H2 Math Binomial Series: Study Guide & Examples

JC · H2 Mathematics · Study guide

Expand recognisable binomial tasks: numerical estimates, products, general coefficients, finite terms and replacements containing more than one power.

Concepts · worked methods · graphs7 min read · practice linked below

Put the constant term into the form 1

For a non-integer or negative power, first write the expression as a constant times (1 + u) raised to the required power. Expand in u, then substitute back. The usual binomial-series condition is |u| < 1; translate it into a condition on x. A positive integer power instead gives a finite expansion valid for every input where the original terms are defined.

Normalise, expand, and state validity

A square root whose constant is not 1

Expand $\sqrt{9+2x}$ in ascending powers of $x$ , up to and including $x^2$ . Hence estimate $\sqrt{10}$ .

\begin{aligned}\sqrt{9+2x}&=3\left(1+\frac{2x}{9}\right)^{1/2},\\&=3\left[1+\frac12\left(\frac{2x}{9}\right)\right.\\&\qquad\left.-\frac18\left(\frac{2x}{9}\right)^2+\cdots\right],\\&=3+\frac{x}{3}-\frac{x^2}{54}+\cdots.\end{aligned}

The binomial-series condition is

\begin{aligned}\left|\frac{2x}{9}\right|&<1,\\-\frac92&<x<\frac92.\end{aligned}

To obtain $\sqrt{10}$ , set $9+2x=10$ , so $x=1/2$ :

\begin{aligned}\sqrt{10}&\approx3+\frac{1/2}{3}-\frac{(1/2)^2}{54},\\&=3+\frac16-\frac1{216},\\&=\frac{683}{216}\\&\approx3.16204.\end{aligned}

The calculator value is $3.16227766\ldots$ , so the three-term approximation is slightly low. Being inside the validity interval permits the infinite expansion; it does not guarantee a chosen short truncation is accurate to any requested number of decimal places.

How far from zero is the substitution?

Choose x:0.5

At x = 0.5: exact value 3.162278; approximation 3.162037; percentage error 0.00761%.

Track every contribution to the required power

A product of two binomial expansions

Expand $\dfrac{\sqrt{1+x}}{1-2x}$ up to and including $x^3$ , stating a validity interval.

Expand each factor far enough to contribute to degree 3:

\begin{aligned}(1+x)^{1/2}&=1+\frac{x}{2}-\frac{x^2}{8}\\&\quad+\frac{x^3}{16}+\cdots,\end{aligned}

\begin{aligned}(1-2x)^{-1}&=1+2x+4x^2\\&\quad+8x^3+\cdots.\end{aligned}

Collect terms by degree. The coefficient of $x^3$ receives four contributions:

\begin{aligned}[x]:\quad2+\frac12&=\frac52,\\{}[x^2]:\quad4+\frac12(2)-\frac18&=\frac{39}{8},\\{}[x^3]:\quad8+\frac12(4)-\frac18(2)\\\quad+\frac1{16}&=\frac{157}{16}.\end{aligned}

Thus

\begin{aligned}\frac{\sqrt{1+x}}{1-2x}&=1+\frac52x+\frac{39}{8}x^2\\&\quad+\frac{157}{16}x^3+\cdots.\end{aligned}

Both expansions must be valid: $|x|<1$ and $|2x|<1$ . Their common interval is $-1/2<x<1/2$ . At $x=0$ , both the original expression and the constant term are 1.

Use partial fractions for a general coefficient

An alternating contribution and a repeated denominator

For $f(x)=\dfrac{2x^2-x+5}{(1+x)(1-x)^2}$ , find the expansion through $x^4$ and the coefficient of $x^r$ .

\begin{aligned}f(x)&=\frac A{1+x}+\frac B{1-x}\\&\quad+\frac C{(1-x)^2}.\end{aligned}

\begin{aligned}2x^2-x+5&=A(1-x)^2\\&\quad+B(1-x^2)\\&\quad+C(1+x).\end{aligned}

Set $x=-1,1,0$ in the polynomial identity:

\begin{aligned}8&=4A\implies A=2,\\6&=2C\implies C=3,\\5&=A+B+C\implies B=0.\end{aligned}

Now expand the two remaining fractions:

\begin{aligned}\frac2{1+x}&=2-2x+2x^2\\&\quad-2x^3+2x^4+\cdots,\end{aligned}

\begin{aligned}\frac3{(1-x)^2}&=3+6x+9x^2\\&\quad+12x^3+15x^4+\cdots,\end{aligned}

\begin{aligned}f(x)&=5+4x+11x^2+10x^3\\&\quad+17x^4+\cdots.\end{aligned}

For integer $r\geq0$ , the general coefficient is

\begin{aligned}[x^r]f(x)&=2(-1)^r+3(r+1).\end{aligned}

The validity interval is $|x|<1$ . For example, $r=4$ gives $2+15=17$ , checking the displayed coefficient. Remember that the coefficient from $(1-x)^{-2}$ is $r+1$ , rather than $r$ .

A finite binomial: find the required term before expanding

The term independent of x

Find the term independent of $x$ in $(2x^2-1/x)^9$ , for $x\ne0$ .

The general term uses $r$ copies of the second term and $9-r$ copies of the first:

\begin{aligned}T_{r+1}&=\binom9r(2x^2)^{9-r}(-x^{-1})^r,\\&=\binom9r2^{9-r}(-1)^rx^{18-3r}.\end{aligned}

For a term independent of $x$ , its exponent must be zero:

\begin{aligned}18-3r&=0,\\r&=6,\\T_7&=\binom96 2^3(-1)^6\\&=84(8)\\&=672.\end{aligned}

The required term is 672. The integer $r=6$ lies between 0 and 9, so it corresponds to an actual term. Had the exponent condition produced a non-integer $r$ , there would be no such term.

When the replacement itself contains several powers

Keep the quadratic contribution from the square

Expand $\sqrt{4+4x+3x^2}$ through $x^2$ .

Put $F(x)=\sqrt{4+4x+3x^2}$ . Then

\begin{aligned}F(x)&=2\left(1+x+\frac34x^2\right)^{1/2}.\end{aligned}

Put $u=x+3x^2/4$ . For terms through $x^2$ , the only required term from $u^2$ is $x^2$ :

\begin{aligned}u^2&=x^2+\frac32x^3+\frac9{16}x^4,\end{aligned}

\begin{aligned}F(x)&=2\left[1+\frac12\left(x+\frac34x^2\right)\right.\\&\qquad\left.-\frac18x^2+\cdots\right],\\&=2+x+\frac12x^2+\cdots.\end{aligned}

The binomial substitution is valid where $|x+3x^2/4|<1$ . Since $x+3x^2/4\geq-1/3$ , the lower bound is automatic. The upper bound gives

\begin{aligned}3x^2+4x-4&<0,\\(3x-2)(x+2)&<0,\\-2&<x<\frac23.\end{aligned}

The constant and first derivative at zero check as 2 and 1. Keep the full squared replacement long enough to decide which powers can contribute.

Before you finish

Is the expanded bracket of the form 1 + u?
Have you intersected the validity intervals of all factors?
Did every product of powers that reaches the required degree contribute?
Is the chosen substitution small enough for a useful short approximation?

Continue with binomial series practice