Reverse of Euler's Homogeneous Function Theorem

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Euler's Homogeneous Function Theorem says,
if $$f(tx,ty)=t^nf(x,y),$$
then one has: $$xfracpartial fpartial x(x,y)+yfracpartial fpartial x(x,y)=nf(x,y)$$



However, can we reverse it? ie.
if $$f(tx,ty)=t^nf(x,y),$$
and there exist 2 functions $g(x,y)$ and $h(x,y)$ such that $$xg(x,y)+yh(x,y)=nf(x,y).$$
then can we say that it must be $$g(x,y)=fracpartial fpartial x(x,y)$$ and $$h(x,y)=fracpartial fpartial x(x,y),?$$




calculus




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  • @GiuseppeNegro updated again.
    – athos
    Jul 31 at 9:36






  • 1




    Not exactly your question, but there is a different way of "reversing" Euler's theorem: en.wikipedia.org/wiki/Homogeneous_function#Positive_homogeneity
    – Theoretical Economist
    Jul 31 at 11:57










  • now i get the whole picture, thanks! @TheoreticalEconomist
    – athos
    Aug 2 at 0:41















up vote
1
down vote

favorite












Euler's Homogeneous Function Theorem says,
if $$f(tx,ty)=t^nf(x,y),$$
then one has: $$xfracpartial fpartial x(x,y)+yfracpartial fpartial x(x,y)=nf(x,y)$$



However, can we reverse it? ie.
if $$f(tx,ty)=t^nf(x,y),$$
and there exist 2 functions $g(x,y)$ and $h(x,y)$ such that $$xg(x,y)+yh(x,y)=nf(x,y).$$
then can we say that it must be $$g(x,y)=fracpartial fpartial x(x,y)$$ and $$h(x,y)=fracpartial fpartial x(x,y),?$$




calculus




share|cite|improve this question





















  • @GiuseppeNegro updated again.
    – athos
    Jul 31 at 9:36






  • 1




    Not exactly your question, but there is a different way of "reversing" Euler's theorem: en.wikipedia.org/wiki/Homogeneous_function#Positive_homogeneity
    – Theoretical Economist
    Jul 31 at 11:57










  • now i get the whole picture, thanks! @TheoreticalEconomist
    – athos
    Aug 2 at 0:41













up vote
1
down vote

favorite









up vote
1
down vote

favorite











Euler's Homogeneous Function Theorem says,
if $$f(tx,ty)=t^nf(x,y),$$
then one has: $$xfracpartial fpartial x(x,y)+yfracpartial fpartial x(x,y)=nf(x,y)$$



However, can we reverse it? ie.
if $$f(tx,ty)=t^nf(x,y),$$
and there exist 2 functions $g(x,y)$ and $h(x,y)$ such that $$xg(x,y)+yh(x,y)=nf(x,y).$$
then can we say that it must be $$g(x,y)=fracpartial fpartial x(x,y)$$ and $$h(x,y)=fracpartial fpartial x(x,y),?$$




calculus




share|cite|improve this question













Euler's Homogeneous Function Theorem says,
if $$f(tx,ty)=t^nf(x,y),$$
then one has: $$xfracpartial fpartial x(x,y)+yfracpartial fpartial x(x,y)=nf(x,y)$$



However, can we reverse it? ie.
if $$f(tx,ty)=t^nf(x,y),$$
and there exist 2 functions $g(x,y)$ and $h(x,y)$ such that $$xg(x,y)+yh(x,y)=nf(x,y).$$
then can we say that it must be $$g(x,y)=fracpartial fpartial x(x,y)$$ and $$h(x,y)=fracpartial fpartial x(x,y),?$$





calculus





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 31 at 9:35
























asked Jul 31 at 9:22









athos

70411236




70411236











  • @GiuseppeNegro updated again.
    – athos
    Jul 31 at 9:36






  • 1




    Not exactly your question, but there is a different way of "reversing" Euler's theorem: en.wikipedia.org/wiki/Homogeneous_function#Positive_homogeneity
    – Theoretical Economist
    Jul 31 at 11:57










  • now i get the whole picture, thanks! @TheoreticalEconomist
    – athos
    Aug 2 at 0:41

















  • @GiuseppeNegro updated again.
    – athos
    Jul 31 at 9:36






  • 1




    Not exactly your question, but there is a different way of "reversing" Euler's theorem: en.wikipedia.org/wiki/Homogeneous_function#Positive_homogeneity
    – Theoretical Economist
    Jul 31 at 11:57










  • now i get the whole picture, thanks! @TheoreticalEconomist
    – athos
    Aug 2 at 0:41
















@GiuseppeNegro updated again.
– athos
Jul 31 at 9:36




@GiuseppeNegro updated again.
– athos
Jul 31 at 9:36




1




1




Not exactly your question, but there is a different way of "reversing" Euler's theorem: en.wikipedia.org/wiki/Homogeneous_function#Positive_homogeneity
– Theoretical Economist
Jul 31 at 11:57




Not exactly your question, but there is a different way of "reversing" Euler's theorem: en.wikipedia.org/wiki/Homogeneous_function#Positive_homogeneity
– Theoretical Economist
Jul 31 at 11:57












now i get the whole picture, thanks! @TheoreticalEconomist
– athos
Aug 2 at 0:41





now i get the whole picture, thanks! @TheoreticalEconomist
– athos
Aug 2 at 0:41











1 Answer
1






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The answer is no. Consider $f(x, y)=x^2+y^2$, which is $2$-homogeneous. It satisfies
$$
2f(x, y)= xg(x, y)+yh(x, y)$$
with $g(x, y)= 2x+ 2y, h(x, y)=2y-2x$, so $gne partial_x f$ and $hnepartial_y f$.






share|cite|improve this answer





















  • How about in general $n$? Eg if $n=1$, does it hold ?
    – athos
    Jul 31 at 10:08










  • No. With $f=(x^2+y^2)^n/2$ and $g=partial_x f +ny, h=partial_y f-nx$ you get a counterexample for all $n$.
    – Giuseppe Negro
    Jul 31 at 10:11











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
3
down vote



accepted










The answer is no. Consider $f(x, y)=x^2+y^2$, which is $2$-homogeneous. It satisfies
$$
2f(x, y)= xg(x, y)+yh(x, y)$$
with $g(x, y)= 2x+ 2y, h(x, y)=2y-2x$, so $gne partial_x f$ and $hnepartial_y f$.






share|cite|improve this answer





















  • How about in general $n$? Eg if $n=1$, does it hold ?
    – athos
    Jul 31 at 10:08










  • No. With $f=(x^2+y^2)^n/2$ and $g=partial_x f +ny, h=partial_y f-nx$ you get a counterexample for all $n$.
    – Giuseppe Negro
    Jul 31 at 10:11















up vote
3
down vote



accepted










The answer is no. Consider $f(x, y)=x^2+y^2$, which is $2$-homogeneous. It satisfies
$$
2f(x, y)= xg(x, y)+yh(x, y)$$
with $g(x, y)= 2x+ 2y, h(x, y)=2y-2x$, so $gne partial_x f$ and $hnepartial_y f$.






share|cite|improve this answer





















  • How about in general $n$? Eg if $n=1$, does it hold ?
    – athos
    Jul 31 at 10:08










  • No. With $f=(x^2+y^2)^n/2$ and $g=partial_x f +ny, h=partial_y f-nx$ you get a counterexample for all $n$.
    – Giuseppe Negro
    Jul 31 at 10:11













up vote
3
down vote



accepted







up vote
3
down vote



accepted






The answer is no. Consider $f(x, y)=x^2+y^2$, which is $2$-homogeneous. It satisfies
$$
2f(x, y)= xg(x, y)+yh(x, y)$$
with $g(x, y)= 2x+ 2y, h(x, y)=2y-2x$, so $gne partial_x f$ and $hnepartial_y f$.






share|cite|improve this answer













The answer is no. Consider $f(x, y)=x^2+y^2$, which is $2$-homogeneous. It satisfies
$$
2f(x, y)= xg(x, y)+yh(x, y)$$
with $g(x, y)= 2x+ 2y, h(x, y)=2y-2x$, so $gne partial_x f$ and $hnepartial_y f$.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 31 at 9:40









Giuseppe Negro

15.6k328117




15.6k328117











  • How about in general $n$? Eg if $n=1$, does it hold ?
    – athos
    Jul 31 at 10:08










  • No. With $f=(x^2+y^2)^n/2$ and $g=partial_x f +ny, h=partial_y f-nx$ you get a counterexample for all $n$.
    – Giuseppe Negro
    Jul 31 at 10:11

















  • How about in general $n$? Eg if $n=1$, does it hold ?
    – athos
    Jul 31 at 10:08










  • No. With $f=(x^2+y^2)^n/2$ and $g=partial_x f +ny, h=partial_y f-nx$ you get a counterexample for all $n$.
    – Giuseppe Negro
    Jul 31 at 10:11
















How about in general $n$? Eg if $n=1$, does it hold ?
– athos
Jul 31 at 10:08




How about in general $n$? Eg if $n=1$, does it hold ?
– athos
Jul 31 at 10:08












No. With $f=(x^2+y^2)^n/2$ and $g=partial_x f +ny, h=partial_y f-nx$ you get a counterexample for all $n$.
– Giuseppe Negro
Jul 31 at 10:11





No. With $f=(x^2+y^2)^n/2$ and $g=partial_x f +ny, h=partial_y f-nx$ you get a counterexample for all $n$.
– Giuseppe Negro
Jul 31 at 10:11













 

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