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I've been interested in representing $\sqrt 2\approx 1.414213562373095$ by an equation which uses each of $1,2,\cdots,9$ once. Suppose that the following conditions must be satisfied. Then, can we get a 'good' approximate value of $\sqrt 2$ ?

Condition 1 : Use each of $1,2,\cdots,9$ once.

Condition 2 : You can use $+,-,\times, \div, (\ ), \{\ \}, [\ ], !$ such as $$\begin{align}1+2-[4\div \{(3+5)\times (6+8)\}+7!].\end{align}$$

Condition 3 : You can use a decimal point. You can represent $0.1234$ as "$.1234$".

Condition 4 : You can use neither the radical sign nor exponential notation such as $\sqrt{3}, 4^5$.

Then, here are my examples.

Example 1 : $.436215+.978=\underline{1.41421}5\rightarrow 6$ digit.

Example 2 : $12653\div 8947=\underline{1.41421}70\cdots\rightarrow 6$ digit.

Example 3 : $3\div 7+.9856421=\underline{1.4142135}28\cdots\rightarrow 8$ digit.

So far I'm facing difficulty for finding an equation which corresponds to $\sqrt 2$ at more than $8$ digit. Can you help?

Update 1 : I've just got the following :

$\{.9\times (5+6)\}\div\{7+(3!\times 2.4)\div (8!+1)\}=\underline{1.4142135623}22\cdots\rightarrow 11$ digit.

Update 2 : I've just got the following :

$(3-.1)\times \{2-4\div (8!\div 6+7)\}\div (5-.9)=\underline{1.4142135623}82\cdots\rightarrow 11$ digit.

It can be said that this example is better than the example of update 1.

mathlove
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  • There are only finitely many expressions you can create of the given form, so you cannot approximate $\sqrt{2}$ with an arbitrary degree of accuracy. I have no idea whether you can do better than $8$ digits of accuracy. – Michael Joyce Oct 23 '13 at 16:43
  • @MichaelJoyce: Yes, I agree with you. I meant that how many digit we can correspond to $\sqrt 2$. – mathlove Oct 23 '13 at 16:46
  • I don't know of any general principle at play here. Seems like you would have to do a brute search of some form or other. – Michael Joyce Oct 23 '13 at 16:47
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    I'm not so sure there's finite many expressions here. Consider the set ${.12345678+9!,.12345678+9!!,.12345678+9!!!,.12345678+9!!!!,...}$ – Chuck Franklin Oct 23 '13 at 16:49
  • @AustinRannow: You mean $9!!!!=(9!!!)!$? Anyway, I think you are right, thanks. – mathlove Oct 23 '13 at 16:51
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    Yes, it was just an example. In theory you could take $.1234567+\frac{9!!!...!}{8!!!...!}$ and find the number of "!" for the numerator and the denominator to get infinite precision for your approximation. – Chuck Franklin Oct 23 '13 at 16:56
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    @austinRannow Is it immediately clear that taking the integer numbers of factorial symbols can get one an arbitrarily close approximation? It seems insertion of one on top drastically overshoots, then one would have to jack up the bottom, etc, but it seems doubtful this process could converge to a given real... (Maybe I just don't see it, it may be true). – coffeemath Oct 23 '13 at 18:11
  • I agree, $\frac{9!!!\cdots}{8!!!\cdots}$ is extremely unlikely to be dense in any finite interval. – Greg Martin Oct 25 '13 at 07:31
  • Note that the ratios of consecutive Pell numbers provide approximations for the silver ratio, $1+\sqrt{2}$, to any arbitrary degree of accuracy. – Jaycob Coleman Oct 25 '13 at 08:14
  • 9!!!.../8!!!... is going to be some factorial over another factorial which, when less than 1 is going to be 1/n(n+1)...(n+m) for some n, m so 9!!!.../8!!!... is going to be at most as dense as 1/i for arbitrary i which means it's dense at 0 and nowhere else. – Darrell Plank Nov 04 '13 at 23:07
  • Can we use Tetrations also? – Neil Apr 11 '15 at 00:14
  • mathlove, can you explain how you found the approximations in the edits ? Is it just trial and error ? And if yes, how long have you spent on it ? – Evargalo Jul 21 '17 at 14:14

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