Every linearly-ordered real-parametrized family of asymptotic classes is nowhere dense?












3












$begingroup$


Upon writing this post, I had the following natural conjecture:
$
defnn{mathbb{N}}
defrr{mathbb{R}}
$




Take any family ${ C_r : rinrr }$ of functions from $nn$ to $rr$ where $0 < C_s(n) ll C_t(n)$ as $n to infty$ for every $s,t in rr$ such that $s < t$. Then for every $r in rr$ there is some function $D_r$ from $nn$ to $rr$ such that for every $ε in rr^+$ we have $C_r(n) ll D_r(n) ll C_{r+ε}(n)$ as $n to infty$. (By considering the reciprocal family this immediately implies the other side as well.)




Intuitively, I claim that there will always be some asymptotic class that falls between the cracks of any linearly-ordered real-parametrized family of asymptotic classes. I think that this is equivalent to the stronger claim of nowhere-denseness, but I am not sure.



For example:




  • If $C_r(n) = n^r$ for every $r in rr$ and $r in nn$, then $D_r$ where $D_r(n) = n^r·ln(n)$ for every $n in nn$ provides a suitable witness, since $n^0 ll ln(n) ll n^ε$ as $n to infty$ for every $ε in rr^+$.


  • If $C_r(n) = r^n$ for every $r in rr$ and $r in nn$, then $D_r$ where $D_r(n) = n^r·n$ for every $n in nn$ provides a suitable witness, since $r^0 ll n ll (1+frac{ε}{r})^n$ as $n to infty$ for every $ε in rr^+$.



It is easy to show that $C$ has strict upper and lower bounds since $C_{-n}(n) ll C_r(n) ll C_n(n)$ as $n to infty$ for every $r in rr$. But I cannot find a general way to construct 'in-between' functions. I know that $C_{r+frac1n}(n) ll C_{r+ε}(n)$ as $n to infty$ for every $r in rr$ and $ε in rr^+$, but it is possible that $C_{r+frac1n}(n) sim C_r(n)$, as is indeed the case in both the above examples.



Is my conjecture true? If so, it suffices to prove that $C_0(n) ll D_0(n) ll C_ε(n)$ as $n to infty$ for some function $D_0$ from $nn$ to $rr$, since the general claim follows by translation. If not, it suffices to prove that for some family $C$ there is no such $D_0$, again due to translation.










share|cite|improve this question









$endgroup$












  • $begingroup$
    You'll want to look at Gordon Fisher's 1981 paper The infinite and infinitesimal quantities of du Bois-Reymond and their reception and Hausdorff gaps (see this survey).
    $endgroup$
    – Dave L. Renfro
    Jan 16 '18 at 17:56










  • $begingroup$
    @DaveL.Renfro: If you have an answer to my question, could you kindly post an answer that doesn't rely on links or articles hidden behind a great paywall? Thank you!
    $endgroup$
    – user21820
    Jan 17 '18 at 2:02










  • $begingroup$
    I've been too busy at work lately (non-academic) to spend time trying to dig into this topic, but maybe the following will help. First, even if you can't visit a library to access the Gordon Fisher paper (my copy is a photocopy I made back in the late 1980s; I don't have online access either), the title of the paper itself provides words and phrases you can use in searches for relevant work and for books/papers that cite Fisher's paper. Second, Math Origins: Orders of Growth at the MAA website should be useful.
    $endgroup$
    – Dave L. Renfro
    Jan 23 '18 at 11:11












  • $begingroup$
    @DaveL.Renfro: Thanks for that! I'm too busy these few weeks to have time for mathematical research, but I'll try to get to them eventually. I took a look at the last link you provided, and I already know everything on that page, I think.
    $endgroup$
    – user21820
    Jan 23 '18 at 14:32


















3












$begingroup$


Upon writing this post, I had the following natural conjecture:
$
defnn{mathbb{N}}
defrr{mathbb{R}}
$




Take any family ${ C_r : rinrr }$ of functions from $nn$ to $rr$ where $0 < C_s(n) ll C_t(n)$ as $n to infty$ for every $s,t in rr$ such that $s < t$. Then for every $r in rr$ there is some function $D_r$ from $nn$ to $rr$ such that for every $ε in rr^+$ we have $C_r(n) ll D_r(n) ll C_{r+ε}(n)$ as $n to infty$. (By considering the reciprocal family this immediately implies the other side as well.)




Intuitively, I claim that there will always be some asymptotic class that falls between the cracks of any linearly-ordered real-parametrized family of asymptotic classes. I think that this is equivalent to the stronger claim of nowhere-denseness, but I am not sure.



For example:




  • If $C_r(n) = n^r$ for every $r in rr$ and $r in nn$, then $D_r$ where $D_r(n) = n^r·ln(n)$ for every $n in nn$ provides a suitable witness, since $n^0 ll ln(n) ll n^ε$ as $n to infty$ for every $ε in rr^+$.


  • If $C_r(n) = r^n$ for every $r in rr$ and $r in nn$, then $D_r$ where $D_r(n) = n^r·n$ for every $n in nn$ provides a suitable witness, since $r^0 ll n ll (1+frac{ε}{r})^n$ as $n to infty$ for every $ε in rr^+$.



It is easy to show that $C$ has strict upper and lower bounds since $C_{-n}(n) ll C_r(n) ll C_n(n)$ as $n to infty$ for every $r in rr$. But I cannot find a general way to construct 'in-between' functions. I know that $C_{r+frac1n}(n) ll C_{r+ε}(n)$ as $n to infty$ for every $r in rr$ and $ε in rr^+$, but it is possible that $C_{r+frac1n}(n) sim C_r(n)$, as is indeed the case in both the above examples.



Is my conjecture true? If so, it suffices to prove that $C_0(n) ll D_0(n) ll C_ε(n)$ as $n to infty$ for some function $D_0$ from $nn$ to $rr$, since the general claim follows by translation. If not, it suffices to prove that for some family $C$ there is no such $D_0$, again due to translation.










share|cite|improve this question









$endgroup$












  • $begingroup$
    You'll want to look at Gordon Fisher's 1981 paper The infinite and infinitesimal quantities of du Bois-Reymond and their reception and Hausdorff gaps (see this survey).
    $endgroup$
    – Dave L. Renfro
    Jan 16 '18 at 17:56










  • $begingroup$
    @DaveL.Renfro: If you have an answer to my question, could you kindly post an answer that doesn't rely on links or articles hidden behind a great paywall? Thank you!
    $endgroup$
    – user21820
    Jan 17 '18 at 2:02










  • $begingroup$
    I've been too busy at work lately (non-academic) to spend time trying to dig into this topic, but maybe the following will help. First, even if you can't visit a library to access the Gordon Fisher paper (my copy is a photocopy I made back in the late 1980s; I don't have online access either), the title of the paper itself provides words and phrases you can use in searches for relevant work and for books/papers that cite Fisher's paper. Second, Math Origins: Orders of Growth at the MAA website should be useful.
    $endgroup$
    – Dave L. Renfro
    Jan 23 '18 at 11:11












  • $begingroup$
    @DaveL.Renfro: Thanks for that! I'm too busy these few weeks to have time for mathematical research, but I'll try to get to them eventually. I took a look at the last link you provided, and I already know everything on that page, I think.
    $endgroup$
    – user21820
    Jan 23 '18 at 14:32
















3












3








3


1



$begingroup$


Upon writing this post, I had the following natural conjecture:
$
defnn{mathbb{N}}
defrr{mathbb{R}}
$




Take any family ${ C_r : rinrr }$ of functions from $nn$ to $rr$ where $0 < C_s(n) ll C_t(n)$ as $n to infty$ for every $s,t in rr$ such that $s < t$. Then for every $r in rr$ there is some function $D_r$ from $nn$ to $rr$ such that for every $ε in rr^+$ we have $C_r(n) ll D_r(n) ll C_{r+ε}(n)$ as $n to infty$. (By considering the reciprocal family this immediately implies the other side as well.)




Intuitively, I claim that there will always be some asymptotic class that falls between the cracks of any linearly-ordered real-parametrized family of asymptotic classes. I think that this is equivalent to the stronger claim of nowhere-denseness, but I am not sure.



For example:




  • If $C_r(n) = n^r$ for every $r in rr$ and $r in nn$, then $D_r$ where $D_r(n) = n^r·ln(n)$ for every $n in nn$ provides a suitable witness, since $n^0 ll ln(n) ll n^ε$ as $n to infty$ for every $ε in rr^+$.


  • If $C_r(n) = r^n$ for every $r in rr$ and $r in nn$, then $D_r$ where $D_r(n) = n^r·n$ for every $n in nn$ provides a suitable witness, since $r^0 ll n ll (1+frac{ε}{r})^n$ as $n to infty$ for every $ε in rr^+$.



It is easy to show that $C$ has strict upper and lower bounds since $C_{-n}(n) ll C_r(n) ll C_n(n)$ as $n to infty$ for every $r in rr$. But I cannot find a general way to construct 'in-between' functions. I know that $C_{r+frac1n}(n) ll C_{r+ε}(n)$ as $n to infty$ for every $r in rr$ and $ε in rr^+$, but it is possible that $C_{r+frac1n}(n) sim C_r(n)$, as is indeed the case in both the above examples.



Is my conjecture true? If so, it suffices to prove that $C_0(n) ll D_0(n) ll C_ε(n)$ as $n to infty$ for some function $D_0$ from $nn$ to $rr$, since the general claim follows by translation. If not, it suffices to prove that for some family $C$ there is no such $D_0$, again due to translation.










share|cite|improve this question









$endgroup$




Upon writing this post, I had the following natural conjecture:
$
defnn{mathbb{N}}
defrr{mathbb{R}}
$




Take any family ${ C_r : rinrr }$ of functions from $nn$ to $rr$ where $0 < C_s(n) ll C_t(n)$ as $n to infty$ for every $s,t in rr$ such that $s < t$. Then for every $r in rr$ there is some function $D_r$ from $nn$ to $rr$ such that for every $ε in rr^+$ we have $C_r(n) ll D_r(n) ll C_{r+ε}(n)$ as $n to infty$. (By considering the reciprocal family this immediately implies the other side as well.)




Intuitively, I claim that there will always be some asymptotic class that falls between the cracks of any linearly-ordered real-parametrized family of asymptotic classes. I think that this is equivalent to the stronger claim of nowhere-denseness, but I am not sure.



For example:




  • If $C_r(n) = n^r$ for every $r in rr$ and $r in nn$, then $D_r$ where $D_r(n) = n^r·ln(n)$ for every $n in nn$ provides a suitable witness, since $n^0 ll ln(n) ll n^ε$ as $n to infty$ for every $ε in rr^+$.


  • If $C_r(n) = r^n$ for every $r in rr$ and $r in nn$, then $D_r$ where $D_r(n) = n^r·n$ for every $n in nn$ provides a suitable witness, since $r^0 ll n ll (1+frac{ε}{r})^n$ as $n to infty$ for every $ε in rr^+$.



It is easy to show that $C$ has strict upper and lower bounds since $C_{-n}(n) ll C_r(n) ll C_n(n)$ as $n to infty$ for every $r in rr$. But I cannot find a general way to construct 'in-between' functions. I know that $C_{r+frac1n}(n) ll C_{r+ε}(n)$ as $n to infty$ for every $r in rr$ and $ε in rr^+$, but it is possible that $C_{r+frac1n}(n) sim C_r(n)$, as is indeed the case in both the above examples.



Is my conjecture true? If so, it suffices to prove that $C_0(n) ll D_0(n) ll C_ε(n)$ as $n to infty$ for some function $D_0$ from $nn$ to $rr$, since the general claim follows by translation. If not, it suffices to prove that for some family $C$ there is no such $D_0$, again due to translation.







real-analysis general-topology asymptotics order-theory






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asked Jan 16 '18 at 11:53









user21820user21820

40.2k544162




40.2k544162












  • $begingroup$
    You'll want to look at Gordon Fisher's 1981 paper The infinite and infinitesimal quantities of du Bois-Reymond and their reception and Hausdorff gaps (see this survey).
    $endgroup$
    – Dave L. Renfro
    Jan 16 '18 at 17:56










  • $begingroup$
    @DaveL.Renfro: If you have an answer to my question, could you kindly post an answer that doesn't rely on links or articles hidden behind a great paywall? Thank you!
    $endgroup$
    – user21820
    Jan 17 '18 at 2:02










  • $begingroup$
    I've been too busy at work lately (non-academic) to spend time trying to dig into this topic, but maybe the following will help. First, even if you can't visit a library to access the Gordon Fisher paper (my copy is a photocopy I made back in the late 1980s; I don't have online access either), the title of the paper itself provides words and phrases you can use in searches for relevant work and for books/papers that cite Fisher's paper. Second, Math Origins: Orders of Growth at the MAA website should be useful.
    $endgroup$
    – Dave L. Renfro
    Jan 23 '18 at 11:11












  • $begingroup$
    @DaveL.Renfro: Thanks for that! I'm too busy these few weeks to have time for mathematical research, but I'll try to get to them eventually. I took a look at the last link you provided, and I already know everything on that page, I think.
    $endgroup$
    – user21820
    Jan 23 '18 at 14:32




















  • $begingroup$
    You'll want to look at Gordon Fisher's 1981 paper The infinite and infinitesimal quantities of du Bois-Reymond and their reception and Hausdorff gaps (see this survey).
    $endgroup$
    – Dave L. Renfro
    Jan 16 '18 at 17:56










  • $begingroup$
    @DaveL.Renfro: If you have an answer to my question, could you kindly post an answer that doesn't rely on links or articles hidden behind a great paywall? Thank you!
    $endgroup$
    – user21820
    Jan 17 '18 at 2:02










  • $begingroup$
    I've been too busy at work lately (non-academic) to spend time trying to dig into this topic, but maybe the following will help. First, even if you can't visit a library to access the Gordon Fisher paper (my copy is a photocopy I made back in the late 1980s; I don't have online access either), the title of the paper itself provides words and phrases you can use in searches for relevant work and for books/papers that cite Fisher's paper. Second, Math Origins: Orders of Growth at the MAA website should be useful.
    $endgroup$
    – Dave L. Renfro
    Jan 23 '18 at 11:11












  • $begingroup$
    @DaveL.Renfro: Thanks for that! I'm too busy these few weeks to have time for mathematical research, but I'll try to get to them eventually. I took a look at the last link you provided, and I already know everything on that page, I think.
    $endgroup$
    – user21820
    Jan 23 '18 at 14:32


















$begingroup$
You'll want to look at Gordon Fisher's 1981 paper The infinite and infinitesimal quantities of du Bois-Reymond and their reception and Hausdorff gaps (see this survey).
$endgroup$
– Dave L. Renfro
Jan 16 '18 at 17:56




$begingroup$
You'll want to look at Gordon Fisher's 1981 paper The infinite and infinitesimal quantities of du Bois-Reymond and their reception and Hausdorff gaps (see this survey).
$endgroup$
– Dave L. Renfro
Jan 16 '18 at 17:56












$begingroup$
@DaveL.Renfro: If you have an answer to my question, could you kindly post an answer that doesn't rely on links or articles hidden behind a great paywall? Thank you!
$endgroup$
– user21820
Jan 17 '18 at 2:02




$begingroup$
@DaveL.Renfro: If you have an answer to my question, could you kindly post an answer that doesn't rely on links or articles hidden behind a great paywall? Thank you!
$endgroup$
– user21820
Jan 17 '18 at 2:02












$begingroup$
I've been too busy at work lately (non-academic) to spend time trying to dig into this topic, but maybe the following will help. First, even if you can't visit a library to access the Gordon Fisher paper (my copy is a photocopy I made back in the late 1980s; I don't have online access either), the title of the paper itself provides words and phrases you can use in searches for relevant work and for books/papers that cite Fisher's paper. Second, Math Origins: Orders of Growth at the MAA website should be useful.
$endgroup$
– Dave L. Renfro
Jan 23 '18 at 11:11






$begingroup$
I've been too busy at work lately (non-academic) to spend time trying to dig into this topic, but maybe the following will help. First, even if you can't visit a library to access the Gordon Fisher paper (my copy is a photocopy I made back in the late 1980s; I don't have online access either), the title of the paper itself provides words and phrases you can use in searches for relevant work and for books/papers that cite Fisher's paper. Second, Math Origins: Orders of Growth at the MAA website should be useful.
$endgroup$
– Dave L. Renfro
Jan 23 '18 at 11:11














$begingroup$
@DaveL.Renfro: Thanks for that! I'm too busy these few weeks to have time for mathematical research, but I'll try to get to them eventually. I took a look at the last link you provided, and I already know everything on that page, I think.
$endgroup$
– user21820
Jan 23 '18 at 14:32






$begingroup$
@DaveL.Renfro: Thanks for that! I'm too busy these few weeks to have time for mathematical research, but I'll try to get to them eventually. I took a look at the last link you provided, and I already know everything on that page, I think.
$endgroup$
– user21820
Jan 23 '18 at 14:32












1 Answer
1






active

oldest

votes


















2





+50







$begingroup$

You can diagonalize to find a sequence asymptotically bigger than $C_{r}$ but asymptotically smaller than $C_{r+epsilon}$ for any $epsilon>0$.



Fix $r$. Define $d_r(n)$ to be the greatest positive integer such that $C_{r+1/j}(n')/C_r(n')ge d_r(n)$ for all positive integer $j le d_r(n)$ and $n'ge n$, or $d_r(n)=1$ if none exists. Take $D_r(n)=C_r(n)d_r(n)$.



For any positive integer $j$, either $j le d_r(n)$ or $d_r(n) le j$, so by construction $D_r(n) le C_{r+1/j}(n)$ or $D_r(n) le j·C_r(n)$ as $n to infty$. Therefore $D_rll C_{r+epsilon}$ for any $epsilon>0.$



I claim that $d_r(n)toinfty$ as $ntoinfty$.
For any fixed positive integer $Delta$, from the conjunction of $C_rll C_{r+1/j}$ for positive integer $j le Delta$, there is a large enough $n$ such that for all $jleDelta$ and $n'ge n$ we have $C_{r+1/j}(n')/C_r(n')geDelta$. For this $n$ we have $d_r(n) ge Delta$. And $d_r$ is clearly non-decreasing. So $d_r$ tends to $infty$. Therefore $C_rll D_r$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks! I edited to add a bit of missing detail. =)
    $endgroup$
    – user21820
    Dec 29 '18 at 17:02












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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









2





+50







$begingroup$

You can diagonalize to find a sequence asymptotically bigger than $C_{r}$ but asymptotically smaller than $C_{r+epsilon}$ for any $epsilon>0$.



Fix $r$. Define $d_r(n)$ to be the greatest positive integer such that $C_{r+1/j}(n')/C_r(n')ge d_r(n)$ for all positive integer $j le d_r(n)$ and $n'ge n$, or $d_r(n)=1$ if none exists. Take $D_r(n)=C_r(n)d_r(n)$.



For any positive integer $j$, either $j le d_r(n)$ or $d_r(n) le j$, so by construction $D_r(n) le C_{r+1/j}(n)$ or $D_r(n) le j·C_r(n)$ as $n to infty$. Therefore $D_rll C_{r+epsilon}$ for any $epsilon>0.$



I claim that $d_r(n)toinfty$ as $ntoinfty$.
For any fixed positive integer $Delta$, from the conjunction of $C_rll C_{r+1/j}$ for positive integer $j le Delta$, there is a large enough $n$ such that for all $jleDelta$ and $n'ge n$ we have $C_{r+1/j}(n')/C_r(n')geDelta$. For this $n$ we have $d_r(n) ge Delta$. And $d_r$ is clearly non-decreasing. So $d_r$ tends to $infty$. Therefore $C_rll D_r$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks! I edited to add a bit of missing detail. =)
    $endgroup$
    – user21820
    Dec 29 '18 at 17:02
















2





+50







$begingroup$

You can diagonalize to find a sequence asymptotically bigger than $C_{r}$ but asymptotically smaller than $C_{r+epsilon}$ for any $epsilon>0$.



Fix $r$. Define $d_r(n)$ to be the greatest positive integer such that $C_{r+1/j}(n')/C_r(n')ge d_r(n)$ for all positive integer $j le d_r(n)$ and $n'ge n$, or $d_r(n)=1$ if none exists. Take $D_r(n)=C_r(n)d_r(n)$.



For any positive integer $j$, either $j le d_r(n)$ or $d_r(n) le j$, so by construction $D_r(n) le C_{r+1/j}(n)$ or $D_r(n) le j·C_r(n)$ as $n to infty$. Therefore $D_rll C_{r+epsilon}$ for any $epsilon>0.$



I claim that $d_r(n)toinfty$ as $ntoinfty$.
For any fixed positive integer $Delta$, from the conjunction of $C_rll C_{r+1/j}$ for positive integer $j le Delta$, there is a large enough $n$ such that for all $jleDelta$ and $n'ge n$ we have $C_{r+1/j}(n')/C_r(n')geDelta$. For this $n$ we have $d_r(n) ge Delta$. And $d_r$ is clearly non-decreasing. So $d_r$ tends to $infty$. Therefore $C_rll D_r$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks! I edited to add a bit of missing detail. =)
    $endgroup$
    – user21820
    Dec 29 '18 at 17:02














2





+50







2





+50



2




+50



$begingroup$

You can diagonalize to find a sequence asymptotically bigger than $C_{r}$ but asymptotically smaller than $C_{r+epsilon}$ for any $epsilon>0$.



Fix $r$. Define $d_r(n)$ to be the greatest positive integer such that $C_{r+1/j}(n')/C_r(n')ge d_r(n)$ for all positive integer $j le d_r(n)$ and $n'ge n$, or $d_r(n)=1$ if none exists. Take $D_r(n)=C_r(n)d_r(n)$.



For any positive integer $j$, either $j le d_r(n)$ or $d_r(n) le j$, so by construction $D_r(n) le C_{r+1/j}(n)$ or $D_r(n) le j·C_r(n)$ as $n to infty$. Therefore $D_rll C_{r+epsilon}$ for any $epsilon>0.$



I claim that $d_r(n)toinfty$ as $ntoinfty$.
For any fixed positive integer $Delta$, from the conjunction of $C_rll C_{r+1/j}$ for positive integer $j le Delta$, there is a large enough $n$ such that for all $jleDelta$ and $n'ge n$ we have $C_{r+1/j}(n')/C_r(n')geDelta$. For this $n$ we have $d_r(n) ge Delta$. And $d_r$ is clearly non-decreasing. So $d_r$ tends to $infty$. Therefore $C_rll D_r$.






share|cite|improve this answer











$endgroup$



You can diagonalize to find a sequence asymptotically bigger than $C_{r}$ but asymptotically smaller than $C_{r+epsilon}$ for any $epsilon>0$.



Fix $r$. Define $d_r(n)$ to be the greatest positive integer such that $C_{r+1/j}(n')/C_r(n')ge d_r(n)$ for all positive integer $j le d_r(n)$ and $n'ge n$, or $d_r(n)=1$ if none exists. Take $D_r(n)=C_r(n)d_r(n)$.



For any positive integer $j$, either $j le d_r(n)$ or $d_r(n) le j$, so by construction $D_r(n) le C_{r+1/j}(n)$ or $D_r(n) le j·C_r(n)$ as $n to infty$. Therefore $D_rll C_{r+epsilon}$ for any $epsilon>0.$



I claim that $d_r(n)toinfty$ as $ntoinfty$.
For any fixed positive integer $Delta$, from the conjunction of $C_rll C_{r+1/j}$ for positive integer $j le Delta$, there is a large enough $n$ such that for all $jleDelta$ and $n'ge n$ we have $C_{r+1/j}(n')/C_r(n')geDelta$. For this $n$ we have $d_r(n) ge Delta$. And $d_r$ is clearly non-decreasing. So $d_r$ tends to $infty$. Therefore $C_rll D_r$.







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edited Dec 29 '18 at 17:02









user21820

40.2k544162




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answered Dec 28 '18 at 8:57









DapDap

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  • $begingroup$
    Thanks! I edited to add a bit of missing detail. =)
    $endgroup$
    – user21820
    Dec 29 '18 at 17:02


















  • $begingroup$
    Thanks! I edited to add a bit of missing detail. =)
    $endgroup$
    – user21820
    Dec 29 '18 at 17:02
















$begingroup$
Thanks! I edited to add a bit of missing detail. =)
$endgroup$
– user21820
Dec 29 '18 at 17:02




$begingroup$
Thanks! I edited to add a bit of missing detail. =)
$endgroup$
– user21820
Dec 29 '18 at 17:02


















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