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by: Cassidy Grimes
Cassidy Grimes

GPA 3.51

M. Girardi

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M. Girardi
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This 18 page Class Notes was uploaded by Cassidy Grimes on Monday October 26, 2015. The Class Notes belongs to MATH 142 at University of South Carolina - Columbia taught by M. Girardi in Fall. Since its upload, it has received 34 views. For similar materials see /class/229534/math-142-university-of-south-carolina-columbia in Mathematics (M) at University of South Carolina - Columbia.

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Date Created: 10/26/15
Math 142 107 Taylor and Maclaurin Polynominals Read this handout thoroughly and then do Homeworks 2 4 5 and 6 on a seperate sheet of paper Let s consider a function y 1 and x a point x0 in the domain of y So the graph of y x goes through the point 9007f900 The equation of the tangent line to the graph of y x at the point x07 x0 is found by 247241 mz1 y i 900 fQEOWE 950 y mo fHBOW 900 So the equation y p1x of the tangent line to the graph of y x at the point x07 x0 is 19190 950 fMW 900 1 Recall that the function y x can be approximated locally near x0 by this tangent line y p1x In other words7 if x is close to x0 then the value x is close to the value 191x7 that is7 if x z x0 then x z p1x Draw yourself a picture Example 1 x sinx near x0 can be approximated by the line lt3 lt3 ltm e 1 Homework 2 Find the equation of the tangent line to the function x 7 at the point x0 2 1 Note that this tangent line approximation works well because the tangent line to the graph of y x at x07 x0 is the only line with slope f x0 passing through the point x07 x0 We can generalize this to second degree approximations by nding the parabola passing through the point x07 x0 with the same slope rst derivative as y x at x0 and the same second derivative as y x at x0 Example 3 Consider x e m 1 at x0 1 Such a parabola as we want looks like p2x co 61x 71 62x 712 and we can nd the constants 00 0102 to make the derivatives of y x and y p2x match up at x0 1 We have P2 co c1x 71 02x 71 and m 57171 PW 61 20225 7 1 and f z 757ml 202 and f m 5 I1 and evaluating these at x0 1 giVes us 1921 Co and l 50 192 61 and f 1 750 1 1921 202 and f 1 5 0 1 and so 1921 f 1 gt co 1 PEG fl gt cl 7 1 2951 MD a 262 1 Q CZ So our parabola is 1 p2x 17 x 71 712 This polynomial is the second order Taylor polynomial of y e m 1 centered at x0 1 Notice that close to x1 this parabola approximates the function rather well Using this as a model we can give a general form of the second order Taylor polynomial for y x at x07 that is the parabola 19290 Co 6195 i 950 6295 i 9602 where we want to nd the constants 00701702 to make the derivatives of y x and y p2x match up at x x0 We have 19290 Co 6195 i 950 0295 i 9602 gt 192950 Co P295 C1 20290 i 900 gt 192900 C1 19295 202 192950 202 and so 192900 mo ltgt Co mo p 20 fmo ltgt C1 fmo jagx0 f x0 ltgt 202 f x0 ltgt 02 z l So our parablola is 7 fQEO 2 WE 7 Hmo f 900W 300 T g00 2 Compare the function y p1x in formula 1 with the function y p2x in formula Starting to see a pattern Homework 4 Find the second order Taylor polynomial for x i at x0 72 Draw a picture Higher order Taylor polynomials are found in the same way For example the third order Taylor polynomial for a function y fz centered at 0 is f10 0 19395 900 fO 900 2 96 7 9602 3 96 i 9003 In general the Nth order Taylor polynomial for y fz at 0 is N mm mo fHEOWE we ltzimogt2 fTWW mow which can also be written as recall that O 1 0 1 2 N pNz f 03 flismkm 7 0 f 22 m 7 mo2 fTWx 7 m0N N open form Formula N open form is in open form It can also be written in closed form by using sigma notation as N 1 900 n 7 N l d f pNm n m mo c ose orm So y pNz is a polynomial of degree at most N and it has the form N pNm 2 90 900 n0 where the Gas TL 35 on f 0 71 are specially chosen so that PN950 900 p gtltz0gt flt1gtltzogt 199900 fame Woo fltNgtltzogt The constant on is called the nth Taylor coefficient of y fz about 0 The Nth order Maclaurin polynomial for y is just the Nth order Taylor polynomial for y fz at 0 0 and so it is N n mm 2 05 n0 39 Homework 5 Compute the fth order Maclaurin polynomial ie y p5z for fz sin3z Why are higher order Taylor polynomials better Let s look at some graphs from Homework 5 To follow are graphs of y sin3m along with its Maclaurin polynomial y pNz for N 13 4 7 9 11 13 FIGURE 1 y sin3m along with its rst order Maclaurin Polynomial y p1z FIGURE 3 y sin3m along with its third order Maclaurin Polynomial y p3z FIGURE 5 y sin3m along with its fth order Maclaurin Polynomial y p5 FIGURE 7 y sin3m along with its 7th order Maclaurin Polynomial y p7z FIGURE 9 y sin3m along with its 9th order Maclaurin Polynomial y pgz FIGURE 11 y sin3m along with its 11th order Maclaurin Polynomial y p11z FIGURE 13 y sin3m along with its 13th order Maclaurin Polynomial y p13z Notice that as N increases the approximation of y sin3m by y pNz gets better and better7 even over a wider and wider interval around the center mo 0 So for a xed z the approximation of y x by y pNz becomes more accurate as N gets bigger Homework 6 Find the Maclaurin polynomials y p m7 y 193m7 y 195m7 y 197m7 y pgm7 y 1911m7 and y p13z above ie7 for y sin3m about 0 0 Just to think about Take another look at Homework 6 Do you notice any pattern in the Taylor coefficients Why did we only use odd order Taylor polynomials Math 142 Techniques of integration Trigonometric Integrals The book gives lots of rules in Tables 831 amp 832 Do NOT memorize these long lists of rules Use logic instead as done below Avoid reduction formulas 17 27 197 20 I ll show you easier ways7 see below Recall the following natural pairings in terms of derivatives sin lt gt cos tan lt gt sec cot lt gt csc If the integrand does not contain natural pairings7 then rewrite it so that it does Eg tanzcos z cos z sinzcosz Make a natural choice for s or t For example7 if the integrand involves sine and cosine7 then try 3 cos z t sin z d8 7sinzdz or dtcosxdx If the integrand involves tangent and secant7 then try 3 tanz or t sec z If the integrand involves cotangent and cosecant7 then try 3 cotz or t csc z Then isolate off your ds or dt and try to express what is left in terms ofjust s or 25 The following will be helpful and you must memorize them 17 cos2z 2 1 cos2z sin2 z f and cos z f half angle formulas 2 A 2 1 cos2zsin2z1 gt COS xsm m7 gt 1tan2zsec2z cosz z cosz z cosz z 2 A 2 1 cos2zsin2z1 gt C szxsn2 2 gt cot2z1csc2z sin z sin z sin z For integrals of the form f sin mz cos nz dz or f sin mz sin nz dz or f cos mz cos nz dz where m 7 71 use the trig identities these 3 you do not have to memorize 1 i sina cos3 SIH04 7 3 sina 3 1 sina sin3 cosa 7 3 7 cosa 3 2 cosa cos3 cosa 7 3 cosa 3 3 Example 4 fsin4zdz Way 1 Integration by Parts7 homework 82 55a Way 2 below 3 If we try 3 cosz or t sin z7 it will not work why fsin4zdz 7 f s z 7 sinzdz Here we use the half angle formulas sin4zdz sin2z2dz 17s22dz i172cos2zcos22z dz Example 1 sin7zdm s cosz t sinz d8 isinzdz or dtcosxdx Sin7dlisinzdxl Now can we express in terms of only 3 cosz 7 Yes7 easily sin6z sinm6 sinzm3 17 cosZz3 So sin7xdm7isinmdx 77sinmdm 717523ds 36733433271 d3 37 335 333 i 7 7 7 7 C 7 5 3 s 7 cos7 x 3 cos5 x 7 7 5 7 cos3micosz0 sin7 zdm would want here cosz dm The substitution with t sinz doesn t work because there is no cosz in the original integrand Example 2 sin417z cos317z dz 3 cos17z or t sin17z ds 717 sin17z dz dt 17 cos17zdz sin417z cos317z dz I771 cos317z sin317z 717 sin17zdz Can we express cos317z sin317z in terms of only 3 cos17z Not easily7 there is an extra sin cos317z sin317z cos317z sin217z sin17z cos317z 17 cos217z sin17z 1 sin417z cos317z dz E sin417z cos217z 17 cos17z dz Can we express sin417m cos217z in terms of only 25 sin17m7 Yes sin417z cos217z sin417z 17 sin217z SO sin417zcos317zdz sin417zcos217z 17cos17zdz sin417z 17 sin217z 17 cos17z dz t417t2dt 1 E t4 7t6dt 1 t5 t7 E E gt t C 7 1 sin517z sin717m 5 7 C 17 Example 3 tan 9 sec4 9 d0 stan0 or tsec0 ds sec2 00 dt sec0tan0d0 tanoseewo seem Can we express in terms of only 3 tan 0 Yes tant9sec2 t9 tan0 1 tan2 0 tan0sec40d0 secz0d0 secz0d0 s132ds s33ds 82 S4 7 7 C 2 4 itan20tan40 7 2 4 SO 0 tan0sec40d0sec0tan0d0 Can we express in terms of only 25 sec 0 Of course So tan 9 sec4 9 d0 sec 0tan 9 d0 3 t4 tdtZk sec4 9 7 k 4 Why are these answers different They re not sec40 1 2 2 2 2 Tk1sec 0 k 1tan 0 k tanzt9tan4t9 kl 2 4 4 l 4 1 Zlt12tan20tan40k Commonly Used Taylor Series SERIES WHEN IS VALIDTRUE 1 2 3 4 NOTE THIS IS THE GEOMETRIC SERIES 1 z z z z 1 7 I JUST THINK OF 95 AS r 00 Z I z E 71 1 710 SO 1 1 1 13 I4 511g e 1 I 7 7 7 V n v v v 17 7 so 171 7 17quot quot 2 3 4 elt x E70 2 mm 710 00 n I Z 7x I E R 710 n NOTE y cos 95 IS AN EVEN FUNCTION 2 I4 15 LE COS7E COSI AND THE 0051 1 7 2y 7 7 a 7 g 7 TAYLOR SERIS OFycosac HAS ONLY EVEN POWERS 00 Zn I 71 z E R 7 2n 3 7 9 NOTE y sinz IS AN ODD FUNCTION 7 I I5 I IE sin7ac 7Sinx AND THE SH 7 I 7 7 5 7 W 7 9 7 TAYLOR SERIS OF y Sinx HAS ONLY ODD POWERS 00 27171 0quot 2n1 I or I 71W 1 7 71 7 z e R n27 2n71 n27 2n1 1 1 I2 13 I4 15 QUESTION IS y 1n1 z EVEN nlt I 7 If 7 ZKi ODDORNEITHER 00 In 00 In 71lt7Hgt7 E 71 7 z e 71 1 Zlt gt n Zlt gt n lt I n1 n1 3 5 7 9 tanil I 7 I7 7 L 7 I7 7 I7 7 QUESTION IS y arctanx EVEN 3 5 7 9 ODD OR NEITHER 00 711 Z1n71 E g Z1n 2n71 2n1 z 6 711 Math 142 107 amp 109 TaylorMaclaurin Polynomials and Series Prof Girardi Fix an interval I in the real line eg I might be 717190 and let 10 be a point in I ie 10 E I Next consider a function Whose domain is I f I 7gt R and Whose derivatives f I 7gt R exist on the interval I for n l 2 3 N De nition 1 The Nthorder Taylor polynomial for y at 10 is N mac 7 fzo f ror 7 me 7 use 7 zogtN open form Which can also be Written as recall that 0 l 0 1 2 N pNz fTWf17WIirofTSIOIiro21710N lt gt a nite sum ie the sum stops Formula open form is in open form It can also be Written in closed form by using sigma notation as N fnI0 pNz 0 z 7 r0 closed form 7 So y pNz is a polynomial of degree at most N and it has the form 71 an fiw N pNz Z on I 7 Ion Where the constants n are specially chosen so that derivatives match up at 10 ie the constants cn7s are chosen so that pNIo fzo 209m 7 flt1gtltzogt p 3gtltzogt 7 flt2gtltzogt Moo 7 fltNgtltzogt The constant on is the nth Taylor coef cient ofy about 10 The Nthorder Maclauriri polynomial for y is just the Nthorder Taylor polynomial for y at 10 0 and so it is N n 0 PNI Z 107 In 710 De nition 2 1 The Taylor series for y at 10 is the power series n 13001 fzo fzoz 7 10 wamp 7 zo2 fTWQ 7 10 open form Which can also be Written as f0 I f1 I 102 I fn I The Taylor series can also be Written in closed form by using sigma notation as I710 lt gt the sum keeps on going and going n 13001 Z 167210 I 7 Io closed form n0 The Maclauriri series for y is just the Taylor series for y at 10 0 1Here We are assuming that the derivatives y fquotm exist for each m in the interval I and for each n 6 N E 1234 5 H Big Questions 3 For what values of I does the power aka Taylor series fn I0 n poem z 7 Io lt1 converge usually the Root or Ratio test helps us out with this question If the powerTaylor series in formula 1 does indeed converge at a point I does the series converge to what we would want it to converge to7 ie7 oes 1W 13001 7 2 Question 2 is going to take some thought De nition 4 The Nthorder Remainder term for y at 10 is d f RNW 5 f1 PNW where y PNI is the Nthorder Taylor polynomial for y at 10 So 1 PNW RNW 3 that is m PNz within an error of RNI We often think of all this as N N fn10 n h N N 7 I 7 10 lt gt a n1te sum7 t e sum stops at We would LIKE TO HAVE THAT n E 167210 I 7 IO lt gt the sum keeps on going and going n 710 In other notation W m PNz and the question is 5 P00 where y PDo is the Taylor series of y at 10 Well7 letls think about what needs to be for 130017 ie7 for f to equal to its Taylor series Notice 5 Taking the limNH00 of both sides in equation 37 we see that n Z 10710 z 7 Ion lt gt the sum keeps on going and going n0 nl if and only if Algnm RNI 0 Recall 6 limNH00 RNI 0 if and only iflimNH00 lRNzl 0 So 7 If ngnm 13 le 0 lt4 then 7010 1710 So we basically want to show that 4 holds true How to do this Well7 this is where Mr Taylor comes to the rescue 2According to Mr Taylor his Remainder Theorem see next page was motivated by coffeehouse conversations about works of Newton on planetary motion and works of Halley of Halley s comet on roots of polynomi s Taylor7s Remainder Theorem Version 1 for a xed point z E I and a xed N E N 3 There exists 5 between I and 10 so that e eorem was f fltzgt 7 PM z ezogtltN1gtA lt5 So either I S c S 10 or 10 S c S I So we do not know exactly what 5 is but atleast we know that c is between I and 10 and so 5 E I Remark This is a by Taylor See the book for the proof The proof uses the Mean Value Theorem Note that formula 5 implies that fN1C N iRNltzgti lawml izizor 1 lt6 Version 2 for the whole interval I and a xed N E N3 Assume we can nd M so that the maximum of fN1 on the interval I S M ie mg fN1c S M Then M N1 lRNIl S lrifol 7 for each I E I Remark This follows from formula Version 3 for the whole interval I and all N E N 4 Now assume that we can nd a sequence MNV 1 so that mg fN1C 3 MN for each N E N and also so that lim lz 7 zolNTl 0 Nam N 1 for each I E I Then by formula 7 and the Squeeze Theorem gymiwzn o for each I E I Thus by So 7 f z n fltzgt 2 were n0 for eacthI 3Here we assume that the N 1ederivative of y fz ive y fltN1m exists for each m 6 I 4Here we assume that y fNm exists for each m 6 I and each N 6 N Indeterminate Forms L7H6pital7s Rulel At z u7 D has the indeterminate form E if lim f l and lim g 1 43 g 0 0 lt2 a oo oo 3 f96 996 0 00 0 00 4 f 95 7 Wt 00 i 00 00 00 5 lf96lgm 00 0 0 6 lf96lgm 000 00 0 7 lf96lgm 1 1 GO HERE u stands for any of the symbols 17 1 7 cf 7 7007 00 1 and 2 If f m i i 0 gm has the interdeterminate form or at u s 0 lim exists ie this limit is a nite number or 700 or 00 M we then i we f x Elm 31 We 39 lf fx g has the interdeterminate form at u7 then rewrite me gltzgt 7 which has the interdeterminate form at u 01 W i i a z z 7 Wthh has the interdeterminate form at u f gt glt gt MM 00 and then apply L7H6pital7s Rule 4 If fx 7 g has the interdeterrninate form at u7 then use algebraic manipulation to convert f 7 g into a form of the type or and then apply L7Hopital7s Rule 5 If fz9w has the interdeterrninate form at u7 then follow these steps Let y lf96lgm So lny ln lt f PM Next7 simplify lny l9l 39 1H WM Note that lny ln has the interdeterrninate form at u Using an appropriate above method ie 37 evaluate lny E L Conclude hm minor L more 5L 6 and 7 If fzgw has the interdeterrninate form or 7 then proceed similarly as in 5 Note that lny ln will have the interdeterrninate form 6 at u 7 at u


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