Maclaurin series Worked Solutions — Further Maths Maths

Fully worked, step-by-step solutions to Further Maths Maclaurin series questions. See exactly how to solve problems on standard-results, substitution, composite-series, interval-of-validity.

standard-resultssubstitutioncomposite-seriesinterval-of-validitycoefficientlimits
Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
Use the standard series expansions to find the series expansion of f(x)=e4xf(x)=e^{4x} in ascending powers of xx, up to and including the term in x3x^{3}.

Worked solution

  1. Write down the function to be expanded

    f(x)=e4xf(x)=e^{4x}

    The expansion is built from the standard series, not from a table of derivatives.

  2. Quote the standard series

    eu=1+u+u22!+u33!+u44!+⋯e^{u}=1+u+\frac{u^{2}}{2!}+\frac{u^{3}}{3!}+\frac{u^{4}}{4!}+\cdots

    This standard result is quoted, not re-derived.

  3. Substitute u=4xu=4x into the standard series

    u=4xu=4x

    The composite series is obtained by substituting the inner function into the standard result.

  4. State the required series expansion

    f(x)≈1+4x+8x2+32x33f(x)\approx 1+4x+8x^{2}+\frac{32x^{3}}{3}

    This is the required series expansion in ascending powers of xx.

Answer
1+4x+8x2+32x331+4x+8x^{2}+\frac{32x^{3}}{3}
Question 2
2 markseasy
Use the standard series expansions to find the series expansion of f(x)=e−3xf(x)=e^{-3x} in ascending powers of xx, up to and including the term in x3x^{3}.

Worked solution

  1. Write down the function to be expanded

    f(x)=e−3xf(x)=e^{-3x}

    The expansion is built from the standard series, not from a table of derivatives.

  2. Quote the standard series

    eu=1+u+u22!+u33!+u44!+⋯e^{u}=1+u+\frac{u^{2}}{2!}+\frac{u^{3}}{3!}+\frac{u^{4}}{4!}+\cdots

    This standard result is quoted, not re-derived.

  3. State the required series expansion

    f(x)≈1−3x+9x22−9x32f(x)\approx 1-3x+\frac{9x^{2}}{2}-\frac{9x^{3}}{2}

    This is the required series expansion in ascending powers of xx.

Answer
1−3x+9x22−9x321-3x+\frac{9x^{2}}{2}-\frac{9x^{3}}{2}
Question 3
2 markseasy
Use the standard series expansions to find the series expansion of f(x)=sin⁡(4x)f(x)=\sin\left(4x\right) in ascending powers of xx, up to and including the term in x3x^{3}.

Worked solution

  1. Write down the function to be expanded

    f(x)=sin⁡(4x)f(x)=\sin\left(4x\right)

    The expansion is built from the standard series, not from a table of derivatives.

  2. Quote the standard series

    sin⁡u=u−u33!+u55!−⋯\sin u=u-\frac{u^{3}}{3!}+\frac{u^{5}}{5!}-\cdots

    This standard result is quoted, not re-derived.

  3. Substitute u=4xu=4x into the standard series

    u=4xu=4x

    The composite series is obtained by substituting the inner function into the standard result.

  4. State the required series expansion

    f(x)≈4x−32x33f(x)\approx 4x-\frac{32x^{3}}{3}

    This is the required series expansion in ascending powers of xx.

Answer
4x−32x334x-\frac{32x^{3}}{3}
Question 4
2 markseasy
Use the standard series expansions to find the series expansion of f(x)=cos⁡(5x)f(x)=\cos\left(5x\right) in ascending powers of xx, up to and including the term in x2x^{2}.

Worked solution

  1. Write down the function to be expanded

    f(x)=cos⁡(5x)f(x)=\cos\left(5x\right)

    The expansion is built from the standard series, not from a table of derivatives.

  2. Quote the standard series

    cos⁡u=1−u22!+u44!−⋯\cos u=1-\frac{u^{2}}{2!}+\frac{u^{4}}{4!}-\cdots

    This standard result is quoted, not re-derived.

  3. Substitute u=5xu=5x into the standard series

    u=5xu=5x

    The composite series is obtained by substituting the inner function into the standard result.

  4. State the required series expansion

    f(x)≈1−25x22f(x)\approx 1-\frac{25x^{2}}{2}

    This is the required series expansion in ascending powers of xx.

Answer
1−25x221-\frac{25x^{2}}{2}
Question 5
2 markseasy
Use the standard series expansions to find the series expansion of f(x)=ln⁡(1+4x)f(x)=\ln\left(1+4x\right) in ascending powers of xx, up to and including the term in x3x^{3}.

Worked solution

  1. Write down the function to be expanded

    f(x)=ln⁡(1+4x)f(x)=\ln\left(1+4x\right)

    The expansion is built from the standard series, not from a table of derivatives.

  2. Quote the standard series

    ln⁡(1+u)=u−u22+u33−u44+⋯\ln\left(1+u\right)=u-\frac{u^{2}}{2}+\frac{u^{3}}{3}-\frac{u^{4}}{4}+\cdots

    This standard result is quoted, not re-derived.

  3. State the required series expansion

    f(x)≈4x−8x2+64x33f(x)\approx 4x-8x^{2}+\frac{64x^{3}}{3}

    This is the required series expansion in ascending powers of xx.

Answer
4x−8x2+64x334x-8x^{2}+\frac{64x^{3}}{3}

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