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• ## Related Books

 1 answerLast reply by: Thomas ZhangWed Mar 19, 2014 4:43 PMPost by Delaram Sadaghdar on January 30, 2014For example 4, how is the answer x^n? 1 answerLast reply by: Jinrong ShiWed Apr 17, 2013 5:27 PMPost by Ahmed Murjis on April 10, 2013which one are you talking about? what do you mean by f'''(x)=3!which = 6 not 3 1 answerLast reply by: Thomas ZhangWed Mar 19, 2014 4:37 PMPost by G Aych on December 30, 2012f'''(x)=3!which = 6 not 3

### Taylor Series

• Approximates value of
• Taylor series:
• MacLaurin series: , simply Taylor series when
• Solving for MacLaurin series:
• Solve for first few derivative terms
• Evaluate derivative terms at 0
• Write out series terms
• Notice pattern and write MacLaurin series proper form

### Taylor Series

Find the MacLaurin Series for f(x) = [1/(2 − x)]
• Find derivatives and fn(0)
•  f(x) = [1/(2 − x)]
 f(0) = [1/2]
 f′(x) = [1/((x − 2)2)]
 f′(0) = [1/4]
 f"(x) = [2/((x − 2)3)]
 f"(0) = [1/8]
 f"′(x) = [6/((x − 2)4)]
 f"′(0) = 116
Apply ∑n = 0 [(fn(0))/n!] ( xn )∑n = 0 [(fn(0))/n!] ( xn ) = [1/2] + [x/4] + [(x2)/8] + [(x4)/32] + ...
What is the MacLaurin Series representaiton for [1/(2 − x)]?
Find General Form of ∑n = 0 [(fn(0))/n!] ( xn )[1/(2 − x)] = [1/2] + [x/4] + [(x2)/8] + [(x4)/32] + ... = ∑n = 0 (x − 1)n
Find the MacLaurin Series for f(x) = [1/(x + 1)]
• Find derivatives and fn(0)
•  f(x) = [1/(x + 1)]
 f(0) = 1
 f′(x) = − [1/(( x + 1 )2)]
 f′(0) = − 1
 f"(x) = [2/(( x + 1 )3)]
 f"(0) = 2
 f"′(x) = − [6/(( x + 1 )4)]
 f"′(0) = − 6
Apply ∑n = 0 [(fn(0))/n!] ( xn )∑n = 0 [(fn(0))/n!] ( xn ) = 1 − [x/1] + [(2x2)/2!] − [(6x3)/3!] + ... = 1 − x + x2 − x3 + ...
What is the MacLaurin Series representaiton for [1/(x + 1)]?
• Find General Form of ∑n = 0 [(fn(0))/n!] ( xn )
[1/(x + 1)] = 1 − x + x2 − x3 + ... = ∑n = 0 ( − 1)nxn
Find the MacLaurin Series for f(x) = e − x
• Find derivatives and fn(0)
•  f(x) = ex
 f(0) = 1
 f′(x) = − ex
 f′(0) = 1
 f"(x) = ex
 f"(0) = 1
 f"′(x) = − ex
 f"′(0) = 1
Apply ∑n = 0 [(fn(0))/n!] ( xn )∑n = 0 [(fn(0))/n!] ( xn ) = 1 − [x/1] + [(x2)/2!] − [(x3)/3!] + ... = 1 − x + [(x2)/2] − [(x3)/6] + ...
What is the MacLaurin Series representaiton for e − x?
Find General Form of ∑n = 0 [(fn(0))/n!] ( xn )e − x = 1 − x + [(x2)/2] − [(x3)/6] + ... = ∑n = 0 [(( − 1)nxn)/n!]
Find the MacLaurin Series for f(x) = [x/(ex)]
• Find derivatives and fn(0)
•  f(x) = [x/(ex)]
 f(0) = 0
 f′(x) = − [(x − 1)/(ex)]
 f′(0) = 1
 f"(x) = [(x − 2)/(ex)]
 f"(0) = − 2
 f"′(x) = − [(x − 3)/(ex)]
 f"′(0) = 3
Apply ∑n = 0 [(fn(0))/n!] ( xn )∑n = 0 [(fn(0))/n!] ( xn ) = 0 + [x/1!] − [(2x2)/2!] + [(3x3)/6!] + ... = x − x2 + [(x3)/2] − [(x4)/6] + ...
What is the MacLaurin Series representaiton for [x/(ex)]?
• Find General Form of ∑n = 0 [(fn(0))/n!] ( xn )
[x/(ex)] = x − x2 + [(x3)/2] − [(x4)/6] + ... = x∑n = 0 [(( − 1)nxn)/n!]
Find the MacLaurin Series for f(x) = ln(x + 1)
• Find derivatives and fn(0)
•  f(x) = ln|x + 1|
 f(0) = 0
 f′(x) = [1/(x + 1)]
 f′(0) = 1
 f"(x) = − [1/(( x + 1 )2)]
 f"(0) = 1
 f"′(x) = [2/(( x + 1 )3)]
 f"′(0) = [1/4]
Apply ∑n = 0 [(fn(0))/n!] ( xn )∑n = 0 [(fn(0))/n!] ( xn ) = 0 + [x/1] − [(x2)/2!] + [(2x3)/3!] − [(6x4)/4!] + ... = x − [(x2)/2] + [(x3)/3] − [(x4)/4] + ...
What is the MacLaurin Series representaiton for ln(x + 1)?
• Find General Form of ∑n = 0 [(fn(0))/n!] ( xn )
ln(x + 1) = x − [(x2)/2] + [(x3)/3] − [(x4)/4] + ... = ∑n = 0 [(( − 1)nxn)/n]

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.