What Maclaurin series should I know for A-Level Further Maths?

You should be confident using the standard expansions for functions such as $e^x$, $\sin x$, $\cos x$, $\ln(1+x)$ and $(1+x)^n$. Check your exam board's current specification for the precise list you are expected to recall or derive.

How do I know how many terms to keep in a Maclaurin expansion?

Work backwards from the highest required power. Keep every term that could contribute to that power after substitution, multiplication, division, differentiation or integration.

Do I need to state the range of validity?

State it whenever the question requests it or when you derive an expansion from a standard series with a restricted interval. Transform the original condition carefully after making a substitution.

Can I earn method marks if my final Maclaurin series is wrong?

Often, yes. Clear use of a correct standard expansion, valid substitution or coefficient-comparison method may earn credit even if later arithmetic is incorrect. Exact marking depends on the paper's official mark scheme.

What is the best way to use a Maclaurin series past paper?

Attempt the question without notes, mark every line against the official mark scheme, and classify the error as recall, substitution, algebra, truncation or validity. Then practise another question testing the same method.

Maclaurin Series Exam Questions A-Level Further Maths

A Maclaurin series question can look harmless. The formula is familiar, the algebra begins neatly, and then one missed factorial changes every coefficient that follows. For many students, the difficulty is not understanding the idea. It is keeping the method organised under exam pressure.

This guide works through three Maclaurin series exam questions for A-Level Further Maths, following a hypothetical student named Maya as she moves from direct expansion to substitution and series division. Each solution includes indicative method marks so you can see where credit is likely to come from. These are teaching mark allocations rather than official exam-board mark schemes.

The method to use in an exam

A Maclaurin series is a Taylor series centred at $x = 0$ :

f(x)=f(0)+xf′(0)+x22!f′′(0)+x33!f′′′(0)+⋯f(x)=f(0)+xf'(0)+\frac{x^2}{2!}f''(0)+\frac{x^3}{3!}f'''(0)+\cdots

Before beginning a question, check four things:

Which power of $x$ is required?
Is the question asking for differentiation, a known expansion or substitution?
Must you state the range of validity?
Will the final expression require multiplication, division or integration of series?

Maya writes the required final power at the top of her page. It is a small habit, but it prevents her from stopping one term too early or producing unnecessary algebra.

You can review the wider course through the Edexcel A-Level Further Maths revision hub, where the matching Maclaurin series lesson sits alongside Further Maths practice resources. If the underlying algebra needs attention, the Edexcel A-Level Maths revision course also provides supporting pure maths material.

Question one: multiplying familiar expansions

Question: Find the Maclaurin expansion of $e^{2x}\cos x$ up to and including the term in $x^3$ .

Maya first decides not to differentiate the product three times. Instead, she uses two standard expansions and keeps only the terms that can contribute up to $x^3$ .

The standard exponential expansion is

eu=1+u+u22!+u33!+⋯e^u=1+u+\frac{u^2}{2!}+\frac{u^3}{3!}+\cdots

Substituting $u = 2 x$ gives

e2x=1+2x+(2x)22!+(2x)33!+⋯e^{2x}=1+2x+\frac{(2x)^2}{2!}+\frac{(2x)^3}{3!}+\cdots

e2x=1+2x+2x2+43x3+⋯e^{2x}=1+2x+2x^2+\frac{4}{3}x^3+\cdots

Indicative marks: M1 for using the exponential series with $u = 2 x$ ; A1 for the correct coefficients through $x^3$ .

For cosine,

cos⁡x=1−x22!+x44!−⋯\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots

Only the constant and $x^2$ terms are needed:

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

Multiplying,

e2xcos⁡x=(1+2x+2x2+43x3)(1−x22)+⋯e^{2x}\cos x=\left(1+2x+2x^2+\frac{4}{3}x^3\right)\left(1-\frac{x^2}{2}\right)+\cdots

Collecting coefficients up to $x^3$ :

e2xcos⁡x=1+2x+(2−12)x2+(43−1)x3+⋯e^{2x}\cos x =1+2x+\left(2-\frac{1}{2}\right)x^2 +\left(\frac{4}{3}-1\right)x^3+\cdots

Therefore,

e2xcos⁡x=1+2x+32x2+13x3+O(x4)\boxed{e^{2x}\cos x=1+2x+\frac{3}{2}x^2+\frac{1}{3}x^3+O(x^4)}

Indicative marks: M1 for multiplying the series; A1 for the $x^2$ coefficient; A1 for the $x^3$ coefficient and final expansion.

The important moment is the $x^3$ term. It comes from both $43x3×1\frac{4}{3}x^3\times 1$ and $2x×(−x22)2x\times\left(-\frac{x^2}{2}\right)$ . Looking only at terms with matching powers would lose the second contribution.

Open the Maclaurin series Lessons and Question Bank from the Further Maths revision hub when you want to practise this multiplication method without having to assemble questions yourself.

Question two: substitution and an approximation

Question: Use a standard Maclaurin series to show that

),\ln\left(\frac{1+x}{1-x}\right)=2\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots\right),

and state the range of validity. Hence estimate $ln⁡3\ln 3$ using terms up to $x^5$ .

Maya begins with

ln⁡(1+x)=x−x22+x33−x44+x55−⋯\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\cdots

Replacing $x$ by $- x$ gives

ln⁡(1−x)=−x−x22−x33−x44−x55−⋯\ln(1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-\frac{x^5}{5}-\cdots

Now use logarithm laws:

ln⁡(1+x1−x)=ln⁡(1+x)−ln⁡(1−x).\ln\left(\frac{1+x}{1-x}\right)=\ln(1+x)-\ln(1-x).

Subtracting the expansions causes the even powers to cancel:

ln⁡(1+x1−x)=2x+2x33+2x55+⋯\ln\left(\frac{1+x}{1-x}\right) =2x+\frac{2x^3}{3}+\frac{2x^5}{5}+\cdots

Hence,

)\boxed{\ln\left(\frac{1+x}{1-x}\right)=2\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots\right)}

Indicative marks: M1 for substituting $- x$ correctly; M1 for subtracting the logarithmic series; A1 for the simplified result.

The original logarithmic expansions are valid for $∣ x ∣ < 1$ , so the derived expansion has range

∣x∣<1.\boxed{|x|<1}.

For the approximation, Maya solves

1+x1−x=3.\frac{1+x}{1-x}=3.

This gives

1 + x = 3 - 3 x,

x=12.x=\frac{1}{2}.

Therefore,

ln⁡3≈2(12+(1/2)33+(1/2)55).\ln 3\approx 2\left(\frac{1}{2}+\frac{(1/2)^3}{3}+\frac{(1/2)^5}{5}\right).

ln⁡3≈2(12+124+1160)=263240.\ln 3\approx 2\left(\frac{1}{2}+\frac{1}{24}+\frac{1}{160}\right) =\boxed{\frac{263}{240}}.

Indicative marks: M1 for finding $x=12x=\frac{1}{2}$ ; M1 for correct substitution; A1 for the approximation.

This is a common Maclaurin series past paper structure: establish an expansion, state its validity, and then use it to approximate a constant.

Question three: finding a series by division

Question: Given the standard expansions of $sin⁡x\sin x$ and $cos⁡x\cos x$ , find the Maclaurin expansion of $tan⁡x\tan x$ up to and including the term in $x^5$ .

Maya could attempt formal long division. Instead, she assumes

tan x=ax+bx^3+cx^5+O(x^7).

Since $tan⁡xcos⁡x=sin⁡x\tan x\cos x=\sin x$ ,

(ax+bx3+cx5)(1−x22+x424)=x−x36+x5120+O(x7).(ax+bx^3+cx^5)\left(1-\frac{x^2}{2}+\frac{x^4}{24}\right) =x-\frac{x^3}{6}+\frac{x^5}{120}+O(x^7).

Comparing coefficients of $x$ gives

a = 1.

Comparing coefficients of $x^3$ gives

b−a2=−16.b-\frac{a}{2}=-\frac{1}{6}.

Using $a = 1$ :

b=13.b=\frac{1}{3}.

Comparing coefficients of $x^5$ gives

c−b2+a24=1120.c-\frac{b}{2}+\frac{a}{24}=\frac{1}{120}.

Therefore,

c=1120+16−124=215.c=\frac{1}{120}+\frac{1}{6}-\frac{1}{24}=\frac{2}{15}.

tan⁡x=x+x33+2x515+O(x7).\boxed{\tan x=x+\frac{x^3}{3}+\frac{2x^5}{15}+O(x^7)}.

Indicative marks: M1 for assuming an appropriate odd-power series; M1 for forming the product; A1 for $a$ and $b$ ; A1 for $c$ ; A1 for the final expansion.

Maya has now moved through three levels of demand: multiplying known series, transforming a standard series, and constructing an unknown series by comparing coefficients. That progression is more useful than completing several nearly identical questions.

For authentic paper practice, use the Edexcel A-Level Further Maths resources and compare your working carefully with the available mark schemes. You can also see how examiners describe method and accuracy marks in this OCR Further Maths Core Pure mark scheme.

Common mistakes in a Maclaurin series markscheme

Forgetting factorials

The coefficient of $x^n$ in the derivative form is

f(n)(0)n!,\frac{f^{(n)}(0)}{n!},

not simply $f^{(n)}(0)$ . Write the factorial before evaluating derivatives.

Substituting into coefficients but not powers

When replacing $x$ by $2 x$ , every occurrence changes. For example,

x33!⟶(2x)33!,\frac{x^3}{3!}\longrightarrow\frac{(2x)^3}{3!},

not $2x33!\frac{2x^3}{3!}$ .

Truncating too early

When two series are multiplied, an apparently unimportant lower-order term can combine with another term to produce the required final power. Write enough terms before multiplying.

Omitting the range of validity

If a question asks for the range, include it explicitly. A correct expansion can still lose an accuracy mark when the interval is absent or copied incorrectly.

Giving only the final line

A Maclaurin series markscheme often separates method marks from accuracy marks. Show the standard expansion, substitution and coefficient comparison so that one arithmetic slip does not hide the whole method.

Exam-style practice routine

Try one direct expansion, one substitution question and one multiplication or division question. Mark each attempt by identifying where the method began correctly, not only whether the boxed answer matched. Then return to the same three question types a few days later without notes.

MathsGenie brings the course materials together, including A-Level Maths revision resources, supporting lessons and Further Maths paper practice. Knowing the expansion is the first step; recognising which method an unfamiliar question needs is what turns that knowledge into marks.

Continue with MathsGenie

Maya's improvement came from giving each question a structure: identify the target power, choose the source series, show the transformation and collect coefficients carefully. You can use the same structure, but it only becomes reliable through focused practice before your exam date. Start the Maclaurin series Lessons and Question Bank in the A-Level Further Maths revision hub.

The method to use in an exam

A Maclaurin series is a Taylor series centred at $x = 0$ :

f(x)=f(0)+xf′(0)+x22!f′′(0)+x33!f′′′(0)+⋯f(x)=f(0)+xf'(0)+\frac{x^2}{2!}f''(0)+\frac{x^3}{3!}f'''(0)+\cdots

Before beginning a question, check four things:

Which power of $x$ is required?
Is the question asking for differentiation, a known expansion or substitution?
Must you state the range of validity?
Will the final expression require multiplication, division or integration of series?

Maya writes the required final power at the top of her page. It is a small habit, but it prevents her from stopping one term too early or producing unnecessary algebra.

Question one: multiplying familiar expansions

Question: Find the Maclaurin expansion of $e^{2x}\cos x$ up to and including the term in $x^3$ .

Maya first decides not to differentiate the product three times. Instead, she uses two standard expansions and keeps only the terms that can contribute up to $x^3$ .

The standard exponential expansion is

eu=1+u+u22!+u33!+⋯e^u=1+u+\frac{u^2}{2!}+\frac{u^3}{3!}+\cdots

Substituting $u = 2 x$ gives

e2x=1+2x+(2x)22!+(2x)33!+⋯e^{2x}=1+2x+\frac{(2x)^2}{2!}+\frac{(2x)^3}{3!}+\cdots

e2x=1+2x+2x2+43x3+⋯e^{2x}=1+2x+2x^2+\frac{4}{3}x^3+\cdots

Indicative marks: M1 for using the exponential series with $u = 2 x$ ; A1 for the correct coefficients through $x^3$ .

For cosine,

cos⁡x=1−x22!+x44!−⋯\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots

Only the constant and $x^2$ terms are needed:

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

Multiplying,

e2xcos⁡x=(1+2x+2x2+43x3)(1−x22)+⋯e^{2x}\cos x=\left(1+2x+2x^2+\frac{4}{3}x^3\right)\left(1-\frac{x^2}{2}\right)+\cdots

Collecting coefficients up to $x^3$ :

e2xcos⁡x=1+2x+(2−12)x2+(43−1)x3+⋯e^{2x}\cos x =1+2x+\left(2-\frac{1}{2}\right)x^2 +\left(\frac{4}{3}-1\right)x^3+\cdots

Therefore,

e2xcos⁡x=1+2x+32x2+13x3+O(x4)\boxed{e^{2x}\cos x=1+2x+\frac{3}{2}x^2+\frac{1}{3}x^3+O(x^4)}

Indicative marks: M1 for multiplying the series; A1 for the $x^2$ coefficient; A1 for the $x^3$ coefficient and final expansion.

Open the Maclaurin series Lessons and Question Bank from the Further Maths revision hub when you want to practise this multiplication method without having to assemble questions yourself.

Question two: substitution and an approximation

Question: Use a standard Maclaurin series to show that

),\ln\left(\frac{1+x}{1-x}\right)=2\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots\right),

and state the range of validity. Hence estimate $ln⁡3\ln 3$ using terms up to $x^5$ .

Maya begins with

ln⁡(1+x)=x−x22+x33−x44+x55−⋯\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\cdots

Replacing $x$ by $- x$ gives

ln⁡(1−x)=−x−x22−x33−x44−x55−⋯\ln(1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-\frac{x^5}{5}-\cdots

Now use logarithm laws:

ln⁡(1+x1−x)=ln⁡(1+x)−ln⁡(1−x).\ln\left(\frac{1+x}{1-x}\right)=\ln(1+x)-\ln(1-x).

Subtracting the expansions causes the even powers to cancel:

ln⁡(1+x1−x)=2x+2x33+2x55+⋯\ln\left(\frac{1+x}{1-x}\right) =2x+\frac{2x^3}{3}+\frac{2x^5}{5}+\cdots

Hence,

)\boxed{\ln\left(\frac{1+x}{1-x}\right)=2\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots\right)}

Indicative marks: M1 for substituting $- x$ correctly; M1 for subtracting the logarithmic series; A1 for the simplified result.

The original logarithmic expansions are valid for $∣ x ∣ < 1$ , so the derived expansion has range

∣x∣<1.\boxed{|x|<1}.

For the approximation, Maya solves

1+x1−x=3.\frac{1+x}{1-x}=3.

This gives

1 + x = 3 - 3 x,

x=12.x=\frac{1}{2}.

Therefore,

ln⁡3≈2(12+(1/2)33+(1/2)55).\ln 3\approx 2\left(\frac{1}{2}+\frac{(1/2)^3}{3}+\frac{(1/2)^5}{5}\right).

ln⁡3≈2(12+124+1160)=263240.\ln 3\approx 2\left(\frac{1}{2}+\frac{1}{24}+\frac{1}{160}\right) =\boxed{\frac{263}{240}}.

Indicative marks: M1 for finding $x=12x=\frac{1}{2}$ ; M1 for correct substitution; A1 for the approximation.

This is a common Maclaurin series past paper structure: establish an expansion, state its validity, and then use it to approximate a constant.

Question three: finding a series by division

Question: Given the standard expansions of $sin⁡x\sin x$ and $cos⁡x\cos x$ , find the Maclaurin expansion of $tan⁡x\tan x$ up to and including the term in $x^5$ .

Maya could attempt formal long division. Instead, she assumes

tan x=ax+bx^3+cx^5+O(x^7).

Since $tan⁡xcos⁡x=sin⁡x\tan x\cos x=\sin x$ ,

(ax+bx3+cx5)(1−x22+x424)=x−x36+x5120+O(x7).(ax+bx^3+cx^5)\left(1-\frac{x^2}{2}+\frac{x^4}{24}\right) =x-\frac{x^3}{6}+\frac{x^5}{120}+O(x^7).

Comparing coefficients of $x$ gives

a = 1.

Comparing coefficients of $x^3$ gives

b−a2=−16.b-\frac{a}{2}=-\frac{1}{6}.

Using $a = 1$ :

b=13.b=\frac{1}{3}.

Comparing coefficients of $x^5$ gives

c−b2+a24=1120.c-\frac{b}{2}+\frac{a}{24}=\frac{1}{120}.

Therefore,

c=1120+16−124=215.c=\frac{1}{120}+\frac{1}{6}-\frac{1}{24}=\frac{2}{15}.

tan⁡x=x+x33+2x515+O(x7).\boxed{\tan x=x+\frac{x^3}{3}+\frac{2x^5}{15}+O(x^7)}.

Indicative marks: M1 for assuming an appropriate odd-power series; M1 for forming the product; A1 for $a$ and $b$ ; A1 for $c$ ; A1 for the final expansion.