CP 4.6 - Standard and compound Maclaurin series | Edexcel A-Level Further Mathematics

Standard expansions let you build new series without repeatedly differentiating a complicated function. You will choose and adapt the five main series, keep their validity conditions and derive expansions of products and functions inside other functions.

The standard series and their ranges

Can a function's series contain a constant, an $x$ term and an $x^2$ term, then skip $x^3$ altogether? The function $\mathrm{e}^{\sin x}$ does exactly that. Its missing cubic term comes from cancellation, and we will uncover it by putting two standard series together.

You already know

A Maclaurin series expands about zero. Its coefficient of $x^r$ is $f^{(r)}(0)/r!$ , and a finite truncation usually approximates the function near zero. From A-level Mathematics, use index laws, logarithm laws and the general binomial expansion. Trigonometric arguments are in radians.

Read a standard series as a pattern: its signs, powers and denominators all matter. The logarithm uses ordinary integer denominators, whereas exponential and trigonometric series use factorials.

The five standard expansions

\mathrm{e}^x=\sum_{r=0}^{\infty}\frac{x^r}{r!} =1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots \qquad\text{all real }x

\ln(1+x)=\sum_{r=1}^{\infty}\frac{(-1)^{r+1}x^r}{r} =x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots \qquad(-1<x\le1)

\sin x=\sum_{r=0}^{\infty}\frac{(-1)^r x^{2r+1}}{(2r+1)!} =x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots \qquad\text{all real }x

\cos x=\sum_{r=0}^{\infty}\frac{(-1)^r x^{2r}}{(2r)!} =1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots \qquad\text{all real }x

For a real constant $a$ ,

(1+x)^a=1+ax+\frac{a(a-1)}{2!}x^2 +\frac{a(a-1)(a-2)}{3!}x^3+\cdots

Its term in $x^r$ , for $r\ge1$ , is

\frac{a(a-1)\cdots(a-r+1)}{r!}x^r

The general binomial series is valid for $|x|<1$ . For a non-negative integer exponent it terminates as the corresponding polynomial, valid for all real $x$ ; endpoint convergence of a non-terminating binomial series depends on the exponent.

In the formula booklet.

“All real $x$ ” belongs to the full exponential, sine and cosine series, not to a claim that a three-term truncation is accurate everywhere. The logarithm's positive endpoint is included: its series at $x=1$ converges to $\ln2$ . The negative endpoint is excluded, and $\ln0$ is undefined.

Key idea

Choose the standard function, identify its entire argument and carry the range condition through that substitution. A formula without its conditions is only half the information.

This check distinguishes the logarithm's exact endpoint rule from the usual open binomial condition.

Substitution and a non-unit constant

Before using a logarithm or binomial series, rewrite its inside as a non-zero constant times $1+u$ , with $u=0$ at $x=0$ .

For example, $4+3x=4(1+3x/4)$ . A logarithm turns the outside factor into an added constant; a power raises that outside factor to the same exponent. Every occurrence of the standard argument must then be replaced by $u$ , including its powers.

Worked example 1: normalising a logarithm

Question. Expand $\ln(4+3x)$ through the term in $x^3$ , and state the range of validity of its full series.

Plan. Extract $4$ , use the logarithm series with $u=3x/4$ and transform $-1<u\le1$ .

\begin{aligned} \ln(4+3x)&=\ln4+\ln\left(1+\frac{3x}{4}\right) &&\text{logarithm product law}\\ &=\ln4+\frac{3x}{4}-\frac12\left(\frac{3x}{4}\right)^2 +\frac13\left(\frac{3x}{4}\right)^3+\cdots &&\text{substitute the whole argument}\\ &=\ln4+\frac34x-\frac9{32}x^2+\frac9{64}x^3+\cdots &&\text{simplify coefficients} \end{aligned}

The range is

-1<\frac{3x}{4}\le1 \quad\Longrightarrow\quad -\frac43<x\le\frac43

Why this step? The standard logarithm series starts at $\ln1=0$ . Extracting the constant preserves the required value $\ln4$ at zero instead of accidentally replacing it by zero.

Answer. $\ln4+3x/4-9x^2/32+9x^3/64+\cdots$ , valid for $-4/3<x\le4/3$ .

Check. The value at zero is $\ln4$ , and the linear coefficient is $3/4$ , agreeing with $f'(0)=3/(4+0)$ .

Worked example 2 changes only the outside function, keeping the same $4+3x$ inside. The binomial exponent now determines both the outside constant and the coefficient pattern.

Worked example 2: a negative fractional exponent

Question. Expand $(4+3x)^{-3/2}$ through $x^3$ , and state its series' range of validity.

Plan. Extract $4^{-3/2}=1/8$ , then use the binomial coefficients with exponent $-3/2$ and argument $3x/4$ .

\begin{aligned} (4+3x)^{-3/2} &=\frac18\left(1+\frac{3x}{4}\right)^{-3/2} &&\text{normalise the constant}\\ &=\frac18\left[1-\frac32u+\frac{15}{8}u^2-\frac{35}{16}u^3+\cdots\right] &&u=\frac{3x}{4}\\ &=\frac18\left[1-\frac98x+\frac{135}{128}x^2-\frac{945}{1024}x^3+\cdots\right] &&\text{substitute powers of }u\\ &=\frac18-\frac9{64}x+\frac{135}{1024}x^2-\frac{945}{8192}x^3+\cdots &&\text{multiply every coefficient by }\frac18 \end{aligned}

Here $|3x/4|<1$ , so $-4/3<x<4/3$ . For this exponent neither endpoint gives a convergent series.

Why this step? The quadratic binomial coefficient is $(-3/2)(-5/2)/2!=15/8$ . Its positive sign comes from two negative factors; the cubic coefficient is negative because it has three.

Answer. The displayed series is valid for $|x|<4/3$ .

Check. At zero the series gives $1/8=4^{-3/2}$ . Its linear coefficient is $-9/64$ , also obtained from $(-3/2)(3)4^{-5/2}$ .

The general logarithm and binomial series are in the formula booklet; these substituted expansions are not printed there. Their coefficients and conditions come from the displayed substitutions.

Watch out

Wrong: $(4+3x)^{-3/2}=4(1+3x/4)^{-3/2}$ . The extracted constant must also be raised to $-3/2$ . Right: the outside factor is $4^{-3/2}=1/8$ .

The next expression changes the constant and the inner coefficient, but its binomial exponent is unchanged. Deal with the outside factor before expanding.

Multiplying series to a specified power

To find the coefficient of $x^k$ in a product, collect every pair of terms whose powers add to $k$ ; a missing pair is a missing contribution.

Keep enough terms in both factors. If one factor has a non-zero constant, its partner's term in $x^k$ contributes directly to the final coefficient. The resulting full product series is used on the common range of the two original series.

Worked example 3: exponential times cosine

Question. Expand $\mathrm{e}^x\cos x$ through $x^4$ .

Plan. Multiply the exponential and cosine expansions, collecting terms by their total power.

\mathrm{e}^x\cos x =\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\cdots\right) \left(1-\frac{x^2}{2}+\frac{x^4}{24}+\cdots\right)

Hence

\begin{aligned} \mathrm{e}^x\cos x &=1+x+\left(\frac12-\frac12\right)x^2 +\left(\frac16-\frac12\right)x^3\\ &\quad+\left(\frac1{24}-\frac14+\frac1{24}\right)x^4+\cdots &&\text{collect equal powers}\\ &=1+x-\frac13x^3-\frac16x^4+\cdots &&\text{simplify, including the zero coefficient} \end{aligned}

Why this step? The quartic coefficient has three contributions: quartic times constant, quadratic times quadratic and constant times quartic. Keeping only the outer two would miss $-1/4$ .

Answer. $1+x-x^3/3-x^4/6+\cdots$ ; the full series is valid for all real $x$ .

Check. The quadratic coefficient is zero because $f''(0)=1-1=0$ , also obtained by differentiating the product twice.

Worked example 4 changes just the second factor, from $\cos x$ to $\ln(1+x)$ . The logarithm has no constant term, and it also restricts the product's range.

Worked example 4: a product with a restricted range

Question. Expand $\mathrm{e}^x\ln(1+x)$ through $x^4$ , and state the full series' range of validity.

Plan. Multiply by the logarithm expansion, keeping all combinations through total degree four, then use the logarithm's range.

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

Thus

\begin{aligned} \mathrm{e}^x\ln(1+x) &=x+\left(1-\frac12\right)x^2 +\left(\frac12-\frac12+\frac13\right)x^3\\ &\quad+\left(\frac16-\frac14+\frac13-\frac14\right)x^4+\cdots &&\text{include every degree-four pair}\\ &=x+\frac12x^2+\frac13x^3+0x^4+\cdots &&\text{the quartic contributions cancel} \end{aligned}

Why this step? The logarithm starts with $x$ , so the exponential's cubic term is enough for a quartic product. Its quartic term would first contribute at degree five.

Answer. $x+x^2/2+x^3/3$ , with zero quartic coefficient and higher terms omitted. The full series is valid for $-1<x\le1$ .

Check. At zero the product vanishes, and its first derivative is $1$ , agreeing with the constant and linear coefficients. All four displayed quartic contributions total zero.

These product expansions are derived results, not printed in the formula booklet. Intersect conditions when both factors impose restrictions; a factor valid for all real $x$ adds no further restriction.

Watch out

Wrong: multiply only matching powers, such as the two quadratic terms. Right: for the cubic coefficient also include constant times cubic and linear times quadratic. Track total degree, not matching position in the two lists.

The next product changes the exponential scale and reverses the logarithm's argument. Both changes affect coefficients, and the latter reverses its endpoint conditions.

Substituting one series inside another

For a function inside another function, first expand the inner expression, then treat it as the whole argument of the outer series.

If the inner expression starts with $x$ , its fourth power starts with $x^4$ . If it starts with $x^2$ , its fourth power starts with $x^8$ . This degree count tells you how many outer terms can affect the requested answer. It assumes the inner expression is zero at $x=0$ ; extract any non-zero constants first, as in the normalisation examples.

Worked example 5: an exponential of a sine

Question. Derive the Maclaurin expansion of $\mathrm{e}^{\sin x}$ through $x^4$ .

Plan. Put $u=\sin x=x-x^3/6+\cdots$ , then use $\mathrm{e}^u=1+u+u^2/2+u^3/6+u^4/24+\cdots$ .

To the required order,

u^2=x^2-\frac{x^4}{3}+\cdots,\quad u^3=x^3+\cdots,\quad u^4=x^4+\cdots

Consequently,

\begin{aligned} \mathrm{e}^{\sin x} &=1+\left(x-\frac{x^3}{6}\right) +\frac12\left(x^2-\frac{x^4}{3}\right) +\frac{x^3}{6}+\frac{x^4}{24}+\cdots\\ &=1+x+\frac{x^2}{2} +\left(-\frac16+\frac16\right)x^3 +\left(-\frac16+\frac1{24}\right)x^4+\cdots\\ &=1+x+\frac12x^2-\frac18x^4+\cdots \end{aligned}

Why this step? The cubic term comes from both the inner sine and the cubic term of the outer exponential. Their equal and opposite coefficients explain the missing power in the opening puzzle.

Answer. $1+x+x^2/2-x^4/8+\cdots$ , with zero cubic coefficient. The full series is valid for all real $x$ .

Check. At zero the function is $1$ . Also $f'(x)=\cos x\,\mathrm{e}^{\sin x}$ , so $f'(0)=1$ , matching the linear coefficient.

Worked example 6 changes only the inner scale, from $x$ to $2x$ . You can repeat the composition or substitute $2x$ into the whole expansion just derived.

Worked example 6: scaling the inner function

Question. Derive the Maclaurin expansion of $\mathrm{e}^{\sin(2x)}$ through $x^4$ .

Plan. Use $u=2x-4x^3/3+\cdots$ , keeping its powers through degree four.

u^2=4x^2-\frac{16}{3}x^4+\cdots,\quad u^3=8x^3+\cdots,\quad u^4=16x^4+\cdots

Hence

\begin{aligned} \mathrm{e}^{\sin(2x)} &=1+2x-\frac43x^3+2x^2-\frac83x^4 +\frac86x^3+\frac{16}{24}x^4+\cdots\\ &=1+2x+2x^2+\left(-\frac43+\frac43\right)x^3 +\left(-\frac83+\frac23\right)x^4+\cdots\\ &=1+2x+2x^2-2x^4+\cdots \end{aligned}

Why this step? The powers of $2$ must follow the powers of $x$ . Substituting into the whole previously derived expression gives the same check: its quartic term becomes $-(2x)^4/8=-2x^4$ .

Answer. $1+2x+2x^2-2x^4+\cdots$ , valid as a full series for all real $x$ .

Check. At $x=0.1$ , the truncation gives $1.2198$ , while the function is approximately $1.21978$ . Agreement near zero is consistent with the expansion; it is not a global error guarantee.

These compound-function expansions are not printed in the formula booklet: derive them from the supplied constituent series. A zero coefficient can arise through cancellation even when a function has both odd and even powers, as the opening example proves.

Watch out

Wrong: $\mathrm{e}^{\sin x}=1+\sin x+x^2/2+x^3/6+\cdots$ . Only the first argument was replaced. Right: use $1+\sin x+(\sin x)^2/2!+(\sin x)^3/3!+\cdots$ , then expand every required power.

The next inner scale gives one final check on the cubic cancellation and the quartic coefficient.

Quick Recap

Use the five standard coefficient patterns with their conditions; the logarithm includes its positive endpoint.
Extract a non-unit constant before a logarithm or binomial expansion, and transform the range with the whole argument.
In a product, collect every pair whose powers add to the requested degree and intersect validity ranges.
In a composition, expand the inner function and all required powers of it; check which coefficients cancel.

The opening missing term is real: $\mathrm{e}^{\sin x}=1+x+x^2/2-x^4/8+\cdots$ , because its two cubic contributions cancel.

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CP 4.6 - Standard and compound Maclaurin series

The standard series and their ranges

Substitution and a non-unit constant

Multiplying series to a specified power

Substituting one series inside another

Quick Recap

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