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Lecture Comments (5)

0 answers

Post by Thomas Zhang on March 19, 2014

For example 5, doesn't the limit not exist because it's +inf x-->0+ and it's -inf when x-->=-

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Post by Akshay Tiwary on February 28, 2013

In the third problem, it would be simpler to substitute '-x' in the Maclauren Expansion for 1/1-x

2 answers

Last reply by: Acme Wang
Sun Jul 10, 2016 3:25 AM

Post by Troy Wingert on February 27, 2013

Just a comment about Example 1 on Power Series Operations .... there's a mistake in the solution....it should be 1 - x^2 + x^4 - x^6 ..... Threw me off for a bit.

Power Series Operations

  • Operating on a power series:
    • For substitution:
      • Find similar and simpler form of function
      • Find series of simpler form
      • Substitute appropriate term
    • For derivatives and integrals:
      • Find integral of function
      • Find series of integral
      • Differentiate each term in series
    • For multiplication and division
      • Find series of known portion
      • Multiply or divide by other variable portion

Power Series Operations

Find the MacLaurin Series for f(x) = [1/(1 + x3)]
Substitute ( − x3) for the MacLaurin series for [1/(1 − x)][1/(1 − x)] = ∑n = 0 (x)n [1/(1 − ( − x3))] = ∑n = 0 ( − x3)n = 1 − x3 + x6 − x9 + ...
Find the MacLaurin Series for f(x) = [1/(1 + 3x4)]
Substitute ( − x2) for the MacLaurin series for [1/(1 − x)][1/(1 − x)] = ∑n = 0 (x)n [1/(1 − ( − 3x4))] = ∑n = 0 ( − 3x4)n = 1 − 3x4 + 9x8 − 27x12 + ...
Find the first 3 non - zero terms of the MacLaurin Series for f(x) = cos(2x)
Substitute (2x) for the MacLaurin series for cosxcosx = ∑n = 0 [(( − 1)nx2n)/(2n)!] cos2x = ∑n = 0 [(( − 1)n(2x)2n)/(2n)!] = 1 − 2x2 + [2x/3]4 − [4x/45]6 + ...
Find the first 3 non - zero terms of the MacLaurin Series for f(x) = sin( − 3x)
Substitute ( − 3x) for the MacLaurin series for sin xsinx = ∑n = 0 [(( − 1)n − 1x2n − 1)/((2n − 1)!)] sin( − 3x) = ∑n = 0 [(( − 1)n − 1( − 3x)2n − 1)/((2n − 1)!)] = − 3x + [(9x3)/2] − [(81x5)/40] + ...
Use a series to solve limx→0 [sinx/x]
Use the MacLaurin Series of sinx to divide
limx → 0 [sinx/x] = limx → 0 [(∑n = 0 [(( − 1)n − 1x2n − 1)/((2n − 1)!)] )/x] = limx → 0 [(x − [(x3)/6] + [(x5)/120] − [x7/5040])/x] = limx → 0 1 − [(x3)/6x] + [(x5)/120x] − [(x7)/5040x] = 1 − 0 + 0 − 0 = 1
Use a series to solve limx → 0 [cos(x)/( − 4x)]
  • Use the MacLaurin Series of cos x to divide
limx → 0 [cos(x)/( − 4x)] = limx → 0 [(∑n = 0 [(( − 1)nx2n)/(2n)!] )/( − 4x)] = limx → 0 [(1 − [(x2)/2] + [(x4)/24] − [(x6)/720] + ...)/( − 4x)] = limx → 0 [1/( − 4x)] − [(x2)/( − 8x)] + [(x4)/(24( − 4x))] − [(x6)/(720( − 4x))] + ... = ∞− 0 + 0 − ... = ∞
Use a series to solve limx → 0 [(ex)/2x]
Use the MacLaurin Series of ex to divide
limx→0 [(ex)/2x] = limx → 0 [(∑n = 0 [(xn)/n!] )/2x] = limx→0 [(1 + x + [(x2)/2] + [(x3)/6] + ...)/2x] = limx→0 [1/2x] + [1/2] + [x/4] + [(x3)/12] + ... = ∞+ [1/2] + 0 + 0 + ... = ∞
Use a series to solve limx → 0 [(ln(x + 1))/x]
Use the MacLaurin Series of ln(x + 1) to divide
limx → 0 [(ln(x + 1))/x] = limx → 0 [(∑n = 0 [(( − 1)nxn)/n] )/x] = limx → 0 [(x − [(x2)/2] + [(x3)/3] − [(x4)/4] + ...)/x] = limx → 0 1 − [x/2] + [(x2)/3] + ... = 1 − + 0 − 0 = 1
Find the MacLaurin series of f(x) = e3x
Substitute (3x) for the MacLaurin series for exex = ∑n = 0 [(xn)/n!] e3x = ∑n = 0 [((3x)n)/n!] = 1 + 3x + [(9x2)/2] + [(9x3)/2] + [(27x4)/8] + ...
Find the first three times of the MacLaurin Series for the f(x) = − [6/((x + 1)4)]
  • Recognize − [6/((x + 1)4)] is the third derivative of [1/(x + 1)]
[(d3)/(dx3)]( [1/(x + 1)] ) = [(d3)/(dx3)]( 1 − x + x2 − x3 + x4 − x5 + ... ) = − 6 + 24x − 60x2 + ...

*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

Power Series Operations

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
  • Operations 0:07
  • Example 1: Substitution 1:05
  • Example 2: Substitution 3:41
  • Example 3: Differentiation/ Integration 5:39
  • Example 4: Differentiation/ Integration 9:55
  • Example 5: Multiplying/ Dividing 12:32
  • Example 6: Multiplying/ Dividing 14:50