Mixing
Digits of a number between 0 and 1.
$begingroup$
I'm currently working on a problem that requires me to know, for some $x=.x_1x_2x_3....x_n in [0,1]$, when the $nth$ digit is equal to $0$ or $1$, in base $2.$ For example the interval where $x_1 = 1$ in base $2$ is $[frac{1}{2}, 1)$, I think. And $x_2=1$ between $0$ and $frac{1}{4}$ and also between $frac{3}{4}$ and $1$. But when the $n$ gets larger, I feel that I should use some kind of pattern, but I'm not sure what to do.
I would like to do the same thing for the numbers in $[0,1]$ in base three.
Any help is appreciated.
binary
asked Jan 29 at 20:14
mXdXmXdX
898
$endgroup$
add a comment |
$begingroup$
I'm currently working on a problem that requires me to know, for some $x=.x_1x_2x_3....x_n in [0,1]$, when the $nth$ digit is equal to $0$ or $1$, in base $2.$ For example the interval where $x_1 = 1$ in base $2$ is $[frac{1}{2}, 1)$, I think. And $x_2=1$ between $0$ and $frac{1}{4}$ and also between $frac{3}{4}$ and $1$. But when the $n$ gets larger, I feel that I should use some kind of pattern, but I'm not sure what to do.
I would like to do the same thing for the numbers in $[0,1]$ in base three.
Any help is appreciated.
binary
asked Jan 29 at 20:14
mXdXmXdX
898
$endgroup$
2
$begingroup$
The region where the $n$th digit in base $b$ takes on a particular value consists of $b^{n-1}$ disjoint intervals. Each of these has a particular sequence of values of all of the first $n$ digits.
$endgroup$
– Ian
Jan 29 at 20:19
2
$begingroup$
You are mistaken with $x_2$: it should be $x_2 = 1$ when $x in (frac14, frac12)$, not $(0, frac14)$.
$endgroup$
– Théophile
Jan 29 at 20:28
$begingroup$
Oh okay, thanks. Is there some kind of quick way that I can find the intervals? So far, it's a little tedious, I'm not sure if there's a better way.
$endgroup$
– mXdX
Jan 29 at 21:05
add a comment |
$begingroup$
I'm currently working on a problem that requires me to know, for some $x=.x_1x_2x_3....x_n in [0,1]$, when the $nth$ digit is equal to $0$ or $1$, in base $2.$ For example the interval where $x_1 = 1$ in base $2$ is $[frac{1}{2}, 1)$, I think. And $x_2=1$ between $0$ and $frac{1}{4}$ and also between $frac{3}{4}$ and $1$. But when the $n$ gets larger, I feel that I should use some kind of pattern, but I'm not sure what to do.
I would like to do the same thing for the numbers in $[0,1]$ in base three.
Any help is appreciated.
binary
asked Jan 29 at 20:14
mXdXmXdX
898
$endgroup$
I'm currently working on a problem that requires me to know, for some $x=.x_1x_2x_3....x_n in [0,1]$, when the $nth$ digit is equal to $0$ or $1$, in base $2.$ For example the interval where $x_1 = 1$ in base $2$ is $[frac{1}{2}, 1)$, I think. And $x_2=1$ between $0$ and $frac{1}{4}$ and also between $frac{3}{4}$ and $1$. But when the $n$ gets larger, I feel that I should use some kind of pattern, but I'm not sure what to do.
I would like to do the same thing for the numbers in $[0,1]$ in base three.
Any help is appreciated.
binary
binary
asked Jan 29 at 20:14
mXdXmXdX
898
asked Jan 29 at 20:14
mXdXmXdX
898
asked Jan 29 at 20:14
mXdXmXdX
898
asked Jan 29 at 20:14
mXdXmXdX
898
asked Jan 29 at 20:14
mXdXmXdX
898
898
2
$begingroup$
The region where the $n$th digit in base $b$ takes on a particular value consists of $b^{n-1}$ disjoint intervals. Each of these has a particular sequence of values of all of the first $n$ digits.
$endgroup$
– Ian
Jan 29 at 20:19
2
$begingroup$
You are mistaken with $x_2$: it should be $x_2 = 1$ when $x in (frac14, frac12)$, not $(0, frac14)$.
$endgroup$
– Théophile
Jan 29 at 20:28
$begingroup$
Oh okay, thanks. Is there some kind of quick way that I can find the intervals? So far, it's a little tedious, I'm not sure if there's a better way.
$endgroup$
– mXdX
Jan 29 at 21:05
add a comment |
2
$begingroup$
The region where the $n$th digit in base $b$ takes on a particular value consists of $b^{n-1}$ disjoint intervals. Each of these has a particular sequence of values of all of the first $n$ digits.
$endgroup$
– Ian
Jan 29 at 20:19
2
$begingroup$
You are mistaken with $x_2$: it should be $x_2 = 1$ when $x in (frac14, frac12)$, not $(0, frac14)$.
$endgroup$
– Théophile
Jan 29 at 20:28
$begingroup$
Oh okay, thanks. Is there some kind of quick way that I can find the intervals? So far, it's a little tedious, I'm not sure if there's a better way.
$endgroup$
– mXdX
Jan 29 at 21:05
2
2
$begingroup$
The region where the $n$th digit in base $b$ takes on a particular value consists of $b^{n-1}$ disjoint intervals. Each of these has a particular sequence of values of all of the first $n$ digits.
$endgroup$
– Ian
Jan 29 at 20:19
$begingroup$
The region where the $n$th digit in base $b$ takes on a particular value consists of $b^{n-1}$ disjoint intervals. Each of these has a particular sequence of values of all of the first $n$ digits.
$endgroup$
– Ian
Jan 29 at 20:19
2
2
$begingroup$
You are mistaken with $x_2$: it should be $x_2 = 1$ when $x in (frac14, frac12)$, not $(0, frac14)$.
$endgroup$
– Théophile
Jan 29 at 20:28
$begingroup$
You are mistaken with $x_2$: it should be $x_2 = 1$ when $x in (frac14, frac12)$, not $(0, frac14)$.
$endgroup$
– Théophile
Jan 29 at 20:28
$begingroup$
Oh okay, thanks. Is there some kind of quick way that I can find the intervals? So far, it's a little tedious, I'm not sure if there's a better way.
$endgroup$
– mXdX
Jan 29 at 21:05
$begingroup$
Oh okay, thanks. Is there some kind of quick way that I can find the intervals? So far, it's a little tedious, I'm not sure if there's a better way.
$endgroup$
– mXdX
Jan 29 at 21:05
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Let $x$ be the number.
Is the integer part of $2^n x$ odd or even?
If it's odd, the $n$th bit after the binary point is $1$.
If it's even, the $n$th bit after the binary point is $0$.
If you're not convinced, try multiplying a decimal number by $10^n$ and looking at the last digit of its integer part. Then remember that an even binary number ends in $0$.
Multiplying $x$ by $2^n$ just shifts its binary representation to the left by $n$ places.
Dividing by $2^n$ again, your intervals will be based on odd and even multiples of $2^{-n}$.
answered Jan 29 at 23:07
timtfjtimtfj
2,503420
$endgroup$
add a comment |
$begingroup$
Is $x$ greater than $2^{-1}$? Yes: $x_1 = 1$; subtract $2^{-1}$ from $x$. No: $x_1 = 0$.
Is $x$ greater than $2^{-2}$? Yes: $x_2 = 1$; subtract $2^{-2}$ from $x$. No: $x_2 = 0$.
$cdots$
Is $x$ greater than $2^{-n}$? Yes: $x_n = 1$; subtract $2^{-n}$ from $x$. No: $x_n = 0$.
answered Jan 29 at 20:20
ThéophileThéophile
20.3k13047
$endgroup$
$begingroup$
This isn't right...you need to subtract off the contribution from the earlier digits as you go. Or follow one of the other answers in looking at $2^n x$.
$endgroup$
– Ian
Jan 30 at 0:57
$begingroup$
@Ian I intended this in an algorithmic sense where $x$ is updated at every step, so that each line should be read as "Is [the current value of] $x$ ...".
$endgroup$
– Théophile
Jan 30 at 16:46
add a comment |
$begingroup$
Sometimes a picture is worth 1000 words.
Regions represented by a $1$ in successive digit places
base 2
base 3
answered Jan 29 at 20:31
David G. StorkDavid G. Stork
11.6k41533
$endgroup$
$begingroup$
Would there be a systematic way for me to find these intervals on my own? I've been simply converting numbers to base 2/base 3 and checking if I need to go up or down to get the digit to be of the value I want. So, I've just been guessing and checking. Is there a more efficient method?
$endgroup$
– mXdX
Jan 29 at 21:09
$begingroup$
See if you can infer the principle from my figures....
$endgroup$
– David G. Stork
Jan 29 at 21:50
$begingroup$
I was looking at the figures but I thought that for base 3, the region where the $nth$ digit takes a particular value is given by $3^{n-1}$ intervals. Shouldn't there be three intervals for $x_2$ on the second picture? I am missing something.
$endgroup$
– mXdX
Jan 30 at 3:05
1
$begingroup$
You are missing something. There are three regions for the first digit, corresponding the the value being $0$, $1$ or $2$. Only one is shown in my figure (for $1$); the region for $0$ is to the left, and for $2$ is to the right. For the second digit, there are .... (can you continue?)....
$endgroup$
– David G. Stork
Jan 30 at 3:11
1
$begingroup$
You were right. Thanks for catching the typo. Now (I think!) fixed.
$endgroup$
– David G. Stork
Jan 30 at 4:42
|
show 1 more comment
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $x$ be the number.
Is the integer part of $2^n x$ odd or even?
If it's odd, the $n$th bit after the binary point is $1$.
If it's even, the $n$th bit after the binary point is $0$.
If you're not convinced, try multiplying a decimal number by $10^n$ and looking at the last digit of its integer part. Then remember that an even binary number ends in $0$.
Multiplying $x$ by $2^n$ just shifts its binary representation to the left by $n$ places.
Dividing by $2^n$ again, your intervals will be based on odd and even multiples of $2^{-n}$.
answered Jan 29 at 23:07
timtfjtimtfj
2,503420
$endgroup$
add a comment |
$begingroup$
Let $x$ be the number.
Is the integer part of $2^n x$ odd or even?
If it's odd, the $n$th bit after the binary point is $1$.
If it's even, the $n$th bit after the binary point is $0$.
If you're not convinced, try multiplying a decimal number by $10^n$ and looking at the last digit of its integer part. Then remember that an even binary number ends in $0$.
Multiplying $x$ by $2^n$ just shifts its binary representation to the left by $n$ places.
Dividing by $2^n$ again, your intervals will be based on odd and even multiples of $2^{-n}$.
answered Jan 29 at 23:07
timtfjtimtfj
2,503420
$endgroup$
add a comment |
$begingroup$
Let $x$ be the number.
Is the integer part of $2^n x$ odd or even?
If it's odd, the $n$th bit after the binary point is $1$.
If it's even, the $n$th bit after the binary point is $0$.
If you're not convinced, try multiplying a decimal number by $10^n$ and looking at the last digit of its integer part. Then remember that an even binary number ends in $0$.
Multiplying $x$ by $2^n$ just shifts its binary representation to the left by $n$ places.
Dividing by $2^n$ again, your intervals will be based on odd and even multiples of $2^{-n}$.
answered Jan 29 at 23:07
timtfjtimtfj
2,503420
$endgroup$
Let $x$ be the number.
Is the integer part of $2^n x$ odd or even?
If it's odd, the $n$th bit after the binary point is $1$.
If it's even, the $n$th bit after the binary point is $0$.
If you're not convinced, try multiplying a decimal number by $10^n$ and looking at the last digit of its integer part. Then remember that an even binary number ends in $0$.
Multiplying $x$ by $2^n$ just shifts its binary representation to the left by $n$ places.
Dividing by $2^n$ again, your intervals will be based on odd and even multiples of $2^{-n}$.
answered Jan 29 at 23:07
timtfjtimtfj
2,503420
edited Jan 29 at 23:18
answered Jan 29 at 23:07
timtfjtimtfj
2,503420
answered Jan 29 at 23:07
timtfjtimtfj
2,503420
answered Jan 29 at 23:07
timtfjtimtfj
2,503420
2,503420
add a comment |
add a comment |
$begingroup$
Is $x$ greater than $2^{-1}$? Yes: $x_1 = 1$; subtract $2^{-1}$ from $x$. No: $x_1 = 0$.
Is $x$ greater than $2^{-2}$? Yes: $x_2 = 1$; subtract $2^{-2}$ from $x$. No: $x_2 = 0$.
$cdots$
Is $x$ greater than $2^{-n}$? Yes: $x_n = 1$; subtract $2^{-n}$ from $x$. No: $x_n = 0$.
answered Jan 29 at 20:20
ThéophileThéophile
20.3k13047
$endgroup$
$begingroup$
This isn't right...you need to subtract off the contribution from the earlier digits as you go. Or follow one of the other answers in looking at $2^n x$.
$endgroup$
– Ian
Jan 30 at 0:57
$begingroup$
@Ian I intended this in an algorithmic sense where $x$ is updated at every step, so that each line should be read as "Is [the current value of] $x$ ...".
$endgroup$
– Théophile
Jan 30 at 16:46
add a comment |
$begingroup$
Is $x$ greater than $2^{-1}$? Yes: $x_1 = 1$; subtract $2^{-1}$ from $x$. No: $x_1 = 0$.
Is $x$ greater than $2^{-2}$? Yes: $x_2 = 1$; subtract $2^{-2}$ from $x$. No: $x_2 = 0$.
$cdots$
Is $x$ greater than $2^{-n}$? Yes: $x_n = 1$; subtract $2^{-n}$ from $x$. No: $x_n = 0$.
answered Jan 29 at 20:20
ThéophileThéophile
20.3k13047
$endgroup$
$begingroup$
This isn't right...you need to subtract off the contribution from the earlier digits as you go. Or follow one of the other answers in looking at $2^n x$.
$endgroup$
– Ian
Jan 30 at 0:57
$begingroup$
@Ian I intended this in an algorithmic sense where $x$ is updated at every step, so that each line should be read as "Is [the current value of] $x$ ...".
$endgroup$
– Théophile
Jan 30 at 16:46
add a comment |
$begingroup$
Is $x$ greater than $2^{-1}$? Yes: $x_1 = 1$; subtract $2^{-1}$ from $x$. No: $x_1 = 0$.
Is $x$ greater than $2^{-2}$? Yes: $x_2 = 1$; subtract $2^{-2}$ from $x$. No: $x_2 = 0$.
$cdots$
Is $x$ greater than $2^{-n}$? Yes: $x_n = 1$; subtract $2^{-n}$ from $x$. No: $x_n = 0$.
answered Jan 29 at 20:20
ThéophileThéophile
20.3k13047
$endgroup$
Is $x$ greater than $2^{-1}$? Yes: $x_1 = 1$; subtract $2^{-1}$ from $x$. No: $x_1 = 0$.
Is $x$ greater than $2^{-2}$? Yes: $x_2 = 1$; subtract $2^{-2}$ from $x$. No: $x_2 = 0$.
$cdots$
Is $x$ greater than $2^{-n}$? Yes: $x_n = 1$; subtract $2^{-n}$ from $x$. No: $x_n = 0$.
answered Jan 29 at 20:20
ThéophileThéophile
20.3k13047
answered Jan 29 at 20:20
ThéophileThéophile
20.3k13047
answered Jan 29 at 20:20
ThéophileThéophile
20.3k13047
answered Jan 29 at 20:20
ThéophileThéophile
20.3k13047
20.3k13047
$begingroup$
This isn't right...you need to subtract off the contribution from the earlier digits as you go. Or follow one of the other answers in looking at $2^n x$.
$endgroup$
– Ian
Jan 30 at 0:57
$begingroup$
@Ian I intended this in an algorithmic sense where $x$ is updated at every step, so that each line should be read as "Is [the current value of] $x$ ...".
$endgroup$
– Théophile
Jan 30 at 16:46
add a comment |
$begingroup$
This isn't right...you need to subtract off the contribution from the earlier digits as you go. Or follow one of the other answers in looking at $2^n x$.
$endgroup$
– Ian
Jan 30 at 0:57
$begingroup$
@Ian I intended this in an algorithmic sense where $x$ is updated at every step, so that each line should be read as "Is [the current value of] $x$ ...".
$endgroup$
– Théophile
Jan 30 at 16:46
$begingroup$
This isn't right...you need to subtract off the contribution from the earlier digits as you go. Or follow one of the other answers in looking at $2^n x$.
$endgroup$
– Ian
Jan 30 at 0:57
$begingroup$
This isn't right...you need to subtract off the contribution from the earlier digits as you go. Or follow one of the other answers in looking at $2^n x$.
$endgroup$
– Ian
Jan 30 at 0:57
$begingroup$
@Ian I intended this in an algorithmic sense where $x$ is updated at every step, so that each line should be read as "Is [the current value of] $x$ ...".
$endgroup$
– Théophile
Jan 30 at 16:46
$begingroup$
@Ian I intended this in an algorithmic sense where $x$ is updated at every step, so that each line should be read as "Is [the current value of] $x$ ...".
$endgroup$
– Théophile
Jan 30 at 16:46
add a comment |
$begingroup$
Sometimes a picture is worth 1000 words.
Regions represented by a $1$ in successive digit places
base 2
base 3
answered Jan 29 at 20:31
David G. StorkDavid G. Stork
11.6k41533
$endgroup$
$begingroup$
Would there be a systematic way for me to find these intervals on my own? I've been simply converting numbers to base 2/base 3 and checking if I need to go up or down to get the digit to be of the value I want. So, I've just been guessing and checking. Is there a more efficient method?
$endgroup$
– mXdX
Jan 29 at 21:09
$begingroup$
See if you can infer the principle from my figures....
$endgroup$
– David G. Stork
Jan 29 at 21:50
$begingroup$
I was looking at the figures but I thought that for base 3, the region where the $nth$ digit takes a particular value is given by $3^{n-1}$ intervals. Shouldn't there be three intervals for $x_2$ on the second picture? I am missing something.
$endgroup$
– mXdX
Jan 30 at 3:05
1
$begingroup$
You are missing something. There are three regions for the first digit, corresponding the the value being $0$, $1$ or $2$. Only one is shown in my figure (for $1$); the region for $0$ is to the left, and for $2$ is to the right. For the second digit, there are .... (can you continue?)....
$endgroup$
– David G. Stork
Jan 30 at 3:11
1
$begingroup$
You were right. Thanks for catching the typo. Now (I think!) fixed.
$endgroup$
– David G. Stork
Jan 30 at 4:42
|
show 1 more comment
$begingroup$
Sometimes a picture is worth 1000 words.
Regions represented by a $1$ in successive digit places
base 2
base 3
answered Jan 29 at 20:31
David G. StorkDavid G. Stork
11.6k41533
$endgroup$
$begingroup$
Would there be a systematic way for me to find these intervals on my own? I've been simply converting numbers to base 2/base 3 and checking if I need to go up or down to get the digit to be of the value I want. So, I've just been guessing and checking. Is there a more efficient method?
$endgroup$
– mXdX
Jan 29 at 21:09
$begingroup$
See if you can infer the principle from my figures....
$endgroup$
– David G. Stork
Jan 29 at 21:50
$begingroup$
I was looking at the figures but I thought that for base 3, the region where the $nth$ digit takes a particular value is given by $3^{n-1}$ intervals. Shouldn't there be three intervals for $x_2$ on the second picture? I am missing something.
$endgroup$
– mXdX
Jan 30 at 3:05
1
$begingroup$
You are missing something. There are three regions for the first digit, corresponding the the value being $0$, $1$ or $2$. Only one is shown in my figure (for $1$); the region for $0$ is to the left, and for $2$ is to the right. For the second digit, there are .... (can you continue?)....
$endgroup$
– David G. Stork
Jan 30 at 3:11
1
$begingroup$
You were right. Thanks for catching the typo. Now (I think!) fixed.
$endgroup$
– David G. Stork
Jan 30 at 4:42
|
show 1 more comment
$begingroup$
Sometimes a picture is worth 1000 words.
Regions represented by a $1$ in successive digit places
base 2
base 3
answered Jan 29 at 20:31
David G. StorkDavid G. Stork
11.6k41533
$endgroup$
Sometimes a picture is worth 1000 words.
Regions represented by a $1$ in successive digit places
base 2
base 3
answered Jan 29 at 20:31
David G. StorkDavid G. Stork
11.6k41533
edited Jan 30 at 4:42
answered Jan 29 at 20:31
David G. StorkDavid G. Stork
11.6k41533
answered Jan 29 at 20:31
David G. StorkDavid G. Stork
11.6k41533
answered Jan 29 at 20:31
David G. StorkDavid G. Stork
11.6k41533
11.6k41533
$begingroup$
Would there be a systematic way for me to find these intervals on my own? I've been simply converting numbers to base 2/base 3 and checking if I need to go up or down to get the digit to be of the value I want. So, I've just been guessing and checking. Is there a more efficient method?
$endgroup$
– mXdX
Jan 29 at 21:09
$begingroup$
See if you can infer the principle from my figures....
$endgroup$
– David G. Stork
Jan 29 at 21:50
$begingroup$
I was looking at the figures but I thought that for base 3, the region where the $nth$ digit takes a particular value is given by $3^{n-1}$ intervals. Shouldn't there be three intervals for $x_2$ on the second picture? I am missing something.
$endgroup$
– mXdX
Jan 30 at 3:05
1
$begingroup$
You are missing something. There are three regions for the first digit, corresponding the the value being $0$, $1$ or $2$. Only one is shown in my figure (for $1$); the region for $0$ is to the left, and for $2$ is to the right. For the second digit, there are .... (can you continue?)....
$endgroup$
– David G. Stork
Jan 30 at 3:11
1
$begingroup$
You were right. Thanks for catching the typo. Now (I think!) fixed.
$endgroup$
– David G. Stork
Jan 30 at 4:42
|
show 1 more comment
$begingroup$
Would there be a systematic way for me to find these intervals on my own? I've been simply converting numbers to base 2/base 3 and checking if I need to go up or down to get the digit to be of the value I want. So, I've just been guessing and checking. Is there a more efficient method?
$endgroup$
– mXdX
Jan 29 at 21:09
$begingroup$
See if you can infer the principle from my figures....
$endgroup$
– David G. Stork
Jan 29 at 21:50
$begingroup$
I was looking at the figures but I thought that for base 3, the region where the $nth$ digit takes a particular value is given by $3^{n-1}$ intervals. Shouldn't there be three intervals for $x_2$ on the second picture? I am missing something.
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– mXdX
Jan 30 at 3:05
1
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You are missing something. There are three regions for the first digit, corresponding the the value being $0$, $1$ or $2$. Only one is shown in my figure (for $1$); the region for $0$ is to the left, and for $2$ is to the right. For the second digit, there are .... (can you continue?)....
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– David G. Stork
Jan 30 at 3:11
1
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You were right. Thanks for catching the typo. Now (I think!) fixed.
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– David G. Stork
Jan 30 at 4:42
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Would there be a systematic way for me to find these intervals on my own? I've been simply converting numbers to base 2/base 3 and checking if I need to go up or down to get the digit to be of the value I want. So, I've just been guessing and checking. Is there a more efficient method?
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– mXdX
Jan 29 at 21:09
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Would there be a systematic way for me to find these intervals on my own? I've been simply converting numbers to base 2/base 3 and checking if I need to go up or down to get the digit to be of the value I want. So, I've just been guessing and checking. Is there a more efficient method?
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– mXdX
Jan 29 at 21:09
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See if you can infer the principle from my figures....
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– David G. Stork
Jan 29 at 21:50
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See if you can infer the principle from my figures....
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– David G. Stork
Jan 29 at 21:50
$begingroup$
I was looking at the figures but I thought that for base 3, the region where the $nth$ digit takes a particular value is given by $3^{n-1}$ intervals. Shouldn't there be three intervals for $x_2$ on the second picture? I am missing something.
$endgroup$
– mXdX
Jan 30 at 3:05
$begingroup$
I was looking at the figures but I thought that for base 3, the region where the $nth$ digit takes a particular value is given by $3^{n-1}$ intervals. Shouldn't there be three intervals for $x_2$ on the second picture? I am missing something.
$endgroup$
– mXdX
Jan 30 at 3:05
1
1
$begingroup$
You are missing something. There are three regions for the first digit, corresponding the the value being $0$, $1$ or $2$. Only one is shown in my figure (for $1$); the region for $0$ is to the left, and for $2$ is to the right. For the second digit, there are .... (can you continue?)....
$endgroup$
– David G. Stork
Jan 30 at 3:11
$begingroup$
You are missing something. There are three regions for the first digit, corresponding the the value being $0$, $1$ or $2$. Only one is shown in my figure (for $1$); the region for $0$ is to the left, and for $2$ is to the right. For the second digit, there are .... (can you continue?)....
$endgroup$
– David G. Stork
Jan 30 at 3:11
1
1
$begingroup$
You were right. Thanks for catching the typo. Now (I think!) fixed.
$endgroup$
– David G. Stork
Jan 30 at 4:42
$begingroup$
You were right. Thanks for catching the typo. Now (I think!) fixed.
$endgroup$
– David G. Stork
Jan 30 at 4:42
|
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The region where the $n$th digit in base $b$ takes on a particular value consists of $b^{n-1}$ disjoint intervals. Each of these has a particular sequence of values of all of the first $n$ digits.
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– Ian
Jan 29 at 20:19
2
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You are mistaken with $x_2$: it should be $x_2 = 1$ when $x in (frac14, frac12)$, not $(0, frac14)$.
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– Théophile
Jan 29 at 20:28
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Oh okay, thanks. Is there some kind of quick way that I can find the intervals? So far, it's a little tedious, I'm not sure if there's a better way.
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– mXdX
Jan 29 at 21:05