Further Maths Maclaurin series Practice Questions

Free Further Maths Maclaurin series practice questions with full step-by-step worked solutions. Covers standard-results, substitution, composite-series, interval-of-validity. Practise exam-style problems and check your method.

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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}.
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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
Which of the following is the series expansion of f(x)=ln⁡(1+5x)f(x)=\ln\left(1+5x\right) in ascending powers of xx, up to and including the term in x2x^{2}?
Show worked solution

Worked solution

  1. Write down the function to be expanded

    f(x)=ln⁡(1+5x)f(x)=\ln\left(1+5x\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. 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. Select the option matching this expansion

    f(x)≈5x−25x22f(x)\approx 5x-\frac{25x^{2}}{2}

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

Answer
5x−25x225x-\frac{25x^{2}}{2}
Question 3
4 marksintermediate
Which of the following is the coefficient of x4x^{4} in the series expansion of f(x)=ln⁡(1+x)1−xf(x)=\frac{\ln\left(1+x\right)}{1-x}?
Show worked solution

Worked solution

  1. Write down the function

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

    Only one coefficient is wanted, but the whole expansion up to x4x^{4} must be built.

  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−x44+⋯\ln\left(1+x\right)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\cdots

    Every power of xx above x4x^{4} 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. Select the correct coefficient

    712\frac{7}{12}

    This is the required coefficient.

Answer
712\frac{7}{12}
Question 4
6 markshard
Using series expansions, which of the following is the value of lim⁡x→0ln⁡(1+3x)−3xx2\lim_{x\to 0}\frac{\ln\left(1+3x\right)-3x}{x^{2}}?
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Worked solution

  1. Write down the limit to be found

    lim⁡x→0ln⁡(1+3x)−3xx2\lim_{x\to0}\frac{\ln\left(1+3x\right)-3x}{x^{2}}

    Both the numerator and the denominator tend to 00, so the standard series are used.

  2. Expand the numerator using the standard series

    ln⁡(1+3x)−3x=−9x22+9x3−81x44+⋯\ln\left(1+3x\right)-3x=-\frac{9x^{2}}{2}+9x^{3}-\frac{81x^{4}}{4}+\cdots

    Each function in the numerator is replaced by its Maclaurin series.

  3. Expand the denominator using the standard series

    x2=x2+⋯x^{2}=x^{2}+\cdots

    The denominator is expanded to the same accuracy.

  4. Identify the lowest power of xx in the denominator

    x2x^{2}

    The lowest power present decides how far the numerator must be expanded.

  5. Form the quotient of the two series

    −9x22+9x3−81x44+⋯x2+⋯\frac{-\frac{9x^{2}}{2}+9x^{3}-\frac{81x^{4}}{4}+\cdots}{x^{2}+\cdots}

    Divide the leading behaviour of the numerator by that of the denominator.

  6. Cancel the common factor of x2x^{2}

    =−92+9x−81x24+⋯1+⋯=\frac{-\frac{9}{2}+9x-\frac{81x^{2}}{4}+\cdots}{1+\cdots}

    Cancelling x2x^{2} leaves an expression with a finite value at x=0x=0.

  7. Let x→0x\to 0

    lim⁡x→0ln⁡(1+3x)−3xx2=−92\lim_{x\to0}\frac{\ln\left(1+3x\right)-3x}{x^{2}}=-\frac{9}{2}

    Every remaining term containing xx tends to 00.

  8. Check numerically at x=1100x=\frac{1}{100}

    ln⁡(1+3x)−3xx2≈−4.4119776\frac{\ln\left(1+3x\right)-3x}{x^{2}}\approx -4.4119776

    The quotient is already close to the limit for small xx.

  9. Check numerically at x=11000x=\frac{1}{1000}

    ln⁡(1+3x)−3xx2≈−4.4910202\frac{\ln\left(1+3x\right)-3x}{x^{2}}\approx -4.4910202

    The quotient is already close to the limit for small xx.

  10. 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.

  11. 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.

  12. Select the value of the limit

    −92-\frac{9}{2}

    This is the value of the limit.

Answer
−92-\frac{9}{2}
Question 5
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}

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