2. Combining Expansions

Video coming soon3 worked examples

Theory

Differentiating a function like $e^{2x}\sin x$ four or five times is punishing, and each derivative gets messier than the last. It is almost always faster to build the series from the standard expansions instead. There are three techniques:

Multiply two series together, collecting like powers
Substitute a function into a standard series, e.g. $u = x^2$ in $\ln(1+u)$
Factorise into standard form, e.g. $\ln(3+x) = \ln 3 + \ln\!\left(1 + \tfrac{x}{3}\right)$

The saving is real: a product that would need four rounds of the product rule reduces to multiplying out a bracket.

The Golden Rule: decide the highest power you need first, then take enough terms from each standard series to reach it — and discard any product that overshoots. If you want the $x^3$ term, a term in $x^2$ multiplied by one in $x^2$ can be ignored entirely.

⚠️ Common Examiner Traps

Taking too few terms: to reach $x^3$ in a product you may still need the $x^3$ term of each factor, since it pairs with the constant term of the other.
Keeping terms you don't need: anything above the required power is wasted work and invites arithmetic slips — cross it out as you go.
Substituting carelessly: replacing $x$ by $2x$ means every $x$ changes, so $\frac{x^2}{2}$ becomes $\frac{4x^2}{2} = 2x^2$ .
Ranges of validity carry over: $\ln(1+u)$ needs $-1 < u \le 1$ , so substituting $u = 2x$ restricts the result to $-\frac{1}{2} < x \le \frac{1}{2}$ .

Worked examples

Example 1

Use the standard expansions to find the series for $e^{2x}\sin x$ as far as the term in $x^3$ .

Step 1: Write down each standard series, substituting $2x$ into the one for $e^x$ . Take terms as far as $x^3$ :

e^{2x} = 1 + 2x + \frac{(2x)^2}{2!} + \frac{(2x)^3}{3!} + \cdots = 1 + 2x + 2x^2 + \frac{4}{3}x^3 + \cdots

\sin x = x - \frac{x^3}{3!} + \cdots = x - \frac{x^3}{6} + \cdots

Step 2: Multiply, keeping only products whose total power is $3$ or less:

\begin{aligned} 1 \times x &= x \\ 1 \times \left(-\tfrac{x^3}{6}\right) &= -\tfrac{1}{6}x^3 \\ 2x \times x &= 2x^2 \\ 2x^2 \times x &= 2x^3 \end{aligned}

Step 3: Every other product reaches $x^4$ or beyond, so discard them. Collect like powers:

e^{2x}\sin x = x + 2x^2 + \left(2 - \tfrac{1}{6}\right)x^3 + \cdots = x + 2x^2 + \frac{11}{6}x^3 + \cdots

Example 2

Find the series for $\ln(1 + x^2)$ as far as the term in $x^6$ , and state the range of values of $x$ for which it is valid.

Step 1: Start from the standard expansion, using $u$ as the variable to keep the substitution clear:

\ln(1+u) = u - \frac{u^2}{2} + \frac{u^3}{3} - \cdots \qquad (-1 < u \le 1)

Step 2: Substitute $u = x^2$ . Each power of $u$ doubles in $x$ , so three terms are enough to reach $x^6$ :

\ln(1+x^2) = x^2 - \frac{(x^2)^2}{2} + \frac{(x^2)^3}{3} - \cdots = x^2 - \frac{x^4}{2} + \frac{x^6}{3} - \cdots

Step 3: Carry the range through the substitution. We need $-1 < x^2 \le 1$ , and since $x^2 \ge 0$ always, this reduces to $x^2 \le 1$ :

-1 \le x \le 1

Example 3

Find the series for $\ln(3 + x)$ as far as the term in $x^3$ .

Step 1: The standard series needs the form $\ln(1 + \ldots)$ , so factor out the $3$ :

\ln(3+x) = \ln\!\left[3\left(1 + \frac{x}{3}\right)\right] = \ln 3 + \ln\!\left(1 + \frac{x}{3}\right)

Step 2: Expand the second term using $u = \dfrac{x}{3}$ :

\ln\!\left(1 + \frac{x}{3}\right) = \frac{x}{3} - \frac{1}{2}\left(\frac{x}{3}\right)^2 + \frac{1}{3}\left(\frac{x}{3}\right)^3 - \cdots

Step 3: Simplify each term carefully — the cube of $\frac{x}{3}$ brings a factor of $27$ :

= \frac{x}{3} - \frac{x^2}{18} + \frac{x^3}{81} - \cdots

Step 4: Add back the constant. Note that $\ln 3$ is the $x^0$ term of the series:

\ln(3+x) = \ln 3 + \frac{x}{3} - \frac{x^2}{18} + \frac{x^3}{81} - \cdots