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Hopping to infinity along a string of digits
Announcing the arrival of Valued Associate #679: Cesar Manara
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9
$begingroup$
Let $s$ be an infinite string of decimal digits, for example:
begin{array}{cccccccccc}
s = 3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 & cdots
end{array}
Consider a marker, the head, pointing to the first digit, $3$ in the above example. Interpret the digit under
the head as an instruction to move the head $3$ digits to the right, i.e., to the $4$th digit. Now the head is pointing to $1$. Interpret this as an instruction to move $1$ place to the left. Continue in this manner, hopping through the string, alternately moving right and left. Think of the head as akin to the head of a Turing machine, and $s$ as the tape of instructions.
There are three possible behaviors.
(1) The head moves off the left end of $s$:
begin{array}{cccccccccc}
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{}
end{array}
(2) The head goes into a cycle, e.g., when the head hits $0$:
begin{array}{cccccccccccccc}
6 & 4 & 5 & 7 & 5 & 1 & 3 & 1 & 1 & 0 & 6 & 4 & 5 & 9 \
text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} \
6 & 4 & 5 & 7 & 5 & 1 & 3 & 1 & 1 & 0 & 6 & 4 & 5 & 9 \
text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} & text{} \
6 & 4 & 5 & 7 & 5 & 1 & 3 & 1 & 1 & 0 & 6 & 4 & 5 & 9 \
text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} \
6 & 4 & 5 & 7 & 5 & 1 & 3 & 1 & 1 & 0 & 6 & 4 & 5 & 9 \
text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} \
6 & 4 & 5 & 7 & 5 & 1 & 3 & 1 & 1 & 0 & 6 & 4 & 5 & 9 \
text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} \
6 & 4 & 5 & 7 & 5 & 1 & 3 & 1 & 1 & 0 & 6 & 4 & 5 & 9 \
text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{}
end{array}
(3) The head moves off rightward to infnity:
begin{array}{ccccccccccccc}
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} \
end{array}
This last string could be viewed as the decimal expansion of $31/99 = 0.3131313131313cdots$.
Q1. What is an example of an irrational number
$0.d_1 d_2 d_3 cdots$ whose string $s=d_1 d_2 d_3 cdots$ causes the head to hop rightward to infinity?
Q1.5. (Added). Is there an explicit irrational algebraic number with the hop-to-$infty$ property?
I'm thinking of something like $sqrt{7}-2$, the
2nd example above (which cycles).
Q2. More generally, which strings
cause the head to hop rightward to infinity?
Update (summarizing answers, 13Apr2019).
Q1. There are irrationals with the hop-to-$infty$ property
(@EthanBolker, @TheSimpliFire),
but explicit construction requires using, e.g., the
Thue-Morse sequence (@Wojowu).
Q1.5. @EthanBolker suggests this may be difficult, and @Wojowu suggests it may be false (b/c: nine consecutive zeros): Perhaps no algebraic irrational has the hop-to-$infty$ property.
Q2. A partial algorithmic characterization by @TheSimpliFire.
sequences-and-series recreational-mathematics irrational-numbers decimal-expansion turing-machines
asked yesterday
Joseph O'RourkeJoseph O'Rourke
18.3k351113
$endgroup$
|
show 3 more comments
9
$begingroup$
Let $s$ be an infinite string of decimal digits, for example:
begin{array}{cccccccccc}
s = 3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 & cdots
end{array}
Consider a marker, the head, pointing to the first digit, $3$ in the above example. Interpret the digit under
the head as an instruction to move the head $3$ digits to the right, i.e., to the $4$th digit. Now the head is pointing to $1$. Interpret this as an instruction to move $1$ place to the left. Continue in this manner, hopping through the string, alternately moving right and left. Think of the head as akin to the head of a Turing machine, and $s$ as the tape of instructions.
There are three possible behaviors.
(1) The head moves off the left end of $s$:
begin{array}{cccccccccc}
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{}
end{array}
(2) The head goes into a cycle, e.g., when the head hits $0$:
begin{array}{cccccccccccccc}
6 & 4 & 5 & 7 & 5 & 1 & 3 & 1 & 1 & 0 & 6 & 4 & 5 & 9 \
text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} \
6 & 4 & 5 & 7 & 5 & 1 & 3 & 1 & 1 & 0 & 6 & 4 & 5 & 9 \
text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} & text{} \
6 & 4 & 5 & 7 & 5 & 1 & 3 & 1 & 1 & 0 & 6 & 4 & 5 & 9 \
text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} \
6 & 4 & 5 & 7 & 5 & 1 & 3 & 1 & 1 & 0 & 6 & 4 & 5 & 9 \
text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} \
6 & 4 & 5 & 7 & 5 & 1 & 3 & 1 & 1 & 0 & 6 & 4 & 5 & 9 \
text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} \
6 & 4 & 5 & 7 & 5 & 1 & 3 & 1 & 1 & 0 & 6 & 4 & 5 & 9 \
text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{}
end{array}
(3) The head moves off rightward to infnity:
begin{array}{ccccccccccccc}
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} \
end{array}
This last string could be viewed as the decimal expansion of $31/99 = 0.3131313131313cdots$.
Q1. What is an example of an irrational number
$0.d_1 d_2 d_3 cdots$ whose string $s=d_1 d_2 d_3 cdots$ causes the head to hop rightward to infinity?
Q1.5. (Added). Is there an explicit irrational algebraic number with the hop-to-$infty$ property?
I'm thinking of something like $sqrt{7}-2$, the
2nd example above (which cycles).
Q2. More generally, which strings
cause the head to hop rightward to infinity?
Update (summarizing answers, 13Apr2019).
Q1. There are irrationals with the hop-to-$infty$ property
(@EthanBolker, @TheSimpliFire),
but explicit construction requires using, e.g., the
Thue-Morse sequence (@Wojowu).
Q1.5. @EthanBolker suggests this may be difficult, and @Wojowu suggests it may be false (b/c: nine consecutive zeros): Perhaps no algebraic irrational has the hop-to-$infty$ property.
Q2. A partial algorithmic characterization by @TheSimpliFire.
sequences-and-series recreational-mathematics irrational-numbers decimal-expansion turing-machines
asked yesterday
Joseph O'RourkeJoseph O'Rourke
18.3k351113
$endgroup$
1
$begingroup$
You want a string such that the $D_1=(1+d_1)$th digit is less than $d_1$, that the $D_2=(1+d_1-D_1)$th digit is greater than $D_1$, that the $D_3=(1+d_1-D_1+D_2)$th digit is less than $D_2$, etc. I think it is possible to generate an algorithm and you can try for some simulations.
$endgroup$
– TheSimpliFire
yesterday
2
$begingroup$
Another question. For sequences that don't hop to infinity behavior is determined by a (finite) initial subsequence. There are only countably many of those. What are they?
$endgroup$
– Ethan Bolker
yesterday
3
$begingroup$
I think Q1.5 is hard since digit sequences for algebraic numbers are hard adamczewski.perso.math.cnrs.fr/Siauliai.pdf
$endgroup$
– Ethan Bolker
yesterday
1
$begingroup$
Followup question: what are the measures of the three sets in the interval $(0,1)$?
$endgroup$
– eyeballfrog
yesterday
2
$begingroup$
@eyeballfrog The set of numbers with hop to infinity property has measure zero, because you can't hop past a string of nine zeros. This also strongly suggests no algebraic irrational has this property.
$endgroup$
– Wojowu
yesterday
|
show 3 more comments
9
9
9
3
$begingroup$
Let $s$ be an infinite string of decimal digits, for example:
begin{array}{cccccccccc}
s = 3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 & cdots
end{array}
Consider a marker, the head, pointing to the first digit, $3$ in the above example. Interpret the digit under
the head as an instruction to move the head $3$ digits to the right, i.e., to the $4$th digit. Now the head is pointing to $1$. Interpret this as an instruction to move $1$ place to the left. Continue in this manner, hopping through the string, alternately moving right and left. Think of the head as akin to the head of a Turing machine, and $s$ as the tape of instructions.
There are three possible behaviors.
(1) The head moves off the left end of $s$:
begin{array}{cccccccccc}
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{}
end{array}
(2) The head goes into a cycle, e.g., when the head hits $0$:
begin{array}{cccccccccccccc}
6 & 4 & 5 & 7 & 5 & 1 & 3 & 1 & 1 & 0 & 6 & 4 & 5 & 9 \
text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} \
6 & 4 & 5 & 7 & 5 & 1 & 3 & 1 & 1 & 0 & 6 & 4 & 5 & 9 \
text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} & text{} \
6 & 4 & 5 & 7 & 5 & 1 & 3 & 1 & 1 & 0 & 6 & 4 & 5 & 9 \
text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} \
6 & 4 & 5 & 7 & 5 & 1 & 3 & 1 & 1 & 0 & 6 & 4 & 5 & 9 \
text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} \
6 & 4 & 5 & 7 & 5 & 1 & 3 & 1 & 1 & 0 & 6 & 4 & 5 & 9 \
text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} \
6 & 4 & 5 & 7 & 5 & 1 & 3 & 1 & 1 & 0 & 6 & 4 & 5 & 9 \
text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{}
end{array}
(3) The head moves off rightward to infnity:
begin{array}{ccccccccccccc}
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} \
end{array}
This last string could be viewed as the decimal expansion of $31/99 = 0.3131313131313cdots$.
Q1. What is an example of an irrational number
$0.d_1 d_2 d_3 cdots$ whose string $s=d_1 d_2 d_3 cdots$ causes the head to hop rightward to infinity?
Q1.5. (Added). Is there an explicit irrational algebraic number with the hop-to-$infty$ property?
I'm thinking of something like $sqrt{7}-2$, the
2nd example above (which cycles).
Q2. More generally, which strings
cause the head to hop rightward to infinity?
Update (summarizing answers, 13Apr2019).
Q1. There are irrationals with the hop-to-$infty$ property
(@EthanBolker, @TheSimpliFire),
but explicit construction requires using, e.g., the
Thue-Morse sequence (@Wojowu).
Q1.5. @EthanBolker suggests this may be difficult, and @Wojowu suggests it may be false (b/c: nine consecutive zeros): Perhaps no algebraic irrational has the hop-to-$infty$ property.
Q2. A partial algorithmic characterization by @TheSimpliFire.
sequences-and-series recreational-mathematics irrational-numbers decimal-expansion turing-machines
asked yesterday
Joseph O'RourkeJoseph O'Rourke
18.3k351113
$endgroup$
Let $s$ be an infinite string of decimal digits, for example:
begin{array}{cccccccccc}
s = 3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 & cdots
end{array}
Consider a marker, the head, pointing to the first digit, $3$ in the above example. Interpret the digit under
the head as an instruction to move the head $3$ digits to the right, i.e., to the $4$th digit. Now the head is pointing to $1$. Interpret this as an instruction to move $1$ place to the left. Continue in this manner, hopping through the string, alternately moving right and left. Think of the head as akin to the head of a Turing machine, and $s$ as the tape of instructions.
There are three possible behaviors.
(1) The head moves off the left end of $s$:
begin{array}{cccccccccc}
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{}
end{array}
(2) The head goes into a cycle, e.g., when the head hits $0$:
begin{array}{cccccccccccccc}
6 & 4 & 5 & 7 & 5 & 1 & 3 & 1 & 1 & 0 & 6 & 4 & 5 & 9 \
text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} \
6 & 4 & 5 & 7 & 5 & 1 & 3 & 1 & 1 & 0 & 6 & 4 & 5 & 9 \
text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} & text{} \
6 & 4 & 5 & 7 & 5 & 1 & 3 & 1 & 1 & 0 & 6 & 4 & 5 & 9 \
text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} \
6 & 4 & 5 & 7 & 5 & 1 & 3 & 1 & 1 & 0 & 6 & 4 & 5 & 9 \
text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} \
6 & 4 & 5 & 7 & 5 & 1 & 3 & 1 & 1 & 0 & 6 & 4 & 5 & 9 \
text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} \
6 & 4 & 5 & 7 & 5 & 1 & 3 & 1 & 1 & 0 & 6 & 4 & 5 & 9 \
text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{}
end{array}
(3) The head moves off rightward to infnity:
begin{array}{ccccccccccccc}
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} & text{} & text{} \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} & text{} & text{} & text{} \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{} & text{ ${}^{wedge}$} & text{} \
end{array}
This last string could be viewed as the decimal expansion of $31/99 = 0.3131313131313cdots$.
Q1. What is an example of an irrational number
$0.d_1 d_2 d_3 cdots$ whose string $s=d_1 d_2 d_3 cdots$ causes the head to hop rightward to infinity?
Q1.5. (Added). Is there an explicit irrational algebraic number with the hop-to-$infty$ property?
I'm thinking of something like $sqrt{7}-2$, the
2nd example above (which cycles).
Q2. More generally, which strings
cause the head to hop rightward to infinity?
Update (summarizing answers, 13Apr2019).
Q1. There are irrationals with the hop-to-$infty$ property
(@EthanBolker, @TheSimpliFire),
but explicit construction requires using, e.g., the
Thue-Morse sequence (@Wojowu).
Q1.5. @EthanBolker suggests this may be difficult, and @Wojowu suggests it may be false (b/c: nine consecutive zeros): Perhaps no algebraic irrational has the hop-to-$infty$ property.
Q2. A partial algorithmic characterization by @TheSimpliFire.
sequences-and-series recreational-mathematics irrational-numbers decimal-expansion turing-machines
sequences-and-series recreational-mathematics irrational-numbers decimal-expansion turing-machines
asked yesterday
Joseph O'RourkeJoseph O'Rourke
18.3k351113
asked yesterday
Joseph O'RourkeJoseph O'Rourke
18.3k351113
edited yesterday
Joseph O'Rourke
asked yesterday
Joseph O'RourkeJoseph O'Rourke
18.3k351113
asked yesterday
Joseph O'RourkeJoseph O'Rourke
18.3k351113
asked yesterday
Joseph O'RourkeJoseph O'Rourke
18.3k351113
18.3k351113
1
$begingroup$
You want a string such that the $D_1=(1+d_1)$th digit is less than $d_1$, that the $D_2=(1+d_1-D_1)$th digit is greater than $D_1$, that the $D_3=(1+d_1-D_1+D_2)$th digit is less than $D_2$, etc. I think it is possible to generate an algorithm and you can try for some simulations.
$endgroup$
– TheSimpliFire
yesterday
2
$begingroup$
Another question. For sequences that don't hop to infinity behavior is determined by a (finite) initial subsequence. There are only countably many of those. What are they?
$endgroup$
– Ethan Bolker
yesterday
3
$begingroup$
I think Q1.5 is hard since digit sequences for algebraic numbers are hard adamczewski.perso.math.cnrs.fr/Siauliai.pdf
$endgroup$
– Ethan Bolker
yesterday
1
$begingroup$
Followup question: what are the measures of the three sets in the interval $(0,1)$?
$endgroup$
– eyeballfrog
yesterday
2
$begingroup$
@eyeballfrog The set of numbers with hop to infinity property has measure zero, because you can't hop past a string of nine zeros. This also strongly suggests no algebraic irrational has this property.
$endgroup$
– Wojowu
yesterday
|
show 3 more comments
1
$begingroup$
You want a string such that the $D_1=(1+d_1)$th digit is less than $d_1$, that the $D_2=(1+d_1-D_1)$th digit is greater than $D_1$, that the $D_3=(1+d_1-D_1+D_2)$th digit is less than $D_2$, etc. I think it is possible to generate an algorithm and you can try for some simulations.
$endgroup$
– TheSimpliFire
yesterday
2
$begingroup$
Another question. For sequences that don't hop to infinity behavior is determined by a (finite) initial subsequence. There are only countably many of those. What are they?
$endgroup$
– Ethan Bolker
yesterday
3
$begingroup$
I think Q1.5 is hard since digit sequences for algebraic numbers are hard adamczewski.perso.math.cnrs.fr/Siauliai.pdf
$endgroup$
– Ethan Bolker
yesterday
1
$begingroup$
Followup question: what are the measures of the three sets in the interval $(0,1)$?
$endgroup$
– eyeballfrog
yesterday
2
$begingroup$
@eyeballfrog The set of numbers with hop to infinity property has measure zero, because you can't hop past a string of nine zeros. This also strongly suggests no algebraic irrational has this property.
$endgroup$
– Wojowu
yesterday
1
1
$begingroup$
You want a string such that the $D_1=(1+d_1)$th digit is less than $d_1$, that the $D_2=(1+d_1-D_1)$th digit is greater than $D_1$, that the $D_3=(1+d_1-D_1+D_2)$th digit is less than $D_2$, etc. I think it is possible to generate an algorithm and you can try for some simulations.
$endgroup$
– TheSimpliFire
yesterday
$begingroup$
You want a string such that the $D_1=(1+d_1)$th digit is less than $d_1$, that the $D_2=(1+d_1-D_1)$th digit is greater than $D_1$, that the $D_3=(1+d_1-D_1+D_2)$th digit is less than $D_2$, etc. I think it is possible to generate an algorithm and you can try for some simulations.
$endgroup$
– TheSimpliFire
yesterday
2
2
$begingroup$
Another question. For sequences that don't hop to infinity behavior is determined by a (finite) initial subsequence. There are only countably many of those. What are they?
$endgroup$
– Ethan Bolker
yesterday
$begingroup$
Another question. For sequences that don't hop to infinity behavior is determined by a (finite) initial subsequence. There are only countably many of those. What are they?
$endgroup$
– Ethan Bolker
yesterday
3
3
$begingroup$
I think Q1.5 is hard since digit sequences for algebraic numbers are hard adamczewski.perso.math.cnrs.fr/Siauliai.pdf
$endgroup$
– Ethan Bolker
yesterday
$begingroup$
I think Q1.5 is hard since digit sequences for algebraic numbers are hard adamczewski.perso.math.cnrs.fr/Siauliai.pdf
$endgroup$
– Ethan Bolker
yesterday
1
1
$begingroup$
Followup question: what are the measures of the three sets in the interval $(0,1)$?
$endgroup$
– eyeballfrog
yesterday
$begingroup$
Followup question: what are the measures of the three sets in the interval $(0,1)$?
$endgroup$
– eyeballfrog
yesterday
2
2
$begingroup$
@eyeballfrog The set of numbers with hop to infinity property has measure zero, because you can't hop past a string of nine zeros. This also strongly suggests no algebraic irrational has this property.
$endgroup$
– Wojowu
yesterday
$begingroup$
@eyeballfrog The set of numbers with hop to infinity property has measure zero, because you can't hop past a string of nine zeros. This also strongly suggests no algebraic irrational has this property.
$endgroup$
– Wojowu
yesterday
|
show 3 more comments
1 Answer
1
active
oldest
votes
12
$begingroup$
$$
x 1^{x-2} y 1^{y-2} z1^{z-2} ldots
$$
moves off to infinity for any sequence of digits $xyzldots$ between $3$ and $9$. Select a sequence that defines an irrational number.
More generally
$$
x 1 ?^{x-1} y 1 ?^{y-1} z 1 ?^{z-1} ldots
$$
works, where $?^n$ is an arbitrary string of $n$ digits, since those spots will never be hopped on.
answered yesterday
Ethan BolkerEthan Bolker
46.2k554121
$endgroup$
$begingroup$
(+1) I was thinking exactly the same thing just as you posted your answer!
$endgroup$
– TheSimpliFire
yesterday
$begingroup$
Nice! Could you be more explicit about how you know the resulting digit string is irrational? Thanks.
$endgroup$
– Joseph O'Rourke
yesterday
4
$begingroup$
Clearly it will be irrational if $t = 0.xyzldots$ is, since then periodicity is impossible. It can be even when $t$ is rational because the $?$'s can force that.
$endgroup$
– Ethan Bolker
yesterday
$begingroup$
I worry about explicitly specifying $t=0.xyzcdots$, excluding digits ${ 0,1,2 }$, guaranteeing $t$ is irrational.
$endgroup$
– Joseph O'Rourke
yesterday
4
$begingroup$
@JosephO'Rourke Take the Thue-Morse sequence, add 3 to each term and take that as a binary sequence. Don't hope for a better sort of answer - our understanding of decimal expansions of "natural" constants is really bad, so you won't be able to exclude 0,1,2 without artificially constructing the decimal expansion.
$endgroup$
– Wojowu
yesterday
|
show 1 more comment
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1 Answer
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1 Answer
1
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active
oldest
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active
oldest
votes
12
$begingroup$
$$
x 1^{x-2} y 1^{y-2} z1^{z-2} ldots
$$
moves off to infinity for any sequence of digits $xyzldots$ between $3$ and $9$. Select a sequence that defines an irrational number.
More generally
$$
x 1 ?^{x-1} y 1 ?^{y-1} z 1 ?^{z-1} ldots
$$
works, where $?^n$ is an arbitrary string of $n$ digits, since those spots will never be hopped on.
answered yesterday
Ethan BolkerEthan Bolker
46.2k554121
$endgroup$
$begingroup$
(+1) I was thinking exactly the same thing just as you posted your answer!
$endgroup$
– TheSimpliFire
yesterday
$begingroup$
Nice! Could you be more explicit about how you know the resulting digit string is irrational? Thanks.
$endgroup$
– Joseph O'Rourke
yesterday
4
$begingroup$
Clearly it will be irrational if $t = 0.xyzldots$ is, since then periodicity is impossible. It can be even when $t$ is rational because the $?$'s can force that.
$endgroup$
– Ethan Bolker
yesterday
$begingroup$
I worry about explicitly specifying $t=0.xyzcdots$, excluding digits ${ 0,1,2 }$, guaranteeing $t$ is irrational.
$endgroup$
– Joseph O'Rourke
yesterday
4
$begingroup$
@JosephO'Rourke Take the Thue-Morse sequence, add 3 to each term and take that as a binary sequence. Don't hope for a better sort of answer - our understanding of decimal expansions of "natural" constants is really bad, so you won't be able to exclude 0,1,2 without artificially constructing the decimal expansion.
$endgroup$
– Wojowu
yesterday
|
show 1 more comment
12
$begingroup$
$$
x 1^{x-2} y 1^{y-2} z1^{z-2} ldots
$$
moves off to infinity for any sequence of digits $xyzldots$ between $3$ and $9$. Select a sequence that defines an irrational number.
More generally
$$
x 1 ?^{x-1} y 1 ?^{y-1} z 1 ?^{z-1} ldots
$$
works, where $?^n$ is an arbitrary string of $n$ digits, since those spots will never be hopped on.
answered yesterday
Ethan BolkerEthan Bolker
46.2k554121
$endgroup$
$begingroup$
(+1) I was thinking exactly the same thing just as you posted your answer!
$endgroup$
– TheSimpliFire
yesterday
$begingroup$
Nice! Could you be more explicit about how you know the resulting digit string is irrational? Thanks.
$endgroup$
– Joseph O'Rourke
yesterday
4
$begingroup$
Clearly it will be irrational if $t = 0.xyzldots$ is, since then periodicity is impossible. It can be even when $t$ is rational because the $?$'s can force that.
$endgroup$
– Ethan Bolker
yesterday
$begingroup$
I worry about explicitly specifying $t=0.xyzcdots$, excluding digits ${ 0,1,2 }$, guaranteeing $t$ is irrational.
$endgroup$
– Joseph O'Rourke
yesterday
4
$begingroup$
@JosephO'Rourke Take the Thue-Morse sequence, add 3 to each term and take that as a binary sequence. Don't hope for a better sort of answer - our understanding of decimal expansions of "natural" constants is really bad, so you won't be able to exclude 0,1,2 without artificially constructing the decimal expansion.
$endgroup$
– Wojowu
yesterday
|
show 1 more comment
12
12
12
$begingroup$
$$
x 1^{x-2} y 1^{y-2} z1^{z-2} ldots
$$
moves off to infinity for any sequence of digits $xyzldots$ between $3$ and $9$. Select a sequence that defines an irrational number.
More generally
$$
x 1 ?^{x-1} y 1 ?^{y-1} z 1 ?^{z-1} ldots
$$
works, where $?^n$ is an arbitrary string of $n$ digits, since those spots will never be hopped on.
answered yesterday
Ethan BolkerEthan Bolker
46.2k554121
$endgroup$
$$
x 1^{x-2} y 1^{y-2} z1^{z-2} ldots
$$
moves off to infinity for any sequence of digits $xyzldots$ between $3$ and $9$. Select a sequence that defines an irrational number.
More generally
$$
x 1 ?^{x-1} y 1 ?^{y-1} z 1 ?^{z-1} ldots
$$
works, where $?^n$ is an arbitrary string of $n$ digits, since those spots will never be hopped on.
answered yesterday
Ethan BolkerEthan Bolker
46.2k554121
edited yesterday
answered yesterday
Ethan BolkerEthan Bolker
46.2k554121
answered yesterday
Ethan BolkerEthan Bolker
46.2k554121
answered yesterday
Ethan BolkerEthan Bolker
46.2k554121
46.2k554121
$begingroup$
(+1) I was thinking exactly the same thing just as you posted your answer!
$endgroup$
– TheSimpliFire
yesterday
$begingroup$
Nice! Could you be more explicit about how you know the resulting digit string is irrational? Thanks.
$endgroup$
– Joseph O'Rourke
yesterday
4
$begingroup$
Clearly it will be irrational if $t = 0.xyzldots$ is, since then periodicity is impossible. It can be even when $t$ is rational because the $?$'s can force that.
$endgroup$
– Ethan Bolker
yesterday
$begingroup$
I worry about explicitly specifying $t=0.xyzcdots$, excluding digits ${ 0,1,2 }$, guaranteeing $t$ is irrational.
$endgroup$
– Joseph O'Rourke
yesterday
4
$begingroup$
@JosephO'Rourke Take the Thue-Morse sequence, add 3 to each term and take that as a binary sequence. Don't hope for a better sort of answer - our understanding of decimal expansions of "natural" constants is really bad, so you won't be able to exclude 0,1,2 without artificially constructing the decimal expansion.
$endgroup$
– Wojowu
yesterday
|
show 1 more comment
$begingroup$
(+1) I was thinking exactly the same thing just as you posted your answer!
$endgroup$
– TheSimpliFire
yesterday
$begingroup$
Nice! Could you be more explicit about how you know the resulting digit string is irrational? Thanks.
$endgroup$
– Joseph O'Rourke
yesterday
4
$begingroup$
Clearly it will be irrational if $t = 0.xyzldots$ is, since then periodicity is impossible. It can be even when $t$ is rational because the $?$'s can force that.
$endgroup$
– Ethan Bolker
yesterday
$begingroup$
I worry about explicitly specifying $t=0.xyzcdots$, excluding digits ${ 0,1,2 }$, guaranteeing $t$ is irrational.
$endgroup$
– Joseph O'Rourke
yesterday
4
$begingroup$
@JosephO'Rourke Take the Thue-Morse sequence, add 3 to each term and take that as a binary sequence. Don't hope for a better sort of answer - our understanding of decimal expansions of "natural" constants is really bad, so you won't be able to exclude 0,1,2 without artificially constructing the decimal expansion.
$endgroup$
– Wojowu
yesterday
$begingroup$
(+1) I was thinking exactly the same thing just as you posted your answer!
$endgroup$
– TheSimpliFire
yesterday
$begingroup$
(+1) I was thinking exactly the same thing just as you posted your answer!
$endgroup$
– TheSimpliFire
yesterday
$begingroup$
Nice! Could you be more explicit about how you know the resulting digit string is irrational? Thanks.
$endgroup$
– Joseph O'Rourke
yesterday
$begingroup$
Nice! Could you be more explicit about how you know the resulting digit string is irrational? Thanks.
$endgroup$
– Joseph O'Rourke
yesterday
4
4
$begingroup$
Clearly it will be irrational if $t = 0.xyzldots$ is, since then periodicity is impossible. It can be even when $t$ is rational because the $?$'s can force that.
$endgroup$
– Ethan Bolker
yesterday
$begingroup$
Clearly it will be irrational if $t = 0.xyzldots$ is, since then periodicity is impossible. It can be even when $t$ is rational because the $?$'s can force that.
$endgroup$
– Ethan Bolker
yesterday
$begingroup$
I worry about explicitly specifying $t=0.xyzcdots$, excluding digits ${ 0,1,2 }$, guaranteeing $t$ is irrational.
$endgroup$
– Joseph O'Rourke
yesterday
$begingroup$
I worry about explicitly specifying $t=0.xyzcdots$, excluding digits ${ 0,1,2 }$, guaranteeing $t$ is irrational.
$endgroup$
– Joseph O'Rourke
yesterday
4
4
$begingroup$
@JosephO'Rourke Take the Thue-Morse sequence, add 3 to each term and take that as a binary sequence. Don't hope for a better sort of answer - our understanding of decimal expansions of "natural" constants is really bad, so you won't be able to exclude 0,1,2 without artificially constructing the decimal expansion.
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– Wojowu
yesterday
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@JosephO'Rourke Take the Thue-Morse sequence, add 3 to each term and take that as a binary sequence. Don't hope for a better sort of answer - our understanding of decimal expansions of "natural" constants is really bad, so you won't be able to exclude 0,1,2 without artificially constructing the decimal expansion.
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– Wojowu
yesterday
|
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1
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You want a string such that the $D_1=(1+d_1)$th digit is less than $d_1$, that the $D_2=(1+d_1-D_1)$th digit is greater than $D_1$, that the $D_3=(1+d_1-D_1+D_2)$th digit is less than $D_2$, etc. I think it is possible to generate an algorithm and you can try for some simulations.
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– TheSimpliFire
yesterday
2
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Another question. For sequences that don't hop to infinity behavior is determined by a (finite) initial subsequence. There are only countably many of those. What are they?
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– Ethan Bolker
yesterday
3
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I think Q1.5 is hard since digit sequences for algebraic numbers are hard adamczewski.perso.math.cnrs.fr/Siauliai.pdf
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– Ethan Bolker
yesterday
1
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Followup question: what are the measures of the three sets in the interval $(0,1)$?
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– eyeballfrog
yesterday
2
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@eyeballfrog The set of numbers with hop to infinity property has measure zero, because you can't hop past a string of nine zeros. This also strongly suggests no algebraic irrational has this property.
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– Wojowu
yesterday