Hard Further Maths Taylor series Questions

Challenging, exam-style Further Maths Taylor series questions with worked solutions. Stretch yourself on the hardest taylor-series, expansion-about-a-point, maclaurin-series, product-of-series problems.

taylor-seriesexpansion-about-a-pointmaclaurin-seriesproduct-of-seriessuccessive-differentiationseries-solution
Further Maths34 questionsStep-by-step solutions
Question 1
9 markschallenging
Which of the following is the Maclaurin expansion of f(x)=x2exf(x)=x^{2}e^{x} up to and including the term in x5x^{5}?
Show worked solution

Worked solution

  1. Expand the first factor as a Maclaurin series

    x2=x2+⋯x^{2}=x^{2}+\cdots

    Only terms up to x5x^{5} can affect the answer.

  2. Expand the second factor as a Maclaurin series

    ex=x5120+x424+x36+x22+x+1+⋯e^{x}=\frac{x^{5}}{120} + \frac{x^{4}}{24} + \frac{x^{3}}{6} + \frac{x^{2}}{2} + x + 1+\cdots

    The second series is truncated at the same power.

  3. Write the product of the two truncated series

    x2ex=(x2)(x5120+x424+x36+x22+x+1)x^{2}e^{x}=\left(x^{2}\right)\left(\frac{x^{5}}{120} + \frac{x^{4}}{24} + \frac{x^{3}}{6} + \frac{x^{2}}{2} + x + 1\right)

    Multiplying the series term by term produces the expansion of the product.

  4. List the products of terms whose degrees add to 0

    (0)(1)\left(0\right)\left(1\right)

    A term in x0x^{0} arises from every pair of degrees summing to 00.

  5. Add the contributions to obtain the coefficient of x0x^{0}

    a0=0a_{0}=0

    This is the coefficient of x0x^{0} in the product.

  6. List the products of terms whose degrees add to 1

    (0)(1)+(0)(1)\left(0\right)\left(1\right)+\left(0\right)\left(1\right)

    A term in x1x^{1} arises from every pair of degrees summing to 11.

  7. Add the contributions to obtain the coefficient of x1x^{1}

    a1=0a_{1}=0

    This is the coefficient of x1x^{1} in the product.

  8. List the products of terms whose degrees add to 2

    (0)(12)+(0)(1)+(1)(1)\left(0\right)\left(\frac{1}{2}\right)+\left(0\right)\left(1\right)+\left(1\right)\left(1\right)

    A term in x2x^{2} arises from every pair of degrees summing to 22.

  9. Add the contributions to obtain the coefficient of x2x^{2}

    a2=1a_{2}=1

    This is the coefficient of x2x^{2} in the product.

  10. List the products of terms whose degrees add to 3

    (0)(16)+(0)(12)+(1)(1)+(0)(1)\left(0\right)\left(\frac{1}{6}\right)+\left(0\right)\left(\frac{1}{2}\right)+\left(1\right)\left(1\right)+\left(0\right)\left(1\right)

    A term in x3x^{3} arises from every pair of degrees summing to 33.

  11. Add the contributions to obtain the coefficient of x3x^{3}

    a3=1a_{3}=1

    This is the coefficient of x3x^{3} in the product.

  12. List the products of terms whose degrees add to 4

    (0)(124)+(0)(16)+(1)(12)+(0)(1)+(0)(1)\left(0\right)\left(\frac{1}{24}\right)+\left(0\right)\left(\frac{1}{6}\right)+\left(1\right)\left(\frac{1}{2}\right)+\left(0\right)\left(1\right)+\left(0\right)\left(1\right)

    A term in x4x^{4} arises from every pair of degrees summing to 44.

  13. Add the contributions to obtain the coefficient of x4x^{4}

    a4=12a_{4}=\frac{1}{2}

    This is the coefficient of x4x^{4} in the product.

  14. List the products of terms whose degrees add to 5

    (0)(1120)+(0)(124)+(1)(16)+(0)(12)+(0)(1)+(0)(1)\left(0\right)\left(\frac{1}{120}\right)+\left(0\right)\left(\frac{1}{24}\right)+\left(1\right)\left(\frac{1}{6}\right)+\left(0\right)\left(\frac{1}{2}\right)+\left(0\right)\left(1\right)+\left(0\right)\left(1\right)

    A term in x5x^{5} arises from every pair of degrees summing to 55.

  15. Add the contributions to obtain the coefficient of x5x^{5}

    a5=16a_{5}=\frac{1}{6}

    This is the coefficient of x5x^{5} in the product.

  16. Select the correct expansion

    x2ex≈x56+x42+x3+x2x^{2}e^{x}\approx\frac{x^{5}}{6} + \frac{x^{4}}{2} + x^{3} + x^{2}

    Collecting the coefficients gives the required expansion.

Answer
x56+x42+x3+x2\frac{x^{5}}{6} + \frac{x^{4}}{2} + x^{3} + x^{2}
Question 2
9 markschallenging
Given that d2ydx2=dydx+xy\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}=\frac{\mathrm{d}y}{\mathrm{d}x}+xy, with y=1y=1 and dydx=1\frac{\mathrm{d}y}{\mathrm{d}x}=1 when x=0x=0, which of the following is the series solution for yy, up to and including the term in x5x^{5}?
Show worked solution

Worked solution

  1. Write down the differential equation

    d2ydx2=dydx+xy\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}=\frac{\mathrm{d}y}{\mathrm{d}x}+xy

    A second-order equation needs two initial values before the series can start.

  2. Write down the two initial conditions

    y(0)=1,y′(0)=1y(0)=1,\qquad y'(0)=1

    These give the first two coefficients of the Maclaurin series directly.

  3. Evaluate y′′y'' at x=0x=0 from the differential equation

    y′′(0)=1y''(0)=1

    Substituting x=0x=0, y=1y=1 and y′=1y'=1 into the right-hand side gives y′′(0)y''(0).

  4. Differentiate the equation to obtain the derivative of order 3

    y′′′=xy′+y+y′′y'''=x y' + y + y''

    Differentiating both sides raises the order of the equation by one.

  5. Evaluate the derivative of order 3 at x=0x=0

    y(3)(0)=2y^{(3)}(0)=2

    All the lower derivatives at 00 are already known.

  6. Differentiate the equation to obtain the derivative of order 4

    y(4)=xy′′+2y′+y′′′y^{(4)}=x y'' + 2 y' + y'''

    Differentiating both sides raises the order of the equation by one.

  7. Evaluate the derivative of order 4 at x=0x=0

    y(4)(0)=4y^{(4)}(0)=4

    All the lower derivatives at 00 are already known.

  8. Differentiate the equation to obtain the derivative of order 5

    y(5)=xy′′′+3y′′+y(4)y^{(5)}=x y''' + 3 y'' + y^{(4)}

    Differentiating both sides raises the order of the equation by one.

  9. Evaluate the derivative of order 5 at x=0x=0

    y(5)(0)=7y^{(5)}(0)=7

    All the lower derivatives at 00 are already known.

  10. Substitute the derivatives into the Maclaurin series for yy

    y≈1+(1)x+12!x2+23!x3+44!x4+75!x5y\approx1+\left(1\right)x+\frac{1}{2!}x^{2}+\frac{2}{3!}x^{3}+\frac{4}{4!}x^{4}+\frac{7}{5!}x^{5}

    Each derivative at 00 is divided by the corresponding factorial.

  11. Check the two initial conditions

    y(0)=1,y′(0)=1y(0)=1,\quad y'(0)=1

    The constant and linear coefficients of the series reproduce the given data.

  12. Check the series in the differential equation

    y′′−(dydx+xy)=0+⋯y''-\left(\frac{\mathrm{d}y}{\mathrm{d}x}+xy\right)=0+\cdots

    The truncated series satisfies the equation to the order calculated.

  13. Recall the Maclaurin series in its general form

    f(x)=f(0)+f′(0)x+f′′(0)2!x2+f′′′(0)3!x3+⋯f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^{2}+\frac{f'''(0)}{3!}x^{3}+\cdots

    The Maclaurin series is the Taylor series taken about the point x=0x=0.

  14. Recall the Taylor series about a general point

    f(x)=∑n=0∞f(n)(a)n!(x−a)nf(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}\left(x-a\right)^{n}

    Every coefficient is a derivative evaluated at the expansion point, divided by n!n!.

  15. Select the correct series solution

    y≈7x5120+x46+x33+x22+x+1y\approx\frac{7 x^{5}}{120} + \frac{x^{4}}{6} + \frac{x^{3}}{3} + \frac{x^{2}}{2} + x + 1

    This is the required series solution for yy.

Answer
7x5120+x46+x33+x22+x+1\frac{7 x^{5}}{120} + \frac{x^{4}}{6} + \frac{x^{3}}{3} + \frac{x^{2}}{2} + x + 1
Question 3
9 markschallenging
Which of the following is the Taylor expansion of f(x)=e2xf(x)=e^{2x} about x=1x=1, up to and including the term in (x−1)4\left(x-1\right)^{4}?
Show worked solution

Worked solution

  1. Write down the Taylor expansion about x=1x=1

    f(x)=∑n=0∞f(n)(1)n!(x−1)nf(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}\left(1\right)}{n!}\left(x-1\right)^{n}

    Each coefficient needs one derivative evaluated at the expansion point.

  2. Evaluate the function at the expansion point

    f(1)=e2f\left(1\right)=e^{2}

    This provides the constant term of the expansion.

  3. Differentiate to obtain the derivative of order 1

    f(1)(x)=2e2xf^{(1)}(x)=2 e^{2 x}

    Differentiating once more gives the next derivative.

  4. Evaluate the derivative of order 1 at the expansion point

    f(1)(1)=2e2f^{(1)}\left(1\right)=2 e^{2}

    This value supplies the coefficient of (x−a)1\left(x-a\right)^{1} after division by 1!1!.

  5. Differentiate to obtain the derivative of order 2

    f(2)(x)=4e2xf^{(2)}(x)=4 e^{2 x}

    Differentiating once more gives the next derivative.

  6. Evaluate the derivative of order 2 at the expansion point

    f(2)(1)=4e2f^{(2)}\left(1\right)=4 e^{2}

    This value supplies the coefficient of (x−a)2\left(x-a\right)^{2} after division by 2!2!.

  7. Differentiate to obtain the derivative of order 3

    f(3)(x)=8e2xf^{(3)}(x)=8 e^{2 x}

    Differentiating once more gives the next derivative.

  8. Evaluate the derivative of order 3 at the expansion point

    f(3)(1)=8e2f^{(3)}\left(1\right)=8 e^{2}

    This value supplies the coefficient of (x−a)3\left(x-a\right)^{3} after division by 3!3!.

  9. Differentiate to obtain the derivative of order 4

    f(4)(x)=16e2xf^{(4)}(x)=16 e^{2 x}

    Differentiating once more gives the next derivative.

  10. Evaluate the derivative of order 4 at the expansion point

    f(4)(1)=16e2f^{(4)}\left(1\right)=16 e^{2}

    This value supplies the coefficient of (x−a)4\left(x-a\right)^{4} after division by 4!4!.

  11. Substitute the derivative values into the Taylor formula

    f(x)\approxe2+2e21!(x−1)+4e22!(x−1)2+8e23!(x−1)3+16e24!(x−1)4f(x)\approxe^{2}+\frac{2 e^{2}}{1!}\left(x - 1\right)+\frac{4 e^{2}}{2!}\left(x - 1\right)^{2}+\frac{8 e^{2}}{3!}\left(x - 1\right)^{3}+\frac{16 e^{2}}{4!}\left(x - 1\right)^{4}

    Each derivative is divided by the corresponding factorial.

  12. Simplify the coefficients

    2(x−1)4e23+4(x−1)3e23+2(x−1)2e2+2(x−1)e2+e2\frac{2 \left(x - 1\right)^{4} e^{2}}{3} + \frac{4 \left(x - 1\right)^{3} e^{2}}{3} + 2 \left(x - 1\right)^{2} e^{2} + 2 \left(x - 1\right) e^{2} + e^{2}

    Cancelling the factorials gives the coefficients in their simplest form.

  13. Check the expansion numerically at x=1110x=\frac{11}{10}

    f(1110)≈9.02501,T(1110)≈9.02499f\left(\frac{11}{10}\right)\approx9.02501,\quad T\left(\frac{11}{10}\right)\approx9.02499

    The truncated series reproduces the function closely near the expansion point, as it must.

  14. Check the expansion numerically at x=2120x=\frac{21}{20}

    f(2120)≈8.16617,T(2120)≈8.16617f\left(\frac{21}{20}\right)\approx8.16617,\quad T\left(\frac{21}{20}\right)\approx8.16617

    The truncated series reproduces the function closely near the expansion point, as it must.

  15. Select the correct expansion

    f(x)≈2(x−1)4e23+4(x−1)3e23+2(x−1)2e2+2(x−1)e2+e2f(x)\approx\frac{2 \left(x - 1\right)^{4} e^{2}}{3} + \frac{4 \left(x - 1\right)^{3} e^{2}}{3} + 2 \left(x - 1\right)^{2} e^{2} + 2 \left(x - 1\right) e^{2} + e^{2}

    This is the Taylor polynomial requested.

Answer
2(x−1)4e23+4(x−1)3e23+2(x−1)2e2+2(x−1)e2+e2\frac{2 \left(x - 1\right)^{4} e^{2}}{3} + \frac{4 \left(x - 1\right)^{3} e^{2}}{3} + 2 \left(x - 1\right)^{2} e^{2} + 2 \left(x - 1\right) e^{2} + e^{2}
Question 4
9 markschallenging
Which of the following gives the first 44 non-zero terms of the Maclaurin expansion of f(x)=cos⁡(x)f(x)=\cos\left(x\right)?
Show worked solution

Worked solution

  1. Write down the Maclaurin expansion

    f(x)=∑n=0∞f(n)(0)n!xnf(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^{n}

    Each coefficient needs one derivative evaluated at x=0x=0.

  2. Evaluate the function at the expansion point

    f(0)=1f\left(0\right)=1

    This provides the constant term of the expansion.

  3. Differentiate to obtain the derivative of order 1

    f(1)(x)=−sin⁡(x)f^{(1)}(x)=- \sin{\left(x \right)}

    Differentiating once more gives the next derivative.

  4. Evaluate the derivative of order 1 at the expansion point

    f(1)(0)=0f^{(1)}\left(0\right)=0

    This value supplies the coefficient of (x−a)1\left(x-a\right)^{1} after division by 1!1!.

  5. Differentiate to obtain the derivative of order 2

    f(2)(x)=−cos⁡(x)f^{(2)}(x)=- \cos{\left(x \right)}

    Differentiating once more gives the next derivative.

  6. Evaluate the derivative of order 2 at the expansion point

    f(2)(0)=−1f^{(2)}\left(0\right)=-1

    This value supplies the coefficient of (x−a)2\left(x-a\right)^{2} after division by 2!2!.

  7. Differentiate to obtain the derivative of order 3

    f(3)(x)=sin⁡(x)f^{(3)}(x)=\sin{\left(x \right)}

    Differentiating once more gives the next derivative.

  8. Evaluate the derivative of order 3 at the expansion point

    f(3)(0)=0f^{(3)}\left(0\right)=0

    This value supplies the coefficient of (x−a)3\left(x-a\right)^{3} after division by 3!3!.

  9. Differentiate to obtain the derivative of order 4

    f(4)(x)=cos⁡(x)f^{(4)}(x)=\cos{\left(x \right)}

    Differentiating once more gives the next derivative.

  10. Evaluate the derivative of order 4 at the expansion point

    f(4)(0)=1f^{(4)}\left(0\right)=1

    This value supplies the coefficient of (x−a)4\left(x-a\right)^{4} after division by 4!4!.

  11. Differentiate to obtain the derivative of order 5

    f(5)(x)=−sin⁡(x)f^{(5)}(x)=- \sin{\left(x \right)}

    Differentiating once more gives the next derivative.

  12. Evaluate the derivative of order 5 at the expansion point

    f(5)(0)=0f^{(5)}\left(0\right)=0

    This value supplies the coefficient of (x−a)5\left(x-a\right)^{5} after division by 5!5!.

  13. Differentiate to obtain the derivative of order 6

    f(6)(x)=−cos⁡(x)f^{(6)}(x)=- \cos{\left(x \right)}

    Differentiating once more gives the next derivative.

  14. Evaluate the derivative of order 6 at the expansion point

    f(6)(0)=−1f^{(6)}\left(0\right)=-1

    This value supplies the coefficient of (x−a)6\left(x-a\right)^{6} after division by 6!6!.

  15. Substitute the derivative values into the Taylor formula

    f(x)≈1+01!x+−12!x2+03!x3+14!x4+05!x5+−16!x6f(x)\approx1+\frac{0}{1!}x+\frac{-1}{2!}x^{2}+\frac{0}{3!}x^{3}+\frac{1}{4!}x^{4}+\frac{0}{5!}x^{5}+\frac{-1}{6!}x^{6}

    Each derivative is divided by the corresponding factorial.

  16. Simplify the coefficients

    −x6720+x424−x22+1- \frac{x^{6}}{720} + \frac{x^{4}}{24} - \frac{x^{2}}{2} + 1

    Cancelling the factorials gives the coefficients in their simplest form.

  17. Count the non-zero terms that have been obtained

    4 non-zero terms, of degrees 0,2,4,6\text{4 non-zero terms, of degrees }0,2,4,6

    Terms whose coefficient is zero do not count, so the expansion must be taken as far as degree 66.

  18. Select the correct expansion

    f(x)≈−x6720+x424−x22+1f(x)\approx- \frac{x^{6}}{720} + \frac{x^{4}}{24} - \frac{x^{2}}{2} + 1

    This is the Taylor polynomial requested.

Answer
−x6720+x424−x22+1- \frac{x^{6}}{720} + \frac{x^{4}}{24} - \frac{x^{2}}{2} + 1
Question 5
9 markschallenging
Find the first 44 non-zero terms of the Maclaurin expansion of f(x)=sin⁡(x)f(x)=\sin\left(x\right).
Show worked solution

Worked solution

  1. Write down the Maclaurin expansion

    f(x)=∑n=0∞f(n)(0)n!xnf(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^{n}

    Each coefficient needs one derivative evaluated at x=0x=0.

  2. Evaluate the function at the expansion point

    f(0)=0f\left(0\right)=0

    This provides the constant term of the expansion.

  3. Differentiate to obtain the derivative of order 1

    f(1)(x)=cos⁡(x)f^{(1)}(x)=\cos{\left(x \right)}

    Differentiating once more gives the next derivative.

  4. Evaluate the derivative of order 1 at the expansion point

    f(1)(0)=1f^{(1)}\left(0\right)=1

    This value supplies the coefficient of (x−a)1\left(x-a\right)^{1} after division by 1!1!.

  5. Differentiate to obtain the derivative of order 2

    f(2)(x)=−sin⁡(x)f^{(2)}(x)=- \sin{\left(x \right)}

    Differentiating once more gives the next derivative.

  6. Evaluate the derivative of order 2 at the expansion point

    f(2)(0)=0f^{(2)}\left(0\right)=0

    This value supplies the coefficient of (x−a)2\left(x-a\right)^{2} after division by 2!2!.

  7. Differentiate to obtain the derivative of order 3

    f(3)(x)=−cos⁡(x)f^{(3)}(x)=- \cos{\left(x \right)}

    Differentiating once more gives the next derivative.

  8. Evaluate the derivative of order 3 at the expansion point

    f(3)(0)=−1f^{(3)}\left(0\right)=-1

    This value supplies the coefficient of (x−a)3\left(x-a\right)^{3} after division by 3!3!.

  9. Differentiate to obtain the derivative of order 4

    f(4)(x)=sin⁡(x)f^{(4)}(x)=\sin{\left(x \right)}

    Differentiating once more gives the next derivative.

  10. Evaluate the derivative of order 4 at the expansion point

    f(4)(0)=0f^{(4)}\left(0\right)=0

    This value supplies the coefficient of (x−a)4\left(x-a\right)^{4} after division by 4!4!.

  11. Differentiate to obtain the derivative of order 5

    f(5)(x)=cos⁡(x)f^{(5)}(x)=\cos{\left(x \right)}

    Differentiating once more gives the next derivative.

  12. Evaluate the derivative of order 5 at the expansion point

    f(5)(0)=1f^{(5)}\left(0\right)=1

    This value supplies the coefficient of (x−a)5\left(x-a\right)^{5} after division by 5!5!.

  13. Differentiate to obtain the derivative of order 6

    f(6)(x)=−sin⁡(x)f^{(6)}(x)=- \sin{\left(x \right)}

    Differentiating once more gives the next derivative.

  14. Evaluate the derivative of order 6 at the expansion point

    f(6)(0)=0f^{(6)}\left(0\right)=0

    This value supplies the coefficient of (x−a)6\left(x-a\right)^{6} after division by 6!6!.

  15. Differentiate to obtain the derivative of order 7

    f(7)(x)=−cos⁡(x)f^{(7)}(x)=- \cos{\left(x \right)}

    Differentiating once more gives the next derivative.

  16. Evaluate the derivative of order 7 at the expansion point

    f(7)(0)=−1f^{(7)}\left(0\right)=-1

    This value supplies the coefficient of (x−a)7\left(x-a\right)^{7} after division by 7!7!.

  17. Substitute the derivative values into the Taylor formula

    f(x)≈0+11!x+02!x2+−13!x3+04!x4+15!x5+06!x6+−17!x7f(x)\approx0+\frac{1}{1!}x+\frac{0}{2!}x^{2}+\frac{-1}{3!}x^{3}+\frac{0}{4!}x^{4}+\frac{1}{5!}x^{5}+\frac{0}{6!}x^{6}+\frac{-1}{7!}x^{7}

    Each derivative is divided by the corresponding factorial.

  18. Simplify the coefficients

    −x75040+x5120−x36+x- \frac{x^{7}}{5040} + \frac{x^{5}}{120} - \frac{x^{3}}{6} + x

    Cancelling the factorials gives the coefficients in their simplest form.

  19. Count the non-zero terms that have been obtained

    4 non-zero terms, of degrees 1,3,5,7\text{4 non-zero terms, of degrees }1,3,5,7

    Terms whose coefficient is zero do not count, so the expansion must be taken as far as degree 77.

  20. State the required expansion

    f(x)≈−x75040+x5120−x36+xf(x)\approx- \frac{x^{7}}{5040} + \frac{x^{5}}{120} - \frac{x^{3}}{6} + x

    This is the Taylor polynomial requested.

Answer
−x75040+x5120−x36+x- \frac{x^{7}}{5040} + \frac{x^{5}}{120} - \frac{x^{3}}{6} + x

Unlock 29 more Taylor series questions

Create a free account to work through every Further Maths Taylor series question with instant step-by-step worked solutions, progress tracking and interactive lessons.

  • Full worked solutions for every question
  • Interactive lessons and instant feedback
  • Track your mastery across every topic
Create a Free Account

No card required · Free forever

More Taylor series practice

Practice QuestionsWorked Solutions

Related Further Pure topics

Conic sections 1Conic sections 2Further complex numbersGroupsInequalitiesFurther integration techniques