AP Calculus Chapter 9: Ifiite Series
9. Sequeces a, a 2, a 3, a 4, a 5,... Sequece: A fuctio whose domai is the set of positive itegers = 2 3 4 a = a a 2 a 3 a 4 terms of the sequece Begi with the patter or th term. What s the sequece of umbers? a = ( ) (2) Graph each sequece o your graphig calculator i SEQ MODE. Use the trace feature to view the terms of the sequece. a = 2 2
Defie a sequece recursively. a = 0 a 2 = 3 a 3 = 6 a 4 = 9 a + = a + 3 Begi with a sequece of umbers. What s the patter or th term? Fidig the th term of a sequece ca sometimes be a little tricky. Recogizig a few commo forms ca be helpful. For example 2 = 2, 4,8,6, 2 = 2, 4,6,8,0, 2 =, 3,5, 7, 2 =, 3, 7,5, ( ) =,2, 3, 4, 5,6,! =,2,6,24,20,., 2, 4, 8, 6,... a = 2, 4 2, 8 6, 6 24, 32 20, 64 720, a = 2 2. 3. 3 7, 5 0, 7 3, 9 6, 9,... a = 4. 3 2, 4 3, 5 4, 6 5, a = 5. 3 4 5 2, 2 3, 3 4, 4 5, 5 6,... a = Graph sequece # ad #4 above ad observe the terms. What s happeig with these sequeces as becomes large?
Covergece ad Divergece: Whe a sequece approaches a particular value for large values of, we say that the sequece coverges. More formally, we use the cocept of a limit. Limit of a Sequece Let L be a real umber. Let f be a fuctio of a real variable such that If a lim f (x) = L. x { } is a sequece such that f () = a for every positive iteger, the lim x a = L Graph u() = / (2) usig sequece mode. Graph the sequece ad trace the poits to see the values of the sequece. What is the limit of this sequece based o the graph? Evaluate the limit aalytically: lim x 2 Whe the limit exists, the sequece CONVERGES. Whe the limit does ot exist, the sequece DIVERGES. Determie the covergece or divergece of the sequece.. a = 3 + ( ) 2. a = 2 3. a = 2 2 4. a = ( ) 4 5. a = ( ) 6. a = 2! 7. a = 3 3 8. a = ( 2)!!
Mootoic Sequece: A sequece is mootoic whe its terms are No- decreasig or o- icreasig. See examples at right. If a sequece is mootoic ad bouded the it coverges! Bouded Sequeces:
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9.2 Series ad Covergece Two basic questios for this chapter: Does a series coverge or diverge? If a series coverges, what is the sum? A series is just the sum of a sequece: a = a + a 2 + a 3 + + a + Ifiite Series A series ca be aalyzed as a sequece of partial sums: S = a S 2 = a + a 2 S 3 = a + a 2 + a 3 S = a + a 2 + a 3 + + a Covergece or Divergece of a Series If the sequece of partial sums, { S }, coverges to S, the the series coverges ad the sum is S. If the sequece of partial sums diverges, the the series diverges. Fid the sum of each ifiite series. Examie the sequece of partial sums.. = 2 2 + 4 + 8 + 6 + 2. = + + + + Geometrically determie the sum from #4, usig the image from Larso, p. 609.
Geometric Series: A geometric series with ratio r: ar = a + ar + ar 2 + ar 3 + + ar + a 0 Coverget: whe 0 < r < ad coverges to S = a r Diverget: whe r Determie whether the geometric series coverges or diverges, ad its sum, if it exists. 3. 0. + 0.0 + 0.00 + 0.000 +... 4. 3 0 + 3 00 + 3,000 + 3 0,000 + 5. 5 00 + 5 0,000 + 5,000,000 + 6. 3 2 7. 3 2
Telescopig Series: ( b b 2 ) + ( b 2 b 3 ) + ( b 3 b 4 ) + A telescopig series will coverge if ad oly if b approaches a fiite umber as. If the series coverges, its sum is S = b lim (b + ). 8. Determie the sum of the telescopig series: + = 2 + 2 3 + 3 4 + 9. Use partial fractios to write this series as a telescopig series ad determie its sum, if it exists. 2 4 2
th Term Test for Divergece If lim a 0, the a diverges, i.e. if the last term does ot approach 0, the series is DIVERGENT. If a coverges, the lim a = 0, i.e. if the series is CONVERGENT, the last term approaches 0. If the last term approaches 0, the series could be CONVERGENT or DIVERGENT. NOTE: The th term test is merely a test for DIVERGENCE!! Apply the th Term Test for Divergece to determie whether or ot a series is DIVERGENT. 0. 2.! 2. 2! +
9.2 Wrap up questios What do you thik?. How would you describe what is meat by coverget ad diverget series? 2. Describe the differece betwee lim a = 5 ad a = 5. 3. Give a geeral form for a geometric series, state whe it coverges, ad give the formula for the sum of a coverget geometric series. 4. State the th- Term Test for Divergece. 5. Explai ay differeces amog the followig series: (a) a (b) a k (c) a k k = True or False. If lim a = 0, the a coverges. 2. If a = L, the a = L + a 0. 3. If r <, the ar = a r. 3. The series diverges. 000( + ) 4. 0.75 = 0.74999999 5. Every decimal with a repeatig patter of digits is a ratioal umber.
9.3 The Itegral Test ad p- Series Cosider the area uder the curve from to. Cosider the sum of the fuctio heights for iteger values from to. Sice the area uder the curve ca be approximated by a series of rectagles havig the same height as the fuctio heights, does it make ituitive sese that the covergece or divergece of these two sums would behave i the same way. That is the essece of the Itegral Test. Either both sums coverge or both sums diverge. 0.4 0.2 2 3 4 5 6 7 8-0.2 The Itegral Test If a fuctio, f is positive, cotiuous ad decreasig for x, ad a = f (), the a ad f (x)dx either both coverge or both diverge. THIS DOES NOT IMPLY THAT THEY CONVERGE OR DIVERGE TO THE SAME VALUE ONLY THAT EITHER BOTH CONVERGE OR BOTH DIVERGE.
Example : Apply the Itegral Test to the series. 2 + 0.4 0.2 The series could be writte 2 + 2 5 + 3 0 + 4 7 + 2 3 4 5 6 7 8 Does this series coverge or diverge? -0.2 x Does the fuctio f (x) = meet the coditios of the Itegral Test? x 2 + a) Is f (x) positive for x >? (Hit: examie the fuctio) b) Is f (x) cotiuous for x >? (Hit: ay udefied values for x? ay breaks i the graph?) c) Is f (x) decreasig for x >? (Hit: is the derivative egative for x >?) Because the fuctio meets of the coditios of the itegral test, we ca apply it. x x 2 + a x dx = lim a x 2 + dx = let u = x 2 + ad du = 2x du 2 = u 2 lim l(x 2 a +) a = 2 lim l a a2 + ( ) l 2 ( ) = Sice the improper itegral is diverget, the series is also diverget.
Example 2: Apply the Itegral Test to the series = 2 + 2 + 5 + 0 + 7 + Does this series coverge or diverge? Does the fuctio f (x) = meet the coditios of the Itegral Test? x 2 + a) Is f (x) positive for x >? b) Is f (x) cotiuous for x >? c) Is f (x) decreasig for x >? Note that eve though the itegral coverges to π / 4, that does ot mea that the series coverges to π / 4. It merely meas that the series coverges by the itegral test because the improper itegral coverges. Example 3: Cofirm that the Itegral Test ca be applied to the series. Use the Itegral Test to determie covergece or divergece. 2 3 + 5 a) Is f (x) positive for x >? b) Is f (x) cotiuous for x >? c) Is f (x) decreasig for x >?
l a) Is f (x) positive for x >? b) Is f (x) cotiuous for x >? c) Is f (x) decreasig for x >? 4 + 2 2 + a) Is f (x) positive for x >? b) Is f (x) cotiuous for x >? c) Is f (x) decreasig for x >? 3 + 5 + 7 + 9 +... a) Is f (x) positive for x >? b) Is f (x) cotiuous for x >? c) Is f (x) decreasig for x >? Example 4: Explai why the Itegral Test does ot apply to the series a) Is f (x) positive for x >? b) Is f (x) cotiuous for x >? c) Is f (x) decreasig for x >? 2 e cos.5 0.5 5 0-0.5
p- Series ad Harmoic Series p- Series: p = + p 2 + +. p p 3.. Coverges if p > Diverges if 0 < p Harmoic Series: = + 2 + 3 +... (special case of the p- Series with p = ) Example 5: Determie the covergece or divergece of the p- Series. 3 5/3.04 0.99 + 3 4 + 3 9 + 3 6 + 3 25 +... 5
Homework 9.3 Cofirm that the itegral test ca be applied to the series ad the use the itegral test to determie the covergece or divergece of the series.. + 3 2. 2 3. e 4. l2 2 + l 3 3 + l 4 4 + l5 5 + l6 6 +!
5. arcta 2 + Explai why the itegral test does ot apply to the series. 6. ( ) 7. 2 + si Use the itegral test to cofirm the covergece or divergece of the p- series. 8. 3
9. /3 Determie the covergece or divergece of the p- series. 0.. 5 3 5/3 2. + 2 + 3 + 4 + l 5 + 6 +! 3. + 4 + 9 + 6 + l 25 +! 4. + 2 2 + 3 3 + 4 4 + l 5 5 + 6 6 +! 5. + 3 4 + 3 9 + 3 6 + l 3 25 +! 6. π
Aswers:. diverget 2. coverget 3. coverget 4. diverget 5. coverget 6. ( ) is ot always positive 2 + si 7. is always positive for >, but f (x) is ot always egative 8. coverges 9. coverges 0. diverges. coverges 2. diverges 3. coverges 4. coverges 5. diverges 6. coverges
9.4 Compariso of Series Direct Compariso Test: This test allows you to compare a series havig complicated terms with a simpler series whose covergece or divergece is kow. For example: is ot a geometric series, but resembles the simpler series which is 2 + 3 3 geometric. The Direct Compariso Test applies oly to series with positive terms. For a give series, if its terms are less tha a coverget series, the it must also coverge. (If it is below the covergig series it must coverge.) For a give series, if its terms are greater tha a diverget series, the it must also diverge. (If it is above a diverget series it must also diverge.) Direct Compariso Test Let 0 < a b for all.. If b coverges, the a The a terms are below the b terms. 2. If a diverges, the b The b terms are above the a terms. If the larger series coverges, the smaller series must also coverge. If the smaller series diverges, the larger series must also diverge. Example : Determie the covergece or divergece of 3 + 2 Compare this series to the coverget geometric series 2 Term by term compariso shows that 0 < Sice the larger series, 2 3 + 2 < 2., coverges, the smaller series, Therefore, by the Direct Compariso Test, the give series, must also coverge. 3 + 2 coverges. 3 + 2
Example 2: Determie the covergece or divergece of Compare this series to the diverget p- series 2 + /2 Term by term compariso shows that 0 < requiremets for divergece. 2 + which does ot meet the /2 Try comparig to aother diverget series, the harmoic series Term by term compariso shows that 0 < which does meet the requiremets 2 + for divergece. Therefore, by the Direct Compariso Test, the give series diverges. 2 + Example 3: Use the Direct Compariso Test to determie the covergece or divergece of the series. 3 2 + 2 4 5 + 3 3 + The Limit Compariso Test is a secod compariso test that may be useful but is ot withi the scope of the AP Calculus BC course ad will ot be dealt with here. You may wish to explore this test o pages 628-629 of your textbook.
Homework 9.4. Use the Direct Compariso Test to determie the covergece or divergece of the series. 2. 2 + 3. 2
4. 4 + 5. =2 l + 6.! Test for covergece or divergece, usig each test at least oce. Idetify which test was used. (a) th- Term Test (b) Geometric Series Test (c) p- Series Test (d) Telescopig Series Test (e) Itegral Test (f) Direct Compariso Test (g) Limit Compariso Test 7. 3
8. 7 7 9. 5 + 0. 2 3 2. + + 2 2. ( 2 +) 2 Aswers: 6. b) is a coverget p- series 3/2 c) Each of the other series are smaller ad therefore each must be coverget. d) The partial sums are all greater tha their ow correspodig terms. AND The greater the idividual terms, the greater the partial sums. 2. coverget 3. Diverget 4. coverget 5. coverget 6. Coverget 7. diverget 8. coverget 9. Coverget 0. diverget. coverget 2. coverget
9.5 Alteratig Series A series with alteratig positive ad egative terms is called a alteratig series. For example: ( ) = + 2 3 + 4 5 +... ( ) + = 2 + 3 4 + 5... odd terms are egative or eve terms are egative The Exploratio below will help you to uderstad what is meat by the covergece of a alteratig series. Cosider the alteratig harmoic series: ( ) + = 2 + 3 4 +... Fid S, S 2, S 3, S 4..., S 0 to two decimal places ad plot each o the lie below. This lie has bee started for you. 0.50.00 S 2 S Describe the patter of the S. Based o this patter, do you thik the series coverges? Where is S? (Write a iequality ivolvig S.) Thaks to Ruth Dover, IMSA for this exploratio. We ca test for the CONVERGENCE of these series based o the Alteratig Series Test.
Alteratig Series Test Let a > 0. The alteratig series ( ) a ad ( ) +! coverge if the followig two coditios are met.. lim a = 0 ad 2. a + a, for all i.e. The limit of the th term is zero ad the terms of the series are ot icreasig. Determie the covergece or divergece of the followig series: a ) ( ) Are the two coditios met? ) lim = 2) Is + for all? Does the series coverge? 2) ( 2) 3) ( ) + ( + ) = 2 3 2 + 4 3 5 4 + 6 5...
Approximatig the Sum of a Alteratig Series Alteratig Series Remaider If a coverget alteratig series satisfies the coditio a + a, the the absolute value of the remaider R N is less tha or equal to the first eglected term. That is, S S N = R N a N + I other words, add up a limited umber of the first terms i the series ad the sum of the series will be differet by o more tha the value of the ext term i the series. Calculator tip: To fid the sum of a sequece. 2 d STAT MATH sum( 2 d STAT OPS seq( sum( seq( fuctio, variable, lower boud, upper boud) Approximate the sum of each series below by usig the first 6 terms. 5) ( ) +! =! 2! + 3! 4! + 5! 6! +... 6) 4( ) + l( + )
More o the error i approximatig the sum of a Alteratig Series The example below aswers three essetial questios. ) What is the magitude of the error i calculatig the sum of a alteratig series based o a give umber of terms? Cosider the alteratig series ( ) + If we approximate the sum, S, of this alteratig series by usig = 5 terms, what will be the magitude of the error? I other words, what is the value of the + term, i this case, the 6 th term? Begi by fidig R5 = a 6 (i.e., the Remaider for a Sum of 5 terms is the 6 th term) 2) Ca we write the sum as a approximatio withi a particular iterval? Fid S 5, ad write a iequality about S. S 5 R 5 S S 5 + R 5 3) How may terms,, must we iclude to make sure that the error is less tha a give value? Begi with a + 0.0
Absolute ad Coditioal Covergece NOT EMPHASIZED ON AP CALCULUS BC EXAM If the series coverges, the the series coverges absolutely. a a If the series coverges but the series diverges, the the series a coverges coditioally. a a Determie whether each series is coverget or diverget. Classify ay coverget series as absolutely or coditioally coverget. 7) ( )! = 0! 2 2! 0 2 + 2! 2 3! 2 3 +... 2 8) ( ) = + 2 3 + 4... 9) ( ) (+)/2 = 3 9 + 27 + 8... 3
Homework 9.5 Determie the covergece or divergece of the series.. ( ) + + 2. ( ) 3 3. ( ) (5 ) 4 + 4. ( ) + 3 + 2 5. si (2 )π 2 ( ) 6. cos π 7. ( )! Approximate the sum of the series by usig the first six terms. 8. 2( )!
Determie the umber of terms required to approximate the sum of the coverget series with a error of less tha 0.00, ad use a graphig utility to approximate the sum of the series with a error of less tha 0.00. 9. ( )! = e 0. ( ) (2 +)! = si Aswers:. coverget 2. coverget 3. diverget 4. coverget 5. diverget 6. diverget 7. coverget 8. 7305 S.736
9.6 The Ratio Test The Ratio Test is particularly useful for series that coverge rapidly such as series ivolvig factorials or expoetials. Ratio Test Let a be a series with ozero terms.. a coverges absolutely if lim a + a <. a 2. a diverges if lim + a > or if lim a + a =. 3. The Ratio Test is icoclusive if lim a + a =. Determie the covergece or divergece of the series: ) 2! 2) 2 2 + 3 ( ) 3) +
Strategies for Testig for Covergece or Divergece: You ow have 8 tests for determiig covergece or divergece. How do you kow which to choose??? Courtesy of Li McMulli Let the followig questios guide your decisio.. Does the th term approach 0? If ot, the series diverges! 2. Is the series oe of the special types? Geometric, p- series, telescopig, or alteratig? 3. Ca the Itegral test (useful whe the th term ca be easily itegrated), or the Ratio Test (useful with fuctios that icrease rapidly like expoetials or factorials) be applied? 4. Ca the series be compared favorably with oe of the special types? Let s look at some examples ad select a approach.
. + 3 + 2. π 6 3. e 2 3 4. ( ) 4 + 5.! 0 6. + 2 + 7. 3 8. 2 + 9. 3 + 0. 3 3 +
Homework 9.6 Day Use the Ratio Test to determie the covergece or divergece of the series.. 2 2.! 3 3. 6 5 4. 4 2
5.! 3 6. 5 2 + 7. ( ) 2 4 ( 2 +)! Aswers:. coverget 2. diverget 3. diverget 4. diverget 5. diverget 6. diverget 7. coverget
Homework 9.6 Day 2 Determie the covergece or divergece of the series usig ay appropriate test from this chapter. Idetify the test used.. ( ) + 5 2. 00 3. 3 4. 2π 3 5. 5 2
6. 2 2 + 7. ( ) 3 2 2 8. 0 3 3 9. 0 + 3 2 0. 2 4 2
. cos 3 2. =2 ( ) l 3. 7! 4. ( ) 3! 5. ( ) 3 2 Aswers:. coverget alteratig series 2. diverget harmoic series 3. coverget p- series 4. diverget by ratio test 5. diverget by th term test 6. diverget by the itegral test 7. diverget by the ratio test 8. coverget p- series 9. coverget by ratio test 0. diverget by ratio test. coverget by compariso test 2. coverget by alteratig series test 3. coverget by ratio test 4. coverget by ratio test 5. diverget by ratio test
9.7 Taylor Polyomials ad Approximatios Our goal is to fid a polyomial fuctio P that approximates aother fuctio f. We begi by choosig some umber c such that the value of f ad P at c are the same, i.e. P(c) = f (c). Not oly must that be true, but at poit c, P'(c) = f '(c). If this is true, we say that the approximatig polyomial P is expaded about or ceter at c. With these two requiremets, we ca obtai a simple liear approximatio of f usig a first degree polyomial fuctio P. Example : For the fuctio f (x) = e x, fid a first- degree polyomial fuctio P(x) = a x + a 0, a lie, whose value ad slope agree with the slope of f at x = 0. Give that P(0) = f (0), determie the costat a 0 i the equatio P(x) = a x + a 0 Give that P'(0) = f '(0), determie the slope a i the equatio P(x) = a x + a 0. Put it all together i a st degree polyomial approximatio: P(x) = The graph of P(x) alog with f (x) = e x is below. You ca tell that at poits ear (0, ) the graph of the fuctio f (x) ad its approximatio P(x) are very close. However, as we move away from (0,), the approximatio is ot good ad the two graphs move further apart.
To improve the approximatio, we ca further require that the value of the secod derivatives of f (x) ad P(x) agree at x = 0. Let s develop a 2 d degree polyomial approximatio P 3 (x) = a 2 x 2 + a x + a 0 for f (x) = e x. Give that P(0) = f (0), fid a 0 Give that P'(0) = f '(0), fid a Give that P''(0) = f ''(0), fid a 2 Put it all together i a 2 d degree polyomial approximatio: P 2 (x) = The graphs of P 2 (x) ad f (x) = e x are show below. Not bad! But we ca do better! Develop a 3 rd degree polyomial approximatio P 3 (x) = a 3 x 3 + a 2 x 2 + a x + a 0 for f (x) = e x. Give that P(0) = f (0), fid a 0 Give that P'(0) = f '(0), fid a Give that P''(0) = f ''(0), fid a 2 Give that P'''(0) = f '''(0), fid a 3 Put it all together i a 3 rd degree polyomial approximatio: P 3 (x)= The graphs of P 3 (x) ad f (x) = e x are show below. We could cotiue creatig higher order polyomial approximatios to get a better fit!
Use your graphig calculator to fid umerical values for the fuctio f (x) = e x ad its 3 rd degree polyomial approximatio P 3 (x) = a 3 x 3 + a 2 x 2 + a x + a 0. x -.0-0.2-0.0 0 0. 0.2 f (x) = e x P 3 (x) It is ot ecessary that the polyomial be cetered at x = 0. It ca also be cetered at some umber c. This work was doe by Eglish mathematicias Brook Taylor (685-73) ad Coli Maclauri (698-746). Defiitio of th Taylor ad Maclauri Polyomials If f has derivatives at c, the the polyomial P (x) = f (c) + f '(c)(x c) + f ''(c) 2! is called the th Taylor polyomial for f at c. (x c) 2 + f '''(c) 3! (x c) 3 +!+ f () (c) (x c)! If c = 0, the P (x) = f (0) + f '(0)x + f ''(0) 2! x 2 + f '''(0) 3! x 3 +!+ f () (0) x! is called the th Maclauri polyomial for f at c.
Example 2: Fid the Taylor polyomials P, P2, P3, P4 for f (x) = l x cetered at c =. To start, fid f () =. Now fid: f '(x) = ad f '() = f ''(x) = ad f ''() = f '''(x) = ad f '''() = f (4) (x) = ad f (4) () = So ow we ca write: P (x) = P 2 (x) = P 3 (x)= P 4 (x) =
Example 3: Fid the Maclauri polyomials P0, P2, P4, P6 for f (x) = cos x. To start, fid f () =. Now fid: f '(x) = ad f '() = f ''(x) = ad f ''() = f '''(x) = ad f '''() = f (4) (x) = ad f (4) () = So ow we ca write: P 0 (x) = P 2 (x) = P 4 (x) = P 6 (x) = Use P 6 (x) to approximate cos 0.2. What is the differece betwee your approximatio for cos 0.2 ad what you get usig the cosie key o your calculator?
Example 4: Fid the Maclauri polyomials P, P3, P5, P7 for f (x) = si x. Use the result to approximate si 0.2. Example 5: Fid the 4 th Maclauri polyomial to approximate the value of l.. You could use f (x) = l x but a polyomial cetered at x = 0 is t goig to give a great approximatio for a value at x =.. So, let s use f (x) = l(x +) cetered at x = 0 ad approximate x = 0.. Example 6: Let f (t) be a fuctio that has derivatives for all orders for all real umbers. Give f (0) = 5, f '(0) = 3, f ''(0) =, f '''(0) = 4, write the 3 rd degree Maclauri polyomial for f (t).
Accuracy ad the Lagrage Remaider A approximatio techique is of little value uless we have some idea of how accurate it is. To measure the accuracy of a Taylor or Maclauri polyomial for a fuctio f (x), we use the cocept of a remaider R (x) defied as follows: f (x) = P (x) + R (x) exact value = approximate value + remaider It follows that R (x) = f (x) P (x) ad we call the absolute value of R (x) the error associated with the approximatio. Lagrage form of the Remaider: R (x) = f (x) P (x) Taylor s Theorem If a fuctio f (x) is differetiable through order + i a iterval I cotaiig c, the for each x i I, there exists a z betwee x ad c such that : f (x) = f (c) + f '(c)(x c) + f ''(c) 2! (x c) 2 + f '''(c) 3! (x c) 3 +!+ f () (c) (x c) + R (x)! exact value = approximate value + remaider R (x) = f (+) (z) (x c)+ ( +)! This is called the Lagrage form of the remaider. This is essetially the ext term i the sequece.
Example 7: Use the 3 rd degree Maclauri polyomial for f (x) = si x (see example 4) to approximate si 0.. Evaluate the Lagrage form of the remaider to determie the error i the approximatio. Recall: P 3 (x) = x x3 6 ad therefore e 0.. This is the estimate for si 0.. Usig Taylor s Theorem we ca write f (x) = P (x) + R (x) for this problem. si x = x x3 6 + f (4) (z) 4! x 4 What is f (4) (z)? What is the Lagrage form of the remaider? To evaluate the Lagrage form of the remaider, we ca t simply plug i 0 for x because the etire remaider would be 0, so we must cosider the rage of values for P 3 (0.) ± R 3 (0.) si z. If we estimate the maximum possible error, the we ca write a rage of values that we kow will iclude the actual value for si 0.. I the iterval [0, 0.] ( i.e. the distace betwee the ceter ad the x- value that we are tryig to approximate), the si fuctio must lie betwee 0 ad. If we let si x i the Lagrage error equal, we are determiig the maximum possible error. For si z 4! (0.)4 lettig si z = gives R 3 (x) = 4! (0.)4 0.0000046 Therefore, Si 0. = P 3 (0.) ± R 3 (0.) = 0.099833 ± 0.0000046 Ad.099829733 < si 0. <.0998374933 That s quite a small error cosiderig we oly used a 3 rd degree approximatio!!
Example 8: Fid the accuracy of the 4 th Maclauri polyomial for e 0.. See Example for the Maclauri polyomial. e x P 4 (x) = R 4 (x) = < e 0. <
Homework 9.7 Day : Taylor ad Maclauri Polyomials For the followig, fid the Maclauri polyomial of degree for the give fuctio ad evaluate at x = 0.. ) f (x) = e x, = 5 2) f (x) = x, = 5 3) f (x) = si(π x), = 3
4) f (x) = xe x, = 3 5) Fid the fourth degree Taylor Polyomial, cetered at c = for f (x) = x. 6) Calculate the error i approximatig cos (0.3) usig cos(0.3) (0.3)2 + (0.3)4, i.e. 2! 4! calculate R4. Write a iequality to idicate the iterval for the approximatio of cos (0.3). Compare your approximatio with the calculator value for cos (0.3).
Homework 9.7 Day 2: Taylor ad Maclauri Polyomials For the followig, fid the Maclauri polyomial of degree for the give fuctio ad evaluate at x = 0... f (x) = ta x, = 3 2. f (x) = ta x, = 3 3. Fid the secod degree Taylor Polyomial, cetered at c = π for f (x) = x cos x.
4. Estimate the error i approximatig e 0.5 by the fifth- degree polyomial by fidig the Lagrage remaider. 5. Fid the fourth degree Taylor polyomial ceter at c = 2 for f (x) = l x. Use the Taylor polyomial to approximate l 2.. 6. Use Taylor s Theorem to obtai a upper boud for the error of the approximatio of : si(0.5) 0.5 (0.5)3 3! + (0.5)5 5!
9.8 Power Series I sectio 9.7 we leared to approximate fuctios by Taylor Polyomials. Taylor Polyomials have a discrete umber of terms. The more terms, the more accurate will be the approximatio. With Power Series, we approximate fuctios exactly as a ifiite series. f (x) = e x e x + x st degree polyomial e x + x + x2 2! e x + x + x2 2! + x3 3! e x + x + x2 2! + x3 3! + x4 4! e x + x + x2 2! + x3 3! + x4 4! + x5 5! 2d degree polyomial 3rd degree polyomial 4th degree polyomial 5th degree polyomial e x + x + x2 2! + x3 3! + x4 4! + x5 x + + + POWER SERIES 5!! Defiitio of Power Series If x is a variable, the a ifiite series of the form a x = a 0 + a x + a 2 x 2 + a 3 x 3 + + a x + is called a power series. I geeral, a ifiite series of the form a (x c) = a 0 + a (x c) + a 2 (x c) 2 + a 3 (x c) 3 + + a (x c) + is called a power series cetered at c, where c is a costat. **To simplify the otatio for power series, we agree that (x c) 0 =, eve if x = c. Note: A Geometric Series is really a special case of the power series with c = 0 ad a = a, a costat. Geometric Series: ax = a + ax + ax 2 + ax 3 +
The mai focus of this lesso is to determie the values of x for which a give power series will coverge. There are 3 possible ways that a power series will coverge.. A power series cetered at c will coverge oly at the poit c, i which case the radius of covergece is 0. 2. A power series will coverge over a iterval that is R uits to the right ad left of c i which case the radius of covergece is R. We ca the talk about a iterval of covergece (c R, c + R). The edpoits may or may ot be part of the iterval of covergece. 3. A power series will coverge for all real umbers, i which case the radius of covergece is R =, ad the iterval of covergece is, These three cases are illustrated i 3 examples below. The Ratio Test works well to determie the iterval of covergece. As a remider, the Ratio Test tells us: lim x a + a < the the series is coverget, ( ) If lim x a + a > the the series is diverget, lim x a + a = the ratio test is icoclusive.. Fid the iterval of covergece of the power series: x 7
2. Fid the radius ad iterval of covergece of!x 3. Fid the radius ad iterval of covergece of 3(x 2)
4. Fid the iterval of covergece of the power series: ( 3x) ( 2)! 5. Fid the radius ad iterval of covergece of ( ) x 2+ (2 + )!
Homework 9.8 Day Fid the iterval of covergece of the power series. Be sure to iclude a check for covergece at the edpoits of the iterval.. x 4 2. ( ) x
3. x 5! 4. (2)! x 3
5. ( ) + (x 4) 9 6. x 3+ (3 +)! Aswers:. Series coverges o (- 4, 4), Radius of covergece = 4 2. Series coverges o (-, ], Radius of covergece = 3. Iterval of covergece (-,), Radius of covergece = 4. he series is diverget for all x except x = 0, the ceter. Radius of covergece = 0 5. The series coverges o (- 5, 3], Radius of covergece = 9 6. Iterval of covergece o (-,), Radius of covergece =
Differetiatio ad Itegratio of Power Series A fuctio defied by a power series ca be differetiated ad itegrated i the same way as ay polyomial fuctio by usig the power rule for derivatives ad itegrals. The iterval of covergece for the derivative ad itegral of the series will be the same as for the origial series, except possibly at the edpoits. So, CHECK THE ENDPOINTS! J Give the fuctio f (x) = a (x c) = a 0 + a (x c) + a 2 (x c) 2 + a 3 (x c) 3 + The derivative is: f '(x) = a (x c) = a + 2a 2 (x c) + 3a 3 (x c) 2 + The itegral is: (x c) + (x c) f (x)dx = C + 2 (x c) 3 a = C + a 0 (x c) + a + a 2 + 2 3 + I the ext sectio of our textbook, we will use differetiatio ad itegratio of kow power series to fid the power series for a ew fuctio. EXAMPLE Let = f (x) = f (x), f '(x), f ''(x) ad fuctio. ( ) + (x ) +. Fid the iterval of covergece of + f (x)dx. Be sure to check the edpoits of the iterval for each
Homework 9.8 Day 2 Fid the itervals of covergece of f (x), f '(x), f ''(x), ad covergece at the edpoits of the iterval. f (x)dx. Iclude a check for. f (x) = x 3
2. f (x) = ( ) + (x ) + + Aswers: f (x) coverges o (-3, 3). 2. f '(x) coverges o (-3, 3) f ''(x) coverges o (-3, 3) f (x)dx coverges o [-3, 3) f (x) coverges o (0, 2] f '(x) coverges o (0, 2) f ''(x) coverges o (0, 2) f (x)dx coverges o [0, 2]
9.9 Represetatio of Fuctios by Power Series DAY I this lesso we will represet a fuctio as a Power Series. I our earlier lesso geometric series we leared that for the series x, we could write the sum of the ifiite umber of terms as S = x Now, let s suppose we begi with a fuctio i the form of the sum, f (x) = x, you ca see that it is a geometric series that is represeted by x = x for x < I the followig examples, we will attempt to rewrite a give expressio i the form x. Example : Fid a power series for f (x) = x + covergece. cetered at 0. Determie the iterval of
Example 2: Fid a power series for f (x) = cetered at. Determie the iterval of x covergece. Example 3: Fid a power series for f (x) = covergece. cetered at 0. Determie the iterval of 3 + x
Example 4: Fid a power series for f (x) = cetered at 0. Determie the iterval of 2 4x covergece. Example 5: Fid a power series for f (x) = covergece. x 7x + cetered at 0. Determie the iterval of
Example 6: Fid a power series for f (x) = covergece. x 2 + 9 cetered at 0. Determie the iterval of Example 7: Fid a power series, cetered at 0, for the fuctio f (x) = 3x x 2 (Hit: First separate the fuctio usig partial fractios. J )
Homework 9.9 Day Fid a geometric power series for the fuctio cetered at 0.. f (x) = 4 x 2. f (x) = 3 4 + x Fid a power series for the fuctio, cetered at c, ad determie the iterval of covergece. 3. f (x) = 3 x, c =
5 4. g(x) = 2x 3, c = - 3 5. g(x) = 4x x 2 + 2x 3, c = 0 Aswers:. 3. x 4 4, (- 4, 4) 2. x 2 2, (-, 3) 4. 5 9 5. x 3 or x, (-, ) 3 ( ) 3 4 x 4, (- 4, 4) 2x + 6 9, (- 5/2, 3/2)
DAY 2 Operatios with Power Series Let f (x) = a x ad g(x) = b x. f (kx) = a k x Multiply withi the fuctio by a costat! 2. f x N ( ) = a x N Expoetiate the variable withi the fuctio. 3. f (x) ± g(x) = (a ± b)x Add/subtract fuctios to form a ew fuctio. Example 8: Fid a power series for f (x) = l x cetered at x =. (Hit: Itegrate the series that we already kow for the fuctio f (x) =. (See example 4) x dx = l x + C x
Example 9: Fid a power series by itegratio for the fuctio f (x) = arcta x, cetered at 0.
Homework 9.9 Day 2 Use the power series + x = ( ) x to determie a power series, cetered at 0, for the fuctio. Idetify the iterval of covergece.. h(x) = 2 x 2 = + x + x 2. f (x) = (x +) = d 2 dx x +
3. f (x) = l(x +) = x + dx
4. g(x) = x 2 + 5. f (x) = l( x 2 +)
Aswers:. 2. 3. 4. 5. ( ) + x, (-, ) ( ) x, (-, ) ( ) (x) + + C +, (-, ] ( ) (x) 2, (-, ) ( ) (x) 2+2 +, [-, ]
9.0 Taylor ad Maclauri Series The Taylor ad Maclauri Series are very similar to the Taylor ad Maclauri polyomials that we geerated earlier i this chapter. The series are a ifiite set of terms while the polyomials are a particular umber of terms with a remaider. Defiitio: If a fuctio f has derivatives of all orders at x = c, the the series f () (c) (x c)! = f (c) + f '(c)(x c) + + f () (c) (x c) +! is called the Taylor Series for f (x) at c. If c = 0, the the series is called the Maclauri Series for f. Example : Use the fuctio f (x) = si x to form the Maclauri series ad determie the iterval of covergece. Note: I this example we have show that the Maclauri series coverges to a sum for all x, but have ot demostrated that it coverges ecessarily to f(x) = si x. Demostratig covergece to the fuctio for all x goes beyod the scope of this course. However, a brief explaatio is that if the Lagrage error goes to 0 as approaches ifiity, the the Maclauri or Taylor series coverges to f(x). f () (z)(x c) i.e. If lim R = lim = 0, the the series coverges to the fuctio.!
Taylor ad Maclauri Series for a Composite Fuctio Use a basic list of Taylor ad Maclauri series for elemetary fuctios to fid series for composite fuctios. For istace, if you wat to fid the Taylor Series for f (x) = cos x 2, write out the Taylor Series for f (x) = cos x ad substitute x 2 for x. Or, if you wat to fid the Taylor Series for f (x) = e x si x, write out the series for f (x) = e x ad for f (x) = si x, ad multiply them together. Below is a list of basic fuctios from the Larso text. For the AP exam, it would be helpful to memorize the series for l x, e x, si x, ad cos x. From this list, we ca determie power series for other fuctios by the operatios of additio, subtractio, multiplicatio, divisio, differetiatio, itegratio, ad compositio with kow power series. This provides a very useful SHORTCUT to fidig the coefficiets of a Maclauri or Taylor series directly. J
Example 2: Fid the Maclauri series for f (x) = si x 2. Example 3: Fid the first five terms of the Maclauri series for f (x) = e x arcta x
Example 4: Use a power series to approximate e x2 dx. Use the first four terms of the series for your approximatio. 0
Homework 9.0 Day Use the defiitio of Taylor series to fid the Taylor series for the fuctio.. f (x) = cos x, c = π / 4 Fid the Maclauri series for the fuctio. Use the table of power series for elemetary fuctios. 2. f (x) = e x2 /2
3. g(x) = si 3x 4. f (x) = xsi x
Aswers:. 2. 3. 4. (+) ( ) ( ) cos x 2 x π / 4 2! coverges o x 2 e x2 /2 + 2! coverges o ( ) 2+ ( ) 3x si 3x ( 2 +)! coverges o ( ) 2 ( ) + x xsi x ( 2 )! coverges o (,) (,) (,) (,)
Homework 9.0 Day 2 Use the defiitio of Taylor series to fid the Taylor series (cetered at c) for the fuctio.. f (x) = l x, c = Fid the Maclauri series for the fuctio. Use the table of power series for elemetary fuctios. 2. f (x) = l(+ x)
Fid the first four ozero terms of the Maclauri series for the fuctio by multiplyig or dividig the appropriate power series. Use the table of power series for elemetary fuctios. Use a graphig utility to graph the fuctio ad its correspodig polyomial approximatio. 3. f (x) = e x si x 4. g(x) = si x + x
Aswers:. 2. 3. 4. l x l( x +) ( ) + ( x ) or ( ) + ( x) P 5 (x) = x + x 2 + x3 3 x5 30 P 4 (x) = x x 2 + 5x3 6 5x4 6 l x ( ) ( x ) + +