Normalize before expanding · source reference

Printed question

It is given that $f(x)=\sqrt{4-3x}$.

(a) Find the binomial expansion for $f(x)$, up to and including the term in $x^2$. Give the coefficients as exact fractions in their simplest form, and state the range of values of $x$ for which this expansion is valid. [4]

(b) By taking $x=\frac14$, show that $\sqrt{13}\approx\frac{1847}{512}$. [3]

(c) We may also substitute $x=\frac1{13}$ to obtain an approximation of $\sqrt{13}$. Without further calculation, explain why this leads to a better approximation of $\sqrt{13}$ than the value shown in (b). [1]

[2022/EJC/Prelim/II/1]

Printed working

$$\begin{aligned}f(x)&=(4-3x)^{1/2}=4^{1/2}(1-3x/4)^{1/2}\\&=2[1+(1/2)(-3x/4)+\tfrac{(1/2)(-1/2)}{2!}(-3x/4)^2+\cdots]\\&=2-3x/4-9x^2/64+\cdots.\end{aligned}$$

$|-3x/4|<1\Rightarrow |x|<4/3\Rightarrow-4/3<x<4/3$.

At $x=1/4$, $\sqrt{13/4}\approx(2048-3\times64-9)/1024=1847/1024$; $\sqrt{13}\approx2\times1847/1024=1847/512$.

Printed (c): the value obtained with $x=1/13$ is a better approximation because it is closer to 0 than $x=1/4$.

Teacher note 1

Need to factorise out $4$ to make the constant term 1. Use the law of indices $(AB)^n=A^nB^n$. Then use the binomial theorem

$$(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\cdots,$$

where $n=1/2$ and replace $x$ by $-3x/4$.

Teacher note 2

The binomial expansion of $(1+f(x))^n$, where $n$ is a negative integer or fraction, is valid for the range of $x$ given by $|f(x)|<1$.

Mathematical review

(a) Using $(1+u)^{1/2}=1+u/2-u^2/8+O(u^3)$ gives $f(x)=2-3x/4-9x^2/64+O(x^3)$. The stated binomial validity condition is $|3x/4|<1$, hence $-4/3<x<4/3$.

(b) $f(1/4)=\sqrt{13}/2\approx2-3/16-9/1024=1847/1024$, so $\sqrt{13}\approx1847/512$.

(c) $f(1/13)=7/\sqrt{13}$, so recover the target using $\sqrt{13}=7/f(1/13)$. The binomial input magnitude is smaller; its omitted higher powers are smaller. This comparison is specific to these substitutions, not a universal accuracy theorem.

Provenance

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