- #1
Contingency
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Homework Statement
[itex]g:ℝ^{ 2 }\rightarrow ℝ[/itex] is everywhere differentiable.
For all (x,y) and for all t: [itex]g\left( tx,ty \right) =tg\left( x,y \right) [/itex].
Prove [itex]g[/itex] is linear (that there exist constants A, B such that for all (x,y): [itex]g\left( x,y \right) =Ax+By[/itex] .
I think my solution is correct, but the solution given after the problem is different. I'd like to know where I'm wrong and why!
Homework Equations
[itex]g\left( tx,ty \right) =tg\left( x,y \right)[/itex]
Chain rule
The Attempt at a Solution
Define [itex]f\left( x,y,t \right) =g\left( tx,ty \right)[/itex].
We then have [itex]f\left( x,y,t \right) =g\left( tx,ty \right) =tg\left( x,y \right)[/itex].
Define [itex]u\left( x,t \right) =tx,\quad v\left( x,y \right) =ty[/itex].
By the chain rule: [itex]\frac { \partial f }{ \partial t } =\frac { \partial g(u,v) }{ \partial u } \frac { \partial u(x,t) }{ \partial t } +\frac { \partial g(u,v) }{ \partial v } \frac { \partial v(y,t) }{ \partial t } =g(x,y)[/itex]
[itex]\Leftrightarrow\quad \frac { \partial f }{ \partial t } =\frac { \partial g(u,v) }{ \partial u } x+\frac { \partial g(u,v) }{ \partial v } y=g\left( x,y \right)[/itex]. This holds for any (x,y) and t.
Now choose [itex]t=0\quad \Rightarrow \quad { \frac { \partial f(x,y,t) }{ \partial t } }_{ (x,y,0) }={ \frac { \partial g(u,v) }{ \partial u } }_{ (0,0) }x+{ \frac { \partial g(u,v) }{ \partial v } }_{ (0,0) }y=g\left( x,y \right)[/itex], but [itex]{ \frac { \partial g(u,v) }{ \partial u } }_{ (0,0) },{ \frac { \partial g(u,v) }{ \partial v } }_{ (0,0) }[/itex] are constants, so we have found proper A,B and therefore the function is linear.
The problem is that in the solution, the differentiation process was different and I have no idea why.. It doesn't make sense. Here's an exact copy of what's written there (u,v are defined the same way):
... By the chain rule: [itex]\frac { \partial f }{ \partial t } =\frac { \partial g(u,v) }{ \partial x } \frac { \partial u(x,t) }{ \partial t } +\frac { \partial g(u,v) }{ \partial y } \frac { \partial v(y,t) }{ \partial t } =g(x,y)[/itex]
[itex]\Leftrightarrow \quad \frac { \partial f }{ \partial t } =\frac { \partial g(u,v) }{ \partial x } x+\frac { \partial g(u,v) }{ \partial y } y=g\left( x,y \right)[/itex]
[itex]...=\frac { \partial f }{ \partial t } =\frac { \partial g(xt,yt) }{ \partial x } x+\frac { \partial g(xt,yt) }{ \partial y } y=g\left( x,y \right)[/itex]. Then they choose t=0 and arrive at:
[itex]{ g }_{ x }(0,0)x+{ g }_{ y }(0,0)y=g\left( x,y \right) [/itex] where [itex]{ g }_{ x }(0,0),\quad { g }_{ y }(0,0)[/itex] are constants.
I don't understand the differentiation process in the solution, and it's really confusing me.
Please let me know where the mistake is!
Thanks!
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