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# Calculus and Analytic Geometry MATH 222

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MATH 222 SECOND SEMESTER CALCULUS Fall 2010 Math 222 2nd Semester Calculus Lecture notes ver on 16FaH 2010 This is a self contained set of lecture notes for Math 222 The notes were written by Sigurd Angenent starting from an extensive collection of notes and problems compiled by Joel Robbin The EM files7 as well as the XFIG and OCTAVE les which were used to produce these notes are available at the following web site wwwmathwisc edu angenentFreeLectureNotes They are meant to be freely available for noncommercial use in the sense that free software77 is free More precisely Copyright c 2006 Sigurd B Angenent Permission is granted to copy distribute or modify this document under the terms of the GNU Free Documentation License Version 12 or any later version published by the Free Software Foundation with no Invariant Sections no FronteCover Texts and no BackeCover Texts A copy of the license is included in the section entitled 7 GNU Free Documentation License CONTENTS Chapter 1 Methods of Integration 5 1 The inde nite integral 5 2 You can always check the answer 6 3 About 0 6 4 Standard Integrals 7 5 Method of substitution 7 6 The double angle trick 9 7 Integration by Parts 9 8 Reduction Formulas 11 9 Partial Fraction Ebrpansion 14 10 PROBLEMS 18 Chapter 2 Taylor s Formulaand In nite Series 26 11 Taylor Polynomials 26 12 Examples 27 13 Some special Taylor polynomials 31 14 The Remainder Term 31 15 Lagrange s Formula for the Remainder Term 33 16 The limit as x a 07 keeping 71 xed 35 17 The limit 71 a 00 keeping x xed 42 18 Convergence of Taylor Series 45 19 Leibniz7 formulas for ln2 and 7r4 47 20 Proof of Lagrange s formula 48 21 Proof of Theorem 168 49 22 PROBLEMS 49 Chapter 3 Complex Numbers and the Complex Exponential 55 23 Complex numbers 55 24 Argument and Absolute Value 56 25 Geometry of Arithmetic 57 26 Applications in Trigonometry 59 27 Calculus of complex valued functions 60 28 The Complex Exponential Function 60 29 Complex solutions of polynomial equations 62 30 Other handy things you can do With complex numbers 64 31 PROBLEMS 66 Chapter 4 Differential Equations 70 32 What is a DiHEq 7O 33 First Order Separable Equations 70 34 First Order Linear Equations 71 35 Dynamical Systems and Determinism 73 36 Higher order equations 75 37 Constant Coef cient Linear Homogeneous Equations 76 38 Inhomogeneous Linear Equations 79 39 Variation of Constants 8O 40 Applications of Second Order Linear Equations 83 41 PROBLEMS 87 Chapter 5 Vectors 93 42 Introduction to Vectors 93 43 Parametric equations for lines and planes 98 44 Vector Bases 100 45 Dot Product 101 46 Cross Product 108 47 A few applications of the cross product 111 48 Notation 114 49 PROBLEMS 114 Chapter 6 Vector Functions and Parametrized Curves 120 50 Parametric CurVes 120 51 Examples of parametrized curVes 121 52 The derivative of a Vector function 123 53 Higher deriVatiVes and product rules 124 54 Interpretation of it as the Velocity Vector 125 55 Acceleration and Force 127 56 Tangents and the unit tangent Vector 129 57 Sketching a parametric curVe 131 58 Length of a curVe 133 59 The arclength function 135 60 Graphs in Cartesian and in Polar Coordinates 136 61 PROBLEMS 137 GNU Free Documentation License 142 1 APPLICABILITY AND DEFINITIONS 142 2 VERBATIM COPYING 143 3 COPYING IN QUANTITY 143 4 MODIFICATIONS 143 5 COMBINING DOCUMENTS 144 6 COLLECTIONS OF DOCUMENTS 144 7 AGGREGATION WITH INDEPENDENT WORKS 144 8 TRANSLATION 144 9 TERMINATION 144 10 FUTURE REVISIONS OF THIS LICENSE 145 11 RELICENSING 145 Chapter 1 Methods of Integration 1 The inde nite integral We recall some facts about integration from rst semester calculus 11 De nition A function y Fz is called an antide39rivative of another function y if F z for all z 12 Example F1 12 is an antiderivative of 21 F2 12 2004 is also an antiderivative of 2x Ct sin2t l is an antiderivative of gt cos2t l The Fundamental Theorem of Calculus states that if a function y is continuous on an interval a g I S 127 then there always exists an antiderivative Fz of f7 and one has I lt1 fltzgt dz F02 e Fltagt The best way of computing an integral is often to nd an antiderivative F of the given function f7 and then to use the Fundamental Theorem How you go about nding an antideiivative F for some given function f is the subject 0f this chapter The following notation is commonly used for antiderivates lt2 Fltzgt fltzgtdm The integral which appears here does not have the integration bounds a and b It is called an inde nite integral7 as opposed to the integral in l which is called a de nite integral lt7s important to distinguish between the two kinds of integrals Here is a list of differences lNDEFlNlTE INTEGRAL DEFlNlTE INTEGRAL f is a function of I f is a number By de nition is any func was de ned in terms of tion ofz whose derivative is f z Riemann sums an can be inter preted as area under the graph of y fz 7 at least when gt 0 z is not a dummy variable7 for exam I is a dummy variable7 for example7 P167 f21d1 12 C and fgtdt f012zdz l7 and 012tdt 17 so t2 C are functions of diffferent vari fol gzdx fol 2tdt ables7 so they are not equa 2 You can always Check the answer Suppose you want to nd an antiderivative of a given function and after a long and messy computation which you don7t really trust you get an answer You can then throw away the dubious computation and differentiate the Fz you had found If Fz turns out to be equal to then your Fz is indeed an antiderivative and your computation isnlt important anymore 21 Example Suppose we want to nd flnzdz My cousin Bruce says it might be Fz zlnz 7 I Lets see if hes right d l Ezlnz7zzgllnz7llnz Who knows how Bruce thought of thisl7 but hes right We now know that flnzdzzlnz7zC 3 About C Let be a function de ned on some interval a S I S b If is an antiderivative of on this interval7 then for any constant C the function Fz C will also be an antiderivative of So one given function has many different antiderivatives7 obtained by adding different constants to one given antiderivative 31 Theorem If F1 and F2z are antiden39vatz39ves 0f the same function on some interval a S I g b then there is a constant C such that F1 F2 C PROOF Consider the difference Cz F1z 7 F2z Then G z 7 7 07 so that Cz must be constant Hence F1 17F2 C D for some constant It follows that there is some ambiguity in the notation f dz Two functions F1z and F2z can both equal dz without equaling each other When this happens7 they F1 and F2 differ by a constant This can sometimes lead to confusing situations7 eg you can check that 2sinzcoszdz SlHQI 2sinzcoszdz 7cos2z are both correct Just differentiate the two functions sin2 I and 7 cos2 I These two answers look different until you realize that because of the trig identity sin2 I cos2 I 1 they really only differ by a constant sin I 7cos I l To avoid this kind of confusion we will from now on never forget to include the arbitrary constant 07 in our answer when we compute an antideriv ative 1He integrated by parts 4 Standard Integrals Here is a list of the standard derivatives and hence the standard integrals everyone should know fzdzFzC zn1 zndzni10 forallny il dzlnlzlC sinzdz7coszC coszdzsinzC tanzdz7lncoszC dzarctanzC lz2 arccoszC 1 l idz arcsinzC W 2 l lsinz 7r 7r 7 C for7 ltzlt dz 7 7 cosz 7 2 nlisinz All ofthese integrals are familiar from rst semester calculus like Math 2217 except 39 1 ln for the last one You can check the last one by differentiation us1ng ln n a7 simpli es things a bit 5 Method of substitution The chain rule says that imm F ltGltzgtgt was dz so that F Gz Gz dz FGz C 51 Example Consider the function 2z sinz2 3 It does not appear in the list of standard integrals we know by heart But we do notice2 that 2z z2 3 So lets call Cz z2 37 and Fu 7cos u7 then FGz 7 cosz2 3 and dF 91 7 2 7 f 7 s1nz 3 2z 7 FGz 01 2 You will start noticing things like this after doing several examples 3 so that 21 sinz2 3 dz 7 cosz2 3 C1 The most transparent way of computing an integral by substitution is by in troducing new variables Thus to do the integral fltGltzgtgtG ltzgt dz where Fu7 we introduce the substitution u Cz7 and agree to write du dGz G dzi Then we get dz du Fu C At the end of the integration we must remember that u really stands for Cz7 so that Gamaz dz M c FltGltzgtgt or For de nite integrals this implies b fGIG I d1 FGb FGaA which you can also write as b Cb lt3 fltGltzgtgtG ltzgtdz A deu 52 Example Substitution in a de nite integral As an example we com pute 1 I 7d 0 1H2 1 using the substitution u Cz 1 121 Since du 21 dz7 the associated inde nite integral is 1 1 1 ngz dui W 1 L Edu To nd the de nite integral you must compute the new integration bounds C0 and C1 see equation If 1 runs between I 0 and z 17 then u Cz 112 runs between u l 0 l and u l 12 2 so the de nite integral we must compute is 1 2 z 1 dz idu 0 11 1 u which is in our list of memorable integralsi So we nd 1 2 I 1 1 1 21 A dziil Eduii nulliiani 6 The double angle trick If an integral contains sin2z or cos2 I then you can remove the squares by using the double angle formulas from trigonometry Recall that cos2 a sin2 a l7 COS 2a 7 sin2a cos2a and Adding these two equations gives 1 cos2 a E cos 2a 1 while substracting them gives 2 1 sin a i 17 cos2a 61 Example The following integral shows up in many contexts7 so it is worth knowing 1 cos2 1 dz E l cos21dz 7 C 2 l i I 1sin210 2 l l 7z sin21 Since sinQI 2 sinz cosz this result can also be written as l cos2zdz 3 sinzcosza If you don7t want to memorize the double angle formulas7 then you can use Complex Exponentials77 to do these and many similar integrals However7 you will have to wait until we are in 28 where this is explained 7 Integration by Parts The product rule states d 7 dF dG EFIGI d1 C95 drip and therefore7 after rearranging terms7 dG 7 d dF 7 altFltacgt0ltacgtgt e Yam F 7dr This implies the formula for integration by parts dG 7 dF dziFzGz 7 dx Czdz 14 Apply the reduction formula again now with n 2 and you get d1 pFhh l 1 22l 7 7 22l122 22 2 NH 7 I 3 1 I 1 7 12Z 1I2 arctan1 7 1 I 3 I 3 7 KW gm garctan1 Cl 9 Partial Fraction Expansion A rational function is one which is a ratio of polynomials fa 131 PM Pn11n71 1011 100 QW QdId qdard l 111 10 Such rational functions can always be integrated and the trick which allows you to do this is called a partial fraction expansion The whole procedure consists of several steps which are explained in this section The procedure itself has nothing to do with integration its just a way of rewriting rational functions lt is in fact useful in other situations such as finding Taylor series see Part 2 of these notes and computing inverse Laplace transforms77 see MATH 319 91 Reduce to a proper rational function A proper rational function is a rational function where the degree of P1 is strictly less than the degree of the method of partial fractions only applies to proper rational functionsi Fortunately therels an additional trick for dealing with rational functions that are not proper lf PQ isn7t proper ie if degreeP 2 degreeQ then you diVide P by Q with result PW MI Q00 Q95 where 51 is the quotient and R1 is the remainder after diVisioni ln practice you would do a long diVision to find 51 and wm 92 Example Consider the rational function 1 Here the numerator has degree 3 which is more than the degree of the denominator which is 2 To apply the method of partial fractions we must first do a diVision with remainderi One has 13212 121 1 1 S1 12l 1 21 2 I3 712 12 21 12 1 1 2 R1 so that 1372z27 712 1 7 x zlI271 When we integrate we get 1372z2 7z2 dezl I271dz 2 7 2 z 1 dzi 12 12 7 l The rational function which still have to integrate namely is proper ie its numerator has lower degree than its denominator 93 Partial Fraction Expansion The Easy Case To compute the par tial fraction expansion of a proper rational function you must factor the denominator Factoring the denominator is a problem as dif cult as nding all of its roots in Math 222 we shall only do problems where the denominator is already factored into linear and quadratic factors or where this factorization is easy to nd In the easiest partial fractions problems all the roots of are real numbers and distinct so the denominator is factored into distinct linear factors say 131 7 131 Q95 95 MW a2quot391 ran To integrate this function we nd constants A1 A2 i i i An so that 131 7 A1 A2 H An Qziz7a1 17a I z7an Then the integral is 131 QW One way to nd the coef cients Ai in is called the method of equating coefficientsi In this method we multiply both sides of with z 7 a1 I 7 an The result is a polynomial of degree n on both sides Equating the coef cients of these polynomial gives a system of n linear equations for A1 i i i Ani You get the Ai by solVing that system of equations dzAllnlz7a1lAglnlz7a2lAnlnlz7anlCi Another much faster way to nd the coef cients Ai is the Heaviside trick Multiply equation by z 7 ai and then plug in4 z ail On the right you are left with Ai so A mm 7 am Pltaigt QW lam W a1 39 39 39 W ai71ai ai1quot 39ai an 3 Named after OLIVER HEAVISIDE a physicist and electrical engineer in the late 19th and early 20ieth century More properly you should take the limit m 7 1239 The problem here is that equation has m 7 mi in the denominator so that it does not hold for m 1239 Therefore you cannot set w equal to mi in any equation derived from but you can take the limit m 7t 1239 which in practice is just as good 95 Partial Fraction Expansion The General Case Buckle up When the denominator contains repeated factors or quadratic factors or both the partial fraction decomposition is more complicated In the most general case the denominator can be factored in the form 4 z 7 my I 7 ankquot12 1211 011 12 bmz 0702quot Here we assume that the factors I 7 a1 7 z 7 an are all different7 and we also assume that the factors 12 1211 cl 12 bmz cm are all different It is a theorem from advanced algebra that you can always write the rational function as a sum of terms like this 131 A B1 C 5 7 Qltzgt wank ltz2bjzcjgt2 How did this sum come about For each linear factor I 7 ak in the denominator 4 you get terms A1 A2 Ak E 39 39 39 in the decomposition There are as many terms as the exponent of the linear factor that generated them For each quadratic factor I2 121 c you get terms B11C1 BgzCg BmICm 12bzc I2bzc2 HI 12bzc Again7 there are as many terms as the exponent E with which the quadratic factor appears in the denominator ln general7 you nd the constants A7 B and C by the method of equating coef cients 96 Example To do the integral 12 3 dz 121 l12 12 apply the method of equating coef cients to the form 12 7A1A2 A3 B11C1 B21C2 12zl12l2 7 z 12 11 12471 12l2i Solving this last problem will require solving a system of seven linear equations in the seven unknowns 1411421437B17 C17B27C2 A computer program like Maple can do this easily7 but it is a lot of work to do it by hand In general7 the method of equating coef cients requires solving n linear equations in n unknowns where n is the degree of the denominator 3 See Problem 98 for a worked example where the coef cients are found 0 39 Unfortunately7 in the presence of quadratic factors or repeated lin 39 39 o ear factors the Heaviside trick does not give the whole answer you 0 0 must use the method of equating coef cients Once you have found the partial fraction decomposition 83C you still have to integrate the terms which appeared The rst three terms are of the form f Az 7 a p dz and they are easy to integrate AlnlzialC a I Adz 7 A I i a 7 1 PXI 174 ifp gt 11 The next fourth term in Ex can be written as B1zC1 7 z dz 7z21 dziBlIQ 1dzClI21 B 1 lnz2 1 C1 aircta nr Cintegration const 0 While these integrals are already not very simple the integrals Bz C dz with gt 1 z2 bz c p which can appear are particularly unpleasantr If you really must compute one of these then complete the square in the denominator so that the integral takes the orm Az B d z b a2 1 After the change of variables u z b and factoring out constants you have to do the integrals du d u du ltu2 H2 a u a2gtp Use the reduction formula we found in example 814 to compute this integral An alternative approach is to use complex numbers which are on the menu for this semester If you allow complex numbers then the quadratic factors z2 bz c can be factored and your partial fraction expansion only contains terms ofthe form A z 7 ap although A and a can now be complex numbers The integrals are then easy but the answer has complex numbers in it and rewriting the answer in terms of real numbers again can be quite involved 10 PROBLEMS DEFINITE VERSUS INDEFINITE INTEGRALS 1 Compute the following three integrals Az 2dz Br2dt 0x 2dt 2 One of the following three integrals is not the salne as the other two 4 4 4 A 2 B 5 0 I m 1 1 1 BASIC INTEGRALS Which one 22 77 Prove the formula 1 1 n7 1 2 cosn1 dx isinx cos x cos x dx n 7 O n n 7r4 and use it to evaluate cos4 x 21 0 78 Prove the formula xm1lnxquot xmlnx dx m1 7771 and use it to evaluate the following integrals 79 lncdx 80 lnx2 dx xmlnx 71dx m 7 71 81 x3lnc2 dx 82 EValuate c71 lnx dx by another method Hint the solution is shortl 83 For an integer n gt 1 derive the formula tan xdc 1 1tannilxitanni2id8 n 7r4 Using this7 nd tan5 x dx by doing just one explicit integration Use the reduztion formula from example 84 to compute these integrals 84 85 lf 8639 Hm Mummy is my 87 117952 dag 88 Group problem The reduction formula from example 84 is valid for all n 7 O In particular 71 does not have to be an integer and it does not have to be positive Find a relation between 1 x2 dx and INTEGRATION OF RATIONAL FUNCTIONS by setting 71 7 Ebrpress each of the following rational functions as a polynomial plus a proper rational function See 91 for de nitions IaiI2iIi5 89 91 x3 74 x3 7 1 90 92 x2 7 1 COMPLETING THE SQUARE and let s see What its derivatives look like They are PW ao aw 1ng agxa I4 Pi a1 2121 3a3x2 4a 3 P2x 12a2 2 3a3c 3 4a4x2 P3x 123a3 2 34m P4x 1234a4 When you set 1 0 all the terms Which have a positive power of x vanish7 and you are left With the rst entry on each line7 ie P0 a0 P 0 a1 Plt2gto 2a2 P30 2 3a3 etc and in general PWO km for 0 g k g n For k 2 n l 1 the derivatives pkx all vanish of course since Pc is a polynomial of degree 71 Therefore7 if we Want P to have the same values and derivatives at x 0 of orders 1 n as the function f then We must have klak Pk0 fk0 for all k g 71 Thus NR0 ak forO kgn 12 Examples Note that the zeroth order Taylor polynomial is just a constant7 TrifW 1 01 While the rst order Taylor polynomial is THW NO f aI a This is exactly the linear approximation of for x close to a which Was derived in 1st semester calculus The Taylor polynomial generalizes this rst order approximation by providing higher order approximations to y TofW Figure 1 The Taylor polynomials of degree 0 1 and 2 of fx e at a 0 The zeroth order Taylor polynomial has the right value at x 0 but it doesn39t know whether or not the function f is increasing at x 0 The first order Taylor polynomial has the right slope at x 0 but it doesn t see if the graph of f is curved up or down at x 0 The second order Taylor polynomial also has the right curvature at x 0 Tofx T1fI T2fx 1 x xZ 1 1 x The graphs are found in Figure 2 As you can see from the graphs the Taylor polynomial Tofx of degree 0 is close to e for small x by Virtue of the continuity of e The Taylor polynomial of degree 0 ie Tofx 1 captures the fact that e by Virtue of its continuity does not change Very much if x stays close to x O The Taylor polynomial of degree 1 ie T1fx 1 x corresponds to the tangent line to the graph of e and so it also captures the fact that the function is increasing near x 0 Clearly T1fx is a better approximation to 5quot than Tofx The graphs of both y Tofx and y T1fx are straight lines while the graph 0 y e is curVed in fact convex The second order Taylor polynomial captures this convexity In fact the graph of y T2fx is a parabola and since it has the same rst and second deriVatiVe at 0 its curVature is the same as the curVature of the graph of y e 0 at x So it seems that y T2fx 1 x x22 is an approximation to y e which beats both Tofx and T1fc 122 Example Find the Taylor polynomials of sinx When you start computing the deriVatiVes of sinx you nd sinx cosx f x isinx f3x icosx and thus f4x sinx So after four deriVatiVes you re back to where you started and the sequence of deriVatiVes of sinx cycles through the pattern sinx cosx isinx 7cos x sinx cos x isinx 7cos x sinx on and on At x 0 you then get the following Values for the deriVatiVes fj0 J 1112131415161718139 flt 010111011101llolill39 This giVes the following Taylor polynomials Tofx O T1fx x T2fx x 3 T3fx x 7 a T4fx x 7 a 5 T5fxxiy Note that since f20 O the Taylor polynomials T1fx and T2fx are the same The second order Taylor polynomial in this example is really only a polynomial of degree 1 In general the Taylor polynomial Tnfx of any function is a polynomial of degree at most 71 and this example shows that the degree can sometimes be strictly less T1f T5f TefW y sin x 7r 27r 727r 77quot T3fx T7fc T11fx Figure 3 Taylor polynomials of fx sinx 123 Example 7 Compute the Taylor polynomials of degree two and three of 1 x x2 x3 at a 3 Solution Remember that our notation for the 71th degree Taylor polynomial of a function f at a is T x and that it is de ned by We have fx1 2x 3x2 f x 2 6x fHx 6 Therefore f3 40 f 3 34 f 3 20 f 3 6 and thus 8 Tme 40 34x e 3 x e 32 40 34x e 3 10x e 3 Why don t we expand the answer You could do this ie replace x 7 32 by x2 7 6x 9 throughout and sort the powers of x but as we will see in this chapter the Taylor polynomial is used as an approximation for when x is close to a In this example is to be used when x is close to 3 If x 7 3 is a small number then the successive powers x 7 37 x 7 327 x 7 337 decrease rapidly7 and so the terms in 8 are arranged in decreasing order We can also compute the third degree Taylor polynomial It is T fac 40 34x e 3 27 e 32 go 7 33 40 34x e 3 10x e 32 x e 3 If you expand this this takes a little work you nd that 40347310I732x7331Ix2x3 So the third degree Taylor polynomial is the function f itself Why is this so Because of Theorem 112 Both sides in the above equation are third degree polynomials and their derivatives of order 07 1 2 and 3 are the same at x 37 so they must be the same polynomial 142 Example If sinx then We have found that T3fx I 7 gas so that R3sinx sinx 7 x gas This is a completely correct formula for the remainder term but it s rather useless there s nothing about this expression that suggests that x 7 133 is a much better approximation to sinx than say x x The usual situation is that there is no simple formula for the remainder term 143 An unusual example in which there is a simple formula for Rnflt Consider 17 x 3x2 7 Then you nd T2fx 1 7 x 3x2 so that R2fx 7 T2fc 715 x3 The moral of this example is this Given a polynomial you nd its nth degree Taylor polynomial by taking all terms of degree g n in the remainder Ruf then consists of the remaining terms 144 Another unusual but important example where you can compute Ruf Consider the function f a Then repeated differentiation giVes 2 7 f am 17w fw my and thus Consequently NR0 n iflt gtltogt 7 1 n and you see that the Taylor polynomials of this function are really simple namely Tnfx 1xx2x3x4 x But this sum should be really familiar it is just the Geometric Sum each term is x times the previous term lts sum is giVen by5 7 n1 Tums1zz2z3z4zquot11 7 x Which We can rewrite as The remainder term therefore is Rnf fI7Tnf 5Multiply both sides with 1 7 m to Verify this in case you had forgotten the formula 16 The limit as x gt O keeping 71 xed 161 Little 0h Lagrange s formula for the remainder term lets us write a function y which is de ned on some interVal containing x O in the following way fWo NR0 n flt 1gtlt5 2v 2m n I n1 1 11 NE f0f 0I The last term contains the 5 from Lagrange s theorem which depends on x and of which you only know that it lies between 0 and x For many purposes it is not necessary to know the last term in this much detail 7 often it is enough to know that in some sense77 the last term is the smallest term in particular as x a 0 it is much smaller than x or 2 or or x 162 Theorem If the 71 1st derivative fltn1 is continuous at x 0 then the remainder term Rnflt13gt fltn1 x 1n 1 satis es lim 7R xHO xk 0 for anyk0l2ri PROOF Since 5 lies between 0 and x one has 11111an fltn1 fltn10 and therefore n1 um RquotfI1inJfquot L k InngmW k flt 1gtltogt o0 36 x we xHO xk So we can rephrase 11 by saying 2 R f John f n50 x remainder fI N f 0I where the remainder is much smaller than N unil x2 x or 1 In order to express the condition that some function is much smaller than xquot at least for Very small x Landau introduced the following notation which many people nd usefu 163 De nition ox is an abbreviation for any function which satis es lim W 0 wHO M So you can rewrite 11 as 2 W M f0f 0x f 25mm f nfohuow The nice thing about Landau s littleoh is that you can compute with it as long as you obey the following at rst sight rather strange rules which will be proved in class in 0xm o IWFPW x ox ifriltni ifriltni for any constant C Figure 4 How the powers stack up All graphs of y x n gt 1 are tangent to the xaxis at the origin But the larger the exponent n the flatter the graph of y xquot is 164 Example rove one of these little oh rules Let s do the rst one ie let s show that xquot 0xm is 0xquotm as x gt 0 Remember if someone Writes xquot 0xm then the 0xm is an abbreviation for some function which satis es 1mx 0hlt xm 0 So the x 0xm we are given here really is an abbreviation for We then have n lim m lim M 0 since 0xm xHO xquot m xHO xm 165 Can you see that x3 0x2 by looking at the graphs of these func tions A picture is of course never a proof but have a look at gure 4 which shows you the graphs of y x x2 x3 x4 x5 and x10 As you see when x approaches 0 the graphs of higher powers of x approach the xaxis much faster than do the graphs of lower powers You should also have a look at gure 5 which exhibits the graphs of y x2 as well as several linear functions y Ox with O g an ii For each of these linear functions one has x2 lt Ox if x is small enough how small is actually small enough depends on O The smaller the constant O the closer you have to keep x to O to be sure that x2 is smaller than Ox Nevertheless no matter how small O is the parabola will eventually always reach the region below the line y Ox 166 Example Little oh arithmetic is a little funny Both x2 and x3 are functions which are 0x ie x2 0x and x3 0x Nevertheless x2 7 x3 So in working with littleoh we are giving up on the principle that says that two things which both equal a third object must themselves be equal in other words a b and b 0 implies a c but not when you re using littleohsl You can also put it like this just because two quantities both are much smaller than x they don t have to be equal In particular you can never cancel little ohs Figure 5 x2 is smaller than any multiple of x if x is small enough Compare the quadratic function y x2 with a linear function y Cm Their graphs are a parabola and a straight line Parts of the parabola may lie above the line but as x 0 the parabola will always duck underneath the line In other words the following is pretty wrong 0x2 7 0x2 0 Why The two 0x2 s both refer to functions which satisfy limw o 07 but there are many such functions7 and the two 0x2 s could be abbreviations for different functions Contrast this with the following computation7 which at rst sight looks wrong even though it is actually right 0132 7 0x2 0x2 ln words if you subtract two quantities both of which are negligible compared to x2 for small x then the result will also be negligible compared to x for sma x 167 Computations with Taylor polynomials The following theorem is Very useful because it lets you compute Taylor polynomials of a function without differentiating it 168 Theorem If and gx are 71 1 times dz erentiable functions than 12 Tnf Tngm ltgt f 996 096quot In other words7 if two functions have the same nth degree Taylor polynomial7 then their difference is much smaller than xquot at least7 if x is smal In principle the de nition of Tnfx lets you compute as many terms of the Taylor polynomial as you want but in many most examples the computations quickly get out of hand To see what can happen go though the following example 169 How NOT to compute the Taylor polynomial of degree 12 of ll x2 Diligently computing derivatives one by one you nd ag 17 so f0 1 fi so fO 0 6 2 7 2 fI1723 sof072 7 3 f3x 24 so f30 0 few 24 so flt4gto 7 24 7 4 f5ac 240Zg3 5 so f40 0 few 420 so flt4gto 720 6 I m getting tired of differentiating 7 can you nd f12x7 After a lot of Work We give up at the sixth derivative and all We have found is 1 7 7 2 47 6 T61271 xx x BytheWay 1 7 7893 715 x4 7 1716906 1287908 7 286mm 139512 12 47nnn12nn f 1 1213 and 479001600 12 1610 The right approach to nding the Taylor polynomial of any degree of ll x2 Start With the Geometric Series if gt 11 7 t then gm1tt2t3t4tquototquot Now substitute t 7x2 in this limit7 97x2 1 7x2 x4 7 x6 71 132 0 7x2 Since 0 7x2n 0amp2quot and We have found 1772 476 7n2n 2 7177271 xx lx 0c By Theorem 168 this implies 1 7 7 2 47 6 7 n 2 T2n1271 xx x 1c 40 Multiply both sides with 1 7 x 7 x2 and you get 1 17x7x2co01x02x2 cnox x70 co cw C2902 auxquot 090quot 7 cox 7 81x2 7 7 Cn71n 0W 00132 7 7 Cn72 0W i H 0 coc17coxcg 7017coI2ca 702 7C1I3 on 7 Cn71 Cn721 n 0x x 7gt 0 Compare the coef cients of powers xk on both sides for k O7 1 n and you nd 001 c17co0 c1coL 027C17coO c2c1co2 and in general on 7 717 cuiz 0 Cn Cn71 Cn72 Therefore the coe cients of the Taylor series Toofx are exactly the Fibonacci numbers on f for 7107 123 Since it is much easier to compute the Fibonacci numbers one by one than it is to compute the derivatives of 11 7 x 7 x2 this is a better way to compute the Taylor series of than just directly from the de nition 1613 More about the Fibonacci numbers In this example you ll see a trick that lets you compute the Taylor series of any rational function You already know the trick nd the partial fraction decomposition of the given rational function Ignoring the case that you have quadratic expressions in the denominator7 this lets you represent your rational function as a sum of terms of the form I i 1 These are easy to differentiate any number of times and thus they allow you to write their Taylor series Let s apply this to the function 11 7 x 7 2 from the example 1612 First we factor the denominator 7 177x20 ltgt x2x710 ltgt 1g The number 4 g x 1618 033988 749 89 is called the Golden Ratio lt sa tis es6 gt xS The roots of our polynomial x2 a 7 1 are therefore 3 7 EPM 2 2 gt and we can factor 1 7 x 7 x2 as follows 17x7z27x2x71 7z7xc7c 7c7 6T thj 1 2 2 17 71 5 o rove suse777 V p lt25 1 1 517 5 2 1616 Example We compute the Taylor polynomial of 117132 by noting that 1 i FI Where 1 7 it Since Tn1Fx1xx2x3cn1 theorem 1615 implies that 1 7 2 3 n THW7121334 n1x 1617 Example Ebrarnplez Taylor polynomials of arctan r Let arctanx Then know that 1 l 7 f as 7 1 By substitution of t 7x2 in the Taylor polynomial of 11 7 t We had found 7 7 1 7 7 2 4 7 6 I I I 7 n 2n T2nfx7T2n1271 xx x 1c This Taylor polynomial must be the derivative of T2n1fx so We have is 5 TI I27 T2n1arctanx 137 g F 71 271 1 17 The limit 71 7gt oo keeping x xed 171 Sequences and their limits We shall call a sequence any ordered sequence of numbers 11 a2 a3 for each positive integer n We have to specify a number a 172 Examples of sequences rst few number in the sequence de nition t anin 1234 bn0 07070707 1 CF was 71777quot 775777717 Eil 1 1 1 1 12212221243 77 gmg 77 57 g7 Q7 737777 a 2n1 a a 5 7 4 7 3 7 7 79L 3 7L Sn7T2n1SLUI7I 3 1 2n1 13 3 x 3 5 The last tWo sequences are derived from the Taylor polynomials of e at x 1 and sin x at any The last example 3 really is a sequence of functions ie for every choice of x you get a di erent sequence 173 De nition A sequence of numbers 1011 converges to a limit L szor every 5 gt 0 there is a number N6 such that for all n gt NE one an 7 L lt 5 One writes lim an L n7 ltgtltgt The Taylor series at this point does 7 eelx not converge to f Figure 6 An innocent looking function with an unexpected Taylor series See example 184 which shows that even when a Taylor series of some function f converges you can t be sure that it converges to f it could converge to a different function i 1 2 If you try to compute the Taylor series of f you need its derivatives at x O of all orders These can be computed not easily and the result turns out to be that all derivatives of f vanish at x O m m m Wmmo The Taylor series of f is therefore 2 3 wax0OO7070 2 3 Clearly this series converges all terms are zero after all but instead of converging to the function we started with it converges to the function 9a 0 What does this mean for the chemical reaction rates and Arrhenius law We wanted to simplify the Arrhenius law by computing the Taylor series of at T 0 but we have just seen that all terms in this series are zero Therefore replacing the Arrhenius reaction rate by its Taylor series at T O has the effect of setting all reaction rates equal to zero 19 Leibniz formulas for ln2 and 7r4 Leibniz showed that 1 1 1 1 1 i 13 39m2 and 1 1 1 1 1 7r i 3 i 39 Z Both formulas arise by setting a 1 in the Taylor series for x x3 x4 ln1x7c7 If x3 x5 x7 arctanx7x7 g7i This is only justi ed if you show that the series actually converge which we ll do here at least for the rst of these two formulas The proof of the second is similar The following is not Leibniz7 original proof You begin with the geometric sum 1 71nixn1 2 a uni 1 xx x 1c 771 1 48 Then you integrate both sides from x O to x 1 and get 1 1 1 1 1dx 1xquot1dx 777 if 71 71n1 3 n1 01x O 1c 1 2 1 n1 7 7 1 x dx iln2 1 O 1 Use 01 xkdx Instead of computing the last integral you estimate it by saying Olt lt 1 Oltl ltlxquotldx 1 71I7 0 1c 0 n2 Hence 1 1 n lim 71 1 7 d7 0 new 0 1 x and We get 4 1 1 1 n 1 7 4 1 1xquot1dx 33201 5 g quot39lt1gt n1 71n211320ltc1gt 0 1H ln2 20 Proof of Lagrange s formula For simplicity assume I gt 0 Consider the function n gm M New f 20 fo W Ktquot1 fa Where W def f0 f 0I 90quot fI 14 K 7 In1 We have chosen this particular K to be sure that 990 0 Just by computing the derivatives you also nd that 90 g lt0gt g lt0 Wm 0 While lt15 9quot1tltn1K 7 fquot1t We noW apply Rolle s Theorem 71 times 0 since gt vanishes at t O and at t x there exists an 1 With 0 lt I1 lt x such that g 1 0 since g t vanishes at t O and at t I1 there exists an 2 With 0 lt I2 lt 1 such that g x2 0 since gIt vanishes at t O and at t 2 there exists an is With 0 lt I3 lt 2 such that g x3 O 0 since gltnt vanishes at t O and at t in there exists an xn1 With 0 lt xn1 lt x such that gltnxn1 0 We noW set 5 In1 and observe that We have shown that gltn1 07 so by 15 We get n1 K f 5 n 1 Apply that to 14 and you nally get NR0 90 n x M f 0x 21 Proof of Theorem 168 211 Lemma If is a k times di erentiable function on some interval can taining O and iffor some integer k lt 71 one has 710 140 Mk7 0 0 then hc 7 hltkgt0 16 215 xk k PROOF Just apply l Hopital s rule k times You get m mh mi y Wm 1m acAO xk xHO kxk l acAO kk 71ck 2 oxo H lim h ankUciDHQIl kk7121 H llolo 3 First de ne the function lf and 915 are n times dilferentiable7 then so is hx The condition Tnfx Tngc means that M 90 f 0 M m NW Wm which says7 in terms of hx f 110 h lt0gt W0 Wm 0 Le Tum 0 We now prove the rst pat of the theorem suppose and gc have the same nth degree Taylor polynomial Then we have just argued that Tnhx O7 and Lemma 211 with k 71 says that limxHo 07 as claimed To conclude we show the conVerse also holds So suppose that limx o O We ll show that follows If were not true then there would be a smallest integer k S n such that h0 h 0 W0 W Wo 0 but hk0 7i 0 This runs into the following contradiction with Lemma 211 Wu 4 he 4 he as 4 k alliltlj xk allgllj xquot 39ITOILII 0 M Here the limit G exists because 71 Z k 22 PROBLEMS TAYLOR S FORMULA

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