Can we find an approximation to $\sqrt{5}+\sqrt{3}$?

Can we find an approximation to $\sqrt{5}+\sqrt{3}$?

Question

If

$\dfrac{b}{a}$

is small enough for powers of

$\dfrac{b}{a}$

higher than the third to be neglected, show that

$$\begin{equation*} (a+b)^\frac{1}{2} - (a-b)^\frac{1}{2} = a^\frac{1}{2} \left( \frac{b}{a} + \frac{b^3}{8a^3} \right). \end{equation*}$$

Use this result to deduce a rational approximation to $\sqrt 5 - \sqrt 3$.

Deduce, or obtain otherwise, a rational approximation to $\sqrt 5 + \sqrt 3$.

Solution

If $a > 0$, we can write $(a+b)^\frac{1}{2} = \sqrt{a} \left( 1 + \frac{b}{a} \right)^{1/2}.$

The

binomial theorem

says that, if

$\big |x \big| < 1$

,

$$\begin{equation*} (1+x)^\alpha = 1 + \alpha x + \frac{\alpha(\alpha-1)}{2!} x^2 + \frac{\alpha(\alpha-1)(\alpha-2)}{3!} x^3 + \dots b \end{equation*}$$

and so, if

$\dfrac{b}{a}$

is small enough,

$$\begin{align*} (a+b)^\frac{1}{2} &= \sqrt{a} \left( 1 + \frac{b}{a} \right)^{1/2} \\ &= \sqrt{a} \left( 1 + \frac{1}{2} \frac{b}{a} + \frac{\frac{1}{2} \times \left(-\frac{1}{2} \right)}{2!} \left( \frac{b}{a} \right)^2 + \frac{\frac{1}{2} \times \left( -\frac{1}{2} \right) \times \left( -\frac{3}{2} \right)}{3!} \left( \frac{b}{a} \right)^3 + \dots b \right) \\ &\approx \sqrt{a} \left( 1 + \frac{b}{2a} - \frac{b^2}{8a^2} + \frac{b^3}{16a^3} \right). \end{align*}$$

Similarly, assuming that

$a > 0$

and

$\dfrac{b}{a}$

is small enough we will get the same result as above replacing

$b$

with

$-b$

,

$$\begin{align*} (a-b)^\frac{1}{2} &\approx \sqrt{a} \left( 1 + \frac{-b}{2a} - \frac{(-b)^2}{8a^2} + \frac{(-b)^3}{16a^3} \right)\\ &= \sqrt{a} \left( 1 - \frac{b}{2a} - \frac{b^2}{8a^2} - \frac{b^3}{16a^3} \right). \end{align*}$$

As a result,

$$\begin{align*} (a+b)^\frac{1}{2} - (a-b)^\frac{1}{2} &\approx \sqrt{a} \left( \left( 1 + \frac{b}{2a} - \frac{b^2}{8a^2} + \frac{b^3}{16a^3} \right) - \left( 1 - \frac{b}{2a} - \frac{b^2}{8a^2} - \frac{b^3}{16a^3} \right) \right) \\ &= \sqrt{a} \left( \frac{b}{a} + \frac{b^3}{8a^3} \right). \end{align*}$$

Let’s put $a + b = 5$ and $a - b = 3$. Now adding gives $2a = 8 \implies a = 4, b = 1$. Thus $a > 0$ and $\dfrac{b}{a}$ is suitably small.

From our work above,

$$\begin{align*} \sqrt 5 - \sqrt 3 = (a+b)^\frac{1}{2} - (a-b)^\frac{1}{2} &\approx \sqrt{a} \left( \frac{b}{a} + \frac{b^3}{8a^3} \right) \\ &= \sqrt{4} \left( \frac{1}{4} + \frac{1}{8 \times 64} \right) \\ &= \frac{1}{2} + \frac{1}{256} \\ &= \frac{129}{256}\\ & \approx 0.5039... \end{align*}$$

The approximation is an accurate one: $\sqrt 5 - \sqrt 3= 0.5040...$

We have

$(\sqrt 5 + \sqrt 3)(\sqrt 5 - \sqrt 3) = 2.$

Thus

$\sqrt 5 + \sqrt 3 \approx 2 \times \dfrac{256}{129} = \dfrac{512}{129} = 3.9690...$

Again, the approximation is a good one: $\sqrt 5 + \sqrt 3 = 3.9681...$