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Does a Bijective Commutative transformation on a vector of angles exist?
Can non-linear transformations be represented as Transformation Matrices?Intuition of Addition Formula for Sine and CosineLooking for peculiar vector transformationfind the angles of a given vector sumHow is $Asintheta +Bcostheta = Csin(theta + phi)$ derived?Inverting a somewhat complicated trig transformationDoes this ray fall inside this angle?Derive the formula for the cosine of the difference of two angles from the dot product formulaSampling the transformation matrixHow to track position using Euler angles?
$begingroup$
I have a problem where I have two vectors a and b representing a list of angles.
I need to find a transformation T where T(a,b) = T(b,a), where T has a distance metric to compare two transformations, and that is inversible : I should be able to retrieve a and b (swapped is ok) from T(a,b)
For example:
a+b, cos(a+b), sin(a)+sin(b), cos(a)*cos(b) or any combinations of these qualify
a.b doesn't qualify because I don't have a metric to compare square angles.
a-b doesn't qualify because it is not commutative
Do you think this is possible?
What I have tried that didn't work:
T(a,b) = cos(a)*cos(b), cos(a)+cos(b), sin(a)*sin(b), sin(a)+sin(b)
And solve the system for cos(a) cos(b) sin(a) sin(b). However because it is a quadratic equation, I get two solutions and my two arrays a and b are not consistent anymore, I got two new vectors that have either values from a or from b depending when the determinant of the equations reaches zero.
This was just one idea of transformation that was symmetric so it would satisfy my conditions, but I couldn't get back my original vectors.
Thank you!
trigonometry transformation angle
asked Mar 14 at 4:30
user5441518user5441518
32
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user5441518 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
$begingroup$
I have a problem where I have two vectors a and b representing a list of angles.
I need to find a transformation T where T(a,b) = T(b,a), where T has a distance metric to compare two transformations, and that is inversible : I should be able to retrieve a and b (swapped is ok) from T(a,b)
For example:
a+b, cos(a+b), sin(a)+sin(b), cos(a)*cos(b) or any combinations of these qualify
a.b doesn't qualify because I don't have a metric to compare square angles.
a-b doesn't qualify because it is not commutative
Do you think this is possible?
What I have tried that didn't work:
T(a,b) = cos(a)*cos(b), cos(a)+cos(b), sin(a)*sin(b), sin(a)+sin(b)
And solve the system for cos(a) cos(b) sin(a) sin(b). However because it is a quadratic equation, I get two solutions and my two arrays a and b are not consistent anymore, I got two new vectors that have either values from a or from b depending when the determinant of the equations reaches zero.
This was just one idea of transformation that was symmetric so it would satisfy my conditions, but I couldn't get back my original vectors.
Thank you!
trigonometry transformation angle
asked Mar 14 at 4:30
user5441518user5441518
32
New contributor
user5441518 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I have a problem where I have two vectors a and b representing a list of angles.
I need to find a transformation T where T(a,b) = T(b,a), where T has a distance metric to compare two transformations, and that is inversible : I should be able to retrieve a and b (swapped is ok) from T(a,b)
For example:
a+b, cos(a+b), sin(a)+sin(b), cos(a)*cos(b) or any combinations of these qualify
a.b doesn't qualify because I don't have a metric to compare square angles.
a-b doesn't qualify because it is not commutative
Do you think this is possible?
What I have tried that didn't work:
T(a,b) = cos(a)*cos(b), cos(a)+cos(b), sin(a)*sin(b), sin(a)+sin(b)
And solve the system for cos(a) cos(b) sin(a) sin(b). However because it is a quadratic equation, I get two solutions and my two arrays a and b are not consistent anymore, I got two new vectors that have either values from a or from b depending when the determinant of the equations reaches zero.
This was just one idea of transformation that was symmetric so it would satisfy my conditions, but I couldn't get back my original vectors.
Thank you!
trigonometry transformation angle
asked Mar 14 at 4:30
user5441518user5441518
32
New contributor
user5441518 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I have a problem where I have two vectors a and b representing a list of angles.
I need to find a transformation T where T(a,b) = T(b,a), where T has a distance metric to compare two transformations, and that is inversible : I should be able to retrieve a and b (swapped is ok) from T(a,b)
For example:
a+b, cos(a+b), sin(a)+sin(b), cos(a)*cos(b) or any combinations of these qualify
a.b doesn't qualify because I don't have a metric to compare square angles.
a-b doesn't qualify because it is not commutative
Do you think this is possible?
What I have tried that didn't work:
T(a,b) = cos(a)*cos(b), cos(a)+cos(b), sin(a)*sin(b), sin(a)+sin(b)
And solve the system for cos(a) cos(b) sin(a) sin(b). However because it is a quadratic equation, I get two solutions and my two arrays a and b are not consistent anymore, I got two new vectors that have either values from a or from b depending when the determinant of the equations reaches zero.
This was just one idea of transformation that was symmetric so it would satisfy my conditions, but I couldn't get back my original vectors.
Thank you!
trigonometry transformation angle
trigonometry transformation angle
asked Mar 14 at 4:30
user5441518user5441518
32
New contributor
user5441518 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 14 at 4:30
user5441518user5441518
32
New contributor
user5441518 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 14 at 4:35
user5441518
asked Mar 14 at 4:30
user5441518user5441518
32
New contributor
user5441518 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 14 at 4:30
user5441518user5441518
32
asked Mar 14 at 4:30
user5441518user5441518
32
32
New contributor
user5441518 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user5441518 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
user5441518 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Notation convention: the $k$th element of an array, such as $a$, will be denoted with a subscript, as in $a_k$.
... I get two solutions and my two arrays a and b are not consistent anymore, I got two new vectors that have either values from a or from b depending ...
If you're working with functions of the form $c_k=f(a_k,b_k)$ for some symmetric $f$, this is unavoidable. We can swap $a_1$ and $b_1$, leave all of the other $a_k$ and $b_k$ alone, and the $c_k$ will still be the same. Repeat with $d_k=g(a_k,b_k)$ and the same thing happens; adding more functions will never solve the problem.
So then, we need functions that cross over, and use more than one $k$. Here's an idea:
beginalign*c_1 = a_1+b_1quad d_1=cos(a_1-b_1) &\
c_2 = a_2+b_2quad d_2=cos(a_2-b_2) &quad e_2=cos(a_1+a_2-b_1-b_2)\
c_3 = a_3+b_3quad d_3=cos(a_3-b_3) &quad e_3=cos(a_1+a_2+a_3-b_1-b_2-b_3)endalign*
and so on. By adding in the third vector there, we know how the angle differences combine, and can tell the difference between the likes of $((x,y),(u,v))$ and $((x,v),(y,u))$ since $|x+y-u-v|neq |x+v-u-y|$ in general.
answered Mar 14 at 5:19
jmerryjmerry
15.2k1632
$endgroup$
$begingroup$
This is what was missing!! Thanks a lot!
$endgroup$
– user5441518
Mar 14 at 6:00
add a comment |
Your Answer
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1 Answer
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1 Answer
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active
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votes
$begingroup$
Notation convention: the $k$th element of an array, such as $a$, will be denoted with a subscript, as in $a_k$.
... I get two solutions and my two arrays a and b are not consistent anymore, I got two new vectors that have either values from a or from b depending ...
If you're working with functions of the form $c_k=f(a_k,b_k)$ for some symmetric $f$, this is unavoidable. We can swap $a_1$ and $b_1$, leave all of the other $a_k$ and $b_k$ alone, and the $c_k$ will still be the same. Repeat with $d_k=g(a_k,b_k)$ and the same thing happens; adding more functions will never solve the problem.
So then, we need functions that cross over, and use more than one $k$. Here's an idea:
beginalign*c_1 = a_1+b_1quad d_1=cos(a_1-b_1) &\
c_2 = a_2+b_2quad d_2=cos(a_2-b_2) &quad e_2=cos(a_1+a_2-b_1-b_2)\
c_3 = a_3+b_3quad d_3=cos(a_3-b_3) &quad e_3=cos(a_1+a_2+a_3-b_1-b_2-b_3)endalign*
and so on. By adding in the third vector there, we know how the angle differences combine, and can tell the difference between the likes of $((x,y),(u,v))$ and $((x,v),(y,u))$ since $|x+y-u-v|neq |x+v-u-y|$ in general.
answered Mar 14 at 5:19
jmerryjmerry
15.2k1632
$endgroup$
$begingroup$
This is what was missing!! Thanks a lot!
$endgroup$
– user5441518
Mar 14 at 6:00
add a comment |
$begingroup$
Notation convention: the $k$th element of an array, such as $a$, will be denoted with a subscript, as in $a_k$.
... I get two solutions and my two arrays a and b are not consistent anymore, I got two new vectors that have either values from a or from b depending ...
If you're working with functions of the form $c_k=f(a_k,b_k)$ for some symmetric $f$, this is unavoidable. We can swap $a_1$ and $b_1$, leave all of the other $a_k$ and $b_k$ alone, and the $c_k$ will still be the same. Repeat with $d_k=g(a_k,b_k)$ and the same thing happens; adding more functions will never solve the problem.
So then, we need functions that cross over, and use more than one $k$. Here's an idea:
beginalign*c_1 = a_1+b_1quad d_1=cos(a_1-b_1) &\
c_2 = a_2+b_2quad d_2=cos(a_2-b_2) &quad e_2=cos(a_1+a_2-b_1-b_2)\
c_3 = a_3+b_3quad d_3=cos(a_3-b_3) &quad e_3=cos(a_1+a_2+a_3-b_1-b_2-b_3)endalign*
and so on. By adding in the third vector there, we know how the angle differences combine, and can tell the difference between the likes of $((x,y),(u,v))$ and $((x,v),(y,u))$ since $|x+y-u-v|neq |x+v-u-y|$ in general.
answered Mar 14 at 5:19
jmerryjmerry
15.2k1632
$endgroup$
$begingroup$
This is what was missing!! Thanks a lot!
$endgroup$
– user5441518
Mar 14 at 6:00
add a comment |
$begingroup$
Notation convention: the $k$th element of an array, such as $a$, will be denoted with a subscript, as in $a_k$.
... I get two solutions and my two arrays a and b are not consistent anymore, I got two new vectors that have either values from a or from b depending ...
If you're working with functions of the form $c_k=f(a_k,b_k)$ for some symmetric $f$, this is unavoidable. We can swap $a_1$ and $b_1$, leave all of the other $a_k$ and $b_k$ alone, and the $c_k$ will still be the same. Repeat with $d_k=g(a_k,b_k)$ and the same thing happens; adding more functions will never solve the problem.
So then, we need functions that cross over, and use more than one $k$. Here's an idea:
beginalign*c_1 = a_1+b_1quad d_1=cos(a_1-b_1) &\
c_2 = a_2+b_2quad d_2=cos(a_2-b_2) &quad e_2=cos(a_1+a_2-b_1-b_2)\
c_3 = a_3+b_3quad d_3=cos(a_3-b_3) &quad e_3=cos(a_1+a_2+a_3-b_1-b_2-b_3)endalign*
and so on. By adding in the third vector there, we know how the angle differences combine, and can tell the difference between the likes of $((x,y),(u,v))$ and $((x,v),(y,u))$ since $|x+y-u-v|neq |x+v-u-y|$ in general.
answered Mar 14 at 5:19
jmerryjmerry
15.2k1632
$endgroup$
Notation convention: the $k$th element of an array, such as $a$, will be denoted with a subscript, as in $a_k$.
... I get two solutions and my two arrays a and b are not consistent anymore, I got two new vectors that have either values from a or from b depending ...
If you're working with functions of the form $c_k=f(a_k,b_k)$ for some symmetric $f$, this is unavoidable. We can swap $a_1$ and $b_1$, leave all of the other $a_k$ and $b_k$ alone, and the $c_k$ will still be the same. Repeat with $d_k=g(a_k,b_k)$ and the same thing happens; adding more functions will never solve the problem.
So then, we need functions that cross over, and use more than one $k$. Here's an idea:
beginalign*c_1 = a_1+b_1quad d_1=cos(a_1-b_1) &\
c_2 = a_2+b_2quad d_2=cos(a_2-b_2) &quad e_2=cos(a_1+a_2-b_1-b_2)\
c_3 = a_3+b_3quad d_3=cos(a_3-b_3) &quad e_3=cos(a_1+a_2+a_3-b_1-b_2-b_3)endalign*
and so on. By adding in the third vector there, we know how the angle differences combine, and can tell the difference between the likes of $((x,y),(u,v))$ and $((x,v),(y,u))$ since $|x+y-u-v|neq |x+v-u-y|$ in general.
answered Mar 14 at 5:19
jmerryjmerry
15.2k1632
answered Mar 14 at 5:19
jmerryjmerry
15.2k1632
answered Mar 14 at 5:19
jmerryjmerry
15.2k1632
answered Mar 14 at 5:19
jmerryjmerry
15.2k1632
15.2k1632
$begingroup$
This is what was missing!! Thanks a lot!
$endgroup$
– user5441518
Mar 14 at 6:00
add a comment |
$begingroup$
This is what was missing!! Thanks a lot!
$endgroup$
– user5441518
Mar 14 at 6:00
$begingroup$
This is what was missing!! Thanks a lot!
$endgroup$
– user5441518
Mar 14 at 6:00
$begingroup$
This is what was missing!! Thanks a lot!
$endgroup$
– user5441518
Mar 14 at 6:00
add a comment |
user5441518 is a new contributor. Be nice, and check out our Code of Conduct.
user5441518 is a new contributor. Be nice, and check out our Code of Conduct.
user5441518 is a new contributor. Be nice, and check out our Code of Conduct.
user5441518 is a new contributor. Be nice, and check out our Code of Conduct.
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