(t) = n+ + t.
3 Using both our inequlities with our choices for, b, f(t) nd g(t) long with the eqution bove, we get the following: ( )n n+ (n + ) ln( + ) = ( ) n t n+ dt + t t n+ + t dt n+ + t dt = n+ ln( + ). This is vlid ssuming. If < < we need to use = nd b =. We cn use the sme f(t), but now use g(t) = n+. Doing this results in the inequlity + t ( )n n+ ln( + ) (n + ) n+ ln( + ) when < < (notice tht the logrithm here is negtive). We re lmost done now. From erlier this semester, we know tht ln(y) y whenever y. Now plugging y = + into our inequlity when shows tht ( )n n+ ln( + ) (n + ) n+ ( + ) whenever nd n is ny nonnegtive integer. For the cse where < < we use the fct tht ln(y) = ln(/y) /y whenever < y <. Plugging y = + into our second inequlity when < < shows tht ( )n n+ (n + ) ln( + ) n+ + whenever < <. We hve now found polynomils tht pproimte ln( + ) when < <. More importntly, the qulity of the pproimtions is quntified. For emple, if we re interested in between nd /2, we see tht ( )n n+ (n + ) ln( + ) n+ ( + ) (/2) n+ ( + /2) = 3 2 n+2.
4 Notice tht 3/2 n+2 is very smll when n is lrge, so our pproimtion gets better when n gets lrger. In prticulr, if we wnt to pproimte ln( + ) to within. = 5 for ll between 3 nd /2, we first check tht 2 < 5. This mens tht we cn use n = 8 nd get ln( + ) 8 < 5. When you compre this with using Simpson s rule (for emple) to pproimte ln( + ) = + with n error less thn 5, you cn strt to see the utility of our polynomil pproimtions. We cn lso tie this in with infinite sequences. Plug = /2 into our polynomils of degree t dt, 2, 3,... bove nd we get n infinite sequence of numbers /2, /2 /8, /2 /8 + /24,... tht converges to ln(3/2). Agin, we not only know tht this sequence converges to ln(3/2), we even hve n upper bound on the bsolute vlue of the difference between ln(3/2) nd ny prticulr number in this sequence. We cn use similr process to produce polynomils tht pproimte the inverse tngent function. Strt with our originl formul nd replce with t 2 to get n = n+ t 2 + t 4 t ( ) n t 2n = + ( )n t 2n+2 + t 2. This is vlid for ll vlues of t, since t 2 cn never equl (becuse it s negtive). As we did before, we cn rewrite this eqution s t 2 + t 4 t ( ) n t 2n + t 2 = ( )n t 2n+2 + t 2. Agin s we did before, we integrte both sides of this eqution from to to get t 2 + t 4 t ( ) n t 2n + t 2 dt = ( ) n t 2n+2 + t 2 dt,
5 so fter integrting n+ + + ( )n 7 2n + rctn() = ( ) n t 2n+2 + t 2 dt. Unlike wht we hd with ln( + ), this eqution is vlid for ll vlues of, not just >. In order for the polynomils to pproimte the rctngent, though, we need the integrl on the righthnd side to be smll. This requires to be less thn. We cn estimte in mnner very similr to wht we did before nd end up with the inequlity n+ + + ( )n 7 2n + rctn() 2n+3 rctn() 2n + 3 Estimting the righthnd side is not so difficult; we just use rctn() < π/2. This gives us the pproimtion n+ + + ( )n 7 2n + rctn() π 2n+3 4n + 6. This is even simpler to use thn our pproimtions for ln( + ). Here it s quite pprent tht the righthnd side is very smll if < nd n is big. For emple, suppose we wnt to pproimte rctn() with n error no lrger thn 5 for ll in the intervl [ /2, /2]. We check tht whenever /2. π < 5, so using n = 6 gives us rctn() < 5. We cn gin tie this into sequences. Plugging = / 3 (which is less thn ) into our polynomils of degree, 3, 5,... gives us sequence tht converges to rctn(/ 3) = π/6: 3, 3 9 3, ,... Agin, we not only know tht this sequence converges to π/6, we lso hve n upper bound on the difference between ny of the numbers in this sequence nd π/6.
The Regulated and Riemann Integrals
Chpter 1 The Regulted nd Riemnn Integrls 1.1 Introduction We will consider severl different pproches to defining the definite integrl f(x) dx of function f(x). These definitions will ll ssign the sme vlue
More informationGoals: Determine how to calculate the area described by a function. Define the definite integral. Explore the relationship between the definite
Unit #8 : The Integrl Gols: Determine how to clculte the re described by function. Define the definite integrl. Eplore the reltionship between the definite integrl nd re. Eplore wys to estimte the definite
More informationGoals: Determine how to calculate the area described by a function. Define the definite integral. Explore the relationship between the definite
Unit #8 : The Integrl Gols: Determine how to clculte the re described by function. Define the definite integrl. Eplore the reltionship between the definite integrl nd re. Eplore wys to estimte the definite
More informationapproaches as n becomes larger and larger. Since e > 1, the graph of the natural exponential function is as below
. Eponentil nd rithmic functions.1 Eponentil Functions A function of the form f() =, > 0, 1 is clled n eponentil function. Its domin is the set of ll rel f ( 1) numbers. For n eponentil function f we hve.
More informationChapter 6 Notes, Larson/Hostetler 3e
Contents 6. Antiderivtives nd the Rules of Integrtion.......................... 6. Are nd the Definite Integrl.................................. 6.. Are............................................ 6. Reimnn
More informationFundamental Theorem of Calculus
Fundmentl Theorem of Clculus Recll tht if f is nonnegtive nd continuous on [, ], then the re under its grph etween nd is the definite integrl A= f() d Now, for in the intervl [, ], let A() e the re under
More informationARITHMETIC OPERATIONS. The real numbers have the following properties: a b c ab ac
REVIEW OF ALGEBRA Here we review the bsic rules nd procedures of lgebr tht you need to know in order to be successful in clculus. ARITHMETIC OPERATIONS The rel numbers hve the following properties: b b
More informationMathematics Number: Logarithms
plce of mind F A C U L T Y O F E D U C A T I O N Deprtment of Curriculum nd Pedgogy Mthemtics Numer: Logrithms Science nd Mthemtics Eduction Reserch Group Supported y UBC Teching nd Lerning Enhncement
More informationMath 131. Numerical Integration Larson Section 4.6
Mth. Numericl Integrtion Lrson Section. This section looks t couple of methods for pproimting definite integrls numericlly. The gol is to get good pproimtion of the definite integrl in problems where n
More information5.7 Improper Integrals
458 pplictions of definite integrls 5.7 Improper Integrls In Section 5.4, we computed the work required to lift pylod of mss m from the surfce of moon of mss nd rdius R to height H bove the surfce of the
More informationand that at t = 0 the object is at position 5. Find the position of the object at t = 2.
7.2 The Fundmentl Theorem of Clculus 49 re mny, mny problems tht pper much different on the surfce but tht turn out to be the sme s these problems, in the sense tht when we try to pproimte solutions we
More informationMath 360: A primitive integral and elementary functions
Mth 360: A primitive integrl nd elementry functions D. DeTurck University of Pennsylvni October 16, 2017 D. DeTurck Mth 360 001 2017C: Integrl/functions 1 / 32 Setup for the integrl prtitions Definition:
More informationSection 5.1 #7, 10, 16, 21, 25; Section 5.2 #8, 9, 15, 20, 27, 30; Section 5.3 #4, 6, 9, 13, 16, 28, 31; Section 5.4 #7, 18, 21, 23, 25, 29, 40
Mth B Prof. Audrey Terrs HW # Solutions by Alex Eustis Due Tuesdy, Oct. 9 Section 5. #7,, 6,, 5; Section 5. #8, 9, 5,, 7, 3; Section 5.3 #4, 6, 9, 3, 6, 8, 3; Section 5.4 #7, 8,, 3, 5, 9, 4 5..7 Since
More informationTopics Covered AP Calculus AB
Topics Covered AP Clculus AB ) Elementry Functions ) Properties of Functions i) A function f is defined s set of ll ordered pirs (, y), such tht for ech element, there corresponds ectly one element y.
More information1 Functions Defined in Terms of Integrals
November 5, 8 MAT86 Week 3 Justin Ko Functions Defined in Terms of Integrls Integrls llow us to define new functions in terms of the bsic functions introduced in Week. Given continuous function f(), consider
More informationThe First Fundamental Theorem of Calculus. If f(x) is continuous on [a, b] and F (x) is any antiderivative. f(x) dx = F (b) F (a).
The Fundmentl Theorems of Clculus Mth 4, Section 0, Spring 009 We now know enough bout definite integrls to give precise formultions of the Fundmentl Theorems of Clculus. We will lso look t some bsic emples
More informationAP Calculus Multiple Choice: BC Edition Solutions
AP Clculus Multiple Choice: BC Edition Solutions J. Slon Mrch 8, 04 ) 0 dx ( x) is A) B) C) D) E) Divergent This function inside the integrl hs verticl symptotes t x =, nd the integrl bounds contin this
More information7.2 The Definite Integral
7.2 The Definite Integrl the definite integrl In the previous section, it ws found tht if function f is continuous nd nonnegtive, then the re under the grph of f on [, b] is given by F (b) F (), where
More informationMORE FUNCTION GRAPHING; OPTIMIZATION. (Last edited October 28, 2013 at 11:09pm.)
MORE FUNCTION GRAPHING; OPTIMIZATION FRI, OCT 25, 203 (Lst edited October 28, 203 t :09pm.) Exercise. Let n be n rbitrry positive integer. Give n exmple of function with exctly n verticl symptotes. Give
More informationMath& 152 Section Integration by Parts
Mth& 5 Section 7.  Integrtion by Prts Integrtion by prts is rule tht trnsforms the integrl of the product of two functions into other (idelly simpler) integrls. Recll from Clculus I tht given two differentible
More informationMath Solutions to homework 1
Mth 75  Solutions to homework Cédric De Groote October 5, 07 Problem, prt : This problem explores the reltionship between norms nd inner products Let X be rel vector spce ) Suppose tht is norm on X tht
More information1 Part II: Numerical Integration
Mth 4 Lb 1 Prt II: Numericl Integrtion This section includes severl techniques for getting pproimte numericl vlues for definite integrls without using ntiderivtives. Mthemticll, ect nswers re preferble
More information1 The Riemann Integral
The Riemnn Integrl. An exmple leding to the notion of integrl (res) We know how to find (i.e. define) the re of rectngle (bse height), tringle ( (sum of res of tringles). But how do we find/define n re
More information2 b. , a. area is S= 2π xds. Again, understand where these formulas came from (pages ).
AP Clculus BC Review Chpter 8 Prt nd Chpter 9 Things to Know nd Be Ale to Do Know everything from the first prt of Chpter 8 Given n integrnd figure out how to ntidifferentite it using ny of the following
More informationImproper Integrals. Type I Improper Integrals How do we evaluate an integral such as
Improper Integrls Two different types of integrls cn qulify s improper. The first type of improper integrl (which we will refer to s Type I) involves evluting n integrl over n infinite region. In the grph
More informationThe Trapezoidal Rule
_.qd // : PM Pge 9 SECTION. Numericl Integrtion 9 f Section. The re of the region cn e pproimted using four trpezoids. Figure. = f( ) f( ) n The re of the first trpezoid is f f n. Figure. = Numericl Integrtion
More informationP 3 (x) = f(0) + f (0)x + f (0) 2. x 2 + f (0) . In the problem set, you are asked to show, in general, the n th order term is a n = f (n) (0)
1 Tylor polynomils In Section 3.5, we discussed how to pproximte function f(x) round point in terms of its first derivtive f (x) evluted t, tht is using the liner pproximtion f() + f ()(x ). We clled this
More informationTaylor Polynomial Inequalities
Tylor Polynomil Inequlities Ben Glin September 17, 24 Abstrct There re instnces where we my wish to pproximte the vlue of complicted function round given point by constructing simpler function such s polynomil
More informationZ b. f(x)dx. Yet in the above two cases we know what f(x) is. Sometimes, engineers want to calculate an area by computing I, but...
Chpter 7 Numericl Methods 7. Introduction In mny cses the integrl f(x)dx cn be found by finding function F (x) such tht F 0 (x) =f(x), nd using f(x)dx = F (b) F () which is known s the nlyticl (exct) solution.
More information6.5 Numerical Approximations of Definite Integrals
Arknss Tech University MATH 94: Clculus II Dr. Mrcel B. Finn 6.5 Numericl Approximtions of Definite Integrls Sometimes the integrl of function cnnot be expressed with elementry functions, i.e., polynomil,
More information63. Representation of functions as power series Consider a power series. ( 1) n x 2n for all 1 < x < 1
3 9. SEQUENCES AND SERIES 63. Representtion of functions s power series Consider power series x 2 + x 4 x 6 + x 8 + = ( ) n x 2n It is geometric series with q = x 2 nd therefore it converges for ll q =
More informationW. We shall do so one by one, starting with I 1, and we shall do it greedily, trying
Vitli covers 1 Definition. A Vitli cover of set E R is set V of closed intervls with positive length so tht, for every δ > 0 nd every x E, there is some I V with λ(i ) < δ nd x I. 2 Lemm (Vitli covering)
More informationn f(x i ) x. i=1 In section 4.2, we defined the definite integral of f from x = a to x = b as n f(x i ) x; f(x) dx = lim i=1
The Fundmentl Theorem of Clculus As we continue to study the re problem, let s think bck to wht we know bout computing res of regions enclosed by curves. If we wnt to find the re of the region below the
More informationf(x i ) x. i=1 In section 4.2, we defined the definite integral of f from x = a to x = b as f(x) dx = lim f(x i ) x; i=1
The Fundmentl Theorem of Clculus As we continue to study the re problem, let s think bck to wht we know bout computing res of regions enclosed by curves. If we wnt to find the re of the region below the
More informationOverview of Calculus I
Overview of Clculus I Prof. Jim Swift Northern Arizon University There re three key concepts in clculus: The limit, the derivtive, nd the integrl. You need to understnd the definitions of these three things,
More informationMath 1B, lecture 4: Error bounds for numerical methods
Mth B, lecture 4: Error bounds for numericl methods Nthn Pflueger 4 September 0 Introduction The five numericl methods descried in the previous lecture ll operte by the sme principle: they pproximte the
More informationUnit #9 : Definite Integral Properties; Fundamental Theorem of Calculus
Unit #9 : Definite Integrl Properties; Fundmentl Theorem of Clculus Gols: Identify properties of definite integrls Define odd nd even functions, nd reltionship to integrl vlues Introduce the Fundmentl
More informationCalculus II: Integrations and Series
Clculus II: Integrtions nd Series August 7, 200 Integrls Suppose we hve generl function y = f(x) For simplicity, let f(x) > 0 nd f(x) continuous Denote F (x) = re under the grph of f in the intervl [,x]
More information1.2. Linear Variable Coefficient Equations. y + b "! = a y + b " Remark: The case b = 0 and a nonconstant can be solved with the same idea as above.
1 12 Liner Vrible Coefficient Equtions Section Objective(s): Review: Constnt Coefficient Equtions Solving Vrible Coefficient Equtions The Integrting Fctor Method The Bernoulli Eqution 121 Review: Constnt
More informationHandout: Natural deduction for first order logic
MATH 457 Introduction to Mthemticl Logic Spring 2016 Dr Json Rute Hndout: Nturl deduction for first order logic We will extend our nturl deduction rules for sententil logic to first order logic These notes
More informationReview of Calculus, cont d
Jim Lmbers MAT 460 Fll Semester 200910 Lecture 3 Notes These notes correspond to Section 1.1 in the text. Review of Clculus, cont d Riemnn Sums nd the Definite Integrl There re mny cses in which some
More information5.4, 6.1, 6.2 Handout. As we ve discussed, the integral is in some way the opposite of taking a derivative. The exact relationship
5.4, 6.1, 6.2 Hnout As we ve iscusse, the integrl is in some wy the opposite of tking erivtive. The exct reltionship is given by the Funmentl Theorem of Clculus: The Funmentl Theorem of Clculus: If f is
More informationTopic 1 Notes Jeremy Orloff
Topic 1 Notes Jerem Orloff 1 Introduction to differentil equtions 1.1 Gols 1. Know the definition of differentil eqution. 2. Know our first nd second most importnt equtions nd their solutions. 3. Be ble
More informationUNIFORM CONVERGENCE. Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3
UNIFORM CONVERGENCE Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3 Suppose f n : Ω R or f n : Ω C is sequence of rel or complex functions, nd f n f s n in some sense. Furthermore,
More informationTHE EXISTENCEUNIQUENESS THEOREM FOR FIRSTORDER DIFFERENTIAL EQUATIONS.
THE EXISTENCEUNIQUENESS THEOREM FOR FIRSTORDER DIFFERENTIAL EQUATIONS RADON ROSBOROUGH https://intuitiveexplntionscom/picrdlindeloftheorem/ This document is proof of the existenceuniqueness theorem
More informationConvergence of Fourier Series and Fejer s Theorem. Lee Ricketson
Convergence of Fourier Series nd Fejer s Theorem Lee Ricketson My, 006 Abstrct This pper will ddress the Fourier Series of functions with rbitrry period. We will derive forms of the Dirichlet nd Fejer
More informationPhysics 116C Solution of inhomogeneous ordinary differential equations using Green s functions
Physics 6C Solution of inhomogeneous ordinry differentil equtions using Green s functions Peter Young November 5, 29 Homogeneous Equtions We hve studied, especilly in long HW problem, second order liner
More informationMath 113 Exam 2 Practice
Mth Em Prctice Februry, 8 Em will cover sections 6.5, 7.7.5 nd 7.8. This sheet hs three sections. The first section will remind you bout techniques nd formuls tht you should know. The second gives number
More informationLecture 14: Quadrature
Lecture 14: Qudrture This lecture is concerned with the evlution of integrls fx)dx 1) over finite intervl [, b] The integrnd fx) is ssumed to be relvlues nd smooth The pproximtion of n integrl by numericl
More information( dg. ) 2 dt. + dt. dt j + dh. + dt. r(t) dt. Comparing this equation with the one listed above for the length of see that
Arc Length of Curves in Three Dimensionl Spce If the vector function r(t) f(t) i + g(t) j + h(t) k trces out the curve C s t vries, we cn mesure distnces long C using formul nerly identicl to one tht we
More informationProperties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives
Block #6: Properties of Integrls, Indefinite Integrls Gols: Definition of the Definite Integrl Integrl Clcultions using Antiderivtives Properties of Integrls The Indefinite Integrl 1 Riemnn Sums  1 Riemnn
More informationMath Calculus with Analytic Geometry II
orem of definite Mth 5.0 with Anlytic Geometry II Jnury 4, 0 orem of definite If < b then b f (x) dx = ( under f bove xxis) ( bove f under xxis) Exmple 8 0 3 9 x dx = π 3 4 = 9π 4 orem of definite Problem
More informationBest Approximation. Chapter The General Case
Chpter 4 Best Approximtion 4.1 The Generl Cse In the previous chpter, we hve seen how n interpolting polynomil cn be used s n pproximtion to given function. We now wnt to find the best pproximtion to given
More informationHow do we solve these things, especially when they get complicated? How do we know when a system has a solution, and when is it unique?
XII. LINEAR ALGEBRA: SOLVING SYSTEMS OF EQUATIONS Tody we re going to tlk bout solving systems of liner equtions. These re problems tht give couple of equtions with couple of unknowns, like: 6 2 3 7 4
More informationA REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007
A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H Thoms Shores Deprtment of Mthemtics University of Nebrsk Spring 2007 Contents Rtes of Chnge nd Derivtives 1 Dierentils 4 Are nd Integrls 5 Multivrite Clculus
More informationMath 1431 Section M TH 4:00 PM 6:00 PM Susan Wheeler Office Hours: Wed 6:00 7:00 PM Online ***NOTE LABS ARE MON AND WED
Mth 43 Section 4839 M TH 4: PM 6: PM Susn Wheeler swheeler@mth.uh.edu Office Hours: Wed 6: 7: PM Online ***NOTE LABS ARE MON AND WED t :3 PM to 3: pm ONLINE Approimting the re under curve given the type
More informationImproper Integrals, and Differential Equations
Improper Integrls, nd Differentil Equtions October 22, 204 5.3 Improper Integrls Previously, we discussed how integrls correspond to res. More specificlly, we sid tht for function f(x), the region creted
More informationAbstract inner product spaces
WEEK 4 Abstrct inner product spces Definition An inner product spce is vector spce V over the rel field R equipped with rule for multiplying vectors, such tht the product of two vectors is sclr, nd the
More informationLogarithmic Functions
Logrithmic Functions Definition: Let > 0,. Then log is the number to which you rise to get. Logrithms re in essence eponents. Their domins re powers of the bse nd their rnges re the eponents needed to
More informationCalculus  Activity 1 Rate of change of a function at a point.
Nme: Clss: p 77 Mths Helper Plus Resource Set. Copright 00 Bruce A. Vughn, Techers Choice Softwre Clculus  Activit Rte of chnge of function t point. ) Strt Mths Helper Plus, then lod the file: Clculus
More informationIntegration. 148 Chapter 7 Integration
48 Chpter 7 Integrtion 7 Integrtion t ech, by supposing tht during ech tenth of second the object is going t constnt speed Since the object initilly hs speed, we gin suppose it mintins this speed, but
More informationdifferent methods (left endpoint, right endpoint, midpoint, trapezoid, Simpson s).
Mth 1A with Professor Stnkov Worksheet, Discussion #41; Wednesdy, 12/6/217 GSI nme: Roy Zho Problems 1. Write the integrl 3 dx s limit of Riemnn sums. Write it using 2 intervls using the 1 x different
More information. Doubleangle formulas. Your answer should involve trig functions of θ, and not of 2θ. sin 2 (θ) =
Review of some needed Trig. Identities for Integrtion. Your nswers should be n ngle in RADIANS. rccos( 1 ) = π rccos(  1 ) = 2π 2 3 2 3 rcsin( 1 ) = π rcsin(  1 ) = π 2 6 2 6 Cn you do similr problems?
More informationLoudoun Valley High School Calculus Summertime Fun Packet
Loudoun Vlley High School Clculus Summertime Fun Pcket We HIGHLY recommend tht you go through this pcket nd mke sure tht you know how to do everything in it. Prctice the problems tht you do NOT remember!
More informationMATH 144: Business Calculus Final Review
MATH 144: Business Clculus Finl Review 1 Skills 1. Clculte severl limits. 2. Find verticl nd horizontl symptotes for given rtionl function. 3. Clculte derivtive by definition. 4. Clculte severl derivtives
More informationWe divide the interval [a, b] into subintervals of equal length x = b a n
Arc Length Given curve C defined by function f(x), we wnt to find the length of this curve between nd b. We do this by using process similr to wht we did in defining the Riemnn Sum of definite integrl:
More informationBest Approximation in the 2norm
Jim Lmbers MAT 77 Fll Semester 111 Lecture 1 Notes These notes correspond to Sections 9. nd 9.3 in the text. Best Approximtion in the norm Suppose tht we wish to obtin function f n (x) tht is liner combintion
More informationHomework 11. Andrew Ma November 30, sin x (1+x) (1+x)
Homewor Andrew M November 3, 4 Problem 9 Clim: Pf: + + d = d = sin b +b + sin (+) d sin (+) d using integrtion by prts. By pplying + d = lim b sin b +b + sin (+) d. Since limits to both sides, lim b sin
More informationAPPROXIMATE INTEGRATION
APPROXIMATE INTEGRATION. Introduction We hve seen tht there re functions whose ntiderivtives cnnot be expressed in closed form. For these resons ny definite integrl involving these integrnds cnnot be
More informationImproper Integrals. The First Fundamental Theorem of Calculus, as we ve discussed in class, goes as follows:
Improper Integrls The First Fundmentl Theorem of Clculus, s we ve discussed in clss, goes s follows: If f is continuous on the intervl [, ] nd F is function for which F t = ft, then ftdt = F F. An integrl
More informationFact: All polynomial functions are continuous and differentiable everywhere.
Dierentibility AP Clculus Denis Shublek ilernmth.net Dierentibility t Point Deinition: ( ) is dierentible t point We write: = i nd only i lim eists. '( ) lim = or '( ) lim h = ( ) ( ) h 0 h Emple: The
More informationODE: Existence and Uniqueness of a Solution
Mth 22 Fll 213 Jerry Kzdn ODE: Existence nd Uniqueness of Solution The Fundmentl Theorem of Clculus tells us how to solve the ordinry differentil eqution (ODE) du = f(t) dt with initil condition u() =
More information5: The Definite Integral
5: The Definite Integrl 5.: Estimting with Finite Sums Consider moving oject its velocity (meters per second) t ny time (seconds) is given y v t = t+. Cn we use this informtion to determine the distnce
More informationChapter 0. What is the Lebesgue integral about?
Chpter 0. Wht is the Lebesgue integrl bout? The pln is to hve tutoril sheet ech week, most often on Fridy, (to be done during the clss) where you will try to get used to the ides introduced in the previous
More informationf a L Most reasonable functions are continuous, as seen in the following theorem:
Limits Suppose f : R R. To sy lim f(x) = L x mens tht s x gets closer n closer to, then f(x) gets closer n closer to L. This suggests tht the grph of f looks like one of the following three pictures: f
More informationImproper Integrals. Introduction. Type 1: Improper Integrals on Infinite Intervals. When we defined the definite integral.
Improper Integrls Introduction When we defined the definite integrl f d we ssumed tht f ws continuous on [, ] where [, ] ws finite, closed intervl There re t lest two wys this definition cn fil to e stisfied:
More informationThe problems that follow illustrate the methods covered in class. They are typical of the types of problems that will be on the tests.
ADVANCED CALCULUS PRACTICE PROBLEMS JAMES KEESLING The problems tht follow illustrte the methods covered in clss. They re typicl of the types of problems tht will be on the tests. 1. Riemnn Integrtion
More informationf(x) dx, If one of these two conditions is not met, we call the integral improper. Our usual definition for the value for the definite integral
Improper Integrls Every time tht we hve evluted definite integrl such s f(x) dx, we hve mde two implicit ssumptions bout the integrl:. The intervl [, b] is finite, nd. f(x) is continuous on [, b]. If one
More informationReview of basic calculus
Review of bsic clculus This brief review reclls some of the most importnt concepts, definitions, nd theorems from bsic clculus. It is not intended to tech bsic clculus from scrtch. If ny of the items below
More informationDisclaimer: This Final Exam Study Guide is meant to help you start studying. It is not necessarily a complete list of everything you need to know.
Disclimer: This is ment to help you strt studying. It is not necessrily complete list of everything you need to know. The MTH 33 finl exm minly consists of stndrd response questions where students must
More informationMath 231E, Lecture 33. Parametric Calculus
Mth 31E, Lecture 33. Prmetric Clculus 1 Derivtives 1.1 First derivtive Now, let us sy tht we wnt the slope t point on prmetric curve. Recll the chin rule: which exists s long s /. = / / Exmple 1.1. Reconsider
More informationWe know that if f is a continuous nonnegative function on the interval [a, b], then b
1 Ares Between Curves c 22 Donld Kreider nd Dwight Lhr We know tht if f is continuous nonnegtive function on the intervl [, b], then f(x) dx is the re under the grph of f nd bove the intervl. We re going
More informationRead section 3.3, 3.4 Announcements:
Dte: 3/1/13 Objective: SWBAT pply properties of exponentil functions nd will pply properties of rithms. Bell Ringer: 1. f x = 3x 6, find the inverse, f 1 x., Using your grphing clcultor, Grph 1. f x,f
More informationInterpreting Integrals and the Fundamental Theorem
Interpreting Integrls nd the Fundmentl Theorem Tody, we go further in interpreting the mening of the definite integrl. Using Units to Aid Interprettion We lredy know tht if f(t) is the rte of chnge of
More informationMath 135, Spring 2012: HW 7
Mth 3, Spring : HW 7 Problem (p. 34 #). SOLUTION. Let N the number of risins per cookie. If N is Poisson rndom vrible with prmeter λ, then nd for this to be t lest.99, we need P (N ) P (N ) ep( λ) λ ln(.)
More informationSection 6.1 INTRO to LAPLACE TRANSFORMS
Section 6. INTRO to LAPLACE TRANSFORMS Key terms: Improper Integrl; diverge, converge A A f(t)dt lim f(t)dt Piecewise Continuous Function; jump discontinuity Function of Exponentil Order Lplce Trnsform
More informationBasic Derivative Properties
Bsic Derivtive Properties Let s strt this section by remining ourselves tht the erivtive is the slope of function Wht is the slope of constnt function? c FACT 2 Let f () =c, where c is constnt Then f 0
More information2.4 Linear Inequalities and Interval Notation
.4 Liner Inequlities nd Intervl Nottion We wnt to solve equtions tht hve n inequlity symol insted of n equl sign. There re four inequlity symols tht we will look t: Less thn , Less thn or
More informationLecture 1: Introduction to integration theory and bounded variation
Lecture 1: Introduction to integrtion theory nd bounded vrition Wht is this course bout? Integrtion theory. The first question you might hve is why there is nything you need to lern bout integrtion. You
More informationCalculus 2: Integration. Differentiation. Integration
Clculus 2: Integrtion The reverse process to differentition is known s integrtion. Differentition f() f () Integrtion As it is the opposite of finding the derivtive, the function obtined b integrtion is
More informationBefore we can begin Ch. 3 on Radicals, we need to be familiar with perfect squares, cubes, etc. Try and do as many as you can without a calculator!!!
Nme: Algebr II Honors PreChpter Homework Before we cn begin Ch on Rdicls, we need to be fmilir with perfect squres, cubes, etc Try nd do s mny s you cn without clcultor!!! n The nth root of n n Be ble
More information5.5 The Substitution Rule
5.5 The Substitution Rule Given the usefulness of the Fundmentl Theorem, we wnt some helpful methods for finding ntiderivtives. At the moment, if n ntiderivtive is not esily recognizble, then we re in
More informationAntiderivatives/Indefinite Integrals of Basic Functions
Antiderivtives/Indefinite Integrls of Bsic Functions Power Rule: In prticulr, this mens tht x n+ x n n + + C, dx = ln x + C, if n if n = x 0 dx = dx = dx = x + C nd x (lthough you won t use the second
More information4 7x =250; 5 3x =500; Read section 3.3, 3.4 Announcements: Bell Ringer: Use your calculator to solve
Dte: 3/14/13 Objective: SWBAT pply properties of exponentil functions nd will pply properties of rithms. Bell Ringer: Use your clcultor to solve 4 7x =250; 5 3x =500; HW Requests: Properties of Log Equtions
More informationx ) dx dx x sec x over the interval (, ).
Curve on 6 For , () Evlute the integrl, n (b) check your nswer by ifferentiting. ( ). ( ). ( ).. 6. sin cos 7. sec csccot 8. sec (sec tn ) 9. sin csc. Evlute the integrl sin by multiplying the numertor
More informationSection 14.3 Arc Length and Curvature
Section 4.3 Arc Length nd Curvture Clculus on Curves in Spce In this section, we ly the foundtions for describing the movement of n object in spce.. Vector Function Bsics In Clc, formul for rc length in
More informationp(t) dt + i 1 re it ireit dt =
Note: This mteril is contined in Kreyszig, Chpter 13. Complex integrtion We will define integrls of complex functions long curves in C. (This is bit similr to [relvlued] line integrls P dx + Q dy in R2.)
More informationNew Expansion and Infinite Series
Interntionl Mthemticl Forum, Vol. 9, 204, no. 22, 06073 HIKARI Ltd, www.mhikri.com http://dx.doi.org/0.2988/imf.204.4502 New Expnsion nd Infinite Series Diyun Zhng College of Computer Nnjing University
More informationExam 2, Mathematics 4701, Section ETY6 6:05 pm 7:40 pm, March 31, 2016, IH1105 Instructor: Attila Máté 1
Exm, Mthemtics 471, Section ETY6 6:5 pm 7:4 pm, Mrch 1, 16, IH115 Instructor: Attil Máté 1 17 copies 1. ) Stte the usul sufficient condition for the fixedpoint itertion to converge when solving the eqution
More information