ADVANCED CALCULUS I
ADVANCED CALCULUS I MATH 409
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Math 409 502 Harold P Boas b0astamuedu Functions Informal de nition A function is given by a rule or by a formula for example f x x2 Formal de nition A function is a set of ordered pairs with the property that no first element appears more than once for example a 512 a 6 1R Math 409502 October 13 2004 i slide 2 Aside on groups In algebra a group is a set equipped with an associative binary operation for which there is an identity element and such that each element has an inverse with respect to the operation Examples The integers form a group under addition The number 0 is the identity element and the additive inverse of n is in The non zero real numbers form a group under multiplication The number 1 is the identity element and the multiplicative inverse of a is 1 u Math 409502 October 13 2004 i slide 3 Groups of functions Do the real valued functions with domain lR form a group under addition Yes the identity element is the function that is constantly equal to 0 and the inverse of f x is fx Do the non zero real valued functions with domain lR form a group under multiplication It depends on what non zero means If non zero means not equal to the function that is constantly equal to 0 then no the function f x 3 does not have a multiplicative inverse that is everywhere defined lf non zero means nowhere equal to zero then yes the function that is constantly equal to 1 is the multiplicative identity and the inverse of f x is 1 f Math 409502 October 13 2004 i slide 4 Composition Do the real valued functions with domain lR form a group under composition The identity function f x 3 serves as identity under composition But some functions lack inverses The function f x x sinx is not one to one so that function does not have an inverse under composition The function gx ex is one to one but not onto so its inverse function lnx is not everywhere defined Math 409502 October 13 2004 i slide 5 Homework 1 Read sections 101 and 102 pages 137 142 2 D0 Exercises 1012 and 1021 on page 148 Math 409502 October 13 2004 i slide 6 Math 409 502 Harold P Boas b0astamuedu Proving existence of lim an nACX 1 If the Value of the limit is not known we can try to use that a bounded monotonic sequence converges 2 If we know or can guess that an 7 L we can try to estimate the error term en 2 an 7 L Math 409502 September 13 2004 7 slide 2 Examples of error term estimation Powers aquot 7 0and m1quot 70when lul lt 1 1quot 71when0ltu 111quot71 Strategy use the binomial theorem to estimate the error term Geometric series an 1 r r2 r 1 irn1 The error term is en 2 an 7 7 7 r 1 7 r 39 so en 7 0 when M lt 1 by the first case above Math 409502 September 13 2004 7 slide 3 Examples continued Newton s method for approximating Here an an i and the error term satis es a quadratic estimate enm kn if e lt 09 1 1 3 Fibonacci fractions I pl I 5 8 WIN U IIUJ NIH The limit of the sequence is the golden ratio 7 1 NE The error term satisfies enH lt Zinle if e lt 02 Math 409502 September 13 2004 i slide 4 A puzzle 1 0 lim ngtoo 1 1 lim n 1xquot x lim n 1xquot dx H00 0 0 HHOC 1 xn1 d 1 0dx o 0 gt O i What went wrong The limit of the integral is not necessarily equal to the integral of the limit One of our goals in proving theorems about limits is to avoid mistakes like the one above Math 409502 September 13 2004 i slide 5 Homework for Wednesday 1 D0 Exercise 361 on page 47 and Exercise 441 on page 58 2 Read section 51 pages 61 64 Math 409502 September 13 2004 i slide 6 Math 409 502 Harold P Boas b0astamuedu Announcement Math Club Meeting Monday October 18 today 600 PM Blocker 156 Undergraduate speakers Dakota Blair Oscillating Patterns in Langtons Ant Ryan Westbrook New Results in Wavelet Set Theory FREE FOOD Math 409502 October 18 2004 7 slide 2 Limits of functions De nition lim f x L means that for every 6 gt 0 there exists a 6 gt 0 such that l f x 7 Ll lt 6 when x 0 lt 13 7 ul lt 6 Note that f a need not be de ned Example Prove that limx i ii 2 Suppose e gt 0 is given Set 6 min 1 If 13 7 1 lt 6 then in particular 3 gt 0 so xl lt 1 Hence ix 7 1 lt 6 implies x3 2 fl7xu 7 lt 7 lt x1 x1 71x 11lt6ie x3 N N Thus m z 2 when x E 1 as required Math 409502 October 18 2004 7 slide 3 Connection with sequences lim f x L if and only if for every sequence xn such that xn a a but xn a we have x fxn a L Consequently all our theorems for limits of sequences carry over to limits of functions Example limx o x sin1x 0 Proof since ilxl xsin1x lxl forxa O the result follows from the squeeze theorem Math 409502 October 18 2004 i slide 4 Continuity A function f is continuous at a point a if limxnu fu The formal de nition For every 6 gt 0 there exists 6 gt 0 such that l f x 7 f 51M lt 6 when ix 7 ul lt 6 The same proof we did to show that limx 1 2 proves that the function f x is continuous at x 1 Math 409502 October 18 2004 i slide 5 Homework 1 Read sections 111 and 112 pages 151 158 2 D0 Exercises 1115 and 1121 on page 167 Math 409502 October 18 2004 i slide 6 Math 409 502 Harold P Boas b0astamuedu Reminder Second examination is Monday November 1 The exam will have a similar format to the format of the first exam The exam covers material through section 134 Math 409502 October 27 2004 i slide 2 Continuity and boundedness A continuous function on an interval need not be bounded Examples 1 x on the bounded interval 0 1 x2 on the closed interval 0 00 Theorem Every continuous function f on a compact interval 51 b is bounded Proof different from the proof in the book Use from Exercise 111 5 that every continuous function is locally bounded Let S c f is bounded on the interval 51 c By S is not empty Let d sup 8 By d E S If d lt b then by some points to the right of d are in S which contradicts that d is an upper bound for 8 Therefore d b and we are done Math 409502 October 27 2004 i slide 3 Continuity and extreme values A continuous function on an interval need not attain a maximum value Examples 32 on the bounded interval 0 1 arctanx on the closed interval 0 00 Theorem Every continuous function f on a compact interval attains a maximum value and also attains a minimum value Proof by contradiction different from the proof in the book By the previous theorem f is bounded Let M sup f Suppose the supremum is not attained Then M 7 f x gt 0 for all 3 so Mil x is continuous By the previous theorem this new function has an upper bound say N Solve Mil x N to get f x g M 7 contradicting that M sup f Math 409502 October 27 2004 i slide 4 Homework 0 Read sections 133 and 134 pages 187 190 0 In preparation for the examination make a list of the main definitions concepts and theorems from sections 75 through 134 Math 409502 October 27 2004 i slide 5 Math 409 502 Harold P Boas b0astamuedu Announcement Math Club Meeting Monday November 15 at 730 PM Blocker 156 Speaker Jeff Nash Aggie former student from Veritas DGC Title Math and Seismic Imaging FREE FOOD Math 409502 November 8 2004 i slide 2 The derivative and applications The definition f a W Some applications 0 the mean value theorem 0 l39H pital39s rule 0 monotonicity extreme values Math 409502 November 8 2004 i slide 3 The meanvalue theorem generalized Cauchy s form of the meanvalue theorem Suppose f and g are continuous functions on 11 b that are differentiable on a h Then there exists a point c in a b for w ic 17240 gm ltng 3a N 7H f0 fOZ NC or I gt if the denominators are nonzero g c Proof Set Mac 7 171310 gx 7 lt8ltbg lt gt A computation shows that 1a 111 so by the ordinary meanValue theorem there is a point c for which h c 0 That reduces to the desired conclusion Math 409502 November 8 2004 7 Slide 4 A bestselling author Guillaume Francois Antoine Marquis de L Hopital 166171704 I Kiri Author of the first calculus textbook Analyse des in niment petits Math 409502 November 8 2004 7 Slide 5 L Hopital s rules If f and g are differentiable functions and if lim 2 is a formally unde ned expression of the x form 9 or E then lim m lim fltxgt if the second limit exists 0 0 xgtu 05 xau 8 x Variations one sided limits 3 a 00 1nx x Example lim is formally L39Hopital39s rule says that the limit equals lim 0 XHOO XHOO Example lirtn x lnx is formally 0 700 Rewrite as lirg L39Hopital39s rule says that the XH 36 136 7 7 szixlgg x limit equals lim xgtOJr Math 409502 November 8 2004 i slide 6 Proof of one version of L Hopital s rule Theorem Suppose f and g are differentiable for large 3 and lim f x 00 and lim gx oo XHOO XHOO lf lim f xg x 00 then lim fxgx oo XHOO XHOO Proof Fix M gt 0 We must find b such that fxgx gt M when x gt b By hypothesis there is a such that f a gt 0 and f c g c gt 2M when c 2 a Moreover there is b gt a such that lgugxl lt 12 whenx gt b Apply Cauchy39s mean Value theorem when x gt b to get fxf amp gltxgtegltugt we gt 2M Then fltx gt fx fW gt 2Mltgx g r SO fXgltx gt 2W1 gt gt 2W1 i i M Math 409502 November 8 2004 i slide 7 Homework 1 Read sections 152 154 pages 212 217 2 D0 Exercise 1432 on page 206 3 D0 Exercise 1542 on page 219 Math 409502 November 8 2004 i slide 8 Math 409 502 Harold P Boas b0astamuedu Summary of convergence tests 0 If an 74gt 0 then En an diverges 0 Comparison tests for positive series if an 3 bn for all large n or alternatively if an bn has a finite limit then convergence of En bn implies convergence of En an Absolute convergence implies convergence if En qu converges then so does En an Ratio and root tests if either lim mill 1 or lim lunH Mnl exists and is strictly less than 1 H00 HHOO then En an converges Special tests for decreasing positive terms an i if i 0 as x a 00 then foofxdx and wam have the same conver gence divergence behavior integral test ii if an i 0 then En71quotun converges al ternating series test Math 409502 October 6 2004 i slide 2 Cauchy s condensation test Another special test for decreasing terms Suppose 0 lt unH an for all large n Then the two series En an and En zn zn either both converge or both diverge Example En m Since MIR is a decreasing function of n the test applies and says that the conver gence divergence behavior is the same as for the series 212quot That simplifies t0 1 2quotln2 39 En 132 which is a multiple of the divergent harmonic series Therefore the original series En m diverges too Math 409502 October 6 2004 i slide 3 Proof of the condensation test sketch usu9u15lt 8amp8 15 1739 31 S 2u4 515 515 517 16 15 S 2ltusu9ms series have the same convergence divergence behavior Adding such inequalities shows that partial sums of En Zn zn are bounded below by partial sums of En an and are bounded above by twice the partial sums of En an Therefore the two Math 409502 October 6 2004 i slide 4 Power series Example E1 237 For which values of 3 does that series converge This is Exercise 811a on page 123 Solution By the root test the series converges absolutely when 171 1 gt 3 lim 34 i 34 271W nHOOZW i 2 11m HHOO that is when lxl lt 2 The series diverges when lxl gt 2 by the proof of the root test A different test is needed to see what happens when x i2 Math 409502 October 6 2004 i slide 5 Homework 0 Read section 81 pages 114 117 0 D0 Exercise 761ac on page 111 0 D0 Exercise 811g on page 123 Math 409502 October 6 2004 i slide 6 Math 409 502 Harold P Boas b0astamuedu Reminders on radius of convergence We can find the radius of convergence of a power series by either the ratio test or the root test but some other test is needed to determine the endpoint behavior Useful tests for endpoint behavior are 0 nth term test 0 comparison tests 0 alternating series test Math 409502 October 11 2004 i slide 2 Followup on endpoint convergence I 2 xquot has radius of convergence equal to 4 and has radius of convergence equal to 2 At the right hand endpoint both 00 Last time we saw by the ratio test that 2 711 n i nix 11352n71 oo n24n series become 2 lt2 That series diverges by the nth term test Indeed 711 4quot lt1 1w 1 21quot 2 2 21quot 1 so 4quot gt 2 E17 Thus quot23 gt 1 so the series cannot converge For the same reason divergence occurs at the left hand endpoint in this example Math 409502 October 11 2004 i slide 3 Operations on power series Addition subtraction multiplication and division of power series work the way you expect Example X2 X4 X6 COSltX1 E n a X3 X5 X7 Slnltxx E 39 so the coef cient of x5 in the product cosx sinx equals i Math 409502 October 11 2004 i slide 4 Remark on the multiplication theorem 00 Theorem page 121 If 2 an and bn both converge absolutely then the product of the two 710 n 0 00 n series equals the absolutely convergent series Z C where cn E ukbnk 710 k0 Counterexample in case of conditional convergence Set an bn 71quotn 1 Then Eu and 2b are conditionally convergent by the alternating series test but the series Zen quot lt71gtklt71gtH is diver ent Indeed c All the terms in this sum have the same g quot kgxkanik 71 sign 71quot so lcnl 2 E k0 n 1 1 Hence Zen diverges Math 409502 October 11 2004 i slide 5 Homework 1 Read Chapter 9 pages 125 134 2 D0 Exercises 923 and 931 pages 134 135 Math 409502 October 11 2004 i slide 6