Hard Further Maths Maclaurin series Questions

Challenging, exam-style Further Maths Maclaurin series questions with worked solutions. Stretch yourself on the hardest standard-results, substitution, composite-series, limits problems.

standard-resultssubstitutioncomposite-serieslimitsseries-expansionapproximation
Further Maths34 questionsStep-by-step solutions
Question 1
9 markschallenging
Which of the following is the series expansion of f(x)=ln⁡(1−x)1+2xf(x)=\frac{\ln\left(1-x\right)}{1+2x} in ascending powers of xx, up to and including the term in x3x^{3}?
Show worked solution

Worked solution

  1. Write down the function to be expanded

    f(x)=ln⁡(1−x)1+2xf(x)=\frac{\ln\left(1-x\right)}{1+2x}

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

  2. Quote the standard series for the first factor

    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. Substitute u=−xu=-x into the standard series for the first factor

    u=−xu=-x

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

  4. Expand ln⁡(1−x)\ln\left(1-x\right) in ascending powers of xx for the first factor

    ln⁡(1−x)=−x−x22−x33+⋯\ln\left(1-x\right)=-x-\frac{x^{2}}{2}-\frac{x^{3}}{3}+\cdots

    Every power of xx above x3x^{3} is discarded straight away.

  5. Quote the standard series for the second factor

    (1+u)n=1+nu+n(n−1)2!u2+n(n−1)(n−2)3!u3+⋯\left(1+u\right)^{n}=1+nu+\frac{n\left(n-1\right)}{2!}u^{2}+\frac{n\left(n-1\right)\left(n-2\right)}{3!}u^{3}+\cdots

    This standard result is quoted, not re-derived.

  6. Substitute u=2xu=2x into the standard series for the second factor

    u=2x,n=−1u=2x,\quad n=-1

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

  7. Expand (1+2x)−1\left(1+2x\right)^{-1} in ascending powers of xx for the second factor

    (1+2x)−1=1−2x+4x2−8x3+⋯\left(1+2x\right)^{-1}=1-2x+4x^{2}-8x^{3}+\cdots

    Every power of xx above x3x^{3} is discarded straight away.

  8. Multiply the series together

    f(x)≈(−x−x22−x33)(1−2x+4x2−8x3)f(x)\approx \left(-x-\frac{x^{2}}{2}-\frac{x^{3}}{3}\right)\left(1-2x+4x^{2}-8x^{3}\right)

    Only the products that give a power of xx at most x3x^{3} are needed.

  9. Expand the product and collect like powers of xx

    f(x)≈−x+3x22−10x33+20x43+8x53+8x63+⋯f(x)\approx -x+\frac{3x^{2}}{2}-\frac{10x^{3}}{3}+\frac{20x^{4}}{3}+\frac{8x^{5}}{3}+\frac{8x^{6}}{3}+\cdots

    Collecting like terms gives the coefficients of the composite series.

  10. Discard the terms of degree greater than x3x^{3}

    f(x)≈−x+3x22−10x33f(x)\approx -x+\frac{3x^{2}}{2}-\frac{10x^{3}}{3}

    The question only asks for the terms up to x3x^{3}.

  11. State the interval of validity of the expansion

    −12<x<12-\frac{1}{2} < x < \frac{1}{2}

    The expansion is only valid where every standard series used converges.

  12. Check the expansion numerically at x=150x=\frac{1}{50}

    f(150)≈−0.01942568,P(150)≈−0.019426667f\left(\frac{1}{50}\right)\approx -0.01942568,\qquad P\left(\frac{1}{50}\right)\approx -0.019426667

    The truncated series and the function agree to several decimal places for small xx, as they must.

  13. Check the expansion numerically at x=1100x=\frac{1}{100}

    f(1100)≈−0.0098532704,P(1100)≈−0.0098533333f\left(\frac{1}{100}\right)\approx -0.0098532704,\qquad P\left(\frac{1}{100}\right)\approx -0.0098533333

    The truncated series and the function agree to several decimal places for small xx, as they must.

  14. Recall the exponential 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 is a standard result; it is valid for every real uu.

  15. Recall the sine series

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

    Only odd powers appear, and the signs alternate.

  16. Recall the cosine series

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

    Only even powers appear, and the signs alternate.

  17. Select the option matching this expansion

    f(x)≈−x+3x22−10x33f(x)\approx -x+\frac{3x^{2}}{2}-\frac{10x^{3}}{3}

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

Answer
−x+3x22−10x33-x+\frac{3x^{2}}{2}-\frac{10x^{3}}{3}
Question 2
9 markschallenging
Which of the following is the series expansion of f(x)=1(1+x)(1−2x)f(x)=\frac{1}{\left(1+x\right)\left(1-2x\right)} in ascending powers of xx, up to and including the term in x3x^{3}?
Show worked solution

Worked solution

  1. Write down the function to be expanded

    f(x)=1(1+x)(1−2x)f(x)=\frac{1}{\left(1+x\right)\left(1-2x\right)}

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

  2. Quote the standard series for the first factor

    (1+u)n=1+nu+n(n−1)2!u2+n(n−1)(n−2)3!u3+⋯\left(1+u\right)^{n}=1+nu+\frac{n\left(n-1\right)}{2!}u^{2}+\frac{n\left(n-1\right)\left(n-2\right)}{3!}u^{3}+\cdots

    This standard result is quoted, not re-derived.

  3. Substitute u=xu=x into the standard series for the first factor

    u=x,n=−1u=x,\quad n=-1

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

  4. Expand (1+x)−1\left(1+x\right)^{-1} in ascending powers of xx for the first factor

    (1+x)−1=1−x+x2−x3+⋯\left(1+x\right)^{-1}=1-x+x^{2}-x^{3}+\cdots

    Every power of xx above x3x^{3} is discarded straight away.

  5. Quote the standard series for the second factor

    (1+u)n=1+nu+n(n−1)2!u2+n(n−1)(n−2)3!u3+⋯\left(1+u\right)^{n}=1+nu+\frac{n\left(n-1\right)}{2!}u^{2}+\frac{n\left(n-1\right)\left(n-2\right)}{3!}u^{3}+\cdots

    This standard result is quoted, not re-derived.

  6. Substitute u=−2xu=-2x into the standard series for the second factor

    u=−2x,n=−1u=-2x,\quad n=-1

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

  7. Expand (1−2x)−1\left(1-2x\right)^{-1} in ascending powers of xx for the second factor

    (1−2x)−1=1+2x+4x2+8x3+⋯\left(1-2x\right)^{-1}=1+2x+4x^{2}+8x^{3}+\cdots

    Every power of xx above x3x^{3} is discarded straight away.

  8. Multiply the series together

    f(x)≈(1−x+x2−x3)(1+2x+4x2+8x3)f(x)\approx \left(1-x+x^{2}-x^{3}\right)\left(1+2x+4x^{2}+8x^{3}\right)

    Only the products that give a power of xx at most x3x^{3} are needed.

  9. Expand the product and collect like powers of xx

    f(x)≈1+x+3x2+5x3−6x4+4x5−8x6+⋯f(x)\approx 1+x+3x^{2}+5x^{3}-6x^{4}+4x^{5}-8x^{6}+\cdots

    Collecting like terms gives the coefficients of the composite series.

  10. Discard the terms of degree greater than x3x^{3}

    f(x)≈1+x+3x2+5x3f(x)\approx 1+x+3x^{2}+5x^{3}

    The question only asks for the terms up to x3x^{3}.

  11. State the interval of validity of the expansion

    −12<x<12-\frac{1}{2} < x < \frac{1}{2}

    The expansion is only valid where every standard series used converges.

  12. Check the expansion numerically at x=150x=\frac{1}{50}

    f(150)≈1.0212418,P(150)≈1.02124f\left(\frac{1}{50}\right)\approx 1.0212418,\qquad P\left(\frac{1}{50}\right)\approx 1.02124

    The truncated series and the function agree to several decimal places for small xx, as they must.

  13. Check the expansion numerically at x=1100x=\frac{1}{100}

    f(1100)≈1.0103051,P(1100)≈1.010305f\left(\frac{1}{100}\right)\approx 1.0103051,\qquad P\left(\frac{1}{100}\right)\approx 1.010305

    The truncated series and the function agree to several decimal places for small xx, as they must.

  14. Recall the exponential 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 is a standard result; it is valid for every real uu.

  15. Recall the sine series

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

    Only odd powers appear, and the signs alternate.

  16. Select the option matching this expansion

    f(x)≈1+x+3x2+5x3f(x)\approx 1+x+3x^{2}+5x^{3}

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

Answer
1+x+3x2+5x31+x+3x^{2}+5x^{3}
Question 3
9 markschallenging
Use the standard series expansions to find the first 44 non-zero terms in the series expansion of f(x)=sin⁡xcos⁡xf(x)=\sin x\cos x, in ascending powers of xx.
Show worked solution

Worked solution

  1. Write down the function to be expanded

    f(x)=sin⁡xcos⁡xf(x)=\sin x\cos x

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

  2. Quote the standard series for the first factor

    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=xu=x into the standard series for the first factor

    u=xu=x

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

  4. Expand sin⁡(x)\sin\left(x\right) in ascending powers of xx for the first factor

    sin⁡(x)=x−x36+x5120−x75040+⋯\sin\left(x\right)=x-\frac{x^{3}}{6}+\frac{x^{5}}{120}-\frac{x^{7}}{5040}+\cdots

    Every power of xx above x7x^{7} is discarded straight away.

  5. Quote the standard series for the second factor

    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.

  6. Substitute u=xu=x into the standard series for the second factor

    u=xu=x

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

  7. Expand cos⁡(x)\cos\left(x\right) in ascending powers of xx for the second factor

    cos⁡(x)=1−x22+x424−x6720+⋯\cos\left(x\right)=1-\frac{x^{2}}{2}+\frac{x^{4}}{24}-\frac{x^{6}}{720}+\cdots

    Every power of xx above x7x^{7} is discarded straight away.

  8. Multiply the series together

    f(x)≈(x−x36+x5120−x75040)(1−x22+x424−x6720)f(x)\approx \left(x-\frac{x^{3}}{6}+\frac{x^{5}}{120}-\frac{x^{7}}{5040}\right)\left(1-\frac{x^{2}}{2}+\frac{x^{4}}{24}-\frac{x^{6}}{720}\right)

    Only the products that give a power of xx at most x7x^{7} are needed.

  9. Expand the product and collect like powers of xx

    f(x)≈x−2x33+2x515−4x7315+41x960480−x1150400+x133628800+⋯f(x)\approx x-\frac{2x^{3}}{3}+\frac{2x^{5}}{15}-\frac{4x^{7}}{315}+\frac{41x^{9}}{60480}-\frac{x^{11}}{50400}+\frac{x^{13}}{3628800}+\cdots

    Collecting like terms gives the coefficients of the composite series.

  10. Discard the terms of degree greater than x7x^{7}

    f(x)≈x−2x33+2x515−4x7315f(x)\approx x-\frac{2x^{3}}{3}+\frac{2x^{5}}{15}-\frac{4x^{7}}{315}

    The question only asks for the terms up to x7x^{7}.

  11. State the interval of validity of the expansion

    x∈Rx\in\mathbb{R}

    The expansion is only valid where every standard series used converges.

  12. Check the expansion numerically at x=150x=\frac{1}{50}

    f(150)≈0.019994667,P(150)≈0.019994667f\left(\frac{1}{50}\right)\approx 0.019994667,\qquad P\left(\frac{1}{50}\right)\approx 0.019994667

    The truncated series and the function agree to several decimal places for small xx, as they must.

  13. Check the expansion numerically at x=1100x=\frac{1}{100}

    f(1100)≈0.0099993333,P(1100)≈0.0099993333f\left(\frac{1}{100}\right)\approx 0.0099993333,\qquad P\left(\frac{1}{100}\right)\approx 0.0099993333

    The truncated series and the function agree to several decimal places for small xx, as they must.

  14. Recall the exponential 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 is a standard result; it is valid for every real uu.

  15. Recall the sine series

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

    Only odd powers appear, and the signs alternate.

  16. Recall the cosine series

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

    Only even powers appear, and the signs alternate.

  17. Recall the logarithmic 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 series is only valid for −1<u≤1-1<u\leq 1.

  18. State the required series expansion

    f(x)≈x−2x33+2x515−4x7315f(x)\approx x-\frac{2x^{3}}{3}+\frac{2x^{5}}{15}-\frac{4x^{7}}{315}

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

Answer
x−2x33+2x515−4x7315x-\frac{2x^{3}}{3}+\frac{2x^{5}}{15}-\frac{4x^{7}}{315}
Question 4
9 markschallenging
Find the coefficient of x5x^{5} in the series expansion of f(x)=e2x1+xf(x)=\frac{e^{2x}}{1+x}.
Show worked solution

Worked solution

  1. Write down the function

    f(x)=e2x1+xf(x)=\frac{e^{2x}}{1+x}

    Only one coefficient is wanted, but the whole expansion up to x5x^{5} must be built.

  2. Quote the standard series for the first factor

    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=2xu=2x into the standard series for the first factor

    u=2xu=2x

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

  4. Expand e2xe^{2x} in ascending powers of xx for the first factor

    e2x=1+2x+2x2+4x33+2x43+4x515+⋯e^{2x}=1+2x+2x^{2}+\frac{4x^{3}}{3}+\frac{2x^{4}}{3}+\frac{4x^{5}}{15}+\cdots

    Every power of xx above x5x^{5} is discarded straight away.

  5. Quote the standard series for the second factor

    (1+u)n=1+nu+n(n−1)2!u2+n(n−1)(n−2)3!u3+⋯\left(1+u\right)^{n}=1+nu+\frac{n\left(n-1\right)}{2!}u^{2}+\frac{n\left(n-1\right)\left(n-2\right)}{3!}u^{3}+\cdots

    This standard result is quoted, not re-derived.

  6. Substitute u=xu=x into the standard series for the second factor

    u=x,n=−1u=x,\quad n=-1

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

  7. Expand (1+x)−1\left(1+x\right)^{-1} in ascending powers of xx for the second factor

    (1+x)−1=1−x+x2−x3+x4−x5+⋯\left(1+x\right)^{-1}=1-x+x^{2}-x^{3}+x^{4}-x^{5}+\cdots

    Every power of xx above x5x^{5} is discarded straight away.

  8. Multiply the expansions together

    f(x)≈1+x+x2+x33+x43−x515f(x)\approx 1+x+x^{2}+\frac{x^{3}}{3}+\frac{x^{4}}{3}-\frac{x^{5}}{15}

    Collect the terms that produce x5x^{5}.

  9. Read off the coefficient of x5x^{5}

    coefficient of x5=−115\text{coefficient of } x^{5} = -\frac{1}{15}

    The coefficient is the number multiplying x5x^{5}.

  10. Check the expansion numerically at x=150x=\frac{1}{50}

    f(150)≈1.0204027,P(150)≈1.0204027f\left(\frac{1}{50}\right)\approx 1.0204027,\qquad P\left(\frac{1}{50}\right)\approx 1.0204027

    The truncated series and the function agree to several decimal places for small xx, as they must.

  11. Check the expansion numerically at x=1100x=\frac{1}{100}

    f(1100)≈1.0101003,P(1100)≈1.0101003f\left(\frac{1}{100}\right)\approx 1.0101003,\qquad P\left(\frac{1}{100}\right)\approx 1.0101003

    The truncated series and the function agree to several decimal places for small xx, as they must.

  12. Recall the exponential 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 is a standard result; it is valid for every real uu.

  13. Recall the sine series

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

    Only odd powers appear, and the signs alternate.

  14. Recall the cosine series

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

    Only even powers appear, and the signs alternate.

  15. Recall the logarithmic 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 series is only valid for −1<u≤1-1<u\leq 1.

  16. Recall the binomial series

    (1+u)n=1+nu+n(n−1)2!u2+n(n−1)(n−2)3!u3+⋯\left(1+u\right)^{n}=1+nu+\frac{n\left(n-1\right)}{2!}u^{2}+\frac{n\left(n-1\right)\left(n-2\right)}{3!}u^{3}+\cdots

    For a non-integer index this is an infinite series, valid only for ∣u∣<1\left|u\right|<1.

  17. State the coefficient of x5x^{5}

    −115-\frac{1}{15}

    This is the required coefficient.

Answer
−115-\frac{1}{15}
Question 5
9 markschallenging
For which set of values of xx is the series expansion of f(x)=1+2x ln⁡(1−4x)f(x)=\sqrt{1+2x}\,\ln\left(1-4x\right) valid?
Show worked solution

Worked solution

  1. Identify the standard series being used (first factor)

    (1+u)n=1+nu+n(n−1)2!u2+n(n−1)(n−2)3!u3+⋯\left(1+u\right)^{n}=1+nu+\frac{n\left(n-1\right)}{2!}u^{2}+\frac{n\left(n-1\right)\left(n-2\right)}{3!}u^{3}+\cdots

    The expansion of 1+2x\sqrt{1+2x} comes from this standard result.

  2. State the condition under which that standard series is valid (first factor)

    valid for ∣u∣<1(n∉Z0+)\text{valid for } \left|u\right| < 1 \quad \left(n \notin \mathbb{Z}^{+}_{0}\right)

    This is the interval of validity of the standard series: the binomial series only converges for ∣u∣<1\left|u\right|<1 when the index is not a non-negative integer.

  3. Substitute u=2xu=2x into the condition (first factor)

    ∣2x∣<1\left|2x\right| < 1

    The condition on uu becomes a condition on xx.

  4. Solve the resulting inequality for xx (first factor)

    −12<x<12-\frac{1}{2} < x < \frac{1}{2}

    This is the set of values of xx for which this series converges.

  5. Identify the standard series being used (second factor)

    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

    The expansion of ln⁡(1−4x)\ln\left(1-4x\right) comes from this standard result.

  6. State the condition under which that standard series is valid (second factor)

    valid for −1<u≤1\text{valid for } -1 < u \leq 1

    This is the interval of validity of the standard series: the logarithmic series only converges for −1<u≤1-1<u\leq 1.

  7. Substitute u=−4xu=-4x into the condition (second factor)

    −1<−4x≤1-1 < -4x \leq 1

    The condition on uu becomes a condition on xx.

  8. Solve the resulting inequality for xx (second factor)

    −14≤x<14-\frac{1}{4} \leq x < \frac{1}{4}

    This is the set of values of xx for which this series converges.

  9. Require both conditions to hold at once

    −12<x<12and−14≤x<14-\frac{1}{2} < x < \frac{1}{2}\quad\text{and}\quad-\frac{1}{4} \leq x < \frac{1}{4}

    A combination of two series only converges where both of them converge.

  10. Take the intersection of the two sets

    {−12<x<12}∩{−14≤x<14}={−14≤x<14}\left\{-\frac{1}{2} < x < \frac{1}{2}\right\}\cap\left\{-\frac{1}{4} \leq x < \frac{1}{4}\right\}=\left\{-\frac{1}{4} \leq x < \frac{1}{4}\right\}

    The tighter of the two conditions wins at each end.

  11. Test x=0x=0 against the condition

    x=0  ⇒  validx = 0 \;\Rightarrow\; \text{valid}

    This value lies inside the interval of validity.

  12. Test x=54x=\frac{5}{4} against the condition

    x=54  ⇒  not validx = \frac{5}{4} \;\Rightarrow\; \text{not valid}

    This value lies outside the interval of validity.

  13. Test x=−54x=-\frac{5}{4} against the condition

    x=−54  ⇒  not validx = -\frac{5}{4} \;\Rightarrow\; \text{not valid}

    This value lies outside the interval of validity.

  14. Recall the exponential 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 is a standard result; it is valid for every real uu.

  15. Recall the sine series

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

    Only odd powers appear, and the signs alternate.

  16. Select the correct interval of validity

    −14≤x<14-\frac{1}{4} \leq x < \frac{1}{4}

    This is the complete set of values of xx for which the expansion is valid.

Answer
−14≤x<14-\frac{1}{4} \leq x < \frac{1}{4}

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