Evaluating surds - different answers using different methods?

Question

I am given that $$ \frac{\sqrt{2}}{\sqrt{5}} \approx \frac{1}{\sqrt{2}} - \frac{1}{8\sqrt{2}} + \frac{3}{128\sqrt{2}} - \frac{5}{1024\sqrt{2}}$$

and asked to show $$ \sqrt{5} \approx \frac{2048}{915}$$.

Evaluating the above in two different ways seems to give slightly different answers, even though there doesn't seem to be any reason it should.

Method 1: $$ \frac{\sqrt{2}}{\sqrt{5}} \approx \frac{1}{\sqrt{2}} - \frac{1}{8\sqrt{2}} + \frac{3}{128\sqrt{2}} - \frac{5}{1024\sqrt{2}}$$ $$ \frac{\sqrt{2}}{\sqrt{5}} \approx \frac{1}{\sqrt{2}}(1 - \frac{1}{8} + \frac{3}{128} - \frac{5}{1024})$$ $$ {\sqrt{2}} \approx \frac{1}{\sqrt{2}}(1 - \frac{1}{8} + \frac{3}{128} - \frac{5}{1024}){\sqrt{5}}$$ $$2 \approx (1 - \frac{1}{8} + \frac{3}{128} - \frac{5}{1024}){\sqrt{5}}$$ $$2 \approx (\frac{915}{1024}){\sqrt{5}}$$ $${\sqrt{5}} \approx \frac{2048}{915}$$

Note: $$\frac{2048}{915} = 2.238251366$$

Method 2: $$ \frac{\sqrt{2}}{\sqrt{5}} \approx \frac{1}{\sqrt{2}} - \frac{1}{8\sqrt{2}} + \frac{3}{128\sqrt{2}} - \frac{5}{1024\sqrt{2}}$$ $$ \frac{\sqrt{2}\sqrt{5}}{\sqrt{5}\sqrt{5}} \approx \frac{1}{\sqrt{2}} - \frac{1}{8\sqrt{2}} + \frac{3}{128\sqrt{2}} - \frac{5}{1024\sqrt{2}}$$ $$ \frac{\sqrt{2}}{{5}}\sqrt{5} \approx \frac{1}{\sqrt{2}} - \frac{1}{8\sqrt{2}} + \frac{3}{128\sqrt{2}} - \frac{5}{1024\sqrt{2}}$$ $$\sqrt{5} \approx (\frac{1}{\sqrt{2}} - \frac{1}{8\sqrt{2}} + \frac{3}{128\sqrt{2}} - \frac{5}{1024\sqrt{2}})\frac{5}{\sqrt{2}}$$ $$\sqrt{5} \approx (\frac{5}{{2}} - \frac{5}{8(2)} + \frac{15}{128(2)} - \frac{25}{1024(2)})$$ $$\sqrt{5} \approx \frac{4575}{{2048}}$$

Note: $$\frac{4575}{2048} = 2.233886719$$

The two methods give different but awfully close answers. Nevertheless, the question asks for an exact term so giving the second answer would be wrong. Am I missing something that is causing a loss of precision in Method 2?

Answer 1

I would explain like this:

When we say $a \approx b$ $(a,b,n\in\mathbb{R})$, this does not necessary mean that $na\approx nb$.

This is because $na-nb=n(a-b)$, this difference can increase dramatically if $|n|$ increases. When both sides of the approximation is multiplied by $n$ times, their difference is also increased by $n$ times.

The second and third steps of method $1$ looks like this:

$$\frac{\sqrt{2}}{\sqrt{5}} \approx \frac{1}{\sqrt{2}}(1 - \frac{1}{8} + \frac{3}{128} - \frac{5}{1024})$$

$${\sqrt{2}} \approx \frac{1}{\sqrt{2}}(1 - \frac{1}{8} + \frac{3}{128} - \frac{5}{1024}){\sqrt{5}}$$

The second line is always true because

$$\frac{1}{\sqrt{2}} - \frac{1}{8\sqrt{2}} + \frac{3}{128\sqrt{2}} - \frac{5}{1024\sqrt{2}}=\frac{1}{\sqrt{2}}\left(1 - \frac{1}{8} + \frac{3}{128} - \frac{5}{1024}\right)$$

However, the third line is not always true, as the difference between the left hand side and the right hand side is increased by $\sqrt{5}$ times (obviously, both sides are positive).

The fourth step also increases the difference between both sides:

$$2 \approx (1 - \frac{1}{8} + \frac{3}{128} - \frac{5}{1024}){\sqrt{5}}$$

This time the difference between both sides is increased again by $\sqrt{2}$ times, which makes the difference from the beginning increased by $\sqrt{5}\times \sqrt{2}=\sqrt{10}$ times overall.

The fifth step in method $1$ looks like this:

$$2 \approx (\frac{915}{1024}){\sqrt{5}}$$

The fifth step is true because the value of the right hand side is not changed.

The sixth step is method $1$ looks like this:

$${\sqrt{5}} \approx \frac{2048}{915}$$

There are two ways to make a conclusion from the fifth step:

• Method $1$ divides both sides by $\dfrac{915}{1024}$, or multiplies both sides by $\dfrac{1024}{915}$, this makes the difference between both sides increased again by $\dfrac{1024}{915}$ times. Because of this, the "beginning" difference has been increased by $\dfrac{1024}{915}\sqrt{10}$ times overall.

• Another method: From the fifth step, multiply both sides by $\dfrac{\sqrt{5}}{2}$ times. This makes the difference increase again by $\dfrac{\sqrt{5}}{2}$ times and the final result is indeed $\sqrt{5} \approx \dfrac{4575}{{2048}}$, this is the answer from method $2$. Doing it this way, the "beginning" difference has been increased by $\sqrt{10}\times\dfrac{\sqrt{5}}{2}=\dfrac{5\sqrt{2}}{2}$ times.

For method $2$, I will do this quickly:

$$ \frac{\sqrt{2}}{\sqrt{5}} \approx \frac{1}{\sqrt{2}} - \frac{1}{8\sqrt{2}} + \frac{3}{128\sqrt{2}} - \frac{5}{1024\sqrt{2}}$$ $$ \frac{\sqrt{2}\sqrt{5}}{\sqrt{5}\sqrt{5}} \approx \frac{1}{\sqrt{2}} - \frac{1}{8\sqrt{2}} + \frac{3}{128\sqrt{2}} - \frac{5}{1024\sqrt{2}}$$ $$ \frac{\sqrt{2}}{{5}}\sqrt{5} \approx \frac{1}{\sqrt{2}} - \frac{1}{8\sqrt{2}} + \frac{3}{128\sqrt{2}} - \frac{5}{1024\sqrt{2}}$$

All three steps do not increase the difference between both sides.

$$\sqrt{5} \approx (\frac{1}{\sqrt{2}} - \frac{1}{8\sqrt{2}} + \frac{3}{128\sqrt{2}} - \frac{5}{1024\sqrt{2}})\frac{5}{\sqrt{2}}$$

The difference between both sides is increased by $\dfrac{5}{\sqrt{2}}=\dfrac{5\sqrt{2}}{2}$ times.

$$\sqrt{5} \approx (\frac{5}{{2}} - \frac{5}{8(2)} + \frac{15}{128(2)} - \frac{25}{1024(2)})$$ $$\sqrt{5} \approx \frac{4575}{{2048}}$$

The difference between both sides is not changed in either of the steps above, so the overall differenced is increased by $\dfrac{5\sqrt{2}}{2}$ times, which is the same as the second method of the last step of method $1$.

Using a calculator:

$\dfrac{1024}{915}\sqrt{10}=3.538986146...$

$\dfrac{5\sqrt{2}}{2}=3.535533906...$

Two values are very close to each other, thus explain why the difference is so small.

Note: I did not say anything about the correctness of either of these two methods, this is just a proof about why two methods are different.

Answer 2

Note that $$ 2048/915 = \frac 5{4575/2048} $$

The mapping $x\mapsto 5/x $ would not change the actual $\sqrt 5$, but it turns your two approximations into each other.

If you assume (counterfactually) that one of the approximations actually equals $\sqrt 5$, you can "prove" that it equals the other.

One is not really more precise than the other -- in fact the the logarithmic distance between each of them and the exact $\sqrt 5$ is the same, just on opposite sides.

Answer 3

I'm putting this here just because I didn't see anyone mention it.

\begin{align} \frac{\sqrt{2}}{\sqrt{5}} &\approx \frac{1}{\sqrt{2}} - \frac{1}{8\sqrt{2}} + \frac{3}{128\sqrt{2}} - \frac{5}{1024\sqrt{2}} \\ \frac{1}{\sqrt{5}} &\approx \frac{1}{2} - \frac{1}{16} + \frac{3}{256} - \frac{5}{2048} \\ \frac{1}{\sqrt{5}} &\approx \frac{915}{2048} \\ \sqrt 5 &\approx \dfrac{2048}{915} \end{align}

If you think of $x \approx y$ to mean that $x = y + \epsilon$ for some "small" value of $\epsilon$, then it should not come as a suprise to find that $f(x) \ne f(y) + \epsilon$ or even that $f(x) \ne f(y) + f(\epsilon)$.

method 1

\begin{align} \frac{\sqrt 2}{\sqrt 5} &= \frac{1} {\sqrt{2}}-\frac{1}{8\sqrt{2}}+\frac{3}{128\sqrt{2}} -\frac{5}{1024\sqrt{2}} + \epsilon \\ \frac{\sqrt 2}{\sqrt 5} &= \frac{1} {\sqrt 2}(1 - \frac{1}{8} + \frac{3}{128} - \frac{5}{1024}) + \epsilon\\ {\sqrt 2} - \sqrt 5 \epsilon &\approx \frac{1} {\sqrt{2}}(1 - \frac{1}{8} + \frac{3}{128} - \frac{5}{1024}){\sqrt 5} \\ 2 - \sqrt{10}\epsilon &= (1 - \frac{1}{8} + \frac{3}{128} - \frac{5}{1024}){\sqrt 5} \\ 2 - \sqrt{10}\epsilon &= (\frac{915}{1024}){\sqrt 5} \\ {\sqrt 5} &= \frac{2048}{915} - \frac{1024}{915}\sqrt{10}\epsilon \end{align}

method 2

\begin{align} \frac{\sqrt 2}{\sqrt 5} &= \frac{1}{\sqrt{2}} - \frac{1}{8\sqrt 2} + \frac{3}{128\sqrt 2} - \frac{5}{1024\sqrt 2} + \epsilon \\ \frac{\sqrt{2}\sqrt{5}}{\sqrt{5}\sqrt{5}} &= \frac{1}{\sqrt 2} - \frac{1}{8\sqrt 2} + \frac{3}{128\sqrt 2} - \frac{5}{1024\sqrt 2} + \epsilon \\ \frac{\sqrt 2}{5} \sqrt{5} &= \frac{1}{\sqrt 2} - \frac{1}{8\sqrt 2} + \frac{3}{128\sqrt 2} - \frac{5}{1024\sqrt 2} + \epsilon \\ \sqrt 5 &= \left(\frac{1}{\sqrt 2} - \frac{1}{8\sqrt 2} + \frac{3}{128\sqrt 2} - \frac{5}{1024\sqrt 2} \right)\frac{5}{\sqrt 2} + \frac{5}{\sqrt 2}\epsilon \\ \sqrt{5} &= \frac 52 - \frac{5}{8(2)} + \frac{15}{128(2)} - \frac{25}{1024(2)} + \frac{5}{\sqrt 2}\epsilon \\ \sqrt{5} &= \frac{4575}{2048} + \frac{5}{\sqrt 2}\epsilon \\ \end{align}

The errors are clearly different.