Continued fractions

Algebra Level 3

Does there exist a real number $0 < x < 1$ with decimal representation $x = 0.\overline{a_1a_2a_3\ldots}$ that can be written as the continued fraction as below?

$\large x = \frac 1{a_1+\frac {a_2}{a_3+\frac {a_4}{a_5+\frac {a_6}{\ddots}}}}$

Yes No

1 solution

Henry U
Nov 25, 2018

$x \approx 0.273944195739271617171$ is one example.

As @Blan Morrison pointed out, the previous example wasn't quite valid; thanks to @Aaghaz Mahajan for showing this page from Wolfram Mathworld where such constants are called Trott constants.

Interesting, I wonder if this constant has a name.

If every second digit is a 1, then how come there is a pair of consecutive 2's and a pair of consecutive 3's?

Blan Morrison - 2 years, 6 months ago

When the continued fraction contained a 23, later a 122 and also a 33, I just added them to the end of the number. Actually, it's not rigorous...

Henry U - 2 years, 6 months ago

@Henry U Hello!!! You can check this out........!!!

Aaghaz Mahajan - 2 years, 6 months ago

Cool, I didn't know that these numbers have a name and that there are a few different ones! It's also interesting to see that the terms of their continued fractions are always less than 10.

Henry U - 2 years, 6 months ago

Makes me wonder if $T_n>1$ and $n$ is finite.

Blan Morrison - 2 years, 6 months ago

Since $x \approx 0.273944195739271617171$ can be written as the continued fraction

$x = \frac 2 {7 + \frac {3} {9 + \frac {4} {4 + \frac {1} {9 + {}_\ddots}}}}$ ,

the continued fraction

$x^\prime = 1+x = 1 + \frac 2 {7 + \frac {3} {9 + \frac {4} {4 + \frac {1} {9 + {}_\ddots}}}}$

will correspond to

$x^\prime = 1+x \approx 1.27394419$ .

The same thing can be done to generate Trott constants of any arbitrary size and therefore there are infinitely many.

However, I don't know if there are finitely or infinitely many Trott constants between 0 and 1.

Henry U - 2 years, 6 months ago

@Henry U – Ah, true. But what about not having a term outside of the fraction; it seems like cheating.

Blan Morrison - 2 years, 6 months ago

@Blan Morrison – This is the only other source I found and it also says that it is unknown how many Trott constants exist. It's probably an open problem.

Henry U - 2 years, 6 months ago

@Henry U – Yeah....this also works.....I had thought about this too......:)
@Blan Morrison But, hey, even Wolfram Alpha cheated........check out the Third Trott Constant.......even tho it is a zero outside the fraction, it still counts...... :p

Aaghaz Mahajan - 2 years, 6 months ago

Note that your format requires that the top most numerator is $1$ (and not the first digit of the decimal), whereas the Trott constants have a slightly different definition.

Calvin Lin Staff - 2 years, 6 months ago

Do you think there arre still numbers that satisfy this?

Henry U - 2 years, 6 months ago

I believe the answer is no, but am not fully certain. This is how I approached it:

First, to determine $a_1$ , observe $\frac{1}{ a_1 + \text{fractional part} }$ is a decreasing sequence and $0.a_1$ is an increasing sequence, hence there is a unique solution. In particular, we must have $a_1 = 3$ , and the decimal will be between $\frac{1}{3} \approx 0.33$ and $\frac{1}{4} \approx 0.25$ . (Alternatively, just list the values of $\frac{1}{1}, \frac{1}{2}, \frac{1}{3} , \ldots$ and compare, as I'm going to do below.

Next, to determine $a_2$ , observe that
$0.25 < \frac{1 } { 3 + \frac{1}{1+ \text{fractional part}}} < 0.2857$
$0.2857 < \frac{1 } { 3 + \frac{1}{2+ \text{fractional part}}} < 0.3$
$0.3 < \frac{1 } { 3 + \frac{1}{3+ \text{fractional part}}} < 0.3077$
$0.3077 < \frac{1 } { 3 + \frac{1}{4+ \text{fractional part}}} < 0.3125$
and clearly $\frac{1 } { 3 + \frac{1}{5+ \text{any part}}} < 0.333$ .
Hence, there is no possible value for $a_2$ .

Let me know if you agree, and I will update the answer to no.

Calvin Lin Staff - 2 years, 6 months ago

@Calvin Lin – I definetly agree, and I also think that this is a better problem than the one about Trott constants because it isn't known much about them. Prooving there are no solutions is better in this case.

Henry U - 2 years, 6 months ago

@Henry U – Oh, I just realized that I misread your problem.

The first Trott constant $T_1 = 0.10841\ldots$ satisfies your problem doesn't it? It starts with a 1 (though $a_2 = 0$ might seem iffy, but that isn't disallowed.)

I had changed the answer to no, and then changed it back to yes.

Can you post the "all numerators = 1" version of this problem?

Calvin Lin Staff - 2 years, 5 months ago

@Calvin Lin – I will right now. Thanks for the help!

Henry U - 2 years, 5 months ago

@Henry U – Argh, I still misread the problem. The 1 in your numerator isn't used in the problem, so $T_1$ doesn't work. At this point in time, I'm not quite sure if a solution exists.

My guess is yes, and it is done by iteratively finding the value that works, similar to what I did for the all 1's version. The crux is to show that this can always be done, which seems somewhat reasonable to me, but I haven't checked the details.

Calvin Lin Staff - 2 years, 5 months ago

@Calvin Lin – I think that you made an error when correcting the problem because, apparently, 105% of people got the problem correct.

Blan Morrison - 2 years, 5 months ago

Continued fractions

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