D1 - General binomial expansions and approximations | AQA A-Level Mathematics

Use rational powers to build useful approximations near a convenient value. You will rewrite brackets into the right form, state a strict validity range and check whether a truncated expansion is accurate enough for its purpose.

Expanding a rational power

Can four simple terms give $\sqrt{1.04}$ correctly to six decimal places? The nearby value $\sqrt1=1$ is easy. We will describe the small change from $1$ , then check how much accuracy survives when we stop the calculation.

For a positive integer power, coefficients eventually stop: in the cubic expansion a fourth selection would contain a factor $3-3=0$ . A power such as $1/2$ never reaches that zero factor. Its expansion generally continues indefinitely.

You already know

For an integer power, the first terms of $(1+t)^n$ are $1+nt+n(n-1)t^2/2!$ . A rational index includes fractional and negative powers, such as $1/2$ and $-2$ .

The falling-product pattern extends to rational powers. You need to use this result, not prove the general theorem.

General binomial expansion

For $\alpha\in\mathbb Q$ ,

(1+t)^\alpha=1+\alpha t+\frac{\alpha(\alpha-1)}{2!}t^2+\frac{\alpha(\alpha-1)(\alpha-2)}{3!}t^3+\cdots

The stated validity condition is $|t|<1$ . A non-negative integer power terminates and gives an identity for all real $t$ ; otherwise use the strict condition. The theorem does not settle the endpoints by this condition alone.
In the formula booklet.

Do not use a factorial such as $(-1/2)!$ to find these coefficients. Use the falling factors. State a truncated result with an ellipsis, or use $\approx$ for an actual numerical approximation.

Worked example 1: a positive fractional power

Question. Expand $(1-2x)^{1/2}$ up to and including the term in $x^3$ , and state the validity range.

Plan. Put $\alpha=1/2$ and $t=-2x$ into the supplied formula.

\begin{aligned} (1-2x)^{1/2}&=1+\frac12(-2x)+\frac{(1/2)(-1/2)}2(-2x)^2\\ &\quad+\frac{(1/2)(-1/2)(-3/2)}6(-2x)^3+\cdots && \text{falling factors}\\ &=1-x-\frac12x^2-\frac12x^3+\cdots && \text{simplify signed powers}\\ |-2x|<1&\ \Longleftrightarrow\ -\frac12<x<\frac12 && \text{apply the validity condition} \end{aligned}

Why this step? The small change is the whole expression $-2x$ ; its square includes both $4$ and $x^2$ .

Answer. $1-x-x^2/2-x^3/2+\cdots$ , valid for $-1/2<x<1/2$ .

Check. At $x=0$ , both sides start at $1$ . Squaring the displayed polynomial gives $1-2x$ through the cubic term.

Change only the power from $1/2$ to $-1/2$ . That changes the falling factors and therefore every non-constant coefficient.

Worked example 2: a negative fractional power

Question. Expand $(1-2x)^{-1/2}$ through $x^3$ , and state its validity range.

Plan. Keep $t=-2x$ , but use $\alpha=-1/2$ .

\begin{aligned} (1-2x)^{-1/2}&=1+\left(-\frac12\right)(-2x)+\frac{(-1/2)(-3/2)}2(-2x)^2\\ &\quad+\frac{(-1/2)(-3/2)(-5/2)}6(-2x)^3+\cdots && \text{use the new power}\\ &=1+x+\frac32x^2+\frac52x^3+\cdots && \text{simplify}\\ |-2x|<1&\ \Longleftrightarrow\ -\frac12<x<\frac12 && \text{same inner change} \end{aligned}

Why this step? A negative power changes the coefficients, while the convergence condition still depends on the inner change.

Answer. $1+x+3x^2/2+5x^3/2+\cdots$ , valid for $-1/2<x<1/2$ .

Check. Multiplying this polynomial by the one in example 1 gives $1$ through $x^3$ , as reciprocal square roots require.

The original function can exist outside the series' interval. For example, $(1-2x)^{-1/2}$ exists at $x=-1$ , but this expansion about zero is outside its stated validity range there.

Watch out

Wrong: the range is $x<1/2$ . That tests the real function's denominator but misses convergence on the negative side. Right: $|-2x|<1$ , giving $-1/2<x<1/2$ .

Change the exponent once more while keeping the same inner change.

Rewriting the bracket before expanding

A bracket such as $4+3x$ does not start with $1$ , but $4(1+3x/4)$ does. Factor out the constant before applying the general expansion.

Normalising a bracket

For $a>0$ ,

(a+bx)^\alpha=a^\alpha\left(1+\frac{bx}{a}\right)^\alpha

Apply the supplied expansion to $t=bx/a$ , requiring $|bx/a|<1$ . In this interval the inner bracket is positive, so the real rational powers are defined. This rewriting uses index laws.
Not in the formula booklet: learn it.

The factor $a^\alpha$ multiplies every retained term. Leaving it out changes the constant term as well as the other coefficients.

Worked example 3: a square root with constant four

Question. Expand $\sqrt{4+3x}$ through $x^3$ , and state the validity range.

Plan. Write the expression as $2(1+3x/4)^{1/2}$ , expand, then multiply by $2$ .

\begin{aligned} \sqrt{4+3x}&=2\left(1+\frac{3x}4\right)^{1/2} && \text{factor out }4\\ &=2\left[1+\frac12\left(\frac{3x}4\right)-\frac18\left(\frac{3x}4\right)^2+\frac1{16}\left(\frac{3x}4\right)^3+\cdots\right]\\ &=2+\frac34x-\frac9{64}x^2+\frac{27}{512}x^3+\cdots && \text{multiply every term by }2 \end{aligned}

Why this step? The binomial formula applies to $1+t$ ; here $t=3x/4$ , rather than $3x$ .

Answer. The displayed expansion is valid for $|3x/4|<1$ , or $-4/3<x<4/3$ .

Check. The constant is $\sqrt4=2$ . Squaring the retained polynomial gives $4+3x$ through $x^3$ .

Change only the power to $-1/2$ . The outside factor becomes $1/2$ , so it must be reconsidered too.

Worked example 4: a reciprocal square root

Question. Expand $(4+3x)^{-1/2}$ through $x^2$ , and state the validity range.

Plan. Factor out $4^{-1/2}$ , then use the negative fractional power.

\begin{aligned} (4+3x)^{-1/2}&=\frac12\left(1+\frac{3x}4\right)^{-1/2} && \text{normalise the bracket}\\ &=\frac12\left[1-\frac12\left(\frac{3x}4\right)+\frac38\left(\frac{3x}4\right)^2+\cdots\right]\\ &=\frac12-\frac3{16}x+\frac{27}{256}x^2+\cdots && \text{include the outside factor} \end{aligned}

Why this step? Taking a reciprocal changes $\sqrt4=2$ into $1/2$ ; retaining the old outside factor would fail even at zero.

Answer. $1/2-3x/16+27x^2/256+\cdots$ , valid for $-4/3<x<4/3$ .

Check. Multiplication by example 3 gives $1$ through $x^2$ .

If the small change contains $x^2$ , track its powers and solve the corresponding inequality: $(1+2x^2)^{-1}=1-2x^2+4x^4-\cdots$ requires $2x^2<1$ , hence $|x|<1/\sqrt2$ . If two expansions are combined, use the intersection of their validity intervals.

Watch out

Wrong: use $|3x|<1$ after factoring $4+3x$ . Right: test the normalised change $|3x/4|<1$ . The range belongs to the expression actually substituted into the formula.

Change the constant under the same reciprocal square root.

Choosing a useful approximation

To estimate $\sqrt{1.04}$ , use the small change $t=0.04$ . Truncation means stopping an infinite expansion; the retained polynomial is generally an approximation, even though its coefficients are exact.

The supplied general binomial formula gives

\sqrt{1+t}=1+\frac t2-\frac{t^2}8+\frac{t^3}{16}+\cdots

This particular square-root expansion is not printed separately in the formula booklet; derive its coefficients from the supplied general formula. Small $|t|$ usually makes a short truncation more useful. Merely satisfying $|t|<1$ does not guarantee a requested number of decimal places.

Worked example 5: a small positive change

Question. Use the cubic binomial approximation to estimate $\sqrt{1.04}$ . Check whether it is correct to six decimal places.

Plan. Substitute $t=0.04$ , then compare the rounded estimate with the calculator value.

\begin{aligned} \sqrt{1.04}&\approx1+\frac{0.04}2-\frac{0.04^2}8+\frac{0.04^3}{16} && \text{retain four terms}\\ &=1.019804 && \text{evaluate without early rounding}\\ \sqrt{1.04}&=1.0198039027\ldots && \text{calculator comparison} \end{aligned}

Why this step? Accuracy is a statement about the omitted tail, so check the actual rounded values rather than counting the retained terms.

Answer. The estimate is $1.019804$ , and both values round to $1.019804$ at six decimal places.

Check. The estimate exceeds the true value by approximately $9.73\times10^{-8}$ . The next term is $-5(0.04)^4/128=-10^{-7}$ , consistent with this sign and scale.

Four rational terms have recovered six correct decimal places of an irrational number. That is the useful surprise behind these expansions: near the centre, simple arithmetic can capture a curved function very closely.

The next example changes only the sign of the small change. The odd-power terms switch sign.

Worked example 6: a small negative change

Question. Use the same cubic approximation to estimate $\sqrt{0.96}$ , and check six-decimal accuracy.

Plan. Use $t=-0.04$ , keeping its sign in the cubic term.

\begin{aligned} \sqrt{0.96}&\approx1-0.02-0.0002-0.000004 && \text{substitute }t=-0.04\\ &=0.979796 && \text{add retained terms}\\ \sqrt{0.96}&=0.9797958971\ldots && \text{calculator comparison} \end{aligned}

Why this step? $(-0.04)^3$ is negative, so the cubic contribution must decrease the estimate.

Answer. $0.979796$ , correct to six decimal places at this input.

Check. The absolute error is approximately $1.03\times10^{-7}$ ; both rounded values are $0.979796$ .

A first omitted term suggests an error's sign and scale in these examples, but it is not automatically the entire error or a rigorous bound. If the question supplies a calculator value, compare with it; if it asks for a guaranteed tolerance, use a justified bound or exact remainder.

Watch out

Wrong: “Four terms means four-decimal accuracy.” Right: the input and the omitted terms determine the error. A value close to a rounding boundary needs particular care.

Change only the size of the positive change.

Quantifying an approximation error

For a reciprocal, multiplication gives an exact remainder. Take $f(x)=1/(1+2x)$ and its quadratic binomial approximation $Q(x)=1-2x+4x^2$ . Then

\begin{aligned} (1+2x)Q(x)&=1+8x^3 && \text{middle powers cancel}\\ f(x)-Q(x)&=\frac{1-(1+8x^3)}{1+2x}=\frac{-8x^3}{1+2x} && \text{use a common denominator} \end{aligned}

This remainder is not in the formula booklet: derive it. The algebraic identity needs $x\ne-1/2$ ; obtaining $Q$ from the binomial series uses $|x|<1/2$ . These are different restrictions.

The reciprocal function and its quadratic approximation, with their signed difference

The curves meet closely near zero. In the lower panel a negative signed error means $Q$ is above the true value. An absolute tolerance uses the magnitude of the error, whatever its sign.

Worked example 7: checking a positive input

Question. Use $Q$ to approximate $f(0.05)$ , and decide whether the absolute error is less than $0.001$ .

Plan. Evaluate the polynomial and then the exact remainder.

\begin{aligned} Q(0.05)&=1-0.1+0.01=\frac{91}{100} && \text{quadratic approximation}\\ f(0.05)-Q(0.05)&=\frac{-8(1/20)^3}{1+2/20}=-\frac1{1100} && \text{exact signed error}\\ \frac1{1100}&<\frac1{1000} && \text{compare positive fractions} \end{aligned}

Why this step? The signed difference identifies overestimation; its absolute value decides the tolerance.

Answer. The estimate is $0.91$ , an overestimate whose error is less than $0.001$ .

Check. $f(0.05)=10/11$ , and $91/100-10/11=1/1100$ .

Change only the input's sign. A similar-looking polynomial calculation now produces an underestimate, and the denominator changes the error's size.

Worked example 8: checking a negative input

Question. Use $Q$ to approximate $f(-0.05)$ , and test the same tolerance $0.001$ .

Plan. Retain the signed cubic in the exact remainder, then compare its modulus.

\begin{aligned} Q(-0.05)&=1+0.1+0.01=\frac{111}{100} && \text{substitute the negative input}\\ f(-0.05)-Q(-0.05)&=\frac{-8(-1/20)^3}{1-2/20}=\frac1{900} && \text{exact remainder}\\ \frac1{900}&>\frac1{1000} && \text{test the tolerance} \end{aligned}

Why this step? A positive signed error means the true value is larger than $Q$ ; it does not mean the tolerance is satisfied.

Answer. The estimate is $1.11$ , an underestimate, and its absolute error exceeds $0.001$ .

Check. $f(-0.05)=10/9$ , and $10/9-111/100=1/900$ .

You can also guarantee accuracy throughout an interval. If $|x|\le\delta<1/2$ , then $|1+2x|\ge1-2\delta$ , so

|f(x)-Q(x)|\le\frac{8\delta^3}{1-2\delta}

This derived bound is not in the formula booklet. For $\delta=0.02$ , it gives $1/15000$ , so the error is at most that value everywhere on the interval, not just at one checked input.

Watch out

Wrong: a negative error automatically meets a positive tolerance. Right: compare $|f-Q|$ with the tolerance; use the sign separately to decide overestimate or underestimate.

Change the magnitude of the positive input in the final check.

Quick Recap

Factor out the constant so the inner bracket starts with $1$ .
Use falling factors for rational powers and state $|bx/a|<1$ .
A truncation's accuracy depends on its omitted tail, not its number of terms.
Use an actual comparison, exact remainder or justified bound to test accuracy.

The opening estimate is $\sqrt{1.04}\approx1.019804$ ; the checked calculator value confirms six-decimal accuracy.

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D1 - General binomial expansions and approximations

Expanding a rational power

Rewriting the bracket before expanding

Choosing a useful approximation

Quantifying an approximation error

Quick Recap

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