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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?

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    $\begingroup$ You did not actually start with an equation, but with an approximation. $\endgroup$ – Peter May 16 '19 at 12:04
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    $\begingroup$ One weird thing that I notice is in the part "rationalize the left hand side". You start with $$ \sqrt{2} = \frac{256}{181} $$ for which the inverse is $$ \frac{1}{\sqrt{2}} = \frac{181}{256} $$ Multiply both sides by $3$ to get $$ \frac{3}{\sqrt{2}} = \frac{543}{256} $$ And now, you have used the relation $\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$, but notice that this is not equivalent to the definition of $\sqrt{2}$ in the beginning. I think this is the source of the error. $\endgroup$ – Matti P. May 16 '19 at 12:08
  • $\begingroup$ Neither is exact: In fact $\sqrt{2}$ is closer to $\dfrac{181.019336}{128}$ and $\dfrac{256}{181.019336}$ since $181.019336 \approx 128 \sqrt{2} = \dfrac{256}{\sqrt2}$ with your approximations rounding ${181.019336}$ to $181$ in both cases $\endgroup$ – Henry May 16 '19 at 12:08
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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.

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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 $\sqrt 2$.

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