! " k x 2k$1 # $ k x 2k. " # p $ 1! px! p " p 1 # !"#$%&'"()'*"+$",&-('./&-/. !"#$%&'()"*#%+!'",' -./#")'.,&'+.0#.1)2,'!%)2%! !"#$%&'"%(")*$+&#,*$,#

Transcription

1 "#$%&'()"*#%+'",' -./#")'.,&'+.0#.1)2,' %)2% "#$%&'"()'*"+$",&-('./&-/. Taylor Series o a unction at x a is " # a k " # " x a# k k0 k It is a Power Series centered at a. Maclaurin Series o a unction is a Taylor Series at x 0. e x "#$%&'"%(")*$+&#,*$,# sin x 1 " x " x2 2 " x3 3 " x k k " # x x3 k0 3 $ x5 # 1 5 cos x 1 x2 2 " x4 1 4 k0 " # p $ 1 px p " p 1 # 1 x 2 " k x 2k$1 # $ k x 2k x 2 2k $ 1" # 2k$ "#$%&'()*$"#*+#" 1 To estimate values o unctions on an interval. 2 To compute limits o unctions. 3 To approximate integrals. 4 To study properties o the unction in question.

2 To ind Taylor series o unctions, we may: 1 Use substitution. 2 Dierentiate known series term by term. 3 Integrate known series term by term. 4 Add, divide, and multiply known series. "#$"#%&'()*+,&-.,".- "#$"%#&'(')$*+#,- Find the Maclaurin Series o the ollowing unctions. sin sin x " x" arctan x x 4 7 cos 2 x" sinh x 5 x 2 e x 1 x 3 " e " x 8 9 x2 arctan x " 3 1 x 6 "#$"%#&'(')$*+#,- Find the Taylor Series o the ollowing unctions at the given value o a. 10 x x3 at a x at a 2 Find the Maclaurin Series o the ollowing unctions. 12 e 2x at a sin x " at a # x at a 1 15 ln 1 x " # at a $ 2

3 "#$%&'"()'*"+,"-&.('/0&.01'2(.3' "#$#%&'()%*+'&,)%,-$.-$/01#2&3')%4)'3$)*-&'),&(+$50-)$ 21#%#&%#6#7#%&'8&59&+%-$:#%-#601&0&%#:;5)**)$<01#3&%#=)%01 #&51?

4 "#$%&'"()'*"+,"-&.('/0&.01'2(.3' 4%&50-5#A4BC4%)D'#*6E

5 "#$%&'"()'*"+,"-&.('/0&.01'2(.3'

6 "#$%&'"()'*"+,"-&.('/0&.01'2(.3'

7 "#$%&'"()'*"+,"-&.('/0&.01'2(.3' I==?'J4'7J,72,2/'K7'LMNN>MN/4OP/N'Q2N/ROP/'

8 "#$%&'"()'*"+,"-&.('/0&.01'2(.3'

9 Problem 1 Substitute x by x 2 Hence sin x 2 "#$"%&'()*+&'+* x" # sin x " 2 in the Maclaurin Series o sine. " x " # 1" 2 2k$1 k # 1 2k $ 1" " k x 4k$2 2k $ 1" Problem 2 "#$"%&'()*+&'+* " x" # sin x x Divide the Maclaurin Series o sine by x. Hence, " sin x x # 1 1" k x 2k$1 # 1 x 2k $ 1" " k x 2k 2k $ 1" Problem 3 Observe that ' x Series o ' x "#$"%&'()*+&'+* Power Series ormula. x" # arctan x" 1 " # 1 $ x " substitute x 2 or x in Basic 2. To ind the Maclaurin Hence ' x "#$"%&'()*+&'+* " # " k 1 1 $ x # 2 x2 # 1. By integrating both sides, we obtain x" # 1" k x 2k dx # 1 # 1 $ C. 2k $ 1 " k x 2k$1 " k x 2k " k x 2k dx

10 "#$"%&'()*+&'+* 0 is in the interval o convergence. Thereore we can insert x 0 to ind that the integration constant c 0. Hence the Maclaurin series o arctan x " # is arctan x " # " 1# k x 2k$1. 2k $ 1 k0 Problem 4 "#$"%&'()*+&'+* x" # cos 2 x" By the trigonometric identity, cos 2 x" # 1 $ cos 2x" " 2. Thereore we start with the Maclaurin Series o cosine. "#$"%&'()*+&'+* Substitute x by 2x incos x Thus cos 2x " # 1" k 2x" 2k and dividing by 2, we obtain " # 1" k x 2k. 2k". Ater adding 1 2k" "#$"%&'()*+&'+* cos 2 x" # $ 1 = 1 2x 1 $ =1+ 1 k#1 " k 2x" 2k " 2 " k 2 2k1 2k" $ 2x " 4 4 2k" x2k "

11 "#$"%&'()*+&'+* "#$"%&'()*+&'+* Problem 5 x" # x 2 e x Problem 6 x" # 1 x 3 Multiply the Maclaurin Seris o e x by x 2. x k Hence, x 2 e x x 2 k k0 k0 x k"2 k. By rewriting "" 12. By substituting x x" # 1 $ x 3 by -x 3 in the binomial ormula with p # 12 we obtain, 1 x 3 # x3 1 8 x6 Problem 7 By rewriting "#$"%&'()*+&'+* the Maclaurin Series o e x x" # sinh x" x" # ex e x. Substitute x by x in 2 # 1 $ x $ x2 2 $ # x k, k " x# k e x k k0 Thus when we add e x 1 x $ x2 2 and e x, the terms with odd power are canceled and the terms with even power are doubled. Ater dividing by 2, we obtain sinh x "#$"%&'()*+&'+* " # 1 $ x2 2 $ x4 4 $ x 2k k0 " 2k#

12 Problem 8 "#$"%&'()*+&'+* x" # ex 1 x We have e x 1 " x " x2 2 " and 1 1 x 1 " x " x2 " To ind the Maclaurin Series o # x$, we multiply these series and group the terms with the same degree. "#$"%&'()*+&'+* " # 1 x x2 2 " 1 x x2 " $ 1 2x x2 higher degree terms $ 1 2x 5 2 x2 higher degree terms Problem 9 "#$"%&'()*+&'+* x" # x 2 arctan x " 3 We have calculated the Maclaurin Series o arctan x arctan x Substituting x by x 3 " # 1" k x 2k$1. 2k $ 1 " in the above ormula, we obtain arctan x 3 "#$"%&'()*+&'+* " x " # 1" 3 2k$1 k # 1. 2k $ 1 2k $ 1 " k x 6k$3 Multiplying by x 2 gives the desired Maclaurin Series x 2 arctan x 3 " # x 2 1" k x 6k$3 # 1 2k $ 1 " k x 6k$4 2k $ 1

13 "#$%&'()&*)( Problem 10 x" # x x 3 at a # 2 Find the Taylor Series o the ollowing unctions at given a. Taylor Series o x" # x x 3 at a # 2 is o the orm 2"$ 1 " 2" x $ 2"$ 2 " 2" x $ 2" 2 2 $ 3 " 2" x $ 2" 3 $ 4 " 2" x $ 2" 4 $ " 3 4 "#$%&'()&*)( Since is a polynominal unction o degree 3, its derivatives o order higher than 3 is 0. Thus Taylor Series is o the orm 2"# 1 " 2" x # 2"# 2 " 2 2 " x # 2" 2 # 3 " 2" 3 x # 2" 3 By direct computation, " 2 2 " # 6, 1 "#$%&'()&*)( " # 11, 2 " 2" # 12, 3 " 2" # 6 So the Taylor Series o x x 3 at a # 2 is 6-11 x $ 2"$ 6 x $ 2" 2 x $ 2" 3

14 Problem 11 "#$%&'()&*)( x" # 1 x at a # 2 " # 1 x at a # 2 is o the orm Taylor Series o x k " 2" x 2" k. We need to ind the general k expression o the k th derivative o 1 x. "#$%&'()&*)( We derive 1 x until a pattern is ound. x" # 1 x # x 1, 1 " x" # 1"x 2 2 " x" # 1" 2"x 3, 3 " x" # 1" 2" 3"x 4 In general, k " " x. Thereore k " " # 1 2" # 1" k k$1" k2. " k k$1 kx "#$%&'()&*)( Ater inserting the general expression o the k th derivative evaluated at 2 we obtain, k " 2" x 2" k 1 # k k 1 " k k$1" k2 x 2" k Hence, the the Taylor Series o 1 x is 1" k x 2 " 2 k$1 " k. Problem 12 Taylor Series o x k " 12" x 1 k 2 "#$%&'()&*)( x" # e 2x at a # 12 " # e 2x at a # 12 is o the orm k. We need to ind the general expression o the k th derivative o e 2x.

15 "#$%&'()&*)( We derive e 2x until a pattern is ound. " " x x x " # e 2x, 1 " # 2e 2x, 2 " # 2 2e 2x " x" # 1" k 2 k e 2x. 12" # 1" k 2 k e # 1 " k 2 k. In general, k Thereore k " e "#$%&'()&*)( Ater inserting the general expression o the k th derivative evaluated at 1 2 we obtain, k " 12" x 1 k 1 1" k 2 k k 2 # x 1 k e 2 Hence, the the Taylor Series o e 2x 1" k 2x 1" k. e k " is k Problem 13 Taylor Series o x orm k " "#$%&'()&*)( " # sin x" at a # 4 is o the k 4" k x 4 general expression o the k th x" # sin x" at a # 4. We need to ind the derivative o sin x". We derive sin x x" # sin x", 1 " In general, k " "#$%&'()&*)( " until a pattern is ound. x" # cos x", 2 " x" # sin x" sin x" i k # 4n cos x" i k # 4n $ 1 x" # sin x" i k # 4n $ 2 cos x" i k # 4n $ 3

16 "#$%&'()&*)( " " and odd order derivatives are either cos x" In other words, even order derivatives are either sin x or -sin x or -cos x". So the Taylor Series at a # 4 can be written as " 1" k sin 4 x 2k" 4 2k " k cos 4" $ 1 x 2k $ 1" 4 2k$1 "#$%&'()&*)( Since, at a 4, sin" 4# cos" 4# 1 2, the Taylor Series can be simpliied to " 1# k x k0 2 " 2k " 1# k 2k# 4 $ x k0 2 " 2k $ 1# 4 2k$1. Problem 14 "#$%&'()&*)( x" # 10 x at a # 1 " # 10 x at a # 1 is o the orm Taylor Series o x k " 1" x 1" k. We need to ind the general k expression o the k th derivative o 10 x. "#$%&'()&*)( We derive 10 x until a pattern is ound. x" # 10 x, 1 " x" # ln 10" 10 x, 2 " x In general, k " x" # ln k 10"10 x. Thereore k " 1" # ln k 10"10. " # ln 2 10 "10 x

17 Ater inserting the general expression o the k th derivative evaluated at 1 we obtain, k " 1" x 1" k ln k 10"10 # x 1 k k "#$%&'()&*)( " k Problem 15 Taylor Series o x "#$%&'()&*)( k " x" # ln 1 $ x" at a # 2 " # ln 1 $ x" at a # 2 is o 2" x $ 2" k the orm. We need to ind the k general expression o the k th derivative o ln 1 $ x". We derive ln x 1 "#$%&'()&*)( " # until a pattern is ound. " " x# $ ln" x 1#, 1 # 1 " x# $ x 1, " 2 # 1 " x# $ " x 1# 2 " x# " 1# k $ " x 1#. Thereore " k# k " In general, k # " 2# $ 1. "#$%&'()&*)( Ater inserting the general expression o the k th derivative evaluated at -2 we obtain k " 2" x # 2" k 1 $ x # 2 k k k$0 k$0 " k.