Binomial expansion
From P1, you are familiar with the binomial expansion for positive integer powers. The coefficients can be expressed using binomial coefficients (nCr notation):
Where $\displaystyle \binom{n}{r} = \frac{n!}{r!\,(n-r)!} = \frac{n(n-1)(n-2)\cdots(n-r+1)}{r!}$
After refreshing your memory with the Quick Recall questions below, we will see how this formula extends to work for any rational $n$ โ not just positive integers.
Quick Recall
From P1, you know the binomial theorem for positive integer exponents. Try these quick questions to refresh your memory!
Expand $(1+x)^3$:
Using Pascal's triangle: row 3 gives coefficients $1, 3, 3, 1$. So $(1+x)^3 = 1 + 3x + 3x^2 + x^3$.
When $n$ is a positive integer, the expansion of $(1+x)^n$ has exactly how many terms?
Since the expansion terminates when the binomial coefficient $\binom{n}{r}$ becomes zero (when $r > n$), the last term is $x^n$, giving $n+1$ terms in total.
The coefficient of $x^2$ in $(1+x)^n$ is:
From the general formula: $\binom{n}{2} = \dfrac{n(n-1)}{2!} = \dfrac{n(n-1)}{2}$. This works for any $n$ (positive, negative, or fractional!)
The series always terminates. For $n = 3$: four terms, then it stops. For $n = 10$: eleven terms, then it stops.
๐งฑ The Wall
A friend challenges you: "Find $\sqrt{1.01}$ to 6 decimal places โ without a calculator."
You know that $\sqrt{1.01} = (1 + 0.01)^{1/2}$. The binomial formula you learned in P1 requires $n$ to be a positive integer โ but $\frac{1}{2}$ is not an integer.
What happens if you plug in $n = \frac{1}{2}$ anyway?
The terms never stop. The formula produces an infinite series. But can infinitely many terms actually add up to $\sqrt{1.01}$?
๐ฏ Today's Mission
The trial widget just revealed something remarkable: the same formula that gives a finite expansion for integer $n$ produces an infinite series for fractional $n$ โ and it converges! This raises four questions we will answer today:
Apply the binomial expansion $(1+x)^n$ for any rational $n$
Discover when the infinite series converges โ and when it breaks
State the validity condition $|kx| < 1$ and find valid ranges
Transform compound forms $(a+bx)^n$ into standard form
By the end of this lesson, you will be able to expand $(1+x)^n$ for any rational $n$, determine exactly when it is valid, and handle real exam expressions like $\dfrac{1}{\sqrt{4+x}}$.
$(1+x)^3 = 1 + 3x + 3x^2 + x^3$
Series terminates after $n+1$ terms. Always valid โ no restriction on $x$.
$(1+x)^{1/2} = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 - \cdots$
Series never terminates โ infinite! Only valid when $|x| < 1$ (we'll prove this in Step 3).
$y = \dfrac{n(n-1)(n-2)\cdots(n-r+1)}{r!}$
The Key Insight above tells us the series is infinite for non-integer $n$. But does an infinite series of terms actually add up to the true value $(1+x)^n$? The Lead In trial showed it works for $x = 0.01$ โ but what about larger $x$? Use the graph tool below to investigate. Drag the $x$ slider toward 1 and beyond โ watch what happens to the amber curve in the shaded divergence zone.
True value: | 3-term approx: | error:
Green curve: true $(1+x)^n$. Amber curve: k-term expansion. Vertical line marks the current $x$. Shaded zone: $|x| > 1$ where the series diverges.
For each scenario, predict the minimum number of terms needed so that the truncated expansion $S_k(x)$ matches the true value $(1+x)^n$ to within $0.001$. But watch out โ one of these scenarios might not converge at all! If you think the series diverges, hit the DIVERGES button instead of guessing a number.
Reflection: Look at your results across all 4 rounds. What pattern do you see? Round 1 ($|x|=0.2$) converged in just 3 terms. Round 2 ($|x|=0.5$) needed 10. Round 3 ($|x|=0.9$) needed 13. And Round 4 ($|x|=1.1$) never converged at all! The key question is: where exactly is the boundary between convergence and divergence? Is it $|x| = 1$? Or somewhere else? That is what we pin down next.
You just discovered the boundary empirically: $|x| = 1$ is where convergence breaks down. Here is the precise mathematical rule โ and it generalises to the compound form $(1+kx)^n$ that you will encounter in exams.
How could you rewrite $(3 - 2x)^{\frac{1}{2}}$ so the bracket starts with 1? And once you do, what does Key relation 2 tell you about the valid range of $x$? Discuss ideas with your partner!
Take turns: one partner explains each step aloud before you reveal the next one.
Factor out 3 from inside the bracket:
$$(3 - 2x)^{\frac{1}{2}} = \left[3\left(1 - \dfrac{2}{3}x\right)\right]^{\frac{1}{2}} = 3^{\frac{1}{2}}\left(1 - \dfrac{2}{3}x\right)^{\frac{1}{2}} = \sqrt{3}\left(1 - \dfrac{2}{3}x\right)^{\frac{1}{2}}$$
Now it's in the standard form $a^n(1 + kx)^n$:
Valid when $\left|kx\right| < 1$, i.e. $\left|-\dfrac{2}{3}x\right| < 1$:
$$|x| < \frac{3}{2} \quad \iff \quad -\frac{3}{2} < x < \frac{3}{2}$$
The valid range is wider than $|x| < 1$ because $|k| = 2/3 < 1$. Smaller $|k|$ means a wider valid range!
(i) Show that $\dfrac{1}{\sqrt{4+x}} = \dfrac{1}{2}\left(1+\dfrac{x}{4}\right)^{-\frac{1}{2}}$.
(ii) Write down the first three terms in the binomial expansion of $\left(1+\dfrac{x}{4}\right)^{-\frac{1}{2}}$ in ascending powers of $x$, stating the range of values of $x$ for which this expansion is valid.
(iii) Find the first three terms in the expansion of $\dfrac{2(1-x)}{\sqrt{4+x}}$ in ascending powers of $x$, for small values of $x$.
For (i), factor $4+x = 4(1 + x/4)$ and take the square root. For (ii), substitute $n = -\tfrac{1}{2}$ and replace $x$ with $\tfrac{x}{4}$ throughout. For (iii), use the result from (i) and expand $(1-x)$ multiplied by the series from (ii), keeping terms up to $x^2$.
Factor out 4 from inside the square root:
$$\dfrac{1}{\sqrt{4+x}} = \dfrac{1}{\sqrt{4\left(1+\dfrac{x}{4}\right)}} = \dfrac{1}{2\sqrt{1+\dfrac{x}{4}}} = \dfrac{1}{2}\left(1+\dfrac{x}{4}\right)^{-\frac{1}{2}} \quad \checkmark$$
Substitute $n = -\tfrac{1}{2}$ and replace $x$ with $\tfrac{x}{4}$:
$$\left(1+\dfrac{x}{4}\right)^{-\frac{1}{2}} = 1 + \left(-\dfrac{1}{2}\right)\!\left(\dfrac{x}{4}\right) + \dfrac{\left(-\dfrac{1}{2}\right)\left(-\dfrac{3}{2}\right)}{2!}\!\left(\dfrac{x}{4}\right)^2 + \cdots$$
Simplify:
$$= 1 - \dfrac{x}{8} + \dfrac{\dfrac{3}{4}}{2}\cdot\dfrac{x^2}{16} + \cdots = 1 - \dfrac{x}{8} + \dfrac{3x^2}{128} + \cdots$$
Valid when $\left|\dfrac{x}{4}\right| < 1$, i.e. $|x| < 4$, i.e. $-4 < x < 4$.
Using parts (i) and (ii):
$$\dfrac{2(1-x)}{\sqrt{4+x}} = 2(1-x) \cdot \dfrac{1}{2}\left(1 - \dfrac{x}{8} + \dfrac{3x^2}{128} + \cdots\right)$$
$$= (1-x)\left(1 - \dfrac{x}{8} + \dfrac{3x^2}{128}\right)$$
Expand (keeping terms up to $x^2$):
$$= 1 - \dfrac{x}{8} + \dfrac{3x^2}{128} - x + \dfrac{x^2}{8} + \cdots$$
$$= 1 + \left(-\dfrac{1}{8} - 1\right)x + \left(\dfrac{3}{128} + \dfrac{1}{8}\right)x^2$$
$$= 1 - \dfrac{9}{8}x + \dfrac{19}{128}x^2$$
(ii) $\left(1+\dfrac{x}{4}\right)^{-\frac{1}{2}} = 1 - \dfrac{x}{8} + \dfrac{3x^2}{128} + \cdots$, valid for $-4 < x < 4$.
(iii) $\dfrac{2(1-x)}{\sqrt{4+x}} = 1 - \dfrac{9}{8}x + \dfrac{19}{128}x^2 + \cdots$
(i) Write down the expansion of $(1-x)^3$.
(ii) Find the first three terms in the expansion of $(1+x)^{-4}$ in ascending powers of $x$. For what values of $x$ is this expansion valid?
(iii) When the expansion is valid, $\dfrac{(1-x)^3}{(1+x)^4}$ can be written as $1 + ax + bx^2 + \cdots$. Find the values of $a$ and $b$.
For (i), use the standard binomial theorem (positive integer $n = 3$). For (ii), substitute $n = -4$ into the general binomial expansion. For (iii), write $\dfrac{(1-x)^3}{(1+x)^4} = (1-x)^3 \cdot (1+x)^{-4}$ and multiply the two expansions, collecting terms up to $x^2$.
Since $n = 3$ is a positive integer, the binomial theorem applies directly:
$$(1-x)^3 = 1 - 3x + 3x^2 - x^3$$
Substitute $n = -4$:
$$(1+x)^{-4} = 1 + (-4)x + \dfrac{(-4)(-5)}{2!}x^2 + \cdots = 1 - 4x + 10x^2 + \cdots$$
Valid when $|x| < 1$ (since $n$ is a negative integer).
$$\dfrac{(1-x)^3}{(1+x)^4} = (1-x)^3 \cdot (1+x)^{-4} = (1 - 3x + 3x^2 - x^3)(1 - 4x + 10x^2 + \cdots)$$
Collecting terms up to $x^2$:
Constant: $1 \cdot 1 = 1$
$x$ term: $1 \cdot (-4x) + (-3x) \cdot 1 = -4x - 3x = -7x$
$x^2$ term: $1 \cdot 10x^2 + (-3x)(-4x) + 3x^2 \cdot 1 = 10x^2 + 12x^2 + 3x^2 = 25x^2$
(ii) $(1+x)^{-4} = 1 - 4x + 10x^2 + \cdots$, valid for $|x| < 1$.
(iii) $a = -7$, $\quad b = 25$.
Test your understanding of all four lesson objectives. Each question targets one objective โ try each before moving on!
What is the coefficient of $x^2$ in the expansion of $(1+x)^{-\frac{1}{2}}$?
Use the $x^2$ term formula: $\dfrac{n(n-1)}{2!}x^2$. With $n = -\tfrac{1}{2}$:
Coefficient = $\dfrac{\left(-\tfrac{1}{2}\right)\left(-\tfrac{3}{2}\right)}{2} = \dfrac{\tfrac{3}{4}}{2} = \dfrac{3}{8}$.
For what values of $x$ is the binomial expansion of $(1 + 3x)^{-2}$ valid?
The expansion of $(1 + kx)^n$ (with $n$ not a positive integer) is valid when $|kx| < 1$. Here $k = 3$, so $|3x| < 1$, giving $|x| < \tfrac{1}{3}$, i.e. $-\tfrac{1}{3} < x < \tfrac{1}{3}$.
Express $(2 + 3x)^{-2}$ in the form $a(1 + bx)^n$, where $a$, $b$, and $n$ are constants.
Factor out the leading coefficient: $(2 + 3x)^{-2} = \left[2\left(1 + \tfrac{3x}{2}\right)\right]^{-2} = 2^{-2}\left(1 + \tfrac{3x}{2}\right)^{-2} = \dfrac{1}{4}\left(1 + \dfrac{3x}{2}\right)^{-2}$. So $a = \tfrac{1}{4}$, $b = \tfrac{3}{2}$, $n = -2$.
What are the first three terms in the binomial expansion of $(1 - 2x)^{\frac{1}{2}}$?
Substitute $n = \tfrac{1}{2}$ and replace $x$ with $-2x$: $(1-2x)^{1/2} = 1 + \tfrac{1}{2}(-2x) + \dfrac{(\tfrac{1}{2})(-\tfrac{1}{2})}{2}(-2x)^2 + \cdots = 1 - x + \dfrac{-\tfrac{1}{4}}{2} \cdot 4x^2 = 1 - x - \dfrac{x^2}{2} + \cdots$ Valid for $|-2x| < 1$, i.e. $|x| < \tfrac{1}{2}$.
Expand $(1+x)^{\frac{1}{2}}$ as a series of ascending powers of $x$ up to and including the term in $x^2$, stating the set of values of $x$ for which the expansion is valid.
Substitute $n = \dfrac{1}{2}$:
$$(1+x)^{\frac{1}{2}} = 1 + \dfrac{1}{2}x + \dfrac{\left(\dfrac{1}{2}\right)\left(-\dfrac{1}{2}\right)}{2!}x^2 + \cdots$$
$$= 1 + \dfrac{x}{2} - \dfrac{x^2}{8} + \cdots$$
Valid when $|x| < 1$.
$$(1+x)^{\frac{1}{2}} = 1 + \dfrac{x}{2} - \dfrac{x^2}{8} + \cdots, \quad |x| < 1$$
Expand $\left(1+\dfrac{x}{3}\right)^{-2}$ as a series of ascending powers of $x$ up to and including the term in $x^3$, stating the set of values of $x$ for which the expansion is valid.
Substitute $n = -2$ and replace $x$ with $\dfrac{x}{3}$:
$$\left(1+\dfrac{x}{3}\right)^{-2} = 1 + (-2)\!\left(\dfrac{x}{3}\right) + \dfrac{(-2)(-3)}{2!}\!\left(\dfrac{x}{3}\right)^2 + \dfrac{(-2)(-3)(-4)}{3!}\!\left(\dfrac{x}{3}\right)^3 + \cdots$$
Simplify each term:
$$= 1 - \dfrac{2x}{3} + 3 \cdot \dfrac{x^2}{9} - 4 \cdot \dfrac{x^3}{27} + \cdots$$
$$= 1 - \dfrac{2x}{3} + \dfrac{x^2}{3} - \dfrac{4x^3}{27} + \cdots$$
Valid when $\left|\dfrac{x}{3}\right| < 1$, i.e. $|x| < 3$, i.e. $-3 < x < 3$.
$$\left(1+\dfrac{x}{3}\right)^{-2} = 1 - \dfrac{2x}{3} + \dfrac{x^2}{3} - \dfrac{4x^3}{27} + \cdots, \quad -3 < x < 3$$
Expand $(1 + x)\sqrt{1 - 2x}$ in ascending powers of $x$, up to and including the term in $x^2$, simplifying the coefficients.
Write $\sqrt{1-2x} = (1-2x)^{1/2}$ and expand using $n = \tfrac{1}{2}$ with $x$ replaced by $-2x$. Then multiply by $(1+x)$ and collect terms up to $x^2$.
Substitute $n = \dfrac{1}{2}$ and replace $x$ with $-2x$:
$$(1-2x)^{\frac{1}{2}} = 1 + \dfrac{1}{2}(-2x) + \dfrac{\left(\dfrac{1}{2}\right)\left(-\dfrac{1}{2}\right)}{2!}(-2x)^2 + \cdots$$
$$= 1 - x + \dfrac{-\dfrac{1}{4}}{2} \cdot 4x^2 + \cdots = 1 - x - \dfrac{x^2}{2} + \cdots$$
$$(1+x)(1 - x - \tfrac{x^2}{2}) = 1 - x - \dfrac{x^2}{2} + x - x^2 + \cdots$$
$$= 1 + (-x + x) + \left(-\dfrac{x^2}{2} - x^2\right) = 1 - \dfrac{3x^2}{2}$$
$$(1+x)\sqrt{1-2x} = 1 - \dfrac{3}{2}x^2 + \cdots$$
(Note: the $x$ term cancels. Valid for $|-2x| < 1$, i.e. $|x| < \tfrac{1}{2}$.)
Expand $\sqrt[3]{1 - 6x}$ in ascending powers of $x$ up to and including the term in $x^3$, simplifying the coefficients.
Write $\sqrt[3]{1-6x} = (1-6x)^{1/3}$ and substitute $n = \tfrac{1}{3}$ with $x$ replaced by $-6x$. Compute the binomial coefficients $\binom{1/3}{1}$, $\binom{1/3}{2}$, $\binom{1/3}{3}$ carefully as fractions.
Substitute $n = \dfrac{1}{3}$ and replace $x$ with $-6x$:
$$(1-6x)^{\frac{1}{3}} = 1 + \dfrac{1}{3}(-6x) + \dfrac{\left(\dfrac{1}{3}\right)\left(-\dfrac{2}{3}\right)}{2!}(-6x)^2 + \dfrac{\left(\dfrac{1}{3}\right)\left(-\dfrac{2}{3}\right)\left(-\dfrac{5}{3}\right)}{3!}(-6x)^3 + \cdots$$
Simplify each term:
Term in $x$: $\quad \dfrac{1}{3}(-6x) = -2x$
Term in $x^2$: $\quad \dfrac{-\tfrac{2}{9}}{2} \cdot 36x^2 = -\dfrac{1}{9} \cdot 36x^2 = -4x^2$
Term in $x^3$: $\quad \dfrac{\tfrac{10}{27}}{6} \cdot (-216x^3) = \dfrac{5}{81} \cdot (-216x^3) = -\dfrac{40}{3}x^3$
$$\sqrt[3]{1-6x} = 1 - 2x - 4x^2 - \dfrac{40}{3}x^3 + \cdots$$
(Valid for $|-6x| < 1$, i.e. $|x| < \tfrac{1}{6}$.)
Exit Ticket
Check the learning objectives you feel confident about: