plotting the remainder r(n,x) as a function of n
0
$begingroup$
I was graphing the remainder of the Euclidean division between an integer $n$ and an irrational $x$: $r(n,x) =n-x cdot lfloor n/x rfloor$
when I stumbled on quasi-linear graph I cannot explain. Any idea how to calculate the slopes of the lines and why there is no lines for $x = phi$, the golden ratio?
:
number-theory
edited Dec 24 '18 at 4:50
gt6989b
34.7k22457
asked Dec 24 '18 at 4:37
ShaktyaiShaktyai
1
$endgroup$
add a comment |
0
$begingroup$
I was graphing the remainder of the Euclidean division between an integer $n$ and an irrational $x$: $r(n,x) =n-x cdot lfloor n/x rfloor$
when I stumbled on quasi-linear graph I cannot explain. Any idea how to calculate the slopes of the lines and why there is no lines for $x = phi$, the golden ratio?
:
number-theory
edited Dec 24 '18 at 4:50
gt6989b
34.7k22457
asked Dec 24 '18 at 4:37
ShaktyaiShaktyai
1
$endgroup$
$begingroup$
You should plot $r(y,x) = y - x lfloor y/x rfloor$ for $y in [0,x]$. Then $frac{r(yx,x)}{x}= y - lfloor y rfloor$ doesn't depend on $x$ so when changing $x$ you obtain the same plot but rescaled differently. Clearly the plot of $y - lfloor y rfloor, y in [0,1]$ is just a line segment from $(0,0)$ to $(1,1)$, and when plotting $y - lfloor y rfloor, y in mathbb{R}$ you get a bunch of line segments from $(n,n)$ to $(n+1,n+1)$. Thus your question now becomes : what happens when sampling the latter plot at discrete values of $y$ ?
$endgroup$
– reuns
Dec 24 '18 at 6:01
$begingroup$
Why is the slope changing from positive for Pi to negative for e? Furthermore, the pattern is filled from the bottom right to up-right, then shift right, goes down, etc...Lines only appear if you plot a lot of points, and points on the line do not correspond to the successive value of n.
$endgroup$
– Shaktyai
Dec 24 '18 at 11:27
add a comment |
0
0
0
$begingroup$
I was graphing the remainder of the Euclidean division between an integer $n$ and an irrational $x$: $r(n,x) =n-x cdot lfloor n/x rfloor$
when I stumbled on quasi-linear graph I cannot explain. Any idea how to calculate the slopes of the lines and why there is no lines for $x = phi$, the golden ratio?
:
number-theory
edited Dec 24 '18 at 4:50
gt6989b
34.7k22457
asked Dec 24 '18 at 4:37
ShaktyaiShaktyai
1
$endgroup$
I was graphing the remainder of the Euclidean division between an integer $n$ and an irrational $x$: $r(n,x) =n-x cdot lfloor n/x rfloor$
when I stumbled on quasi-linear graph I cannot explain. Any idea how to calculate the slopes of the lines and why there is no lines for $x = phi$, the golden ratio?
:
number-theory
number-theory
edited Dec 24 '18 at 4:50
gt6989b
34.7k22457
asked Dec 24 '18 at 4:37
ShaktyaiShaktyai
1
edited Dec 24 '18 at 4:50
gt6989b
34.7k22457
asked Dec 24 '18 at 4:37
ShaktyaiShaktyai
1
edited Dec 24 '18 at 4:50
gt6989b
34.7k22457
edited Dec 24 '18 at 4:50
gt6989b
34.7k22457
edited Dec 24 '18 at 4:50
gt6989b
34.7k22457
34.7k22457
asked Dec 24 '18 at 4:37
ShaktyaiShaktyai
1
asked Dec 24 '18 at 4:37
ShaktyaiShaktyai
1
asked Dec 24 '18 at 4:37
ShaktyaiShaktyai
1
1
$begingroup$
You should plot $r(y,x) = y - x lfloor y/x rfloor$ for $y in [0,x]$. Then $frac{r(yx,x)}{x}= y - lfloor y rfloor$ doesn't depend on $x$ so when changing $x$ you obtain the same plot but rescaled differently. Clearly the plot of $y - lfloor y rfloor, y in [0,1]$ is just a line segment from $(0,0)$ to $(1,1)$, and when plotting $y - lfloor y rfloor, y in mathbb{R}$ you get a bunch of line segments from $(n,n)$ to $(n+1,n+1)$. Thus your question now becomes : what happens when sampling the latter plot at discrete values of $y$ ?
$endgroup$
– reuns
Dec 24 '18 at 6:01
$begingroup$
Why is the slope changing from positive for Pi to negative for e? Furthermore, the pattern is filled from the bottom right to up-right, then shift right, goes down, etc...Lines only appear if you plot a lot of points, and points on the line do not correspond to the successive value of n.
$endgroup$
– Shaktyai
Dec 24 '18 at 11:27
add a comment |
$begingroup$
You should plot $r(y,x) = y - x lfloor y/x rfloor$ for $y in [0,x]$. Then $frac{r(yx,x)}{x}= y - lfloor y rfloor$ doesn't depend on $x$ so when changing $x$ you obtain the same plot but rescaled differently. Clearly the plot of $y - lfloor y rfloor, y in [0,1]$ is just a line segment from $(0,0)$ to $(1,1)$, and when plotting $y - lfloor y rfloor, y in mathbb{R}$ you get a bunch of line segments from $(n,n)$ to $(n+1,n+1)$. Thus your question now becomes : what happens when sampling the latter plot at discrete values of $y$ ?
$endgroup$
– reuns
Dec 24 '18 at 6:01
$begingroup$
Why is the slope changing from positive for Pi to negative for e? Furthermore, the pattern is filled from the bottom right to up-right, then shift right, goes down, etc...Lines only appear if you plot a lot of points, and points on the line do not correspond to the successive value of n.
$endgroup$
– Shaktyai
Dec 24 '18 at 11:27
$begingroup$
You should plot $r(y,x) = y - x lfloor y/x rfloor$ for $y in [0,x]$. Then $frac{r(yx,x)}{x}= y - lfloor y rfloor$ doesn't depend on $x$ so when changing $x$ you obtain the same plot but rescaled differently. Clearly the plot of $y - lfloor y rfloor, y in [0,1]$ is just a line segment from $(0,0)$ to $(1,1)$, and when plotting $y - lfloor y rfloor, y in mathbb{R}$ you get a bunch of line segments from $(n,n)$ to $(n+1,n+1)$. Thus your question now becomes : what happens when sampling the latter plot at discrete values of $y$ ?
$endgroup$
– reuns
Dec 24 '18 at 6:01
$begingroup$
You should plot $r(y,x) = y - x lfloor y/x rfloor$ for $y in [0,x]$. Then $frac{r(yx,x)}{x}= y - lfloor y rfloor$ doesn't depend on $x$ so when changing $x$ you obtain the same plot but rescaled differently. Clearly the plot of $y - lfloor y rfloor, y in [0,1]$ is just a line segment from $(0,0)$ to $(1,1)$, and when plotting $y - lfloor y rfloor, y in mathbb{R}$ you get a bunch of line segments from $(n,n)$ to $(n+1,n+1)$. Thus your question now becomes : what happens when sampling the latter plot at discrete values of $y$ ?
$endgroup$
– reuns
Dec 24 '18 at 6:01
$begingroup$
Why is the slope changing from positive for Pi to negative for e? Furthermore, the pattern is filled from the bottom right to up-right, then shift right, goes down, etc...Lines only appear if you plot a lot of points, and points on the line do not correspond to the successive value of n.
$endgroup$
– Shaktyai
Dec 24 '18 at 11:27
$begingroup$
Why is the slope changing from positive for Pi to negative for e? Furthermore, the pattern is filled from the bottom right to up-right, then shift right, goes down, etc...Lines only appear if you plot a lot of points, and points on the line do not correspond to the successive value of n.
$endgroup$
– Shaktyai
Dec 24 '18 at 11:27
add a comment |
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$begingroup$
You should plot $r(y,x) = y - x lfloor y/x rfloor$ for $y in [0,x]$. Then $frac{r(yx,x)}{x}= y - lfloor y rfloor$ doesn't depend on $x$ so when changing $x$ you obtain the same plot but rescaled differently. Clearly the plot of $y - lfloor y rfloor, y in [0,1]$ is just a line segment from $(0,0)$ to $(1,1)$, and when plotting $y - lfloor y rfloor, y in mathbb{R}$ you get a bunch of line segments from $(n,n)$ to $(n+1,n+1)$. Thus your question now becomes : what happens when sampling the latter plot at discrete values of $y$ ?
$endgroup$
– reuns
Dec 24 '18 at 6:01
$begingroup$
Why is the slope changing from positive for Pi to negative for e? Furthermore, the pattern is filled from the bottom right to up-right, then shift right, goes down, etc...Lines only appear if you plot a lot of points, and points on the line do not correspond to the successive value of n.
$endgroup$
– Shaktyai
Dec 24 '18 at 11:27