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Defining a Neighborhood Around a Point Whose Elements Have Different Units
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Suppose I have some n-dimensional point
$ mathbfx = [x_1,...,x_n] $
where some of the $x_i$ have different units (for example, $x_1$ in Ohms and $x_5$ in seconds). Now if all the elements had the same units I could define a ball $B$ of radius $r$ around $mathbfx$ by
$ B_r(mathbfx) = $
But because the elements of the point do not have the same units, the norm $||mathbfx-mathbfy||$ is not meaningful if it is taken to be the Euclidean norm or something similar.
How could one define a meaningful distance in this space in order to define a ball of a particular radius around a point? My first approach is to consider that each element $x_i$ is bounded by some interval with a minimum $m_i$ and a maximum $M_i$ such that
$x_i in [m_i, M_i]$
and then define a radius for each element by
$r^*_i = frac1(M_i-m_i)r$
Then I could define a "ball" of points around the point in question by
$ B_r(mathbfx) = leq r^*_i$
But I am not sure that this is meaningful. Ultimately I am trying to define a ball of parameters for a dynamical model around a particular parameter, and the model is to be evaluated at each point of this ball (a "parameter hypersphere" instead of a parameter grid). Any help here is appreciated.
general-topology geometry mathematical-modeling computational-mathematics
asked Jul 31 at 20:41
nguzman
295
add a comment |Â
up vote
1
down vote
favorite
Suppose I have some n-dimensional point
$ mathbfx = [x_1,...,x_n] $
where some of the $x_i$ have different units (for example, $x_1$ in Ohms and $x_5$ in seconds). Now if all the elements had the same units I could define a ball $B$ of radius $r$ around $mathbfx$ by
$ B_r(mathbfx) = $
But because the elements of the point do not have the same units, the norm $||mathbfx-mathbfy||$ is not meaningful if it is taken to be the Euclidean norm or something similar.
How could one define a meaningful distance in this space in order to define a ball of a particular radius around a point? My first approach is to consider that each element $x_i$ is bounded by some interval with a minimum $m_i$ and a maximum $M_i$ such that
$x_i in [m_i, M_i]$
and then define a radius for each element by
$r^*_i = frac1(M_i-m_i)r$
Then I could define a "ball" of points around the point in question by
$ B_r(mathbfx) = leq r^*_i$
But I am not sure that this is meaningful. Ultimately I am trying to define a ball of parameters for a dynamical model around a particular parameter, and the model is to be evaluated at each point of this ball (a "parameter hypersphere" instead of a parameter grid). Any help here is appreciated.
general-topology geometry mathematical-modeling computational-mathematics
asked Jul 31 at 20:41
nguzman
295
If I'm interpreting your question correctly, at some point you will need to decide how to compare electrical current to time (how 1 Ohm compares to 1 second). This requires physical knowledge/intuition. I don't think there's any way around that. Once you decide on the scalings, you can nondimensionalise everything and can forget about the physical units/interpretation when calculating distances. I don't understand what you mean by a "parameter hypersphere" and a "parameter grid", are you just saying you don't want to allow all combinations of the variables?
– David
Jul 31 at 22:37
@David A parameter grid is just a colloquial term among people who do computational modeling in my field, called as such because it is usually a grid of points $(x,y)$ of two parameters for a model. I thought that by extending the notion to a hypersphere (since I'm dealing with about nine parameters instead of two and I want parameter combinations equidistant from a point) might make my question clearer to those who do computational modeling.
– nguzman
Aug 1 at 2:44
Once you nondimensionalise, using $x_irightarrow x_i/(M_i-m_i)$ or some other scale (maybe $x_irightarrow(x_i-m_i)/(M_i-m_i)$ so $0leq x_ileq 1$ if you want different quantities weighted evenly), you can use whatever distance metric you want on the $x_i$'s. The hardest part will be working out what value of $r$ to use, and what it means physically.
– David
Aug 1 at 20:11
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Suppose I have some n-dimensional point
$ mathbfx = [x_1,...,x_n] $
where some of the $x_i$ have different units (for example, $x_1$ in Ohms and $x_5$ in seconds). Now if all the elements had the same units I could define a ball $B$ of radius $r$ around $mathbfx$ by
$ B_r(mathbfx) = $
But because the elements of the point do not have the same units, the norm $||mathbfx-mathbfy||$ is not meaningful if it is taken to be the Euclidean norm or something similar.
How could one define a meaningful distance in this space in order to define a ball of a particular radius around a point? My first approach is to consider that each element $x_i$ is bounded by some interval with a minimum $m_i$ and a maximum $M_i$ such that
$x_i in [m_i, M_i]$
and then define a radius for each element by
$r^*_i = frac1(M_i-m_i)r$
Then I could define a "ball" of points around the point in question by
$ B_r(mathbfx) = leq r^*_i$
But I am not sure that this is meaningful. Ultimately I am trying to define a ball of parameters for a dynamical model around a particular parameter, and the model is to be evaluated at each point of this ball (a "parameter hypersphere" instead of a parameter grid). Any help here is appreciated.
general-topology geometry mathematical-modeling computational-mathematics
asked Jul 31 at 20:41
nguzman
295
Suppose I have some n-dimensional point
$ mathbfx = [x_1,...,x_n] $
where some of the $x_i$ have different units (for example, $x_1$ in Ohms and $x_5$ in seconds). Now if all the elements had the same units I could define a ball $B$ of radius $r$ around $mathbfx$ by
$ B_r(mathbfx) = $
But because the elements of the point do not have the same units, the norm $||mathbfx-mathbfy||$ is not meaningful if it is taken to be the Euclidean norm or something similar.
How could one define a meaningful distance in this space in order to define a ball of a particular radius around a point? My first approach is to consider that each element $x_i$ is bounded by some interval with a minimum $m_i$ and a maximum $M_i$ such that
$x_i in [m_i, M_i]$
and then define a radius for each element by
$r^*_i = frac1(M_i-m_i)r$
Then I could define a "ball" of points around the point in question by
$ B_r(mathbfx) = leq r^*_i$
But I am not sure that this is meaningful. Ultimately I am trying to define a ball of parameters for a dynamical model around a particular parameter, and the model is to be evaluated at each point of this ball (a "parameter hypersphere" instead of a parameter grid). Any help here is appreciated.
general-topology geometry mathematical-modeling computational-mathematics
asked Jul 31 at 20:41
nguzman
295
asked Jul 31 at 20:41
nguzman
295
asked Jul 31 at 20:41
nguzman
295
asked Jul 31 at 20:41
nguzman
295
295
If I'm interpreting your question correctly, at some point you will need to decide how to compare electrical current to time (how 1 Ohm compares to 1 second). This requires physical knowledge/intuition. I don't think there's any way around that. Once you decide on the scalings, you can nondimensionalise everything and can forget about the physical units/interpretation when calculating distances. I don't understand what you mean by a "parameter hypersphere" and a "parameter grid", are you just saying you don't want to allow all combinations of the variables?
– David
Jul 31 at 22:37
@David A parameter grid is just a colloquial term among people who do computational modeling in my field, called as such because it is usually a grid of points $(x,y)$ of two parameters for a model. I thought that by extending the notion to a hypersphere (since I'm dealing with about nine parameters instead of two and I want parameter combinations equidistant from a point) might make my question clearer to those who do computational modeling.
– nguzman
Aug 1 at 2:44
Once you nondimensionalise, using $x_irightarrow x_i/(M_i-m_i)$ or some other scale (maybe $x_irightarrow(x_i-m_i)/(M_i-m_i)$ so $0leq x_ileq 1$ if you want different quantities weighted evenly), you can use whatever distance metric you want on the $x_i$'s. The hardest part will be working out what value of $r$ to use, and what it means physically.
– David
Aug 1 at 20:11
add a comment |Â
If I'm interpreting your question correctly, at some point you will need to decide how to compare electrical current to time (how 1 Ohm compares to 1 second). This requires physical knowledge/intuition. I don't think there's any way around that. Once you decide on the scalings, you can nondimensionalise everything and can forget about the physical units/interpretation when calculating distances. I don't understand what you mean by a "parameter hypersphere" and a "parameter grid", are you just saying you don't want to allow all combinations of the variables?
– David
Jul 31 at 22:37
@David A parameter grid is just a colloquial term among people who do computational modeling in my field, called as such because it is usually a grid of points $(x,y)$ of two parameters for a model. I thought that by extending the notion to a hypersphere (since I'm dealing with about nine parameters instead of two and I want parameter combinations equidistant from a point) might make my question clearer to those who do computational modeling.
– nguzman
Aug 1 at 2:44
Once you nondimensionalise, using $x_irightarrow x_i/(M_i-m_i)$ or some other scale (maybe $x_irightarrow(x_i-m_i)/(M_i-m_i)$ so $0leq x_ileq 1$ if you want different quantities weighted evenly), you can use whatever distance metric you want on the $x_i$'s. The hardest part will be working out what value of $r$ to use, and what it means physically.
– David
Aug 1 at 20:11
If I'm interpreting your question correctly, at some point you will need to decide how to compare electrical current to time (how 1 Ohm compares to 1 second). This requires physical knowledge/intuition. I don't think there's any way around that. Once you decide on the scalings, you can nondimensionalise everything and can forget about the physical units/interpretation when calculating distances. I don't understand what you mean by a "parameter hypersphere" and a "parameter grid", are you just saying you don't want to allow all combinations of the variables?
– David
Jul 31 at 22:37
If I'm interpreting your question correctly, at some point you will need to decide how to compare electrical current to time (how 1 Ohm compares to 1 second). This requires physical knowledge/intuition. I don't think there's any way around that. Once you decide on the scalings, you can nondimensionalise everything and can forget about the physical units/interpretation when calculating distances. I don't understand what you mean by a "parameter hypersphere" and a "parameter grid", are you just saying you don't want to allow all combinations of the variables?
– David
Jul 31 at 22:37
@David A parameter grid is just a colloquial term among people who do computational modeling in my field, called as such because it is usually a grid of points $(x,y)$ of two parameters for a model. I thought that by extending the notion to a hypersphere (since I'm dealing with about nine parameters instead of two and I want parameter combinations equidistant from a point) might make my question clearer to those who do computational modeling.
– nguzman
Aug 1 at 2:44
@David A parameter grid is just a colloquial term among people who do computational modeling in my field, called as such because it is usually a grid of points $(x,y)$ of two parameters for a model. I thought that by extending the notion to a hypersphere (since I'm dealing with about nine parameters instead of two and I want parameter combinations equidistant from a point) might make my question clearer to those who do computational modeling.
– nguzman
Aug 1 at 2:44
Once you nondimensionalise, using $x_irightarrow x_i/(M_i-m_i)$ or some other scale (maybe $x_irightarrow(x_i-m_i)/(M_i-m_i)$ so $0leq x_ileq 1$ if you want different quantities weighted evenly), you can use whatever distance metric you want on the $x_i$'s. The hardest part will be working out what value of $r$ to use, and what it means physically.
– David
Aug 1 at 20:11
Once you nondimensionalise, using $x_irightarrow x_i/(M_i-m_i)$ or some other scale (maybe $x_irightarrow(x_i-m_i)/(M_i-m_i)$ so $0leq x_ileq 1$ if you want different quantities weighted evenly), you can use whatever distance metric you want on the $x_i$'s. The hardest part will be working out what value of $r$ to use, and what it means physically.
– David
Aug 1 at 20:11
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
Any metric formed in the way you speak of will be topologically equivalent to the standard Euclidean one whose coordinates will be your 'meaningless' Ohms-seconds-etc. This is because in finite dimensions over $mathbbR$ the normal Euclidean metric is equivalent to the 'taxi-cab metric' which considers the sums of the distances in each variable. Scaling each one-dimensional distance by the (presumed non-zero) factors $(M_i - m_i)^-1$ will again give you an equivalent metric. This is a basic exercise in topology: You can easily show that for any point, in either metric you can find neighborhoods (e.g. n-dimensional rectangular prisms centered around the point with side lengths converging to zero) that are bases for the different topologies but which are nested in each other.
answered Aug 1 at 1:55
John Samples
989415
Thanks, I don't know much topology and had completely forgotten about the $L_1$ distance. This definitely solves the problem.
– nguzman
Aug 1 at 3:11
add a comment |Â
up vote
0
down vote
If for example, x is measured in ft and y in lbs,
then (x,y) = x×y is measured in ft-lbs.
So for |x×y| < r, either consider |x×y| dimensionaless
in the physics sense or require r to measured in ft-lbs.
answered Aug 1 at 1:06
William Elliot
5,0722414
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Any metric formed in the way you speak of will be topologically equivalent to the standard Euclidean one whose coordinates will be your 'meaningless' Ohms-seconds-etc. This is because in finite dimensions over $mathbbR$ the normal Euclidean metric is equivalent to the 'taxi-cab metric' which considers the sums of the distances in each variable. Scaling each one-dimensional distance by the (presumed non-zero) factors $(M_i - m_i)^-1$ will again give you an equivalent metric. This is a basic exercise in topology: You can easily show that for any point, in either metric you can find neighborhoods (e.g. n-dimensional rectangular prisms centered around the point with side lengths converging to zero) that are bases for the different topologies but which are nested in each other.
answered Aug 1 at 1:55
John Samples
989415
Thanks, I don't know much topology and had completely forgotten about the $L_1$ distance. This definitely solves the problem.
– nguzman
Aug 1 at 3:11
add a comment |Â
up vote
2
down vote
accepted
Any metric formed in the way you speak of will be topologically equivalent to the standard Euclidean one whose coordinates will be your 'meaningless' Ohms-seconds-etc. This is because in finite dimensions over $mathbbR$ the normal Euclidean metric is equivalent to the 'taxi-cab metric' which considers the sums of the distances in each variable. Scaling each one-dimensional distance by the (presumed non-zero) factors $(M_i - m_i)^-1$ will again give you an equivalent metric. This is a basic exercise in topology: You can easily show that for any point, in either metric you can find neighborhoods (e.g. n-dimensional rectangular prisms centered around the point with side lengths converging to zero) that are bases for the different topologies but which are nested in each other.
answered Aug 1 at 1:55
John Samples
989415
Thanks, I don't know much topology and had completely forgotten about the $L_1$ distance. This definitely solves the problem.
– nguzman
Aug 1 at 3:11
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Any metric formed in the way you speak of will be topologically equivalent to the standard Euclidean one whose coordinates will be your 'meaningless' Ohms-seconds-etc. This is because in finite dimensions over $mathbbR$ the normal Euclidean metric is equivalent to the 'taxi-cab metric' which considers the sums of the distances in each variable. Scaling each one-dimensional distance by the (presumed non-zero) factors $(M_i - m_i)^-1$ will again give you an equivalent metric. This is a basic exercise in topology: You can easily show that for any point, in either metric you can find neighborhoods (e.g. n-dimensional rectangular prisms centered around the point with side lengths converging to zero) that are bases for the different topologies but which are nested in each other.
answered Aug 1 at 1:55
John Samples
989415
Any metric formed in the way you speak of will be topologically equivalent to the standard Euclidean one whose coordinates will be your 'meaningless' Ohms-seconds-etc. This is because in finite dimensions over $mathbbR$ the normal Euclidean metric is equivalent to the 'taxi-cab metric' which considers the sums of the distances in each variable. Scaling each one-dimensional distance by the (presumed non-zero) factors $(M_i - m_i)^-1$ will again give you an equivalent metric. This is a basic exercise in topology: You can easily show that for any point, in either metric you can find neighborhoods (e.g. n-dimensional rectangular prisms centered around the point with side lengths converging to zero) that are bases for the different topologies but which are nested in each other.
answered Aug 1 at 1:55
John Samples
989415
answered Aug 1 at 1:55
John Samples
989415
answered Aug 1 at 1:55
John Samples
989415
answered Aug 1 at 1:55
John Samples
989415
989415
Thanks, I don't know much topology and had completely forgotten about the $L_1$ distance. This definitely solves the problem.
– nguzman
Aug 1 at 3:11
add a comment |Â
Thanks, I don't know much topology and had completely forgotten about the $L_1$ distance. This definitely solves the problem.
– nguzman
Aug 1 at 3:11
Thanks, I don't know much topology and had completely forgotten about the $L_1$ distance. This definitely solves the problem.
– nguzman
Aug 1 at 3:11
Thanks, I don't know much topology and had completely forgotten about the $L_1$ distance. This definitely solves the problem.
– nguzman
Aug 1 at 3:11
add a comment |Â
up vote
0
down vote
If for example, x is measured in ft and y in lbs,
then (x,y) = x×y is measured in ft-lbs.
So for |x×y| < r, either consider |x×y| dimensionaless
in the physics sense or require r to measured in ft-lbs.
answered Aug 1 at 1:06
William Elliot
5,0722414
add a comment |Â
up vote
0
down vote
If for example, x is measured in ft and y in lbs,
then (x,y) = x×y is measured in ft-lbs.
So for |x×y| < r, either consider |x×y| dimensionaless
in the physics sense or require r to measured in ft-lbs.
answered Aug 1 at 1:06
William Elliot
5,0722414
add a comment |Â
up vote
0
down vote
up vote
0
down vote
If for example, x is measured in ft and y in lbs,
then (x,y) = x×y is measured in ft-lbs.
So for |x×y| < r, either consider |x×y| dimensionaless
in the physics sense or require r to measured in ft-lbs.
answered Aug 1 at 1:06
William Elliot
5,0722414
If for example, x is measured in ft and y in lbs,
then (x,y) = x×y is measured in ft-lbs.
So for |x×y| < r, either consider |x×y| dimensionaless
in the physics sense or require r to measured in ft-lbs.
answered Aug 1 at 1:06
William Elliot
5,0722414
answered Aug 1 at 1:06
William Elliot
5,0722414
answered Aug 1 at 1:06
William Elliot
5,0722414
answered Aug 1 at 1:06
William Elliot
5,0722414
5,0722414
add a comment |Â
add a comment |Â
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If I'm interpreting your question correctly, at some point you will need to decide how to compare electrical current to time (how 1 Ohm compares to 1 second). This requires physical knowledge/intuition. I don't think there's any way around that. Once you decide on the scalings, you can nondimensionalise everything and can forget about the physical units/interpretation when calculating distances. I don't understand what you mean by a "parameter hypersphere" and a "parameter grid", are you just saying you don't want to allow all combinations of the variables?
– David
Jul 31 at 22:37
@David A parameter grid is just a colloquial term among people who do computational modeling in my field, called as such because it is usually a grid of points $(x,y)$ of two parameters for a model. I thought that by extending the notion to a hypersphere (since I'm dealing with about nine parameters instead of two and I want parameter combinations equidistant from a point) might make my question clearer to those who do computational modeling.
– nguzman
Aug 1 at 2:44
Once you nondimensionalise, using $x_irightarrow x_i/(M_i-m_i)$ or some other scale (maybe $x_irightarrow(x_i-m_i)/(M_i-m_i)$ so $0leq x_ileq 1$ if you want different quantities weighted evenly), you can use whatever distance metric you want on the $x_i$'s. The hardest part will be working out what value of $r$ to use, and what it means physically.
– David
Aug 1 at 20:11