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

Worked Example: Using the Binomial Expansion

Question: Use the Maclaurin expansion of $(1+x)^n$ to find the first five terms of the series for $f(x) = \frac{1}{(1-x)^2}$.

Solution:

We start with the general binomial expansion: $(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \ldots + \frac{n(n-1)\ldots(n-1+r)}{r!}x^r + \ldots$

For the function $\frac{1}{(1-x)^2}$, we identify that $n = -2$ and we replace $x$ with $-x$:

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

Simplifying:

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

Validity: Since $n = -2$ is not a positive integer, the expansion of $(1+x)^n$ is valid for $-1 < x < 1$.

In our case, we replaced $x$ with $-x$, so the expansion for $(1-x)^{-2}$ is valid when $-1 < -x < 1$, which can be rearranged to $-1 < x < 1$.

Worked Example: Finding Validity Ranges

Question: Write down the range of values of $x$ that make the Maclaurin series for these functions valid:

a) $\ln(1-3x)$

b) $\ln(1+x^2)$

c) $\frac{1}{3+x}$

Solution:

a) For $\ln(1-3x)$:

The standard series $\ln(1+x)$ is valid for $-1 < x < 1$.

We replace $x$ with $-3x$ in the standard expansion, so we need:

$-1 < -3x \leq 1$

Dividing by $-3$ (and reversing the inequality signs):

$\frac{1}{3} > x \geq -\frac{1}{3}$

Rearranging: $-\frac{1}{3} \leq x < \frac{1}{3}$

b) For $\ln(1+x^2)$:

We replace $x$ with $x^2$ in the standard expansion, so we need:

$-1 < x^2 \leq 1$

Since $x^2 \geq 0$ always, the left inequality is automatically satisfied.

From $x^2 \leq 1$, we get: $-1 \leq x \leq 1$

c) For $\frac{1}{3+x}$:

First, rewrite this in a form we can use:

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

The expansion $(1+x)^{-1}$ is valid when $-1 < x < 1$.

We replace $x$ with $\frac{x}{3}$, so we need:

$-1 < \frac{x}{3} < 1$

Multiplying by 3: $-3 < x < 3$

Therefore, $\frac{1}{3+x}$ is valid when $-3 < x < 3$.

Worked Example: Expanding Compound Functions

Question: Expand $\frac{\sin x}{\sqrt{(1+x)}}$ as far as the term in $x^5$ and give the range of $x$ for which the series is valid.

Solution:

Step 1: Rewrite the function in a suitable form.

$\frac{1}{\sqrt{(1+x)}} = (1+x)^{-\frac{1}{2}}$

Step 2: Use the binomial expansion with $n = -\frac{1}{2}$:

$(1+x)^{-\frac{1}{2}} = 1 + \left(-\frac{1}{2}\right)x + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2!}x^2 + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{3!}x^3 + \ldots$

Simplifying:

$= 1 - \frac{1}{2}x + \frac{3x^2}{8} - \frac{5x^3}{16} + \frac{35x^4}{128} - \frac{63x^5}{256} + \ldots$

Step 3: Use the standard expansion for $\sin x$:

$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \ldots = x - \frac{x^3}{6} + \frac{x^5}{120} - \ldots$

Step 4: Multiply the two expansions together. Only multiply out terms that give powers up to $x^5$:

$\frac{\sin x}{\sqrt{(1+x)}} = (\sin x)(1+x)^{-\frac{1}{2}}$

$= \left(x - \frac{x^3}{6} + \frac{x^5}{120}\right)\left(1 - \frac{1}{2}x + \frac{3x^2}{8} - \frac{5x^3}{16} + \frac{35x^4}{128} - \frac{63x^5}{256}\right)$

Multiplying out (keeping only terms up to $x^5$):

$= x - \frac{1}{2}x^2 + \frac{3x^3}{8} - \frac{5x^4}{16} + \frac{35x^5}{128} + \ldots - \frac{x^3}{6} + \frac{x^4}{12} - \frac{x^5}{16} + \ldots$

Collecting like terms:

$= x - \frac{1}{2}x^2 + \frac{5x^3}{24} - \frac{11x^4}{48} + \frac{42x^5}{1920} + \ldots$

Step 5: Determine validity.

The expansion for $(1+x)^{-\frac{1}{2}}$ is valid for $-1 < x < 1$.

The expansion for $\sin x$ is valid for all values of $x$.

Therefore, the expansion for $\frac{\sin x}{\sqrt{(1+x)}}$ is valid for $-1 < x < 1$ (we choose the range that is valid for both series).