Series Revision Notes for AQA A-Level Further Maths

Maclaurin series 2

Introduction to Maclaurin series expansion

The Maclaurin series is a powerful tool for representing functions as infinite series of terms. When working with these series, we need to understand when they can be applied and how to use them effectively.

A convergent series is one where an infinite number of terms adds up to a finite sum. This is a crucial property because not all infinite series converge. For a function $f(x)$ to be expanded using the Maclaurin series, three conditions must be met:

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Conditions for Maclaurin expansion:

The function can be expressed as a convergent infinite series
Each term in $f(x)$ can be differentiated
Each differentiated term has a finite value when $x = 0$

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Maclaurin series formula

The Maclaurin series formula for any suitable function $f(x)$ is:

$f(x) = f(0) + xf'(0) + \frac{x^2}{2!}f''(0) + \frac{x^3}{3!}f'''(0) + \frac{x^4}{4!}f''''(0) + \ldots + \frac{x^r}{r!}f^{(r)}(0) + \ldots$

This formula is sometimes called the expansion of $f(x)$ about $x = 0$ . Notice that each term involves evaluating a derivative at $x = 0$ and multiplying by the appropriate power of $x$ divided by a factorial.

Range of validity for standard functions

Not all Maclaurin series are valid for all values of $x$ . Different functions have different domains where their series expansions converge. Here are the ranges for commonly used functions:

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Key observations from the range of validity table:

The exponential function $e^x$ and trigonometric functions $\sin x$ and $\cos x$ have series that work for all real values of $x$
The binomial series $(1+x)^n$ converges only when $-1 < x < 1$ (provided $n \in \mathbb{R}$ )
The natural logarithm series $\ln(1+x)$ converges for $-1 < x \leq 1$

Exam tip: These ranges of validity are crucial. You should be able to derive these standard series yourself, not just memorise them.

Deriving Maclaurin series from first principles

To derive a Maclaurin series, we follow a systematic process of finding successive derivatives and evaluating them at $x = 0$ .

Why some functions cannot be expanded

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Functions without Maclaurin series

Not every function has a Maclaurin series. Consider $f(x) = \ln(x)$ . When we try to find $f(0)$ , we get $\ln(0) = -\infty$ , which is not finite. Therefore, a Maclaurin series for $\ln(x)$ is not possible.

However, $\ln(1+x)$ works because $f(0) = \ln(1) = 0$ , which is finite.

Worked example: deriving $\ln(1+x)$

Let's derive the Maclaurin series for $f(x) = \ln(1+x)$ step by step.

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Worked Example: Deriving the Maclaurin series for $\ln(1+x)$

First, we need $f(0)$ : $f(x) = \ln(1+x), \text{ so } f(0) = \ln(1) = 0$

Next, find the first derivative using the chain rule: $f'(x) = (1+x)^{-1} = \frac{1}{1+x}, \text{ so } f'(0) = 1$

Continue with higher derivatives: $f''(x) = -(1+x)^{-2}, \text{ so } f''(0) = -1$

$f'''(x) = 2(1+x)^{-3}, \text{ so } f'''(0) = 2$

$f^{(4)}(x) = -6(1+x)^{-4}, \text{ so } f^{(4)}(0) = -6 = -3!$

$f^{(5)}(x) = 24(1+x)^{-5}, \text{ so } f^{(5)}(0) = 4!$

Exam tip: The pattern becomes clear after a few terms and can be proved by induction if needed.

Now substitute these values into the Maclaurin formula:

$f(x) = f(0) + xf'(0) + \frac{x^2}{2!}f''(0) + \frac{x^3}{3!}f'''(0) + \frac{x^4}{4!}f''''(0) + \ldots$

$f(x) = \ln(1+x) = x - \frac{x^2}{2!} + \frac{2x^3}{3!} - \frac{6x^4}{4!} + \frac{24x^5}{5!} - \ldots$

Simplifying the factorials:

$f(x) = \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \ldots$

This shows the alternating pattern in the series for $\ln(1+x)$ .

Multiplying Maclaurin series

Sometimes you need to find the expansion of a product of two functions by multiplying their individual Maclaurin series. When doing this, you only need to multiply terms up to the required power of $x$ .

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Technique for multiplying series:

When multiplying two Maclaurin series, you only need to multiply terms up to the required power of $x$ . This saves time and reduces the complexity of calculations. Focus on collecting like powers of $x$ after multiplication.

Worked example: expanding $\tan(2x)\ln(1+x)$

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Worked Example: Expanding $\tan(2x)\ln(1+x)$

Given that the first three terms of the series for $\tan x$ are $x + \frac{x^3}{3} + \frac{2x^5}{15}$ , we can find the expansion of $\tan(2x)\ln(1+x)$ up to the term in $x^5$ .

The method involves:

Writing the series for $\tan(2x)$ by replacing $x$ with $2x$ in the given series
Using the series for $\ln(1+x)$ that we derived earlier
Multiplying the two series term by term, collecting like powers of $x$

The final result is: $\tan(2x)\ln(1+x) = 2x^2 - x^3 + \frac{10x^4}{3} - \frac{11x^5}{6} + \ldots$

Exam tip: You only need to multiply terms as far as the highest power requested in the question. Don't waste time computing unnecessary higher-order terms.

Adapting standard series

You can create new series by adapting standard ones using algebraic manipulation and substitution. This is particularly useful for logarithmic functions.

Worked example: deriving $\ln(2+x)$

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Worked Example: Deriving the series for $\ln(2+x)$

To find the series for $\ln(2+x)$ , we adapt the known series for $\ln(1+x)$ .

The key steps are:

Factor out the constant: $\ln(2+x) = \ln\left[2\left(1+\frac{x}{2}\right)\right]$
Use laws of logarithms: $\ln(2+x) = \ln 2 + \ln\left(1+\frac{x}{2}\right)$
Expand $\ln\left(1+\frac{x}{2}\right)$ by replacing $x$ with $\frac{x}{2}$ in the expansion of $\ln(1+x)$

This gives: $\ln(2+x) = \ln 2 + \frac{x}{2} - \frac{x^2}{8} + \frac{x^3}{24} - \frac{x^4}{64} + \frac{x^5}{160} - \ldots$

Exam tip: When adapting series, always check that the substitution you make keeps the series within its range of validity.

General terms

Sequences and series can be represented by a formula for the general term, sometimes called the $r$ th term or $n$ th term. This provides a concise way to express patterns in sequences.

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Understanding general terms:

For example, the formula $3r - 2$ represents the sequence $1, 4, 7, 10, \ldots$

When you know only the first few terms of a sequence, you need to find the $r$ th term by identifying patterns in the sequence structure. This skill is essential for working with Maclaurin series efficiently.

Worked example: finding general terms

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Worked Example: Finding general terms for series

For $e^{2x}$ :

Starting with $e^x = 1 + x + \frac{x^2}{2!} + \ldots + \frac{x^r}{r!} + \ldots$

We replace $x$ with $2x$ to get $e^{2x}$ :

The general term of the sequence for $e^{2x}$ is $\frac{(2x)^r}{r!}$ or equivalently $\frac{2^r x^r}{r!}$

For $\ln(1+3x)$ :

Starting with $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \ldots + (-1)^{r+1}\frac{x^r}{r} + \ldots$

We replace $x$ with $3x$ :

The $r$ th term for $\ln(1+3x)$ is $(-1)^{r+1}\frac{(3x)^r}{r}$ or equivalently $(-1)^{r+1}\frac{3^r x^r}{r}$

Exam tip: The general term formula must work for all values of $r$ from the starting index onwards. Test your formula with the first few terms to verify it's correct.

Using series to expand composite functions

When a function is the product of two functions, you can find its Maclaurin series by first finding the series for each component, then multiplying them together.

Worked example: expanding $f(x) = e^x \cos x$

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Worked Example: Expanding $f(x) = e^x \cos x$

To find the first four non-zero terms:

Step 1: Find the required derivatives using the product rule.

Starting with $f(x) = e^x \cos x$ :

$f'(x) = e^x(\cos x - \sin x)$

$f''(x) = e^x(\cos x - \sin x - \sin x - \cos x) = -2e^x \sin x$

Step 2: Continue finding derivatives.

$f'''(x) = -2e^x(\sin x + \cos x)$

$f^{(4)}(x) = -2e^x(\sin x + \cos x + \cos x - \sin x) = -4e^x \cos x$

Step 3: Substitute into the Maclaurin formula.

Evaluate at $x = 0$ :

$f(0) = e^0 \cos 0 = 1$
$f'(0) = 1(1-0) = 1$
$f''(0) = -2e^0 \sin 0 = 0$
$f'''(0) = -2e^0(0+1) = -2$
$f^{(4)}(0) = -4e^0 \cos 0 = -4$

Therefore: $f(x) = 1 + x(1) + \frac{x^2}{2!}(0) + \frac{x^3}{3!}(-2) + \frac{x^4}{4!}(-4) + \ldots$

$f(x) = 1 + x - \frac{x^3}{3} - \frac{x^4}{6} + \ldots$

Exam tip: When evaluating derivatives at $x=0$ , remember that $\sin 0 = 0$ and $\cos 0 = 1$ . This simplifies many calculations.

Limits

A limiting value, or limit, is a specific value that a function approaches or tends towards as the variable approaches a particular value. This concept is fundamental in calculus and is used extensively in differentiation from first principles.

The derivative of a function $f(x)$ is defined as: $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

This gives the gradient of the curve at any value of $x$ .

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In the definition, $f'(x)$ is the limit as $h$ tends to zero. However, you can also find limits as variables approach other numbers, such as $0$ , $1$ , or $\frac{\pi}{2}$ .

Evaluating simple limits

Sometimes you can work out limits directly by substitution:

$\lim_{n \to 0}(2 + e^n) = 2 + e^0 = 2 + 1 = 3$

$\lim_{n \to \frac{\pi}{2}}(1 - \cos n) = 1 - \cos \frac{\pi}{2} = 1 - 0 = 1$

Manipulating expressions before finding limits

In other cases, you need to manipulate the function before finding the limit. For example, when finding $\lim_{n \to \infty} \frac{n+10}{n}$ , you must manipulate the expression first:

$\lim_{n \to \infty} \frac{n+10}{n} = \lim_{n \to \infty} \frac{n}{n} + \frac{10}{n} = \lim_{n \to \infty} 1 + \frac{10}{n} = 1 + 0 = 1$

Worked example: evaluating limits

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Worked Example: Evaluating limits as $n \to \infty$

Part a: $\lim_{n \to \infty} \frac{2n+5}{n}$

Divide through by $n$ before finding the limit: $\lim_{n \to \infty} \frac{2n+5}{n} = \lim_{n \to \infty} 2 + \frac{5}{n}$

As $n \to \infty$ , $\frac{5}{n} \to 0$

Therefore: $\lim_{n \to \infty} \frac{2n+5}{n} = 2$

Part b: $\lim_{n \to \infty} \frac{2n^2+5}{n}$

$\lim_{n \to \infty} \frac{2n^2+5}{n} = \lim_{n \to \infty} 2n + \frac{5}{n} = \infty$

Part c: $\lim_{n \to \infty} \frac{2n+5}{n^2}$

$\lim_{n \to \infty} \frac{2n+5}{n^2} = \lim_{n \to \infty} \frac{2}{n} + \frac{5}{n^2} = 0$

Part d: $\lim_{n \to \infty} \frac{2n^2 + n - 35}{5n^2 - 3n + 7}$

Divide each term in numerator and denominator by $n^2$ :

$\lim_{n \to \infty} \frac{2 + \frac{1}{n} - \frac{35}{n^2}}{5 - \frac{3}{n} + \frac{7}{n^2}} = \frac{2}{5}$

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Exam tips for working with limits:

Divide through by $n$ (or the highest power) before finding the limit
Never write " $\frac{5}{n} = 0$ if $n = \infty$ " or " $\frac{1}{\infty} = 0$ "
Instead, write " $\frac{5}{n}$ tends to $0$ as $n$ tends to $\infty$ "
The symbol $\infty$ is not a number; " $n$ tends to $\infty$ " is a mathematical way of saying there is no limit to the size $n$ can take

Indeterminate forms

The expression $\frac{\sin x}{x}$ is known as an indeterminate form when $x = 0$ , because both numerator and denominator equal zero at this point, giving $\frac{0}{0}$ .

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When indeterminate forms occur

Indeterminate forms occur when, for an expression of the form $\frac{f(x)}{g(x)}$ in the limit as $x \to a$ :

$\lim_{x \to a} f(x) = \lim_{x \to a} g(x) = 0$

$\lim_{x \to a} f(x) = \pm \infty \text{ and } \lim_{x \to a} g(x) = \pm \infty$

These forms cannot be evaluated directly because the result is ambiguous. Different functions giving rise to $\frac{0}{0}$ can have completely different limits.

For example:

$\lim_{x \to 0} \frac{\sin x}{x} = 1$ (not $0$ or undefined!)
$\lim_{x \to 0} \frac{x}{x} = 1$
$\lim_{x \to 0} \frac{2x}{x} = 2$

All three expressions give $\frac{0}{0}$ when $x = 0$ , but have different limits.

L'Hopital's rule

The French mathematician L'Hopital published a rule which provides a systematic method for evaluating limits of indeterminate forms.

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L'Hopital's rule

If $\lim_{x \to a} \frac{f(x)}{g(x)}$ is indeterminate, where $a$ is any real number or $\pm \infty$ , then:

$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$

This method can be repeated as many times as necessary if $\lim_{x \to a} \frac{f'(x)}{g'(x)}$ is still indeterminate.

The rule says that when the limit of $\frac{f(x)}{g(x)}$ is indeterminate, it can usually be found by evaluating the limit of $\frac{f'(x)}{g'(x)}$ instead. In other words, differentiate the numerator and denominator separately, then find the limit of the resulting expression.

Worked example: using L'Hopital's rule

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Worked Example: Applying L'Hopital's rule

Part a: $\lim_{x \to 1} \frac{x^3-1}{x^2-1}$

This gives $\frac{0}{0}$ , so we apply L'Hopital's rule:

$\lim_{x \to 1} \frac{f'(x)}{g'(x)} = \lim_{x \to 1} \frac{3x^2}{2x} = \lim_{x \to 1} \frac{3x}{2} = \frac{3 \times 1}{2} = \frac{3}{2}$

Therefore: $\lim_{x \to 1} \frac{x^3-1}{x^2-1} = \frac{3}{2}$

Part b: $\lim_{x \to 0} \frac{2-2\cos x}{x^2}$

This gives $\frac{0}{0}$ , so apply L'Hopital's rule:

$\lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} \frac{2\sin x}{2x} = \lim_{x \to 0} \frac{\sin x}{x}$

This is still $\frac{0}{0}$ , so apply L'Hopital's rule again:

$\lim_{x \to 0} \frac{f''(x)}{g''(x)} = \lim_{x \to 0} \frac{\cos x}{1} = \lim_{x \to 0} \cos x = 1$

Therefore: $\lim_{x \to 0} \frac{2-2\cos x}{x^2} = 1$

Exam tip: Sometimes you need to apply L'Hopital's rule more than once. Keep differentiating until you get an expression that can be evaluated directly.

Using Maclaurin series to evaluate limits

When you encounter an indeterminate limit such as $\frac{0}{0}$ or $\frac{\infty}{\infty}$ , you can use Maclaurin expansions as an alternative to L'Hopital's rule. This method is particularly elegant for certain types of problems.

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Strategy for solving problems involving Maclaurin series and limits:

Derive the Maclaurin series from first principles or adapt a standard Maclaurin series
Substitute the series into the expression
Use the Maclaurin series or L'Hopital's rule to work out the limit of the function

Worked example: using series to find a limit

Consider $f(x) = \frac{4x^4}{3\ln(1+x^2) - 3x^2}$

We need to adapt the known Maclaurin series for $\ln(1+x)$ to calculate $\lim_{x \to 0} f(x)$ .

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Worked Example: Using Maclaurin series to evaluate $\lim_{x \to 0} \frac{4x^4}{3\ln(1+x^2) - 3x^2}$

First, check if direct substitution works:

$\lim_{x \to 0} \frac{4x^4}{3\ln(1+x^2) - 3x^2} = \frac{0}{0}$

This is indeterminate, so we must use another method.

The standard Maclaurin series is:

$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \frac{x^6}{6} + \ldots$

Replace $x$ with $x^2$ to get $\ln(1+x^2)$ :

$\ln(1+x^2) = x^2 - \frac{x^4}{2} + \frac{x^6}{3} - \frac{x^8}{4} + \frac{x^{10}}{5} - \frac{x^{12}}{6} + \ldots$

Now substitute this series into the expression for $f(x)$ :

$f(x) = \frac{4x^4}{3\left(x^2 - \frac{x^4}{2} + \frac{x^6}{3} - \frac{x^8}{4} + \frac{x^{10}}{5} - \frac{x^{12}}{6} + \ldots\right) - 3x^2}$

Simplify the denominator:

$f(x) = \frac{4x^4}{3x^2 - \frac{3x^4}{2} + \frac{x^6}{1} - \frac{3x^8}{4} + \ldots - 3x^2}$

$f(x) = \frac{4x^4}{-\frac{3x^4}{2} + x^6 - \frac{3x^8}{4} + \ldots}$

Divide each term in numerator and denominator by $x^4$ :

$f(x) = \frac{4}{-\frac{3}{2} + x^2 - \frac{3x^4}{4} + \ldots}$

Now find the limit:

$\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{4}{-\frac{3}{2} + x^2 - \frac{3x^4}{4} + \ldots} = \frac{4}{-\frac{3}{2}} = \frac{8}{-3} = -\frac{8}{3}$

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Exam tips for using Maclaurin series with limits:

Substitute the series carefully, paying attention to signs
Divide through by the lowest power of $x$ that appears in both numerator and denominator
The Maclaurin series method often provides cleaner algebra than repeated application of L'Hopital's rule

Worked example: comparing methods

Consider $f(x) = \frac{e^{2x} - e^x}{x}$ and finding $\lim_{x \to 0} f(x)$ .

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Worked Example: Comparing Maclaurin series and L'Hopital's rule

Method 1: Using Maclaurin series

We know that $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \ldots + \frac{x^r}{r!} + \ldots$

Replace $x$ with $2x$ to get: $e^{2x} = 1 + 2x + \frac{(2x)^2}{2!} + \frac{(2x)^3}{3!} + \frac{(2x)^4}{4!} + \ldots$

$e^{2x} = 1 + 2x + 2x^2 + \frac{4x^3}{3} + \frac{2x^4}{3} + \ldots$

Therefore: $e^{2x} - e^x = x + \frac{3x^2}{2} + \frac{7x^3}{6} + \frac{5x^4}{8} + \ldots$

So: $\frac{e^{2x} - e^x}{x} = 1 + \frac{3x}{2} + \frac{7x^2}{6} + \frac{5x^3}{8} + \ldots$

Taking the limit: $\lim_{x \to 0} \frac{e^{2x} - e^x}{x} = \lim_{x \to 0} \left(1 + \frac{3x}{2} + \frac{7x^2}{6} + \frac{5x^3}{8} + \ldots\right) = 1$

Method 2: Using L'Hopital's rule

Direct substitution gives $\frac{0}{0}$ , so apply L'Hopital's rule:

$\lim_{x \to 0} \frac{e^{2x} - e^x}{x} = \lim_{x \to 0} \frac{2e^{2x} - e^x}{1} = \frac{2e^0 - e^0}{1} = \frac{2-1}{1} = 1$

Both methods give the same answer. The choice of method depends on the specific problem and personal preference.

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Exam tip: When asked to "use" a specific method (Maclaurin series or L'Hopital's rule), you must use that method. However, if asked to "find" a limit, you can choose the most efficient approach.

Remember!

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Key Points to Remember:

Convergent series: An infinite series is convergent if the sum of all its terms equals a finite value. Only convergent series can be used as Maclaurin expansions.
Maclaurin series formula: $f(x) = f(0) + xf'(0) + \frac{x^2}{2!}f''(0) + \frac{x^3}{3!}f'''(0) + \ldots + \frac{x^r}{r!}f^{(r)}(0) + \ldots$ This formula requires evaluating the function and all its derivatives at $x = 0$ .
Range of validity: Each standard Maclaurin series has a specific range where it converges. For example, $e^x$ works for all $x$ , but $(1+x)^n$ only works for $-1 < x < 1$ .
Multiplying series: When multiplying two Maclaurin series, multiply term by term and only calculate up to the required power of $x$ .
L'Hopital's rule for indeterminate forms: If $\lim_{x \to a} \frac{f(x)}{g(x)}$ gives $\frac{0}{0}$ or $\frac{\infty}{\infty}$ , then: $\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$ Differentiate numerator and denominator separately, not as a quotient.
Alternative methods for limits: Maclaurin series provide an alternative to L'Hopital's rule. Substitute the series expansion, simplify, then take the limit. Both methods should give the same result.

Series (AQA A-Level Further Maths): Revision Notes

Maclaurin series 2

Introduction to Maclaurin series expansion

Range of validity for standard functions

Deriving Maclaurin series from first principles

Why some functions cannot be expanded

Worked example: deriving ln⁡(1+x)\ln(1+x)ln(1+x)

Multiplying Maclaurin series

Worked example: expanding tan⁡(2x)ln⁡(1+x)\tan(2x)\ln(1+x)tan(2x)ln(1+x)

Adapting standard series

Worked example: deriving ln⁡(2+x)\ln(2+x)ln(2+x)

General terms

Worked example: finding general terms

Using series to expand composite functions

Worked example: expanding f(x)=excos⁡xf(x) = e^x \cos xf(x)=excosx

Limits

Evaluating simple limits

Manipulating expressions before finding limits

Worked example: evaluating limits

Indeterminate forms

L'Hopital's rule

Worked example: using L'Hopital's rule

Using Maclaurin series to evaluate limits

Worked example: using series to find a limit

Worked example: comparing methods

Remember!

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Worked example: deriving $\ln(1+x)$

Worked example: expanding $\tan(2x)\ln(1+x)$

Worked example: deriving $\ln(2+x)$

Worked example: expanding $f(x) = e^x \cos x$