Taylor series Worked Solutions — Further Maths Maths

Fully worked, step-by-step solutions to Further Maths Taylor series questions. See exactly how to solve problems on maclaurin-series, standard-series, substitution, non-zero-terms.

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}.

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
Find the Maclaurin expansion of f(x)=e2xf(x)=e^{2x} up to and including the term in x3x^{3}.

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=2xX=2x into the standard series

    e2x=1+(2x)+(2x)22!+(2x)33!+(2x)44!+(2x)55!+⋯e^{2x}=1+\left(2x\right)+\frac{\left(2x\right)^{2}}{2!}+\frac{\left(2x\right)^{3}}{3!}+\frac{\left(2x\right)^{4}}{4!}+\frac{\left(2x\right)^{5}}{5!}+\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

    e2x=4x33+2x2+2x+1+⋯e^{2x}=\frac{4 x^{3}}{3} + 2 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

    e2x≈4x33+2x2+2x+1e^{2x}\approx\frac{4 x^{3}}{3} + 2 x^{2} + 2 x + 1

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

Answer
4x33+2x2+2x+1\frac{4 x^{3}}{3} + 2 x^{2} + 2 x + 1
Question 3
2 markseasy
Find the Maclaurin expansion of f(x)=sin⁡(x)f(x)=\sin\left(x\right) up to and including the term in x3x^{3}.

Worked solution

  1. Quote the standard Maclaurin series for \sin X

    sin⁡X=X−X33!+X55!−⋯\sin X=X-\frac{X^{3}}{3!}+\frac{X^{5}}{5!}-\cdots

    This is the sine series (odd powers only), quoted from the formula book.

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

    sin⁡(x)=(x)−(x)33!+(x)55!−⋯\sin\left(x\right)=\left(x\right)-\frac{\left(x\right)^{3}}{3!}+\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

    sin⁡(x)=−x36+x+⋯\sin\left(x\right)=- \frac{x^{3}}{6} + x+\cdots

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

  4. State the required expansion

    sin⁡(x)≈−x36+x\sin\left(x\right)\approx- \frac{x^{3}}{6} + x

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

Answer
−x36+x- \frac{x^{3}}{6} + x
Question 4
2 markseasy
Find the Maclaurin expansion of f(x)=cos⁡(x)f(x)=\cos\left(x\right) up to and including the term in x3x^{3}.

Worked solution

  1. Quote the standard Maclaurin series for \cos X

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

    This is the cosine series (even powers only), quoted from the formula book.

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

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

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

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

  4. State the required expansion

    cos⁡(x)≈1−x22\cos\left(x\right)\approx1 - \frac{x^{2}}{2}

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

Answer
1−x221 - \frac{x^{2}}{2}
Question 5
2 markseasy
Find the Maclaurin expansion of f(x)=ln⁡(1+x)f(x)=\ln\left(1+x\right) up to and including the term in x3x^{3}.

Worked solution

  1. Quote the standard Maclaurin series for \ln\left(1+X\right)

    ln⁡(1+X)=X−X22+X33−X44+X55−⋯\ln\left(1+X\right)=X-\frac{X^{2}}{2}+\frac{X^{3}}{3}-\frac{X^{4}}{4}+\frac{X^{5}}{5}-\cdots

    This is the logarithmic series, valid for ∣@∣<1\left|@\right|<1, quoted from the formula book.

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

    ln⁡(1+x)=(x)−(x)22+(x)33−(x)44+(x)55−⋯\ln\left(1+x\right)=\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

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

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

  4. State the required expansion

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

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

Answer
x33−x22+x\frac{x^{3}}{3} - \frac{x^{2}}{2} + x

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