Mixing
Heuristics as to when $frac{x-c_1}{c_2-c_1}$ and similar forms are linear and form a line?
$begingroup$
Heuristics as to when $f(x)=frac{x-c_1}{c_2-c_1}$ and similar forms are linear and form a line?
By similar forms I mean that I started to speculate as to how much one can "add stuff" to the above simple case and still know that the function is a linear function and its graph is a line.
Intuitively one can add, subtract and do multiplications by constants. One could also add linear functions to it and still retain linearity. But linearity would surely be broken if one adds a nonlinear function $eta(x)$ like $h(x)=f(x)+eta(x)$?
So additivity and homogeneity do not suffice, but they have some constraints as to what can be added and what can be used as multiplier. What are these constraints?
linear-transformations
asked Jan 9 at 13:04
mavaviljmavavilj
2,75011035
$endgroup$
add a comment |
$begingroup$
Heuristics as to when $f(x)=frac{x-c_1}{c_2-c_1}$ and similar forms are linear and form a line?
By similar forms I mean that I started to speculate as to how much one can "add stuff" to the above simple case and still know that the function is a linear function and its graph is a line.
Intuitively one can add, subtract and do multiplications by constants. One could also add linear functions to it and still retain linearity. But linearity would surely be broken if one adds a nonlinear function $eta(x)$ like $h(x)=f(x)+eta(x)$?
So additivity and homogeneity do not suffice, but they have some constraints as to what can be added and what can be used as multiplier. What are these constraints?
linear-transformations
asked Jan 9 at 13:04
mavaviljmavavilj
2,75011035
$endgroup$
$begingroup$
What do you call "similar forms" and what is $n$ ?
$endgroup$
– Yves Daoust
Jan 9 at 13:10
$begingroup$
@YvesDaoust More clear?
$endgroup$
– mavavilj
Jan 9 at 13:13
$begingroup$
As clear as stuff can be.
$endgroup$
– Yves Daoust
Jan 9 at 13:13
$begingroup$
Use the term 'linear functional' very carefully. $f(x)=frac{x-c_1}{c_2-c_1}$ is a linear functional only for $c_1=0$
$endgroup$
– Shubham Johri
Jan 9 at 13:16
add a comment |
$begingroup$
Heuristics as to when $f(x)=frac{x-c_1}{c_2-c_1}$ and similar forms are linear and form a line?
By similar forms I mean that I started to speculate as to how much one can "add stuff" to the above simple case and still know that the function is a linear function and its graph is a line.
Intuitively one can add, subtract and do multiplications by constants. One could also add linear functions to it and still retain linearity. But linearity would surely be broken if one adds a nonlinear function $eta(x)$ like $h(x)=f(x)+eta(x)$?
So additivity and homogeneity do not suffice, but they have some constraints as to what can be added and what can be used as multiplier. What are these constraints?
linear-transformations
asked Jan 9 at 13:04
mavaviljmavavilj
2,75011035
$endgroup$
Heuristics as to when $f(x)=frac{x-c_1}{c_2-c_1}$ and similar forms are linear and form a line?
By similar forms I mean that I started to speculate as to how much one can "add stuff" to the above simple case and still know that the function is a linear function and its graph is a line.
Intuitively one can add, subtract and do multiplications by constants. One could also add linear functions to it and still retain linearity. But linearity would surely be broken if one adds a nonlinear function $eta(x)$ like $h(x)=f(x)+eta(x)$?
So additivity and homogeneity do not suffice, but they have some constraints as to what can be added and what can be used as multiplier. What are these constraints?
linear-transformations
linear-transformations
asked Jan 9 at 13:04
mavaviljmavavilj
2,75011035
asked Jan 9 at 13:04
mavaviljmavavilj
2,75011035
edited Jan 9 at 13:18
mavavilj
asked Jan 9 at 13:04
mavaviljmavavilj
2,75011035
asked Jan 9 at 13:04
mavaviljmavavilj
2,75011035
asked Jan 9 at 13:04
mavaviljmavavilj
2,75011035
2,75011035
$begingroup$
What do you call "similar forms" and what is $n$ ?
$endgroup$
– Yves Daoust
Jan 9 at 13:10
$begingroup$
@YvesDaoust More clear?
$endgroup$
– mavavilj
Jan 9 at 13:13
$begingroup$
As clear as stuff can be.
$endgroup$
– Yves Daoust
Jan 9 at 13:13
$begingroup$
Use the term 'linear functional' very carefully. $f(x)=frac{x-c_1}{c_2-c_1}$ is a linear functional only for $c_1=0$
$endgroup$
– Shubham Johri
Jan 9 at 13:16
add a comment |
$begingroup$
What do you call "similar forms" and what is $n$ ?
$endgroup$
– Yves Daoust
Jan 9 at 13:10
$begingroup$
@YvesDaoust More clear?
$endgroup$
– mavavilj
Jan 9 at 13:13
$begingroup$
As clear as stuff can be.
$endgroup$
– Yves Daoust
Jan 9 at 13:13
$begingroup$
Use the term 'linear functional' very carefully. $f(x)=frac{x-c_1}{c_2-c_1}$ is a linear functional only for $c_1=0$
$endgroup$
– Shubham Johri
Jan 9 at 13:16
$begingroup$
What do you call "similar forms" and what is $n$ ?
$endgroup$
– Yves Daoust
Jan 9 at 13:10
$begingroup$
What do you call "similar forms" and what is $n$ ?
$endgroup$
– Yves Daoust
Jan 9 at 13:10
$begingroup$
@YvesDaoust More clear?
$endgroup$
– mavavilj
Jan 9 at 13:13
$begingroup$
@YvesDaoust More clear?
$endgroup$
– mavavilj
Jan 9 at 13:13
$begingroup$
As clear as stuff can be.
$endgroup$
– Yves Daoust
Jan 9 at 13:13
$begingroup$
As clear as stuff can be.
$endgroup$
– Yves Daoust
Jan 9 at 13:13
$begingroup$
Use the term 'linear functional' very carefully. $f(x)=frac{x-c_1}{c_2-c_1}$ is a linear functional only for $c_1=0$
$endgroup$
– Shubham Johri
Jan 9 at 13:16
$begingroup$
Use the term 'linear functional' very carefully. $f(x)=frac{x-c_1}{c_2-c_1}$ is a linear functional only for $c_1=0$
$endgroup$
– Shubham Johri
Jan 9 at 13:16
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The affine functions of one variable are of the form
$$ax+b$$ where $a$ and $b$ are constants.
Nothing more to say.
answered Jan 9 at 13:12
Yves DaoustYves Daoust
126k672226
$endgroup$
$begingroup$
This is all there is to functions that are linear and form a line? So then one only needs to be able to factor the function into that form. Well makes sense, since that's the line equation.
$endgroup$
– mavavilj
Jan 9 at 13:13
$begingroup$
@mavavilj: that's it. Strictly speaking, a linear function is of the form $ax$. With an added constant, it is said affine.
$endgroup$
– Yves Daoust
Jan 9 at 13:15
add a comment |
$begingroup$
Generalization
Consider the function $f:Bbb R^nto Bbb R$ as follows $$f(x)={a^Tx+bover c^Tx+d}$$where $a,cin Bbb R^n$ and $b,dinBbb R$. The case where $a=0$ , $b=0$ is trivial, so we investigate the rest. If $f(x)$ is affine, then $ein Bbb R^n$ , $fin Bbb R$ exist such that $$f(x)=e^Tx+f$$by substitution we obtain $${a^Tx+bover c^Tx+d}=e^Tx+f\a^Tx+b=x^Tce^Tx+df+(fc+de)^Tx$$which yields to $$b=df\ce^T=0\fc+de=a$$some cases are worth studying:
Case 1 $d=0$
In this case $b=0$ and we obtain $a=fc$ with $fne 0$ and we have $$f(x)={a^Txover c^Tx}=f{c^Txover c^Tx}$$ which is not affine over $Bbb R^n$ since the function is not defined for $c^Tx=0$. Therefore the case $d=0$ doesn't lead to affinity.
Case 2 $dne 0$ , $c=0$
We obtain$$f={bover d}\de=a$$which means that affinity happens in this case without extra assumptions.
Case 3 $dne0$ and $cne 0$
The equations lead to $$f={bover d}\ec^T=0\fc+de=a$$from $$ec^T=0$$ we obtain $ec^Tc=ecdot ||c||^2=0$ and since $cne 0$ we have $e=0$ therefore $$f={bover d}\fc=a$$ and since $c^Tx+d=0$ has infinitely many roots, then this case either doesn't lead to affinity.
Conclusion
The only case leading to affinity is $c=0$ and $dne 0$.
answered Jan 9 at 15:45
Mostafa AyazMostafa Ayaz
15.4k3939
$endgroup$
$begingroup$
The equivalence is not perfect, as the denominator can vanish.
$endgroup$
– Yves Daoust
Jan 9 at 21:43
$begingroup$
In which case you mean? It's not needed to study such cases since the equivalence must hold on infinitely many points. More precisely, the points for which the equivalence is eliminated, form a zero-measure subset of $Bbb R^n$
$endgroup$
– Mostafa Ayaz
Jan 10 at 10:13
$begingroup$
What you consider "cases not needed" in fact completely destroys the affinity property. At every point.
$endgroup$
– Yves Daoust
Jan 10 at 10:18
$begingroup$
Yes you are right. I mean the cases that the function is well defined (not $pminfty$ valued) e.g for the case $dne 0$ and $c=0$ the domain is $Bbb R^n$ but I will edit my answer.
$endgroup$
– Mostafa Ayaz
Jan 10 at 10:22
$begingroup$
For the function to be defined everywhere, you need to explicitly state its value at the singular points (taking the limit for proper continuation), and this is absolutely equivalent to simplifying the form to $ax+b$. An homographic form is inappropriate.
$endgroup$
– Yves Daoust
Jan 10 at 10:26
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
The affine functions of one variable are of the form
$$ax+b$$ where $a$ and $b$ are constants.
Nothing more to say.
answered Jan 9 at 13:12
Yves DaoustYves Daoust
126k672226
$endgroup$
$begingroup$
This is all there is to functions that are linear and form a line? So then one only needs to be able to factor the function into that form. Well makes sense, since that's the line equation.
$endgroup$
– mavavilj
Jan 9 at 13:13
$begingroup$
@mavavilj: that's it. Strictly speaking, a linear function is of the form $ax$. With an added constant, it is said affine.
$endgroup$
– Yves Daoust
Jan 9 at 13:15
add a comment |
$begingroup$
The affine functions of one variable are of the form
$$ax+b$$ where $a$ and $b$ are constants.
Nothing more to say.
answered Jan 9 at 13:12
Yves DaoustYves Daoust
126k672226
$endgroup$
$begingroup$
This is all there is to functions that are linear and form a line? So then one only needs to be able to factor the function into that form. Well makes sense, since that's the line equation.
$endgroup$
– mavavilj
Jan 9 at 13:13
$begingroup$
@mavavilj: that's it. Strictly speaking, a linear function is of the form $ax$. With an added constant, it is said affine.
$endgroup$
– Yves Daoust
Jan 9 at 13:15
add a comment |
$begingroup$
The affine functions of one variable are of the form
$$ax+b$$ where $a$ and $b$ are constants.
Nothing more to say.
answered Jan 9 at 13:12
Yves DaoustYves Daoust
126k672226
$endgroup$
The affine functions of one variable are of the form
$$ax+b$$ where $a$ and $b$ are constants.
Nothing more to say.
answered Jan 9 at 13:12
Yves DaoustYves Daoust
126k672226
answered Jan 9 at 13:12
Yves DaoustYves Daoust
126k672226
answered Jan 9 at 13:12
Yves DaoustYves Daoust
126k672226
answered Jan 9 at 13:12
Yves DaoustYves Daoust
126k672226
126k672226
$begingroup$
This is all there is to functions that are linear and form a line? So then one only needs to be able to factor the function into that form. Well makes sense, since that's the line equation.
$endgroup$
– mavavilj
Jan 9 at 13:13
$begingroup$
@mavavilj: that's it. Strictly speaking, a linear function is of the form $ax$. With an added constant, it is said affine.
$endgroup$
– Yves Daoust
Jan 9 at 13:15
add a comment |
$begingroup$
This is all there is to functions that are linear and form a line? So then one only needs to be able to factor the function into that form. Well makes sense, since that's the line equation.
$endgroup$
– mavavilj
Jan 9 at 13:13
$begingroup$
@mavavilj: that's it. Strictly speaking, a linear function is of the form $ax$. With an added constant, it is said affine.
$endgroup$
– Yves Daoust
Jan 9 at 13:15
$begingroup$
This is all there is to functions that are linear and form a line? So then one only needs to be able to factor the function into that form. Well makes sense, since that's the line equation.
$endgroup$
– mavavilj
Jan 9 at 13:13
$begingroup$
This is all there is to functions that are linear and form a line? So then one only needs to be able to factor the function into that form. Well makes sense, since that's the line equation.
$endgroup$
– mavavilj
Jan 9 at 13:13
$begingroup$
@mavavilj: that's it. Strictly speaking, a linear function is of the form $ax$. With an added constant, it is said affine.
$endgroup$
– Yves Daoust
Jan 9 at 13:15
$begingroup$
@mavavilj: that's it. Strictly speaking, a linear function is of the form $ax$. With an added constant, it is said affine.
$endgroup$
– Yves Daoust
Jan 9 at 13:15
add a comment |
$begingroup$
Generalization
Consider the function $f:Bbb R^nto Bbb R$ as follows $$f(x)={a^Tx+bover c^Tx+d}$$where $a,cin Bbb R^n$ and $b,dinBbb R$. The case where $a=0$ , $b=0$ is trivial, so we investigate the rest. If $f(x)$ is affine, then $ein Bbb R^n$ , $fin Bbb R$ exist such that $$f(x)=e^Tx+f$$by substitution we obtain $${a^Tx+bover c^Tx+d}=e^Tx+f\a^Tx+b=x^Tce^Tx+df+(fc+de)^Tx$$which yields to $$b=df\ce^T=0\fc+de=a$$some cases are worth studying:
Case 1 $d=0$
In this case $b=0$ and we obtain $a=fc$ with $fne 0$ and we have $$f(x)={a^Txover c^Tx}=f{c^Txover c^Tx}$$ which is not affine over $Bbb R^n$ since the function is not defined for $c^Tx=0$. Therefore the case $d=0$ doesn't lead to affinity.
Case 2 $dne 0$ , $c=0$
We obtain$$f={bover d}\de=a$$which means that affinity happens in this case without extra assumptions.
Case 3 $dne0$ and $cne 0$
The equations lead to $$f={bover d}\ec^T=0\fc+de=a$$from $$ec^T=0$$ we obtain $ec^Tc=ecdot ||c||^2=0$ and since $cne 0$ we have $e=0$ therefore $$f={bover d}\fc=a$$ and since $c^Tx+d=0$ has infinitely many roots, then this case either doesn't lead to affinity.
Conclusion
The only case leading to affinity is $c=0$ and $dne 0$.
answered Jan 9 at 15:45
Mostafa AyazMostafa Ayaz
15.4k3939
$endgroup$
$begingroup$
The equivalence is not perfect, as the denominator can vanish.
$endgroup$
– Yves Daoust
Jan 9 at 21:43
$begingroup$
In which case you mean? It's not needed to study such cases since the equivalence must hold on infinitely many points. More precisely, the points for which the equivalence is eliminated, form a zero-measure subset of $Bbb R^n$
$endgroup$
– Mostafa Ayaz
Jan 10 at 10:13
$begingroup$
What you consider "cases not needed" in fact completely destroys the affinity property. At every point.
$endgroup$
– Yves Daoust
Jan 10 at 10:18
$begingroup$
Yes you are right. I mean the cases that the function is well defined (not $pminfty$ valued) e.g for the case $dne 0$ and $c=0$ the domain is $Bbb R^n$ but I will edit my answer.
$endgroup$
– Mostafa Ayaz
Jan 10 at 10:22
$begingroup$
For the function to be defined everywhere, you need to explicitly state its value at the singular points (taking the limit for proper continuation), and this is absolutely equivalent to simplifying the form to $ax+b$. An homographic form is inappropriate.
$endgroup$
– Yves Daoust
Jan 10 at 10:26
add a comment |
$begingroup$
Generalization
Consider the function $f:Bbb R^nto Bbb R$ as follows $$f(x)={a^Tx+bover c^Tx+d}$$where $a,cin Bbb R^n$ and $b,dinBbb R$. The case where $a=0$ , $b=0$ is trivial, so we investigate the rest. If $f(x)$ is affine, then $ein Bbb R^n$ , $fin Bbb R$ exist such that $$f(x)=e^Tx+f$$by substitution we obtain $${a^Tx+bover c^Tx+d}=e^Tx+f\a^Tx+b=x^Tce^Tx+df+(fc+de)^Tx$$which yields to $$b=df\ce^T=0\fc+de=a$$some cases are worth studying:
Case 1 $d=0$
In this case $b=0$ and we obtain $a=fc$ with $fne 0$ and we have $$f(x)={a^Txover c^Tx}=f{c^Txover c^Tx}$$ which is not affine over $Bbb R^n$ since the function is not defined for $c^Tx=0$. Therefore the case $d=0$ doesn't lead to affinity.
Case 2 $dne 0$ , $c=0$
We obtain$$f={bover d}\de=a$$which means that affinity happens in this case without extra assumptions.
Case 3 $dne0$ and $cne 0$
The equations lead to $$f={bover d}\ec^T=0\fc+de=a$$from $$ec^T=0$$ we obtain $ec^Tc=ecdot ||c||^2=0$ and since $cne 0$ we have $e=0$ therefore $$f={bover d}\fc=a$$ and since $c^Tx+d=0$ has infinitely many roots, then this case either doesn't lead to affinity.
Conclusion
The only case leading to affinity is $c=0$ and $dne 0$.
answered Jan 9 at 15:45
Mostafa AyazMostafa Ayaz
15.4k3939
$endgroup$
$begingroup$
The equivalence is not perfect, as the denominator can vanish.
$endgroup$
– Yves Daoust
Jan 9 at 21:43
$begingroup$
In which case you mean? It's not needed to study such cases since the equivalence must hold on infinitely many points. More precisely, the points for which the equivalence is eliminated, form a zero-measure subset of $Bbb R^n$
$endgroup$
– Mostafa Ayaz
Jan 10 at 10:13
$begingroup$
What you consider "cases not needed" in fact completely destroys the affinity property. At every point.
$endgroup$
– Yves Daoust
Jan 10 at 10:18
$begingroup$
Yes you are right. I mean the cases that the function is well defined (not $pminfty$ valued) e.g for the case $dne 0$ and $c=0$ the domain is $Bbb R^n$ but I will edit my answer.
$endgroup$
– Mostafa Ayaz
Jan 10 at 10:22
$begingroup$
For the function to be defined everywhere, you need to explicitly state its value at the singular points (taking the limit for proper continuation), and this is absolutely equivalent to simplifying the form to $ax+b$. An homographic form is inappropriate.
$endgroup$
– Yves Daoust
Jan 10 at 10:26
add a comment |
$begingroup$
Generalization
Consider the function $f:Bbb R^nto Bbb R$ as follows $$f(x)={a^Tx+bover c^Tx+d}$$where $a,cin Bbb R^n$ and $b,dinBbb R$. The case where $a=0$ , $b=0$ is trivial, so we investigate the rest. If $f(x)$ is affine, then $ein Bbb R^n$ , $fin Bbb R$ exist such that $$f(x)=e^Tx+f$$by substitution we obtain $${a^Tx+bover c^Tx+d}=e^Tx+f\a^Tx+b=x^Tce^Tx+df+(fc+de)^Tx$$which yields to $$b=df\ce^T=0\fc+de=a$$some cases are worth studying:
Case 1 $d=0$
In this case $b=0$ and we obtain $a=fc$ with $fne 0$ and we have $$f(x)={a^Txover c^Tx}=f{c^Txover c^Tx}$$ which is not affine over $Bbb R^n$ since the function is not defined for $c^Tx=0$. Therefore the case $d=0$ doesn't lead to affinity.
Case 2 $dne 0$ , $c=0$
We obtain$$f={bover d}\de=a$$which means that affinity happens in this case without extra assumptions.
Case 3 $dne0$ and $cne 0$
The equations lead to $$f={bover d}\ec^T=0\fc+de=a$$from $$ec^T=0$$ we obtain $ec^Tc=ecdot ||c||^2=0$ and since $cne 0$ we have $e=0$ therefore $$f={bover d}\fc=a$$ and since $c^Tx+d=0$ has infinitely many roots, then this case either doesn't lead to affinity.
Conclusion
The only case leading to affinity is $c=0$ and $dne 0$.
answered Jan 9 at 15:45
Mostafa AyazMostafa Ayaz
15.4k3939
$endgroup$
Generalization
Consider the function $f:Bbb R^nto Bbb R$ as follows $$f(x)={a^Tx+bover c^Tx+d}$$where $a,cin Bbb R^n$ and $b,dinBbb R$. The case where $a=0$ , $b=0$ is trivial, so we investigate the rest. If $f(x)$ is affine, then $ein Bbb R^n$ , $fin Bbb R$ exist such that $$f(x)=e^Tx+f$$by substitution we obtain $${a^Tx+bover c^Tx+d}=e^Tx+f\a^Tx+b=x^Tce^Tx+df+(fc+de)^Tx$$which yields to $$b=df\ce^T=0\fc+de=a$$some cases are worth studying:
Case 1 $d=0$
In this case $b=0$ and we obtain $a=fc$ with $fne 0$ and we have $$f(x)={a^Txover c^Tx}=f{c^Txover c^Tx}$$ which is not affine over $Bbb R^n$ since the function is not defined for $c^Tx=0$. Therefore the case $d=0$ doesn't lead to affinity.
Case 2 $dne 0$ , $c=0$
We obtain$$f={bover d}\de=a$$which means that affinity happens in this case without extra assumptions.
Case 3 $dne0$ and $cne 0$
The equations lead to $$f={bover d}\ec^T=0\fc+de=a$$from $$ec^T=0$$ we obtain $ec^Tc=ecdot ||c||^2=0$ and since $cne 0$ we have $e=0$ therefore $$f={bover d}\fc=a$$ and since $c^Tx+d=0$ has infinitely many roots, then this case either doesn't lead to affinity.
Conclusion
The only case leading to affinity is $c=0$ and $dne 0$.
answered Jan 9 at 15:45
Mostafa AyazMostafa Ayaz
15.4k3939
edited Jan 10 at 10:39
answered Jan 9 at 15:45
Mostafa AyazMostafa Ayaz
15.4k3939
answered Jan 9 at 15:45
Mostafa AyazMostafa Ayaz
15.4k3939
answered Jan 9 at 15:45
Mostafa AyazMostafa Ayaz
15.4k3939
15.4k3939
$begingroup$
The equivalence is not perfect, as the denominator can vanish.
$endgroup$
– Yves Daoust
Jan 9 at 21:43
$begingroup$
In which case you mean? It's not needed to study such cases since the equivalence must hold on infinitely many points. More precisely, the points for which the equivalence is eliminated, form a zero-measure subset of $Bbb R^n$
$endgroup$
– Mostafa Ayaz
Jan 10 at 10:13
$begingroup$
What you consider "cases not needed" in fact completely destroys the affinity property. At every point.
$endgroup$
– Yves Daoust
Jan 10 at 10:18
$begingroup$
Yes you are right. I mean the cases that the function is well defined (not $pminfty$ valued) e.g for the case $dne 0$ and $c=0$ the domain is $Bbb R^n$ but I will edit my answer.
$endgroup$
– Mostafa Ayaz
Jan 10 at 10:22
$begingroup$
For the function to be defined everywhere, you need to explicitly state its value at the singular points (taking the limit for proper continuation), and this is absolutely equivalent to simplifying the form to $ax+b$. An homographic form is inappropriate.
$endgroup$
– Yves Daoust
Jan 10 at 10:26
add a comment |
$begingroup$
The equivalence is not perfect, as the denominator can vanish.
$endgroup$
– Yves Daoust
Jan 9 at 21:43
$begingroup$
In which case you mean? It's not needed to study such cases since the equivalence must hold on infinitely many points. More precisely, the points for which the equivalence is eliminated, form a zero-measure subset of $Bbb R^n$
$endgroup$
– Mostafa Ayaz
Jan 10 at 10:13
$begingroup$
What you consider "cases not needed" in fact completely destroys the affinity property. At every point.
$endgroup$
– Yves Daoust
Jan 10 at 10:18
$begingroup$
Yes you are right. I mean the cases that the function is well defined (not $pminfty$ valued) e.g for the case $dne 0$ and $c=0$ the domain is $Bbb R^n$ but I will edit my answer.
$endgroup$
– Mostafa Ayaz
Jan 10 at 10:22
$begingroup$
For the function to be defined everywhere, you need to explicitly state its value at the singular points (taking the limit for proper continuation), and this is absolutely equivalent to simplifying the form to $ax+b$. An homographic form is inappropriate.
$endgroup$
– Yves Daoust
Jan 10 at 10:26
$begingroup$
The equivalence is not perfect, as the denominator can vanish.
$endgroup$
– Yves Daoust
Jan 9 at 21:43
$begingroup$
The equivalence is not perfect, as the denominator can vanish.
$endgroup$
– Yves Daoust
Jan 9 at 21:43
$begingroup$
In which case you mean? It's not needed to study such cases since the equivalence must hold on infinitely many points. More precisely, the points for which the equivalence is eliminated, form a zero-measure subset of $Bbb R^n$
$endgroup$
– Mostafa Ayaz
Jan 10 at 10:13
$begingroup$
In which case you mean? It's not needed to study such cases since the equivalence must hold on infinitely many points. More precisely, the points for which the equivalence is eliminated, form a zero-measure subset of $Bbb R^n$
$endgroup$
– Mostafa Ayaz
Jan 10 at 10:13
$begingroup$
What you consider "cases not needed" in fact completely destroys the affinity property. At every point.
$endgroup$
– Yves Daoust
Jan 10 at 10:18
$begingroup$
What you consider "cases not needed" in fact completely destroys the affinity property. At every point.
$endgroup$
– Yves Daoust
Jan 10 at 10:18
$begingroup$
Yes you are right. I mean the cases that the function is well defined (not $pminfty$ valued) e.g for the case $dne 0$ and $c=0$ the domain is $Bbb R^n$ but I will edit my answer.
$endgroup$
– Mostafa Ayaz
Jan 10 at 10:22
$begingroup$
Yes you are right. I mean the cases that the function is well defined (not $pminfty$ valued) e.g for the case $dne 0$ and $c=0$ the domain is $Bbb R^n$ but I will edit my answer.
$endgroup$
– Mostafa Ayaz
Jan 10 at 10:22
$begingroup$
For the function to be defined everywhere, you need to explicitly state its value at the singular points (taking the limit for proper continuation), and this is absolutely equivalent to simplifying the form to $ax+b$. An homographic form is inappropriate.
$endgroup$
– Yves Daoust
Jan 10 at 10:26
$begingroup$
For the function to be defined everywhere, you need to explicitly state its value at the singular points (taking the limit for proper continuation), and this is absolutely equivalent to simplifying the form to $ax+b$. An homographic form is inappropriate.
$endgroup$
– Yves Daoust
Jan 10 at 10:26
add a comment |
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$begingroup$
What do you call "similar forms" and what is $n$ ?
$endgroup$
– Yves Daoust
Jan 9 at 13:10
$begingroup$
@YvesDaoust More clear?
$endgroup$
– mavavilj
Jan 9 at 13:13
$begingroup$
As clear as stuff can be.
$endgroup$
– Yves Daoust
Jan 9 at 13:13
$begingroup$
Use the term 'linear functional' very carefully. $f(x)=frac{x-c_1}{c_2-c_1}$ is a linear functional only for $c_1=0$
$endgroup$
– Shubham Johri
Jan 9 at 13:16