AS & A Level · Practice questions · Maclaurin series

Differentiate at zero, divide by the factorial.

Six original Cambridge-style questions on Maclaurin series: deriving series from the formula, successive differentiation, combining standard series, validity, approximations and limits.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Derivatives at \(0\), over \(r!\).

Formula: \(f(x) = f(0) + x f'(0) + \dfrac{x^{2}}{2!} f''(0) + \dfrac{x^{3}}{3!} f'''(0) + \dots\)
Table first: write \(f, f', f'', \dots\) in general, THEN put \(x = 0\) in each.
Messy functions: find a relation such as \(y' = 1 + y^{2}\) or \(y' = y\cos x\) and differentiate that.
Combine: substitute into standard series (bracket the new variable) and multiply, keeping every cross term up to the power asked.
Use it: small \(x\) for approximations, expand and divide by the lowest power for limits, and stay inside the range of validity.

Recall

[3 marks]

Use the Maclaurin formula to show that \(\cos x = 1 - \dfrac{x^{2}}{2} + \dfrac{x^{4}}{24} - \dots\), giving the derivatives you use.

\(f = \cos x\), \(f' = -\sin x\), \(f'' = -\cos x\), \(f''' = \sin x\), \(f^{(4)} = \cos x\). ✓
At \(x = 0\): \(f(0) = 1\), \(f'(0) = 0\), \(f''(0) = -1\), \(f'''(0) = 0\), \(f^{(4)}(0) = 1\). ✓
\[\cos x = 1 + 0 + \frac{x^{2}}{2!}(-1) + 0 + \frac{x^{4}}{4!}(1) = 1 - \frac{x^{2}}{2} + \frac{x^{4}}{24}.\] ✓

Calculation

[5 marks]

Let \(f(x) = \ln(1 + \sin x)\).

(a) Show that \(f''(x) = -\dfrac{1}{1 + \sin x}\).

(b) Hence find the Maclaurin series for \(f(x)\) up to and including the term in \(x^{3}\).

(a) \(f'(x) = \dfrac{\cos x}{1 + \sin x}\). Quotient rule: ✓
\[f''(x) = \frac{-\sin x\,(1 + \sin x) - \cos^{2} x}{(1 + \sin x)^{2}} = \frac{-\sin x - (\sin^{2} x + \cos^{2} x)}{(1 + \sin x)^{2}} = \frac{-(1 + \sin x)}{(1 + \sin x)^{2}}\] ✓
so \(f''(x) = -\dfrac{1}{1 + \sin x}\). ✓
(b) \(f'''(x) = \dfrac{\cos x}{(1 + \sin x)^{2}}\). At \(x = 0\): \(f(0) = \ln 1 = 0\), \(f'(0) = 1\), \(f''(0) = -1\), \(f'''(0) = 1\). ✓
\[\ln(1 + \sin x) = x - \frac{x^{2}}{2} + \frac{x^{3}}{6} + \dots\] ✓

Calculation

[4 marks]

(a) Write down the Maclaurin series for \(\ln(1 + 2x)\) up to and including the term in \(x^{3}\), and state the set of values of \(x\) for which the full series is valid.

(b) Use your answer to estimate \(\ln 1.2\), and compare it with the calculator value.

(a) Replace \(x\) by \(2x\) in \(\ln(1 + x) = x - \dfrac{x^{2}}{2} + \dfrac{x^{3}}{3} - \dots\): ✓
\[\ln(1 + 2x) = 2x - \frac{(2x)^{2}}{2} + \frac{(2x)^{3}}{3} - \dots = 2x - 2x^{2} + \frac{8x^{3}}{3} - \dots\] ✓
Valid for \(-1 \lt 2x \le 1\), that is \(-\tfrac{1}{2} \lt x \le \tfrac{1}{2}\). ✓
(b) \(x = 0.1\) (inside the range): \(0.2 - 0.02 + \dfrac{0.008}{3} = 0.182\,67\). ✓
The calculator gives \(\ln 1.2 = 0.182\,32\), so the estimate is good to about 3 decimal places; the next term, \(-\dfrac{(0.2)^{4}}{4} = -0.0004\), accounts for most of the gap. ✓

Analysis

[5 marks]

Using standard series, find the Maclaurin series for \(e^{2x}\sqrt{1 + x}\) up to and including the term in \(x^{2}\).

\(e^{2x} = 1 + 2x + \dfrac{(2x)^{2}}{2} + \dots = 1 + 2x + 2x^{2} + \dots\) ✓
\((1 + x)^{\frac{1}{2}} = 1 + \tfrac{1}{2}x + \dfrac{\tfrac{1}{2}\left(-\tfrac{1}{2}\right)}{2}x^{2} + \dots = 1 + \dfrac{x}{2} - \dfrac{x^{2}}{8} + \dots\) (valid for \(|x| \lt 1\)). ✓
Multiply, keeping powers up to \(x^{2}\). \(x\): \(2x + \dfrac{x}{2} = \dfrac{5x}{2}\). ✓
\(x^{2}\): \(2x^{2} + 2x \cdot \dfrac{x}{2} - \dfrac{x^{2}}{8} = 2x^{2} + x^{2} - \dfrac{x^{2}}{8} = \dfrac{23x^{2}}{8}\). ✓
\[e^{2x}\sqrt{1 + x} = 1 + \frac{5x}{2} + \frac{23x^{2}}{8} + \dots\] ✓

Analysis

[5 marks]

(a) Find the Maclaurin series for \(e^{x}\sin x\) up to and including the term in \(x^{3}\).

(b) Hence find \(\displaystyle\lim_{x \to 0} \frac{e^{x}\sin x - x - x^{2}}{x^{3}}\).

(a) \(\left(1 + x + \dfrac{x^{2}}{2} + \dots\right)\left(x - \dfrac{x^{3}}{6} + \dots\right)\). ✓
\(x\): \(x\). \(x^{2}\): \(x \cdot x = x^{2}\). \(x^{3}\): \(\dfrac{x^{2}}{2}\cdot x - \dfrac{x^{3}}{6} = \dfrac{x^{3}}{2} - \dfrac{x^{3}}{6} = \dfrac{x^{3}}{3}\). ✓
So \(e^{x}\sin x = x + x^{2} + \dfrac{x^{3}}{3} + \dots\) ✓
(b) The numerator is \(\dfrac{x^{3}}{3} + (\text{terms in } x^{4} \text{ and higher})\). ✓
Dividing by \(x^{3}\): \(\dfrac{1}{3} + (\text{terms in } x \text{ and higher}) \to \dfrac{1}{3}\) as \(x \to 0\). ✓

Proof

[6 marks]

Let \(f(x) = \ln(1 + e^{x})\).

(a) Show that \(f'(x) = \dfrac{1}{1 + e^{-x}}\) and that \(f''(x) = f'(x)\big(1 - f'(x)\big)\).

(b) Hence find the Maclaurin series for \(f(x)\) up to and including the term in \(x^{3}\).

(a) \(f'(x) = \dfrac{e^{x}}{1 + e^{x}}\); divide top and bottom by \(e^{x}\): \(f'(x) = \dfrac{1}{1 + e^{-x}}\). ✓
Then \(f''(x) = \dfrac{e^{-x}}{(1 + e^{-x})^{2}} = \dfrac{1}{1 + e^{-x}} \cdot \dfrac{e^{-x}}{1 + e^{-x}}\), and \(\dfrac{e^{-x}}{1 + e^{-x}} = 1 - \dfrac{1}{1 + e^{-x}} = 1 - f'(x)\). So \(f'' = f'(1 - f')\). ✓
(b) Differentiate the relation: \(f''' = f''(1 - f') + f'(-f'') = f''(1 - 2f')\). ✓
At \(x = 0\): \(f(0) = \ln 2\), \(f'(0) = \tfrac{1}{2}\), \(f''(0) = \tfrac{1}{2}\cdot\tfrac{1}{2} = \tfrac{1}{4}\), \(f'''(0) = \tfrac{1}{4}(1 - 1) = 0\). ✓
\[\ln(1 + e^{x}) = \ln 2 + \frac{x}{2} + \frac{x^{2}}{8} + 0x^{3} + \dots\] ✓
(c) \(x = 0.2\): \(\ln 2 + 0.1 + 0.005 = 0.693\,15 + 0.105 = 0.7981\) (4 d.p.), matching the calculator value \(0.798\,14\). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.