Mixing
Of which continuous line in $Bbb RtoBbb C$ is this sequence a subset?
$begingroup$
The first 250 terms of an $Bbb N$-indexed sequence in $Bbb C$, are shown here.
What's the function $h(x):Bbb RtoBbb C$ of which this is a subset, and what's $frac{dr}{dtheta}$ as a function of either $x$ or $h(x)$ (setting $re^{ipitheta}=h(x)$)?
The sequence $S_n$ is given by the map $S_n=h(x):xin2Bbb N-1$, where $h$ is defined:
$f:2Bbb N-1tofrac12+Bbb Z[frac12]/frac12Bbb Z$
$f(x)=frac x{2^p}inleft[frac12,1right)$
$h(x)=xcdot exp{(4pi icdot f(x))}$
The function's effectively a sequence of segments glued together at $x=1, Im(h(x))=0$.
It seems obvious there must be a fairly simple continuous function $h^times:Bbb RtoBbb C$ of which this is a subset (give or take possible singularities at $0,1$). What is it? Obviously there are infinitely many, but there should be an obvious choice - I speculate the rule that chooses the canonical extension would be something like there will be an $n^{th}$ derivative of $r$ with respect to $theta$ that's fixed or monotonic at any given point on the line.
What's the function $h^times(x):Bbb RtoBbb C$, and what's $frac{dr}{dtheta}$ as a function of either $x$ or $h^times(x)$?
It may be worth pointing out that $f(x) mapsto theta(h(x))$ sends (at least a subset of) the Prufer 2-group in its dyadic form to its 2nd roots of unity form.
complex-analysis algebra-precalculus functions complex-numbers
asked Jan 6 at 16:04
user334732user334732
4,26311240
$endgroup$
add a comment |
$begingroup$
The first 250 terms of an $Bbb N$-indexed sequence in $Bbb C$, are shown here.
What's the function $h(x):Bbb RtoBbb C$ of which this is a subset, and what's $frac{dr}{dtheta}$ as a function of either $x$ or $h(x)$ (setting $re^{ipitheta}=h(x)$)?
The sequence $S_n$ is given by the map $S_n=h(x):xin2Bbb N-1$, where $h$ is defined:
$f:2Bbb N-1tofrac12+Bbb Z[frac12]/frac12Bbb Z$
$f(x)=frac x{2^p}inleft[frac12,1right)$
$h(x)=xcdot exp{(4pi icdot f(x))}$
The function's effectively a sequence of segments glued together at $x=1, Im(h(x))=0$.
It seems obvious there must be a fairly simple continuous function $h^times:Bbb RtoBbb C$ of which this is a subset (give or take possible singularities at $0,1$). What is it? Obviously there are infinitely many, but there should be an obvious choice - I speculate the rule that chooses the canonical extension would be something like there will be an $n^{th}$ derivative of $r$ with respect to $theta$ that's fixed or monotonic at any given point on the line.
What's the function $h^times(x):Bbb RtoBbb C$, and what's $frac{dr}{dtheta}$ as a function of either $x$ or $h^times(x)$?
It may be worth pointing out that $f(x) mapsto theta(h(x))$ sends (at least a subset of) the Prufer 2-group in its dyadic form to its 2nd roots of unity form.
complex-analysis algebra-precalculus functions complex-numbers
asked Jan 6 at 16:04
user334732user334732
4,26311240
$endgroup$
add a comment |
$begingroup$
The first 250 terms of an $Bbb N$-indexed sequence in $Bbb C$, are shown here.
What's the function $h(x):Bbb RtoBbb C$ of which this is a subset, and what's $frac{dr}{dtheta}$ as a function of either $x$ or $h(x)$ (setting $re^{ipitheta}=h(x)$)?
The sequence $S_n$ is given by the map $S_n=h(x):xin2Bbb N-1$, where $h$ is defined:
$f:2Bbb N-1tofrac12+Bbb Z[frac12]/frac12Bbb Z$
$f(x)=frac x{2^p}inleft[frac12,1right)$
$h(x)=xcdot exp{(4pi icdot f(x))}$
The function's effectively a sequence of segments glued together at $x=1, Im(h(x))=0$.
It seems obvious there must be a fairly simple continuous function $h^times:Bbb RtoBbb C$ of which this is a subset (give or take possible singularities at $0,1$). What is it? Obviously there are infinitely many, but there should be an obvious choice - I speculate the rule that chooses the canonical extension would be something like there will be an $n^{th}$ derivative of $r$ with respect to $theta$ that's fixed or monotonic at any given point on the line.
What's the function $h^times(x):Bbb RtoBbb C$, and what's $frac{dr}{dtheta}$ as a function of either $x$ or $h^times(x)$?
It may be worth pointing out that $f(x) mapsto theta(h(x))$ sends (at least a subset of) the Prufer 2-group in its dyadic form to its 2nd roots of unity form.
complex-analysis algebra-precalculus functions complex-numbers
asked Jan 6 at 16:04
user334732user334732
4,26311240
$endgroup$
The first 250 terms of an $Bbb N$-indexed sequence in $Bbb C$, are shown here.
What's the function $h(x):Bbb RtoBbb C$ of which this is a subset, and what's $frac{dr}{dtheta}$ as a function of either $x$ or $h(x)$ (setting $re^{ipitheta}=h(x)$)?
The sequence $S_n$ is given by the map $S_n=h(x):xin2Bbb N-1$, where $h$ is defined:
$f:2Bbb N-1tofrac12+Bbb Z[frac12]/frac12Bbb Z$
$f(x)=frac x{2^p}inleft[frac12,1right)$
$h(x)=xcdot exp{(4pi icdot f(x))}$
The function's effectively a sequence of segments glued together at $x=1, Im(h(x))=0$.
It seems obvious there must be a fairly simple continuous function $h^times:Bbb RtoBbb C$ of which this is a subset (give or take possible singularities at $0,1$). What is it? Obviously there are infinitely many, but there should be an obvious choice - I speculate the rule that chooses the canonical extension would be something like there will be an $n^{th}$ derivative of $r$ with respect to $theta$ that's fixed or monotonic at any given point on the line.
What's the function $h^times(x):Bbb RtoBbb C$, and what's $frac{dr}{dtheta}$ as a function of either $x$ or $h^times(x)$?
It may be worth pointing out that $f(x) mapsto theta(h(x))$ sends (at least a subset of) the Prufer 2-group in its dyadic form to its 2nd roots of unity form.
complex-analysis algebra-precalculus functions complex-numbers
complex-analysis algebra-precalculus functions complex-numbers
asked Jan 6 at 16:04
user334732user334732
4,26311240
asked Jan 6 at 16:04
user334732user334732
4,26311240
edited Jan 6 at 16:30
user334732
asked Jan 6 at 16:04
user334732user334732
4,26311240
asked Jan 6 at 16:04
user334732user334732
4,26311240
asked Jan 6 at 16:04
user334732user334732
4,26311240
4,26311240
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The function
$$h(x) = 2^xexpleft(2pi i cdot2^{x-lfloor xrfloor}right)$$
is continuous, relatively simple, and obviously fits the sequence at $x = log_2(m)$. Since $|h(x)| = 2^x$ and $mathrm{Arg}[h(x)] = 2pi 2^{x-lfloor x rfloor}$,
$$
frac{dr}{dtheta} = frac{dr/dx}{dtheta/dx} = frac{2^xln 2}{2pi 2^{x-lfloor x rfloor} ln 2} = frac{2^{lfloor x rfloor}}{2pi}
$$
As an aside, this function seems to be an approximation of the logarithmic spiral
$$
z(x) = 2^xexp(2pi i x),
$$
which has
$$
frac{dr}{dtheta} = frac{2^x}{2pi}.
$$
answered Jan 6 at 16:53
eyeballfrogeyeballfrog
6,103629
$endgroup$
$begingroup$
Is this your function $z(x)$?: desmos.com/calculator/r8xi2zvogc or have I translated it wrongly?
$endgroup$
– user334732
Jan 6 at 17:01
$begingroup$
Do I understand correctly; the crucial part of this is that $2^{x}$ is even for whole numbers $x$ so $2^{x-lfloor xrfloor}cong2^{-lfloor xrfloor}$ - is that right?
$endgroup$
– user334732
Jan 6 at 17:06
1
$begingroup$
@user334732 Yeah, you got the translation wrong. $h(x)$ is parameterized so that it matches $S(m)$ when $x = log_2(m)$. So you should have $(m cos[2pi log_2(m)],m sin[2pi log_2(m)])$ if you want $z(x)$ to look like your function.
$endgroup$
– eyeballfrog
Jan 6 at 17:20
$begingroup$
thanks. I was sure it would be me! I'll look again at that.
$endgroup$
– user334732
Jan 6 at 17:31
$begingroup$
Right, thanks to your help I have simply $z(x)=xcdot x^{(2pi i/log 2)}$ which I think preserves all the properties I want, and has some added properties w.r.t. continuity etc. Does that look right?
$endgroup$
– user334732
Jan 6 at 23:02
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The function
$$h(x) = 2^xexpleft(2pi i cdot2^{x-lfloor xrfloor}right)$$
is continuous, relatively simple, and obviously fits the sequence at $x = log_2(m)$. Since $|h(x)| = 2^x$ and $mathrm{Arg}[h(x)] = 2pi 2^{x-lfloor x rfloor}$,
$$
frac{dr}{dtheta} = frac{dr/dx}{dtheta/dx} = frac{2^xln 2}{2pi 2^{x-lfloor x rfloor} ln 2} = frac{2^{lfloor x rfloor}}{2pi}
$$
As an aside, this function seems to be an approximation of the logarithmic spiral
$$
z(x) = 2^xexp(2pi i x),
$$
which has
$$
frac{dr}{dtheta} = frac{2^x}{2pi}.
$$
answered Jan 6 at 16:53
eyeballfrogeyeballfrog
6,103629
$endgroup$
$begingroup$
Is this your function $z(x)$?: desmos.com/calculator/r8xi2zvogc or have I translated it wrongly?
$endgroup$
– user334732
Jan 6 at 17:01
$begingroup$
Do I understand correctly; the crucial part of this is that $2^{x}$ is even for whole numbers $x$ so $2^{x-lfloor xrfloor}cong2^{-lfloor xrfloor}$ - is that right?
$endgroup$
– user334732
Jan 6 at 17:06
1
$begingroup$
@user334732 Yeah, you got the translation wrong. $h(x)$ is parameterized so that it matches $S(m)$ when $x = log_2(m)$. So you should have $(m cos[2pi log_2(m)],m sin[2pi log_2(m)])$ if you want $z(x)$ to look like your function.
$endgroup$
– eyeballfrog
Jan 6 at 17:20
$begingroup$
thanks. I was sure it would be me! I'll look again at that.
$endgroup$
– user334732
Jan 6 at 17:31
$begingroup$
Right, thanks to your help I have simply $z(x)=xcdot x^{(2pi i/log 2)}$ which I think preserves all the properties I want, and has some added properties w.r.t. continuity etc. Does that look right?
$endgroup$
– user334732
Jan 6 at 23:02
add a comment |
$begingroup$
The function
$$h(x) = 2^xexpleft(2pi i cdot2^{x-lfloor xrfloor}right)$$
is continuous, relatively simple, and obviously fits the sequence at $x = log_2(m)$. Since $|h(x)| = 2^x$ and $mathrm{Arg}[h(x)] = 2pi 2^{x-lfloor x rfloor}$,
$$
frac{dr}{dtheta} = frac{dr/dx}{dtheta/dx} = frac{2^xln 2}{2pi 2^{x-lfloor x rfloor} ln 2} = frac{2^{lfloor x rfloor}}{2pi}
$$
As an aside, this function seems to be an approximation of the logarithmic spiral
$$
z(x) = 2^xexp(2pi i x),
$$
which has
$$
frac{dr}{dtheta} = frac{2^x}{2pi}.
$$
answered Jan 6 at 16:53
eyeballfrogeyeballfrog
6,103629
$endgroup$
$begingroup$
Is this your function $z(x)$?: desmos.com/calculator/r8xi2zvogc or have I translated it wrongly?
$endgroup$
– user334732
Jan 6 at 17:01
$begingroup$
Do I understand correctly; the crucial part of this is that $2^{x}$ is even for whole numbers $x$ so $2^{x-lfloor xrfloor}cong2^{-lfloor xrfloor}$ - is that right?
$endgroup$
– user334732
Jan 6 at 17:06
1
$begingroup$
@user334732 Yeah, you got the translation wrong. $h(x)$ is parameterized so that it matches $S(m)$ when $x = log_2(m)$. So you should have $(m cos[2pi log_2(m)],m sin[2pi log_2(m)])$ if you want $z(x)$ to look like your function.
$endgroup$
– eyeballfrog
Jan 6 at 17:20
$begingroup$
thanks. I was sure it would be me! I'll look again at that.
$endgroup$
– user334732
Jan 6 at 17:31
$begingroup$
Right, thanks to your help I have simply $z(x)=xcdot x^{(2pi i/log 2)}$ which I think preserves all the properties I want, and has some added properties w.r.t. continuity etc. Does that look right?
$endgroup$
– user334732
Jan 6 at 23:02
add a comment |
$begingroup$
The function
$$h(x) = 2^xexpleft(2pi i cdot2^{x-lfloor xrfloor}right)$$
is continuous, relatively simple, and obviously fits the sequence at $x = log_2(m)$. Since $|h(x)| = 2^x$ and $mathrm{Arg}[h(x)] = 2pi 2^{x-lfloor x rfloor}$,
$$
frac{dr}{dtheta} = frac{dr/dx}{dtheta/dx} = frac{2^xln 2}{2pi 2^{x-lfloor x rfloor} ln 2} = frac{2^{lfloor x rfloor}}{2pi}
$$
As an aside, this function seems to be an approximation of the logarithmic spiral
$$
z(x) = 2^xexp(2pi i x),
$$
which has
$$
frac{dr}{dtheta} = frac{2^x}{2pi}.
$$
answered Jan 6 at 16:53
eyeballfrogeyeballfrog
6,103629
$endgroup$
The function
$$h(x) = 2^xexpleft(2pi i cdot2^{x-lfloor xrfloor}right)$$
is continuous, relatively simple, and obviously fits the sequence at $x = log_2(m)$. Since $|h(x)| = 2^x$ and $mathrm{Arg}[h(x)] = 2pi 2^{x-lfloor x rfloor}$,
$$
frac{dr}{dtheta} = frac{dr/dx}{dtheta/dx} = frac{2^xln 2}{2pi 2^{x-lfloor x rfloor} ln 2} = frac{2^{lfloor x rfloor}}{2pi}
$$
As an aside, this function seems to be an approximation of the logarithmic spiral
$$
z(x) = 2^xexp(2pi i x),
$$
which has
$$
frac{dr}{dtheta} = frac{2^x}{2pi}.
$$
answered Jan 6 at 16:53
eyeballfrogeyeballfrog
6,103629
answered Jan 6 at 16:53
eyeballfrogeyeballfrog
6,103629
answered Jan 6 at 16:53
eyeballfrogeyeballfrog
6,103629
answered Jan 6 at 16:53
eyeballfrogeyeballfrog
6,103629
6,103629
$begingroup$
Is this your function $z(x)$?: desmos.com/calculator/r8xi2zvogc or have I translated it wrongly?
$endgroup$
– user334732
Jan 6 at 17:01
$begingroup$
Do I understand correctly; the crucial part of this is that $2^{x}$ is even for whole numbers $x$ so $2^{x-lfloor xrfloor}cong2^{-lfloor xrfloor}$ - is that right?
$endgroup$
– user334732
Jan 6 at 17:06
1
$begingroup$
@user334732 Yeah, you got the translation wrong. $h(x)$ is parameterized so that it matches $S(m)$ when $x = log_2(m)$. So you should have $(m cos[2pi log_2(m)],m sin[2pi log_2(m)])$ if you want $z(x)$ to look like your function.
$endgroup$
– eyeballfrog
Jan 6 at 17:20
$begingroup$
thanks. I was sure it would be me! I'll look again at that.
$endgroup$
– user334732
Jan 6 at 17:31
$begingroup$
Right, thanks to your help I have simply $z(x)=xcdot x^{(2pi i/log 2)}$ which I think preserves all the properties I want, and has some added properties w.r.t. continuity etc. Does that look right?
$endgroup$
– user334732
Jan 6 at 23:02
add a comment |
$begingroup$
Is this your function $z(x)$?: desmos.com/calculator/r8xi2zvogc or have I translated it wrongly?
$endgroup$
– user334732
Jan 6 at 17:01
$begingroup$
Do I understand correctly; the crucial part of this is that $2^{x}$ is even for whole numbers $x$ so $2^{x-lfloor xrfloor}cong2^{-lfloor xrfloor}$ - is that right?
$endgroup$
– user334732
Jan 6 at 17:06
1
$begingroup$
@user334732 Yeah, you got the translation wrong. $h(x)$ is parameterized so that it matches $S(m)$ when $x = log_2(m)$. So you should have $(m cos[2pi log_2(m)],m sin[2pi log_2(m)])$ if you want $z(x)$ to look like your function.
$endgroup$
– eyeballfrog
Jan 6 at 17:20
$begingroup$
thanks. I was sure it would be me! I'll look again at that.
$endgroup$
– user334732
Jan 6 at 17:31
$begingroup$
Right, thanks to your help I have simply $z(x)=xcdot x^{(2pi i/log 2)}$ which I think preserves all the properties I want, and has some added properties w.r.t. continuity etc. Does that look right?
$endgroup$
– user334732
Jan 6 at 23:02
$begingroup$
Is this your function $z(x)$?: desmos.com/calculator/r8xi2zvogc or have I translated it wrongly?
$endgroup$
– user334732
Jan 6 at 17:01
$begingroup$
Is this your function $z(x)$?: desmos.com/calculator/r8xi2zvogc or have I translated it wrongly?
$endgroup$
– user334732
Jan 6 at 17:01
$begingroup$
Do I understand correctly; the crucial part of this is that $2^{x}$ is even for whole numbers $x$ so $2^{x-lfloor xrfloor}cong2^{-lfloor xrfloor}$ - is that right?
$endgroup$
– user334732
Jan 6 at 17:06
$begingroup$
Do I understand correctly; the crucial part of this is that $2^{x}$ is even for whole numbers $x$ so $2^{x-lfloor xrfloor}cong2^{-lfloor xrfloor}$ - is that right?
$endgroup$
– user334732
Jan 6 at 17:06
1
1
$begingroup$
@user334732 Yeah, you got the translation wrong. $h(x)$ is parameterized so that it matches $S(m)$ when $x = log_2(m)$. So you should have $(m cos[2pi log_2(m)],m sin[2pi log_2(m)])$ if you want $z(x)$ to look like your function.
$endgroup$
– eyeballfrog
Jan 6 at 17:20
$begingroup$
@user334732 Yeah, you got the translation wrong. $h(x)$ is parameterized so that it matches $S(m)$ when $x = log_2(m)$. So you should have $(m cos[2pi log_2(m)],m sin[2pi log_2(m)])$ if you want $z(x)$ to look like your function.
$endgroup$
– eyeballfrog
Jan 6 at 17:20
$begingroup$
thanks. I was sure it would be me! I'll look again at that.
$endgroup$
– user334732
Jan 6 at 17:31
$begingroup$
thanks. I was sure it would be me! I'll look again at that.
$endgroup$
– user334732
Jan 6 at 17:31
$begingroup$
Right, thanks to your help I have simply $z(x)=xcdot x^{(2pi i/log 2)}$ which I think preserves all the properties I want, and has some added properties w.r.t. continuity etc. Does that look right?
$endgroup$
– user334732
Jan 6 at 23:02
$begingroup$
Right, thanks to your help I have simply $z(x)=xcdot x^{(2pi i/log 2)}$ which I think preserves all the properties I want, and has some added properties w.r.t. continuity etc. Does that look right?
$endgroup$
– user334732
Jan 6 at 23:02
add a comment |
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