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Jacobian

Anonymous Sat 03 Aug 18:37:07 2019 No.10861153 View Reply Original Report

Quoted By: >>10861163 >>10861176 >>10863165

Is it possible to find the Jacobian $J=J(u,v)$ if $u=(x^3)/y$ and $v=3x+2y$ ? I can't find the way to get rid of those $x$ and $y$ in the result.

Anonymous

Anonymous Sat 03 Aug 2019 18:42:44 No.10861163 Report

Quoted By: >>10861172 >>10861185 >>10861188

>>10861153
Since when does the Jacobian ever take the form 1/det(A)?

Anonymous

Anonymous Sat 03 Aug 2019 18:46:41 No.10861172 Report

Quoted By: >>10861188

>>10861163
I think he wants to find the jacobian of the inverse map.

Anonymous

Anonymous Sat 03 Aug 2019 18:47:49 No.10861176 Report

Quoted By: >>10861185 >>10863093

>>10861153
The Jacobean should be:
$\frac{\partial (u,v)}{\partial (x,y)} = \begin{vmatrix}<br />\frac{\partial u}{\partial x} & \frac{\partial v}{\partial x}\\ <br />\frac{\partial u}{\partial y} & \frac{\partial v}{\partial y}<br />\end{vmatrix}$ .

Anonymous

Anonymous Sat 03 Aug 2019 18:53:25 No.10861185 Report

Quoted By:

>>10861163
that's right
>>10861176
there are two ways, but that's not the problem.

Anonymous

Anonymous Sat 03 Aug 2019 18:54:30 No.10861188 Report

Quoted By: >>10861194

>>10861172
>>>10861163
>that's right
shit, i meant to quote>>10861172

Anonymous

Anonymous Sat 03 Aug 2019 18:55:22 No.10861194 Report

Quoted By:

>>10861188
I see.

Anonymous

Anonymous Sun 04 Aug 2019 10:42:52 No.10862754 Report

Quoted By:

Bump

Anonymous

Anonymous Sun 04 Aug 2019 10:59:16 No.10862778 Report

Quoted By: >>10863093

Im not a mathematician, but are you sure the function is invertable?

Anonymous

Anonymous Sun 04 Aug 2019 15:11:46 No.10863093 Report

Quoted By: >>10863105

>>10862778
That' not the point. You can do it like >>10861176 said, but the problem is how to get rid of x and y in the final result. I need the J to be the function of u and v only.

Anonymous

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Anonymous Sun 04 Aug 2019 15:19:53 No.10863105 Report

Quoted By: >>10863127

>>10863093
But you cannot get rid of x and y if an inverse function doesn't exist. You basically start with the function F(x,y) = (u(x,y), v(x,y)). But you can't get
$F^{-1} (u,v) = (x(u.v), y(u,v))$ .

Anonymous

Anonymous Sun 04 Aug 2019 15:31:16 No.10863127 Report

Quoted By: >>10863171

>>10863105
Okay, I see now. Basically, the problem is unsolvable.

Anonymous

Anonymous Sun 04 Aug 2019 15:50:16 No.10863165 Report

Quoted By: >>10863175

>>10861153
You’re an idiot.
your substitution converts a coordinate system in u and v to a system of x and y, so the jacobian of this should be in terms of x and y. If you want to go from x and y to u and v, your equations should compute x and y in terms of u and v.

Anonymous

Anonymous Sun 04 Aug 2019 15:52:56 No.10863171 Report

Quoted By: >>10863204

>>10863127
It may just possibly be. And probably is most likely, given that you spent so much time on it. But, like I said, I'm not a mathematician, so I don't know how to invert a function like that....of even check to see if it's invertable.

Anonymous

Anonymous Sun 04 Aug 2019 15:55:49 No.10863175 Report

Quoted By: >>10863185

>>10863165
I bet you think your post was smart. That's impossible to do, unless you want to do Newton-Raphson on four pages.

Anonymous

Anonymous Sun 04 Aug 2019 15:59:03 No.10863180 Report

Quoted By: >>10863185 >>10863204

I don't think it is impossible to express x and y as functions of u and v over the specified region, but it looks like it requires solving a cubic equation.

Anonymous

Anonymous Sun 04 Aug 2019 16:02:49 No.10863185 Report

Quoted By: >>10863204

>>10863180
>it looks like it requires solving a cubic equation
it does and it's as tedious as it gets, almost impossible to solve. see the other half of post>>10863175

Anonymous

Anonymous Sun 04 Aug 2019 16:16:05 No.10863204 Report

Quoted By: >>10863209

>>10863171
>>10863180
>>10863185
But all you need to find is one u and v in the defined region such that there is no unique x,y in the preimage of the defined region. Then you'll know for sure that the function not invertable.

Anonymous

Anonymous Sun 04 Aug 2019 16:19:19 No.10863209 Report

Quoted By: >>10863211

>>10863204
And I found one!
Let u = v = 1.
This is on the defined domain, but provides no unique solution to x.

Anonymous

Anonymous Sun 04 Aug 2019 16:21:20 No.10863211 Report

Quoted By:

>>10863209
No, fug, I messed up false alarm

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