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Maclaurin Series — Question Types

Type 1: Standard Maclaurin from First Principles (5–7 marks)​

Obtain the Maclaurin series for a given function by successive differentiation.

Example: s20/21 Q2 — Use the Maclaurin series to find the first three non-zero terms in the expansion of 2x2^x.

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Let f(x)=2x=exln⁡2f(x) = 2^x = e^{x\ln 2}

f′(x)=2xln⁡2f'(x) = 2^x \ln 2, f′′(x)=2x(ln⁡2)2f''(x) = 2^x (\ln 2)^2, f′′′(x)=2x(ln⁡2)3f'''(x) = 2^x (\ln 2)^3

f(0)=1f(0) = 1, f′(0)=ln⁡2f'(0) = \ln 2, f′′(0)=(ln⁡2)2f''(0) = (\ln 2)^2, f′′′(0)=(ln⁡2)3f'''(0) = (\ln 2)^3

2x=1+(ln⁡2)x+(ln⁡2)22!x2+(ln⁡2)33!x3+⋯2^x = 1 + (\ln 2)x + \frac{(\ln 2)^2}{2!}x^2 + \frac{(\ln 2)^3}{3!}x^3 + \cdots

Marks scheme:

  • M1 — Write 2x=exln⁡22^x = e^{x\ln 2} or attempt to differentiate
  • A1 — Correct derivatives (at least 2)
  • A1 — Correct first term 11
  • A1 — Correct (ln⁡2)x(\ln 2)x term
  • A1 — Correct (ln⁡2)22x2\frac{(\ln 2)^2}{2}x^2 term
  • A1 — Correct (ln⁡2)36x3\frac{(\ln 2)^3}{6}x^3 term

Example: s23/23 Q1 — Find the Maclaurin series for sin⁡−1x\sin^{-1} x up to the term in x3x^3.

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Let f(x)=sin⁡−1xf(x) = \sin^{-1} x

f′(x)=(1−x2)−1/2f'(x) = (1 - x^2)^{-1/2}, f′′(x)=x(1−x2)−3/2f''(x) = x(1 - x^2)^{-3/2}, f′′′(x)=(1−x2)−3/2+3x2(1−x2)−5/2f'''(x) = (1 - x^2)^{-3/2} + 3x^2(1 - x^2)^{-5/2}

f(0)=0f(0) = 0, f′(0)=1f'(0) = 1, f′′(0)=0f''(0) = 0, f′′′(0)=1f'''(0) = 1

sin⁡−1x=x+x36+⋯\sin^{-1} x = x + \frac{x^3}{6} + \cdots

Marks scheme:

  • M1 — Differentiate to find f′(x)=(1−x2)−1/2f'(x) = (1-x^2)^{-1/2}
  • A1 — Correct f′(0)=1f'(0) = 1, f′′(0)=0f''(0) = 0
  • A1 — Correct f′′′(0)=1f'''(0) = 1 and final series

Type 2: Composite Functions / Log Diff (4–7 marks)​

Example: w20/21 Q1 — Find the Maclaurin series for e−x2e^{-x^2} up to the term in x4x^4.

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Substitute u=−x2u = -x^2 into eu=1+u+u22!+u33!+⋯e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \cdots

e−x2=1+(−x2)+(−x2)22!+(−x2)33!+⋯e^{-x^2} = 1 + (-x^2) + \frac{(-x^2)^2}{2!} + \frac{(-x^2)^3}{3!} + \cdots

e−x2=1−x2+x42−x66+⋯e^{-x^2} = 1 - x^2 + \frac{x^4}{2} - \frac{x^6}{6} + \cdots

Marks scheme:

  • M1 — Use substitution u=−x2u=-x^2 in standard eue^u series
  • A1 — Correct 1−x21 - x^2 term
  • A1 — Correct x42\frac{x^4}{2} term

Example: s21/23 Q2 — Find the Maclaurin series for ln⁡(cosh⁡x)\ln(\cosh x) up to the term in x4x^4.

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Method 1: Direct differentiation.

cosh⁡x=1+x22!+x44!+⋯\cosh x = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots

ln⁡(cosh⁡x)=ln⁡(1+x22+x424+⋯ )\ln(\cosh x) = \ln\left(1 + \frac{x^2}{2} + \frac{x^4}{24} + \cdots\right)

Let u=x22+x424+⋯u = \frac{x^2}{2} + \frac{x^4}{24} + \cdots, use ln⁡(1+u)=u−u22+u33−⋯\ln(1+u)=u-\frac{u^2}{2}+\frac{u^3}{3}-\cdots

ln⁡(cosh⁡x)=(x22+x424)−12(x22)2+⋯=x22−x412+⋯\ln(\cosh x) = \left(\frac{x^2}{2} + \frac{x^4}{24}\right) - \frac{1}{2}\left(\frac{x^2}{2}\right)^2 + \cdots = \frac{x^2}{2} - \frac{x^4}{12} + \cdots

Marks scheme:

  • B1 — Correct cosh⁡x\cosh x expansion (at least 2 terms)
  • M1 — Use ln⁡(1+u)\ln(1+u) expansion with correct substitution
  • A1 — Correct x22\frac{x^2}{2} term
  • A1 — Correct −x412-\frac{x^4}{12} term

Example: s24/21 Q2 — Find the Maclaurin series for e1+x2+e1−xe^{1+x^2} + e^{1-x} up to the term in x3x^3.

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e1+x2=e⋅ex2=e(1+x2+x42!+⋯ )e^{1+x^2} = e \cdot e^{x^2} = e\left(1 + x^2 + \frac{x^4}{2!} + \cdots\right)

e1−x=e⋅e−x=e(1−x+x22!−x33!+⋯ )e^{1-x} = e \cdot e^{-x} = e\left(1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \cdots\right)

Sum: e(2−x+32x2−16x3+⋯ )e(2 - x + \frac{3}{2}x^2 - \frac{1}{6}x^3 + \cdots)

Marks scheme:

  • M1 — Factor ee from both expressions
  • A1 — Correct expansion of e1+x2e^{1+x^2} up to x3x^3 terms
  • A1 — Correct expansion of e1−xe^{1-x} up to x3x^3 terms
  • A1 — Correct sum: 2e−ex+32ex2−16ex32e - ex + \frac{3}{2}ex^2 - \frac{1}{6}ex^3

Type 3: Approximation of Integrals (2 marks)​

Example: w22/21 Q1 (part) — Use the Maclaurin series for ln⁡(1+ex)\ln(1+e^x) to approximate ∫00.1ln⁡(1+ex) dx\int_0^{0.1} \ln(1+e^x)\,dx.

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First expand ln⁡(1+ex)\ln(1+e^x).

ex=1+x+x22+x36+⋯e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots

1+ex=2+x+x22+x36+⋯=2(1+x2+x24+x312+⋯ )1+e^x = 2 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots = 2\left(1 + \frac{x}{2} + \frac{x^2}{4} + \frac{x^3}{12} + \cdots\right)

ln⁡(1+ex)=ln⁡2+ln⁡(1+x2+x24+x312+⋯ )\ln(1+e^x) = \ln 2 + \ln\left(1 + \frac{x}{2} + \frac{x^2}{4} + \frac{x^3}{12} + \cdots\right)

Use ln⁡(1+u)=u−u22+⋯\ln(1+u)=u-\frac{u^2}{2}+\cdots with u=x2+x24+x312+⋯u = \frac{x}{2} + \frac{x^2}{4} + \frac{x^3}{12}+\cdots

ln⁡(1+ex)=ln⁡2+x2+x28+⋯\ln(1+e^x) = \ln 2 + \frac{x}{2} + \frac{x^2}{8} + \cdots

Integrate term by term:

∫00.1ln⁡(1+ex) dx≈∫00.1(ln⁡2+x2+x28)dx\int_0^{0.1} \ln(1+e^x)\,dx \approx \int_0^{0.1} \left(\ln 2 + \frac{x}{2} + \frac{x^2}{8}\right)dx

=[ln⁡2⋅x+x24+x324]00.1= \left[\ln 2 \cdot x + \frac{x^2}{4} + \frac{x^3}{24}\right]_0^{0.1}

=0.1ln⁡2+0.014+0.00124≈0.0697= 0.1\ln 2 + \frac{0.01}{4} + \frac{0.001}{24} \approx 0.0697

Marks scheme (integral part):

  • M1 — Integrate series term-by-term
  • A1 — Correct numerical answer