How can this binomial expansion result in two different approximations of root 2?

Question

I have been working on a problem on approximating $\sqrt{2}$ using the first three terms of the following binomial expansion, and a substitution of $x = -\frac{1}{10}$ :

$$(4 - 5x)^.5 = 2 - \frac{5x}{4} - \frac{25x^2}{64}$$

After substituting, I got to the stage:

$ \frac{3}{\sqrt{2}}= \frac{543}{256}$

Now what is bizarre, is that if I solve this equation for $\sqrt{2}$, there are two routes I could take, but they give slightly different answers.

Route 1:

Reciprocate both sides, and multiply both sides by 3 to get:

$ \sqrt{2}= \frac{256}{181} $

Route 2:

Rationalize the left hand side to make the equation:

$ \frac{3\sqrt{2}}{2}= \frac{543}{256}$

Divide both sides by $\frac{3}{2}$:

$ \sqrt{2}= \frac{181}{128}$

So we end up with two approximations of $ \sqrt{2}$: $\frac{256}{181}$ and $\frac{181}{128}$

I am really struggling to understand how this has happened? Why is the same equation leading to two different solutions?

Answer 1

If we denote the approximation with $\ c\ $, we have two equations that we can solve for $\ x\ $ :

• $\ \frac{3}{x}=c\ $ giving $\ x=\frac{3}{c}\ $
• $\ \frac{3x}{2}=c\ $ giving $\ x=\frac{2c}{3}\ $

If $\ c\ $ were exactly $\ \sqrt{2}\ $, both solutions would coincide. But $\ c\ $ is only an approximation, hence the values cannot coincide excactly. Their product , however , is exactly $\ 2\ $ , so one approximation is too small and the other too large.

Answer 2

Let $f(x) = 2 - \frac{5x}{4} - \frac{25x^2}{64}$.

Let $c$ denote the value $f(-\frac{1}{10}) = \frac{543}{256}$.

We can write

$$\tag 1 \frac{3}{\sqrt 2} = c + \varepsilon$$

If we take the OP's route 1 (but keeping $\varepsilon$), then

$$\tag 2 \sqrt 2 = \frac{3}{c+\varepsilon}$$

We can check that the 'route 1 approximation', $\frac{3}{c}$, is strictly greater than $\sqrt 2$. With $\varepsilon$ playing a part in the denominator of the rhs of $\text{(2)}$, to 'fix things up' it must be true that $\varepsilon \gt 0$.

If we take the OP's route 2 (but keeping $\varepsilon$), then

$$\tag 3 \sqrt 2 = \frac{2}{3}\,(c + \varepsilon)$$

Since $\varepsilon \gt 0$, the 'route 2 approximation' $\frac{2c}{3}$ is strictly less than $\sqrt 2$.