Using Binomial Expansions | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Use an expansion for an approximation

A positive-integer binomial expansion is exact for every allowed value of its inputs. Keeping only its first few terms gives an approximation, not an exact equality.

Worked example

Use the first three terms of $(1+x)^5$ to estimate $1.02^5$ .

The required expansion starts $1+5x+10x^2+\cdots$ . To make $1+x=1.02$ , substitute $x=0.02$ :

\begin{aligned}1.02^5&\approx1+5(0.02)\\&\qquad+10(0.02)^2\\&=1.104\end{aligned}

The next omitted term is $10(0.02)^3=0.00008$ . It is small here; all omitted terms are positive, so the estimate is low. A calculator gives $1.1040808032$ as a check.

Small $x$ often makes higher powers small, but check the coefficients and requested accuracy too. The full finite expansion has no small- $x$ restriction.

Replace a whole expression, not one symbol

Worked example

Find the coefficient of $x^2$ in $(1+2x-x^2)^4$ .

Set $y=2x-x^2$ . Use $(1+y)^4=1+4y+6y^2+\cdots$ .

\begin{aligned}y^2&=(2x-x^2)^2\\&=4x^2-4x^3+x^4\end{aligned}

The $x^2$ contributions are $4(-x^2)$ and $6(4x^2)$ , giving coefficient 20. Powers $y^3$ and $y^4$ start at $x^3$ and $x^4$ , so cannot contribute.

Keep every term for an exact substitution

The full expansion $(1+x)^3=1+3x+3x^2+x^3$ can also evaluate a surd exactly. Substitute $x=\sqrt2$ , using $(\sqrt2)^2=2$ and $(\sqrt2)^3=2\sqrt2$ :

(1+\sqrt2)^3=7+5\sqrt2

No terms were omitted, so this is equality, not an approximation.

Now you try

Your turnIndependent practice

(a) Use three terms to estimate $0.99^6$ .

(b) Find the coefficient of $x^2$ in $(1+3x-2x^2)^5$ .

Show worked answer

(a) Substitute $x=-0.01$ into $1+6x+15x^2$ , giving $0.9415$ . The calculator check is $0.941480149401$ .

(b) Use $1+5y+10y^2+\cdots$ , with $y=3x-2x^2$ . The coefficient is $5(-2)+10(3^2)=80$ .

Use an expansion for an approximation

A positive-integer binomial expansion is exact for every allowed value of its inputs. Keeping only its first few terms gives an approximation, not an exact equality.

Worked example

Use the first three terms of $(1+x)^5$ to estimate $1.02^5$ .

The required expansion starts $1+5x+10x^2+\cdots$ . To make $1+x=1.02$ , substitute $x=0.02$ :

\begin{aligned}1.02^5&\approx1+5(0.02)\\&\qquad+10(0.02)^2\\&=1.104\end{aligned}

The next omitted term is $10(0.02)^3=0.00008$ . It is small here; all omitted terms are positive, so the estimate is low. A calculator gives $1.1040808032$ as a check.

Small $x$ often makes higher powers small, but check the coefficients and requested accuracy too. The full finite expansion has no small- $x$ restriction.

Replace a whole expression, not one symbol

Worked example

Find the coefficient of $x^2$ in $(1+2x-x^2)^4$ .

Set $y=2x-x^2$ . Use $(1+y)^4=1+4y+6y^2+\cdots$ .

\begin{aligned}y^2&=(2x-x^2)^2\\&=4x^2-4x^3+x^4\end{aligned}

The $x^2$ contributions are $4(-x^2)$ and $6(4x^2)$ , giving coefficient 20. Powers $y^3$ and $y^4$ start at $x^3$ and $x^4$ , so cannot contribute.

Keep every term for an exact substitution

The full expansion $(1+x)^3=1+3x+3x^2+x^3$ can also evaluate a surd exactly. Substitute $x=\sqrt2$ , using $(\sqrt2)^2=2$ and $(\sqrt2)^3=2\sqrt2$ :

(1+\sqrt2)^3=7+5\sqrt2

No terms were omitted, so this is equality, not an approximation.

Now you try

Your turnIndependent practice

(a) Use three terms to estimate $0.99^6$ .

(b) Find the coefficient of $x^2$ in $(1+3x-2x^2)^5$ .

Show worked answer

(a) Substitute $x=-0.01$ into $1+6x+15x^2$ , giving $0.9415$ . The calculator check is $0.941480149401$ .

(b) Use $1+5y+10y^2+\cdots$ , with $y=3x-2x^2$ . The coefficient is $5(-2)+10(3^2)=80$ .