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Lecture Comments (6)

1 answer

Last reply by: Thomas Zhang
Wed Mar 19, 2014 4:43 PM

Post by Delaram Sadaghdar on January 30, 2014

For example 4, how is the answer x^n?

1 answer

Last reply by: Jinrong Shi
Wed Apr 17, 2013 5:27 PM

Post by Ahmed Murjis on April 10, 2013

which one are you talking about? what do you mean by f'''(x)=3!which = 6 not 3

1 answer

Last reply by: Thomas Zhang
Wed Mar 19, 2014 4:37 PM

Post by G Aych on December 30, 2012

f'''(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.

Answer

Taylor Series

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Taylor Series 0:06
    • MacLaurin Series
  • Example 1 1:02
  • Example 2 2:45
  • Example 3 6:04
  • Example 4 8:23
  • Example 5 11:49