AP Calculus BC Review Chapter 12 (Sequences and Series), Part Two. n n th derivative of f at x = 5 is given by = x = approximates ( 6)

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1 AP Calculus BC Review Chapter (Sequeces a Series), Part Two Thigs to Kow a Be Able to Do Uersta the meaig of a power series cetere at either or a arbitrary a Uersta raii a itervals of covergece, a kow how to fi each Uersta that fuctios ca be represete by power series, how to o this, a why it might be esirable Perform operatios o fuctios represete by power series Uersta Taylor series (a Maclauri series), be able to fi their epasios to a arbitrary umber of terms a i the geeral case, a uersta Taylor s Theorem for the remaier estimate Practice Problems These problems shoul be oe without a calculator. The origial test, of course, require that you show relevat work for free-respose problems. The Taylor series about = for a certai fuctio f coverges to f ( ) for all i the iterval of covergece. The ( ) ( + ) f! th erivative of f at = is give by =, a f ( ) =. = a Write the thir-egree Taylor polyomial for f about =. b Show that the sith-egree Taylor polyomial for f about Fi the sum of the series Let f be give by f ( ) a Fi P( ). b Fi the coefficiet of k k+ ( ) k+ ( k + ) 6!!! = approimates ( 6) 6 + L + + L. Justify your aswer. = si +, a let ( ) f with error less tha P be the thir-egree Taylor polyomial for f about =. i the Taylor series for f about =. c Use the Lagrage error bou to show that f ( ) P( ) <. Let G be give by G( ) = f ( t) t. Write the thir-egree Taylor polyomial for G about =.. + A fuctio f is efie by f ( ) = L + + L for all i the iterval of covergece of the + give power series. a Fi the iterval of covergece for this power series. Show the work that leas to your aswer. f ( ) b Fi lim. c Write the first three ozero terms a the geeral term for a ifiite series that represets f ( ). Fi the sum of the series etermie i part c.

2 ( ) For what values of oes the series coverge? = a b < < c < e < 6 The first four terms of the Maclauri series for f ( ) = + are a + + b + + c e The coefficiet of i the Taylor series about a b c = for f ( ) cos = is 6 e ( ) 8 The iterval of covergece of is = a < b c < < < e 9 The coefficiet of a b i the Maclauri series for ( ) f e c 96 = is e 8 8 For what iteger k, k >, will both ( ) k a k = = coverge? a 6 b c e

3 Aswers 6 6 a ( ) ( ) ( ) a ( ) + P=+ 6 b G( )= + 6! a (,) b L + + L c + e, 6 b, 7 a, 8 c, 9 e, Solutios a ( ) ( )! ( )! ( )! ( + ) ( + ) ( + ) P = + ( ) + ( ) + = +!!! 6 6 = = = b The coefficiet of the et (seveth-egree) term of the Taylor series about = is Recall that the Maclauri series for si is k ( ) k+. k= k k ( ) ( k + ) + ( ) ( ) ( ) ( ) ( )! ( + )! = 7 = <. a ote that this series kth term ca be rewritte as! Igore for a momet that the summa is multiplie by ; we will eal with that later. ( k + )! Now the summa has a very similar form to the series for si : a alteratig factor ( ) k +, a a eomiator of ( k + )!. Sice the costat that is raise to the ( ) posit that is the argumet for the sie fuctio. Sice ( ), a costat raise to k + th power is, we si =, that is the series sum. Do t forget =. about the factor of from the begiig; the sum of the origial series is ( )( ) factorial gives, respectively,,, P ( ) = +. a f ( ) =, f ( ) =, f ( ) =, a ( ) f =. Diviig each of these by the erivative s orer s, a. These are the coefficiets of P( ), so b There will be a factor of i the aswer because evaluatig either cos( + ) or si( + ) = will = give cos( ) = si( ) =. The Chai Rule will have to be applie times, so we also have a factor of. Due to alteratig sigs i the power series for the sie fuctio, the power s term will be egative, a we must also ivie by!. Therefore the coefficiet is.! k

4 c Note that f ( ) P( ) M where M is a upper bou o f ( ) o,.! We have 6 M = 6, so the error is ε = <. (The formula for the Lagrage error bou, fou o page! 8 M of the tetbook, is R ( ) a where M is as escribe, a is the ceter of the power series epasio, a is the orer of the power ( +! ) series. = + The last term may be ig- G. 8 ore because the problem requires oly the thir-egree Taylor polyomial. Atiifferetiate each term of P( ) to get ( ) + a Lettig a =, cosier + b a + + = = =. a ( ) Thus we are itereste i < so <, meaig that the raius of covergece R is R =. We must ow test the epoits, a = the we have ( ) = ( ) ( + ) : if, show that whe +. This oes ot coverge, a it ca be similarly + = = = the series also oes ot coverge. Therefore the iterval of covergece I is I = ( ) + f / / / / lim = lim = lim = lim + + / / + = =. 9 ( ) ( ) c Atiifferetiatig each term of f ( ) gives ( ) = (sice all the terms for + = ,. f / /. Evaluatig this at 9 7 = are themselves zero) gives / + + / This is a geometric series with first term a ratio, so its sum is ( ) f = = =. We ca immeiately elimiate choices a a b; sice the series is cetere at =, so must the iterval of covergece be. It will suffice to test the epoits = a =. At =, the series is which coverges by the Alteratig Series Test. At =, the series is = = = = which is the harmoic se- ries, so it iverges. Therefore the iterval is <, choice e. 6 f ( ) =, f ( ) =, f ( ) =, a ( ) f = 8. Diviig by the erivatives orers factorials gives, respectively,,, 8, a 6, so the first four terms are 7 Several erivatives of ( ) cos = = ( ) ( ) + +, which is b. 8 6 f = at,, f =. The last oe just give is the relevat erivative, but we must first ivie by! to get the correct coefficiet. Thus we are itereste i = =, optio.! 6 6 = are f ( ) = f ( ) = a ( )

5 8 As i problem, we ca elimiate a a b because the iterval of covergece must be symmetric arou the ceter poit,. = We fi the raius of covergece by cosierig ( ) ( ) + + =, so we are itereste i < or <, meaig the raius of covergece is. Therefore e is wrog, because it ivolves a raius of. Give the two remaiig optios, we ee oly test the epoit = because both iicate that = is ( ) iverget. So we look at = ( ) which iverges. The iterval is therefore < <, choice c. = = 9 Several erivatives of f ( ) at = are f ( ) =, f ( ) =, f ( ) = 8, a ( ) oe just give is relevat, but we must ivie by! To get ( ) 6 6 f = = =, choice e.! 8 f = 6. The last We ca immeiately elimiate a, b, a c because these woul prouce i the seco series a geometric series with a ratio too large to allow for covergece. e, too, is impossible because the first series woul the have a factor of ( ) which for iteger is always eve a thus always. That, the, is the harmoic series; the harmoic series is kow to iverge. Therefore is the correct aswer.