Further Maths Taylor series Practice Questions

Free Further Maths Taylor series practice questions with full step-by-step worked solutions. Covers maclaurin-series, standard-series, substitution, non-zero-terms. Practise exam-style problems and check your method.

maclaurin-seriesstandard-seriessubstitutionnon-zero-termstruncationbinomial-series
Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
Find the Maclaurin expansion of f(x)=exf(x)=e^{x} up to and including the term in x3x^{3}.
Show worked solution

Worked solution

  1. Quote the standard Maclaurin series for e^{X}

    eX=1+X+X22!+X33!+X44!+X55!+⋯e^{X}=1+X+\frac{X^{2}}{2!}+\frac{X^{3}}{3!}+\frac{X^{4}}{4!}+\frac{X^{5}}{5!}+\cdots

    This is the exponential series, quoted from the formula book.

  2. Substitute X=xX=x into the standard series

    ex=1+(x)+(x)22!+(x)33!+(x)44!+(x)55!+⋯e^{x}=1+\left(x\right)+\frac{\left(x\right)^{2}}{2!}+\frac{\left(x\right)^{3}}{3!}+\frac{\left(x\right)^{4}}{4!}+\frac{\left(x\right)^{5}}{5!}+\cdots

    The standard series holds for any argument, so the whole of xx may be substituted for XX.

  3. Simplify the powers and the factorials

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

    Each bracket is raised to the required power and the factorial denominators are evaluated.

  4. State the required expansion

    ex≈x36+x22+x+1e^{x}\approx\frac{x^{3}}{6} + \frac{x^{2}}{2} + x + 1

    This is the expansion asked for, and no further terms are required.

Answer
x36+x22+x+1\frac{x^{3}}{6} + \frac{x^{2}}{2} + x + 1
Question 2
2 markseasy
Which of the following is the Maclaurin expansion of f(x)=cosh⁡(2x)f(x)=\cosh\left(2x\right) up to and including the term in x2x^{2}?
Show worked solution

Worked solution

  1. Quote the standard Maclaurin series for \cosh X

    cosh⁡X=1+X22!+X44!+⋯\cosh X=1+\frac{X^{2}}{2!}+\frac{X^{4}}{4!}+\cdots

    This is the hyperbolic cosine series, quoted from the formula book.

  2. Substitute X=2xX=2x into the standard series

    cosh⁡(2x)=1+(2x)22!+(2x)44!+⋯\cosh\left(2x\right)=1+\frac{\left(2x\right)^{2}}{2!}+\frac{\left(2x\right)^{4}}{4!}+\cdots

    The standard series holds for any argument, so the whole of 2x2x may be substituted for XX.

  3. Simplify the powers and the factorials

    cosh⁡(2x)=2x2+1+⋯\cosh\left(2x\right)=2 x^{2} + 1+\cdots

    Each bracket is raised to the required power and the factorial denominators are evaluated.

  4. Select the correct expansion

    cosh⁡(2x)≈2x2+1\cosh\left(2x\right)\approx2 x^{2} + 1

    This is the expansion asked for, and no further terms are required.

Answer
2x2+12 x^{2} + 1
Question 3
4 marksintermediate
Which of the following is the coefficient of x2x^{2} in the Maclaurin expansion of f(x)=ln⁡(1+2x)f(x)=\ln\left(1+2x\right)?
Show worked solution

Worked solution

  1. Write down the coefficient in terms of a derivative

    a2=f(2)(0)2!a_{2}=\frac{f^{(2)}(0)}{2!}

    The coefficient of x2x^{2} in a Maclaurin series is the 2th derivative at 00 divided by 2!2!.

  2. Differentiate to obtain the derivative of order 1

    f(1)(x)=22x+1f^{(1)}(x)=\frac{2}{2 x + 1}

    Each differentiation moves one step closer to the required order.

  3. Differentiate to obtain the derivative of order 2

    f(2)(x)=−4(2x+1)2f^{(2)}(x)=- \frac{4}{\left(2 x + 1\right)^{2}}

    Each differentiation moves one step closer to the required order.

  4. Evaluate the derivative of order 2 at x=0x=0

    f(2)(0)=−4f^{(2)}(0)=-4

    Only the value at the expansion point is needed.

  5. Divide by 2!2!

    a2=−42=−2a_{2}=\frac{-4}{2}=-2

    The factorial denominator is what turns the derivative into a series coefficient.

  6. Select the correct coefficient

    a2=−2a_{2}=-2

    This is the required coefficient of x2x^{2}.

Answer
−2-2
Question 4
6 markshard
Given that dydx=x+y2\frac{\mathrm{d}y}{\mathrm{d}x}=x+y^{2} and that y=1y=1 when x=0x=0, which of the following is the series solution for yy, up to and including the term in x4x^{4}?
Show worked solution

Worked solution

  1. Write down the differential equation and the initial condition

    dydx=x+y2,y=1 at x=0\frac{\mathrm{d}y}{\mathrm{d}x}=x+y^{2},\qquad y=1\text{ at }x=0

    The Taylor-series method builds the solution from the derivatives of yy at 00.

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

    y′(0)=1y'(0)=1

    Substituting x=0x=0 and y=1y=1 into the right-hand side gives the first derivative.

  3. Differentiate the equation to obtain the derivative of order 2

    y′′=2yy′+1y''=2 y y' + 1

    The right-hand side is differentiated with respect to xx, using the chain rule on every yy.

  4. Evaluate the derivative of order 2 at x=0x=0

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

    Every derivative already found is substituted into this expression.

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

    y′′′=2yy′′+2y′2y'''=2 y y'' + 2 y'^{2}

    The right-hand side is differentiated with respect to xx, using the chain rule on every yy.

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

    y(3)(0)=8y^{(3)}(0)=8

    Every derivative already found is substituted into this expression.

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

    y(4)=2yy′′′+6y′y′′y^{(4)}=2 y y''' + 6 y' y''

    The right-hand side is differentiated with respect to xx, using the chain rule on every yy.

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

    y(4)(0)=34y^{(4)}(0)=34

    Every derivative already found is substituted into this expression.

  9. Substitute the derivatives into the Maclaurin series for yy

    y≈1+(1)x+32!x2+83!x3+344!x4y\approx1+\left(1\right)x+\frac{3}{2!}x^{2}+\frac{8}{3!}x^{3}+\frac{34}{4!}x^{4}

    The Maclaurin series for yy uses the derivatives of yy at x=0x=0.

  10. Select the correct series solution

    y≈17x412+4x33+3x22+x+1y\approx\frac{17 x^{4}}{12} + \frac{4 x^{3}}{3} + \frac{3 x^{2}}{2} + x + 1

    This is the required series solution for yy.

Answer
17x412+4x33+3x22+x+1\frac{17 x^{4}}{12} + \frac{4 x^{3}}{3} + \frac{3 x^{2}}{2} + x + 1
Question 5
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}

Unlock 65 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

Hard QuestionsWorked Solutions

Related Further Pure topics

Conic sections 1Conic sections 2Further complex numbersGroupsInequalitiesFurther integration techniques