Conjecture. For any real number $x \in (0,1]$ there exists a unique expansion in the form $x=-2+\sqrt{a_1+\sqrt{a_2+\sqrt{a_3+\cdots}}}$ with $a_k$ being natural numbers from the set $(2,3,4,5,6)$. This expansion always converges to $x$.

We can make a simple continued fraction expansion of any real number in a unified way, using a variation of Euclidean algorithm. For irrational numbers this expansion would be infinite. Sometimes it has a pattern (quadratic irrationals, $e$), most times it doesn't.

For quadratic irrationals, however, we can also make an infinite nested radical expansion, which is connected to its continued fraction expansion in a clear way, take Golden ratio for example.


But in principle, we can make an infinite nested radical expansion for any real number. The 'best' rule for this is not clear, however, this seems doable:






Or, using the notation I just invented (similar to continued fraction notation):

$$\pi=\sqrt{7+}\sqrt{6+}\sqrt{2+}\sqrt{6+}\sqrt{6+}\sqrt{5+}\sqrt{5+}\sqrt{2+}\sqrt{4+}\sqrt{6+}\sqrt{3+}\sqrt{4+}\sqrt{2+}\sqrt{4+}\sqrt{6+} \cdots$$

Now, why am I always taking $2$ as a whole part of the number I square? That's because if I take $1$ I quickly get a number in the form $1. \cdots$, so I can't go on further. And if I take $3$ I get bigger numbers under the root. So this seems optimal. In a similar fashion:

$$e=\sqrt{5+}\sqrt{3+}\sqrt{5+}\sqrt{3+}\sqrt{3+}\sqrt{6+}\sqrt{4+}\sqrt{4+}\sqrt{2+}\sqrt{4+}\sqrt{3+}\sqrt{2+}\sqrt{4+}\sqrt{4+}\sqrt{5+} \cdots$$

Here are plots for the convergence of above expansions for $\pi$ and $e$. You can see that the convergence is very good.

enter image description here

In fact, there are some interesting general properties of this expansion. If we only consider the numbers with the whole part $2$, then we get only numbers from $2$ to $6$ under the radical, with their mean value being around $4$.

How do I prove the observed properties of this expansion? Would this expansion always converge to the number? Is it possible that we get a pattern for some transcendental number in this expansion?

This is related to Bolyai Expansion, as was suggested by @Watson in a comment, however every $a_k$ here is not zero, which makes the expansion more 'orderly' in my opinion.

Also, I conjecture that taking $2$ as a whole part to square is the best way in the sense that if we take $1$ we may get $0$ as the next $a_k$, but if we take $3$ or more the set of $a_k$ becomes much larger, than just $(2,3,4,5,6)$.

Some more examples (I will use a more simple notations for the expansion as a set of numbers):




$$R(\phi+1)=(4,6,2,4,5,5,4,5,6,3,2,3,3,2,3,5,4,2,2,2,2,5,2,2,4,6,3,2,5,6,2,6,2,4,6,6,3,4,6,3,2,6,6,4,3,3,2,2,4,4, \dots)$$

We recover the well known identities and some other similar expansions:

$$R(2)=(2,2,2,2, \dots)$$

$$R(3)=(6,6,6,6,, \dots)$$

$$R(1+\sqrt{2})=(3,6,2,2,2,2, \dots)$$

$$R(\sqrt{5})=(3,2,2,2,2,2, \dots)$$

For some rational numbers (and quadratic irrationals) this expansion seems random:

$$R(2.5)=(4,3,2,3,2,3,3,5,6,6,5,4,2,6,6,2,6,4,2,2,2,4,2,4,5,2,5,5,2,2,6,2,6,3,4,2,3,6,2,4,4,3,2,3,2,2,5,4,3,2, \dots)$$

$$R(1+\sqrt{3})=(5,4,2,3,3,2,3,4,2,6,5,3,4,6,3,3,4,6,5,5,3,4,4,5,5,2,4,5,4,5,3,6,2,2,3,4,6,4,4,5,3,4,2,3,3,4,3,5,6,2, \dots)$$

By the way, this makes a beautiful (if inaccurate) approximations to some constants (see above):

$$\gamma \approx \sqrt{4+\sqrt{4+\sqrt{6+\sqrt{6+\sqrt{3+\sqrt{3}}}}}}-2=0.577$$

$$e \approx \sqrt{5+\sqrt{3+\sqrt{5+\sqrt{3+\sqrt{5}}}}}=2.718$$

And (cheating a little):

$$\pi \approx \sqrt{7+\sqrt{6+\sqrt{5}}}=3.1416$$

  • 1
    $\begingroup$ Related : mathworld.wolfram.com/BolyaiExpansion.html $\endgroup$ – Watson Mar 13 '16 at 21:29
  • $\begingroup$ Thank you. The first reference there seems quite illuminating $\endgroup$ – Yuriy S Mar 13 '16 at 21:35
  • 1
    $\begingroup$ Your last formula ascribes to a movement of research on closed forms giving approximations of $\pi$. You have, for example, $\pi=\sqrt[9]{\frac{34041350274878}{1141978491}}$ giving $15$ exact decimal digits and $\pi= \frac{ln((640320)^3+744)}{\sqrt{163}}$ that gives accuracy of $30$ correct digits. $\endgroup$ – Piquito Mar 15 '16 at 23:09
  • $\begingroup$ It's not about accuracy at all. The expansion I offered is extremely inaccurate for most numbers (it requires more digits than the decimal expansion of the number) $\endgroup$ – Yuriy S Mar 15 '16 at 23:14
  • $\begingroup$ Obviously: your parenthesis is eloquent enough about it (furthermore, four digits of approximation is nothing in this kind of research). $\endgroup$ – Piquito Mar 16 '16 at 1:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.