Maclaurin Series

Overview

A power series in $x$ is an infinite polynomial-like expression

$$\sum _{r=0}^{\infty }{a}_r{x}^r=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+\cdots ,$$

where $a_0,a_1,a_2,\dots$ are constants.

The central idea is:

many non-polynomial functions can be represented near a suitable point by an infinite polynomial.

For H2 Mathematics, the main centre is $x=0$. A power series expanded about $x=0$ is called a Maclaurin series.

This topic sits naturally after sequences and series because it changes the question from:

What does this infinite sum do?

to:

Can this infinite sum represent a familiar function?

Why Maclaurin Series Matter

Maclaurin series are useful because polynomials are easy to handle.

They allow us to:

• approximate functions such as $e^x$, $\ln (1+x)$, $\sin x$, and $\cos x$
• estimate values such as $\ln (1.01)$ or $\sqrt{0.96}$
• justify small-angle approximations
• expand products, quotients, and composite functions
• differentiate or integrate suitable functions term by term
• understand local behaviour near $x=0$

The word local matters. A low-order Maclaurin polynomial usually approximates a function best near the centre of expansion.

Caption: A Maclaurin polynomial uses derivative information at $x=0$. Higher-degree truncations match more derivatives there and can improve the local approximation, but accuracy at a particular nonzero $x$ must still be checked rather than assumed.

The Maclaurin Formula

Suppose

$$f(x)=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+\cdots .$$

Substituting $x=0$ gives

$$a_0=f(0).$$

Differentiating once gives

$$f^{\prime }(x)=a_1+2a_{2}x+3a_3{x}^2+\cdots ,$$

so

$$a_1=f^{\prime }(0).$$

Repeating this process gives

$$a_2=\frac{f^{\prime \prime }(0)}{2!},a_3=\frac{f^{\prime \prime \prime }(0)}{3!},a_r=\frac{f^{(r)}(0)}{r!}.$$

Hence the Maclaurin series is

$$\boxed{f(x)=f(0)+f^{\prime }(0)x+\frac{f^{\prime \prime }(0)}{2!}{x}^2+\frac{f^{\prime \prime \prime }(0)}{3!}{x}^3+\cdots +\frac{f^{(r)}(0)}{r!}{x}^r+\cdots }$$

provided the resulting series converges to $f(x)$ in the range being used.

What The Formula Means

The Maclaurin series builds a polynomial approximation from derivative information at the origin.

• $f(0)$ fixes the value at the origin.
• $f^{\prime }(0)$ fixes the tangent slope at the origin.
• $f^{\prime \prime }(0)$ fixes the local curvature.
• Higher derivatives refine the local shape.

The first few Maclaurin polynomials are

$$P_0(x)=f(0),$$ $$P_1(x)=f(0)+f^{\prime }(0)x,$$ $$P_2(x)=f(0)+f^{\prime }(0)x+\frac{f^{\prime \prime }(0)}{2!}{x}^2,$$

and

$$P_3(x)=f(0)+f^{\prime }(0)x+\frac{f^{\prime \prime }(0)}{2!}{x}^2+\frac{f^{\prime \prime \prime }(0)}{3!}{x}^3.$$

For small $|x|$, higher powers such as $x^3,x^4,x^5$ are usually small. This is why a low-order Maclaurin polynomial can be accurate near $x=0$ but poor far away.

Existence And Convergence

It is not enough for a function to have many derivatives. A Maclaurin series is useful only when the series converges to the original function in the range considered.

At H2 level, the practical checks are:

• the needed derivatives at $x=0$ must exist
• the resulting power series must be used inside its validity range
• the approximation is usually most accurate near $x=0$
• different standard expansions have different validity ranges

For example,

$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$$

is valid for all real $x$, but

$$\frac{1}{1-x}=1+x+x^2+x^3+\cdots$$

is valid only when $|x|<1$.

Standard Maclaurin Expansions

The following standard expansions are the working toolkit.

Exponential Function

$$\boxed{e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots +\frac{x^r}{r!}+\cdots }$$

This is valid for all real $x$.

Replacing $x$ by $kx$ gives

$$e^{kx}=1+kx+\frac{k^2{x}^2}{2!}+\frac{k^3{x}^3}{3!}+\cdots .$$

For $a>0$, since $a^x=e^{x\ln a}$,

$$a^x=1+x\ln a+\frac{x^2(\ln a{)}^2}{2!}+\frac{x^3(\ln a{)}^3}{3!}+\cdots .$$

Sine Function

$$\boxed{\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots }$$

This is valid for all real $x$, with $x$ measured in radians.

Replacing $x$ by $kx$ gives

$$\sin (kx)=kx-\frac{k^3{x}^3}{3!}+\frac{k^5{x}^5}{5!}-\cdots .$$

Cosine Function

$$\boxed{\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots }$$

This is valid for all real $x$, with $x$ measured in radians.

Replacing $x$ by $kx$ gives

$$\cos (kx)=1-\frac{k^2{x}^2}{2!}+\frac{k^4{x}^4}{4!}-\cdots .$$

Logarithmic Function

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

This is valid for

$$-1<x\le 1.$$

Replacing $x$ by $u$ gives

$$\ln (1+u)=u-\frac{u^2}{2}+\frac{u^3}{3}-\cdots ,$$

with validity determined by

$$-1<u\le 1.$$

Binomial Series

For rational $n$,

$$\boxed{(1+x{)}^n=1+nx+\frac{n(n-1)}{2!}{x}^2+\frac{n(n-1)(n-2)}{3!}{x}^3+\cdots }$$

More generally,

$$(1+x{)}^n=1+nx+\frac{n(n-1)}{2!}{x}^2+\cdots +\frac{n(n-1)\cdots (n-r+1)}{r!}{x}^r+\cdots .$$

If $n$ is a non-negative integer, the series terminates and becomes the ordinary binomial theorem.

If $n$ is not a non-negative integer, the series is infinite and is valid for

$$|x|<1.$$

Validity Ranges

A common exam error is to expand correctly but state the wrong validity range.

The reliable question is:

What expression replaced the standard variable?

For example,

$$(1+u{)}^{-1}=1-u+u^2-u^3+\cdots ,|u|<1.$$

If $u=3x$, then

$$(1+3x{)}^{-1}=1-3x+9x^2-27x^3+\cdots ,$$

and the validity condition is

$$|3x|<1,$$

so

$$|x|<\frac{1}{3}.$$

For

$$\ln (1-2x),$$

use $u=-2x$ in $\ln (1+u)$. The validity condition is

$$-1<-2x\le 1.$$

Solving gives

$$-\frac{1}{2}\le x<\frac{1}{2}.$$

The endpoint direction changes because the coefficient of $x$ is negative.

Caption: Validity ranges must be transformed from the standard variable into the actual variable used in the question.

Ascending And Descending Powers

For expressions such as $(a+bx{)}^n$, the factorisation determines the form and validity range.

Ascending Powers Of $x$

For $a\ne 0$ and $b\ne 0$, factor out the constant term only where the chosen real branch is represented consistently:

$$(a+bx{)}^n=a^n{\left(1+\frac{bx}{a}\right)}^n.$$

Then use the binomial series on

$$1+\frac{bx}{a}.$$

The validity condition is

$$\left|\frac{bx}{a}\right|<1,$$

so

$$|x|<\left|\frac{a}{b}\right|.$$

The binomial interval must also be intersected with the original real domain. For a non-integer rational power, choose a real factorisation that preserves the required branch; a positive constant factor is the usual first-learner route.

This gives a series in ascending powers of $x$:

$$A_0+A_{1}x+A_2{x}^2+\cdots .$$

Descending Powers Of $x$

Support/enrichment. This form is useful for large-$|x|$ reasoning, but the real domain and branch of the original function must be preserved.

For $a\ne 0$ and $b\ne 0$, factor out the $x$-term only where the real powers involved are defined consistently:

$$(a+bx{)}^n=(bx{)}^n{\left(1+\frac{a}{bx}\right)}^n.$$

The binomial-series condition is

$$\left|\frac{a}{bx}\right|<1,$$

so

$$|x|>\left|\frac{a}{b}\right|.$$

This gives a series in powers of $1/x$, but the condition must be intersected with the real domain and branch of the original expression. It is not automatically valid on both sides of the number line for a non-integer rational power.

For example,

$$(2+3x{)}^{1/2}=(3x{)}^{1/2}{\left(1+\frac{2}{3x}\right)}^{1/2}$$

on the usual real square-root branch requires $x>0$ as well as $\left|\frac{2}{3x}\right|<1$. Hence this descending expansion is valid for

$$x>\frac{2}{3},$$

not for $x<-\frac{2}{3}$.

Use ascending powers when values near $x=0$ are relevant. Use descending powers when $|x|$ is large and the expression is naturally controlled by powers of $1/x$.

Manipulating Series

Most exam questions require adapting standard expansions rather than deriving every series from first principles.

Substitution

To expand $e^{2x}$, replace $x$ by $2x$:

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

To expand $\cos (3x)$:

$$\cos (3x)=1-\frac{(3x{)}^2}{2!}+\frac{(3x{)}^4}{4!}-\cdots =1-\frac{9}{2}{x}^2+\frac{27}{8}{x}^4-\cdots .$$

Sums And Differences

Expand each part to the required order, then combine like powers.

For example,

$$\ln \left(\frac{1+x}{1-x}\right)=\ln (1+x)-\ln (1-x).$$

Now

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

and

$$\ln (1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\cdots .$$

Therefore

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

The validity range is the intersection of the two ranges:

$$-1<x<1.$$

Products

When multiplying series, keep only the terms needed up to the required order.

For example, expand $e^x\ln (1+x)$ up to and including the term in $x^2$.

Use

$$e^x=1+x+\frac{x^2}{2}+\cdots$$

and

$$\ln (1+x)=x-\frac{x^2}{2}+\cdots .$$

Then

$$e^x\ln (1+x)=\left(1+x+\frac{x^2}{2}+\cdots \right)\left(x-\frac{x^2}{2}+\cdots \right).$$

Keeping terms up to $x^2$,

$$e^x\ln (1+x)=x+\frac{x^2}{2}+\cdots .$$

The important discipline is not to over-expand blindly. Keep enough terms, but only enough terms.

Partial Fractions

For rational functions, partial fractions often make expansion easier.

For example,

$$\frac{x+5}{(1+3x)(2-x)}$$

can be decomposed into simple fractions of the form

$$\frac{A}{1+3x}+\frac{B}{2-x}.$$

Equating numerators gives

$$x+5=A(2-x)+B(1+3x),$$

so $A=2$ and $B=1$. Hence

$$\frac{x+5}{(1+3x)(2-x)}=\frac{2}{1+3x}+\frac{1}{2-x}.$$

Now

$$\frac{2}{1+3x}=2(1-3x+9x^2-\cdots ),|x|<\frac{1}{3},$$

and

$$\frac{1}{2-x}=\frac{1}{2}\left(1+\frac{x}{2}+\frac{x^2}{2^2}+\cdots \right),|x|<2.$$

Therefore, up to $x^2$,

$$\frac{x+5}{(1+3x)(2-x)}=\frac{5}{2}-\frac{23}{4}x+\frac{145}{8}{x}^2+\cdots ,$$

with the intersected validity range

$$|x|<\frac{1}{3}.$$

Each part can then be expanded using a binomial or geometric series. The final validity range is the intersection of the validity ranges of all parts used.

Deriving Series By Differentiation

If a power series for $f(x)$ is known, then a power series for $f^{\prime }(x)$ can often be obtained by differentiating term by term within the interval of validity.

For example,

$$\sin (2x)=2x-\frac{(2x{)}^3}{3!}+\frac{(2x{)}^5}{5!}-\cdots .$$

Differentiating both sides gives

$$2\cos (2x)=2-\frac{3(2x{)}^2\cdot 2}{3!}+\frac{5(2x{)}^4\cdot 2}{5!}-\cdots .$$

Thus

$$\cos (2x)=1-2x^2+\frac{2}{3}{x}^4-\cdots .$$

This agrees with the direct expansion

$$\cos (2x)=1-\frac{(2x{)}^2}{2!}+\frac{(2x{)}^4}{4!}-\cdots .$$

Deriving Series By Integration

Term-by-term integration can also create a new series. Start from the geometric expansion

$$\frac{1}{1+x}=1-x+x^2-x^3+\cdots ,|x|<1.$$

Integrate from $0$ to $x$ so that the constant is fixed automatically:

$${\int }_0^x\frac{1}{1+t}dt={\int }_0^x(1-t+t^2-t^3+\cdots )dt.$$

Thus

$$\ln (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots ,|x|<1.$$

Term-by-term integration preserves the open interval of convergence used in the derivation. An endpoint such as $x=1$ must be checked separately; it should not be included merely because the antiderivative exists there.

At $x=1$, the resulting alternating harmonic series converges, so the familiar final validity range is

$$-1<x\le 1.$$

At $x=-1$, $\ln (1+x)$ is undefined and the series is excluded.

Deriving Series By Implicit Differentiation

Sometimes direct differentiation becomes messy. An implicit relation can make the repeated derivatives at $x=0$ easier to obtain.

For example, let

$$y=\sqrt{5+4e^x}.$$

Then

$$y^2=5+4e^x.$$

Differentiating gives

$$2y\frac{dy}{dx}=4e^x,$$

so

$$y\frac{dy}{dx}=2e^x.$$

At $x=0$,

$$y(0)=3,y^{\prime }(0)=\frac{2}{3}.$$

Further differentiation gives higher derivatives at $x=0$. Substituting into the Maclaurin formula gives

$$y=3+\frac{2}{3}x+\frac{7}{27}{x}^2+\frac{13}{243}{x}^3+\cdots .$$

Implicit differentiation is useful when a function is defined through an equation involving $y$, $e^x$, trigonometric functions, or inverse trigonometric functions.

Approximations Using Maclaurin Series

To approximate a number using a Maclaurin series:

1. Choose a suitable standard expansion.
2. Rewrite the number as a function value near $x=0$.
3. Check the validity range.
4. Substitute the chosen value of $x$.
5. Keep enough terms for the requested accuracy.

Example: Approximate $\ln (1.01)$

Use

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

Put $x=0.01$:

$$\ln (1.01)=0.01-\frac{(0.01{)}^2}{2}+\frac{(0.01{)}^3}{3}-\cdots .$$

To five decimal places,

$$\ln (1.01)\approx \mathrm{0.00995.}$$

This works well because $0.01$ is close to $0$.

Example: Approximate $\sqrt{0.96}$

Write

$$\sqrt{0.96}=\sqrt{1-0.04}.$$

Use

$$(1+x{)}^{1/2}=1+\frac{1}{2}x-\frac{1}{8}{x}^2+\frac{1}{16}{x}^3-\cdots .$$

Put $x=-0.04$:

$$\sqrt{0.96}=1-0.02-0.0002-0.000004-\cdots \approx \mathrm{0.97980.}$$

Accuracy And Number Of Terms

The accuracy of a truncated Maclaurin series depends mainly on:

• how far $x$ is from $0$
• how many terms are kept
• whether the chosen $x$ lies in the validity range

For example,

$$e^x\approx 1+x$$

is a good linear approximation only near $x=0$, while

$$e^x\approx 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}$$

often gives a better approximation near $0$, but a higher truncation order does not guarantee a smaller error at every fixed point.

Caption: For $e^x$, successive Maclaurin polynomials match more derivatives at $x=0$ and generally track the curve well over a larger nearby region. The figure illustrates local improvement; it is not a universal theorem that every added term reduces pointwise error everywhere.

Small-Angle Approximations

Small-angle approximations come directly from the Maclaurin expansions of trigonometric functions.

For $x$ small and measured in radians,

$$\sin x=x-\frac{x^3}{3!}+\cdots \approx x,$$ $$\cos x=1-\frac{x^2}{2!}+\cdots \approx 1-\frac{x^2}{2},$$

and

$$\tan x\approx x.$$

Thus

$$\boxed{\sin x\approx x,\cos x\approx 1-\frac{x^2}{2},\tan x\approx x.}$$

The simpler approximation

$$\cos x\approx 1$$

is sometimes enough, but $1-\frac{x^2}{2}$ is more accurate when terms up to $x^2$ are needed.

For example, if $x$ is small enough to neglect $x^3$ and higher powers, then

$$\sin (2x)\approx 2x,$$

and

$$\cos (3x)\approx 1-\frac{(3x{)}^2}{2}=1-\frac{9}{2}{x}^2.$$

Worked Example: Deriving $e^x$

Let

$$f(x)=e^x.$$

Then

$$f^{\prime }(x)=e^x,f^{\prime \prime }(x)=e^x,f^{\prime \prime \prime }(x)=e^x,$$

and in general

$$f^{(r)}(x)=e^x.$$

At $x=0$,

$$f(0)=f^{\prime }(0)=f^{\prime \prime }(0)=\cdots =1.$$

Therefore

$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots +\frac{x^r}{r!}+\cdots .$$

Worked Example: Deriving $\tan x$ Up To $x^3$

Let

$$f(x)=\tan x.$$

Then

$$f(0)=0.$$

Since

$$f^{\prime }(x)={\sec }^{2}x=1+{\tan }^{2}x,$$

we get

$$f^{\prime }(0)=1.$$

Differentiating again,

$$f^{\prime \prime }(x)=2\tan x{\sec }^{2}x,$$

so

$$f^{\prime \prime }(0)=0.$$

Differentiating once more gives

$$f^{\prime \prime \prime }(0)=2.$$

Therefore

$$\tan x=0+x+\frac{0}{2!}{x}^2+\frac{2}{3!}{x}^3+\cdots =x+\frac{x^3}{3}+\cdots .$$

Thus, for small $x$,

$$\tan x\approx x.$$

Worked Example: Expanding A Binomial Expression

Expand $(1-4x{)}^{1/2}$ up to and including the term in $x^3$.

Use

$$(1+u{)}^n=1+nu+\frac{n(n-1)}{2!}{u}^2+\frac{n(n-1)(n-2)}{3!}{u}^3+\cdots .$$

Here

$$n=\frac{1}{2},u=-4x.$$

Therefore

$$(1-4x{)}^{1/2}=1+\frac{1}{2}(-4x)+\frac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2!}(-4x{)}^2+\frac{\frac{1}{2}\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{3!}(-4x{)}^3+\cdots .$$

Simplifying,

$$(1-4x{)}^{1/2}=1-2x-2x^2-4x^3+\cdots .$$

The validity condition is

$$|-4x|<1,$$

so

$$|x|<\frac{1}{4}.$$

Worked Example: Expanding A Product

Find the expansion of $e^x\ln (1+x)$ up to and including the term in $x^3$.

Use

$$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots ,$$

and

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

Then

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

Collecting terms gives

$$e^x\ln (1+x)=x+\frac{1}{2}{x}^2+\frac{1}{3}{x}^3+\cdots .$$

The validity range comes from $\ln (1+x)$:

$$-1<x\le 1.$$

Worked Example: Coefficient Of $x^r$

Find the coefficient of $x^r$ in

$$(2+3x{)}^{-3}.$$

First factor out $2$:

$$(2+3x{)}^{-3}=2^{-3}{\left(1+\frac{3x}{2}\right)}^{-3}.$$

The general term is

$$2^{-3}\cdot \frac{(-3)(-4)(-5)\cdots (-3-r+1)}{r!}{\left(\frac{3x}{2}\right)}^r.$$

Since

$$(-3)(-4)\cdots (-r-2)=(-1{)}^r(3)(4)\cdots (r+2),$$

and

$$(3)(4)\cdots (r+2)=\frac{(r+2)!}{2!},$$

the coefficient of $x^r$ is

$$\boxed{2^{-3}\cdot (-1{)}^r\frac{(r+2)!}{2!r!}{\left(\frac{3}{2}\right)}^r}$$

or

$$\boxed{(-1{)}^r\frac{(r+1)(r+2)}{16}{\left(\frac{3}{2}\right)}^r}.$$

The validity condition is

$$\left|\frac{3x}{2}\right|<1,$$

so

$$|x|<\frac{2}{3}.$$

Enrichment: Taylor Expansion Around $x=a$

The Maclaurin series is a special case of Taylor series.

A Maclaurin series expands a function around

$$x=0.$$

A Taylor series around $x=a$ expands the function around

$$x=a.$$

The Taylor series of $f(x)$ about $x=a$ is

$$f(x)=f(a)+f^{\prime }(a)(x-a)+\frac{f^{\prime \prime }(a)}{2!}(x-a{)}^2+\frac{f^{\prime \prime \prime }(a)}{3!}(x-a{)}^3+\cdots .$$

If $a=0$, this becomes the Maclaurin series.

So:

$$\boxed{\text{Maclaurin series}=\text{Taylor series centred at }0.}$$

This is enrichment because the current H2 core emphasis is Maclaurin series, but the wider idea helps explain why the centre of expansion matters.

Caption: Maclaurin expansion is centred at $x=0$; Taylor expansion shifts the local centre to $x=a$.

Common Mistakes

• Forgetting to state the validity range.
• Using degrees instead of radians in trigonometric expansions.
• Assuming every Maclaurin series is valid for all real $x$.
• Keeping too few terms before multiplying series.
• Support/enrichment: choosing an invalid real branch when forming a descending-power expansion.
• Mishandling endpoint inequalities for $\ln (1+x)$.
• Treating the formula as memorised decoration rather than derivative information at $x=0$.
• Forgetting that a truncated series is an approximation, not usually an exact equality.

Problem-Solving Checklist

1. Identify whether the target resembles $e^x$, $\sin x$, $\cos x$, $\ln (1+x)$, or $(1+x{)}^n$.
2. Rewrite or factor the expression into a standard form.
3. For core questions, form the required ascending-power expansion. Support/enrichment: use powers of $1/x$ only when a descending expansion is explicitly required, checking the real branch.
4. Substitute carefully into the standard expansion.
5. Keep enough terms for the requested order.
6. Simplify coefficients and combine like powers.
7. State the validity range.
8. For approximations, check that the chosen value lies within the validity range.
9. For small-angle approximations, check that angles are in radians.
10. For non-standard functions, consider using the derivative formula or implicit differentiation.

Summary Table

Function	Maclaurin expansion	Validity
$e^x$	$1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$	all real $x$
$\sin x$	$x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$	all real $x$, radians
$\cos x$	$1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots$	all real $x$, radians
$\ln (1+x)$	$x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots$	$-1<x\le 1$
$(1+x{)}^n$	$1+nx+\frac{n(n-1)}{2!}{x}^2+\cdots$	$

Practice Questions

Basic Fluency

1. Expand $e^{3x}$ up to and including the term in $x^3$.
2. Expand $\sin (2x)$ up to and including the term in $x^5$.
3. Expand $\cos (4x)$ up to and including the term in $x^4$.
4. Expand $\ln (1+5x)$ up to and including the term in $x^3$, and state the validity range.
5. Expand $(1-2x{)}^{-1}$ up to and including the term in $x^4$, and state the validity range.

Exam-Style Practice

6. Expand $(2+3x{)}^{1/2}$ in ascending powers of $x$ up to and including the term in $x^3$.
7. Expand $e^x\cos x$ up to and including the term in $x^3$.
8. Expand $\ln \left(\frac{1+x}{1-x}\right)$ up to and including the term in $x^5$.
9. Use partial fractions to expand

$$\frac{x+5}{(1+3x)(2-x)}$$

in ascending powers of $x$ up to and including the term in $x^2$, and state the validity range.

Enrichment Practice

10. Use implicit differentiation to find the Maclaurin expansion of $y=\sqrt{5+4e^x}$ up to and including the term in $x^3$.
11. Find the Taylor expansion of $\sqrt{x}$ about $x=4$ up to and including the term in $(x-4{)}^2$.
12. Use a suitable series to approximate $\sqrt{0.98}$ correct to five decimal places.

Related

• Series and Sequences
• Differentiation
• Complex Numbers