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Not homework!

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Thread replies: 15
Thread images: 4

File: stirner.gif (17KB, 186x200px) Image search: [Google]
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If you can't solve this problem, then you are a brainlet who needs to get off this board.

Consider a function [math] F:\mathbb{C}^2\to\mathbb{C}^2 [/math] of the form F(x,y)=(p(x,y), q(x,y)) where the components p and q are polynomials.

Suppose the Jacobian determinant [math] p_xq_y-p_yq_x [/math] is a non-zero complex number.

Show that the components of G^{-1} are polynomials.

If you've taken any real analysis course this should not be difficult.
>>
>>8419362
>G
>>
>>8419375
Whoops. Meant to write

>Show that the components of F^{-1} are polynomials
>>
>Linear algebra, complex analysis

>"If you've taken any real analysis course this should not be difficult."

I don't even know why I'm responding to this. OP is a faggot. Do your own damn homework OP.
>>
>OP is baiting his question from the SQT thread because he couldn't figure out the case for n = 2
>>8417567
>>8417634
>>
Isn't this an unsolved math problem?
>>
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>>8419362
polynomials are a spook
>>
File: 1441994157075.png (52KB, 700x419px) Image search: [Google]
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>>8420556
>>
>>8419362
>is a non-zero complex number.
you mean it's constant?
>>
>>8421101
Yes
>>
>>8419362
8
>>
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>>8420556
>>8420701
>>
https://en.wikipedia.org/wiki/Jacobian_conjecture
>>
>>8419362
I have not taken a real analysis course, I am still working on its prerequisites. Not being at your level in mathematics does not make me a brainlet. Intelligence is not a function of knowledge, it is the capacity to gain, retain and use it.
>>
>>8419362
Let F be a polynomial.

Then F^-1 is 1 divided by that polynomial and therefore also a polynomial.

QED

Now hand over the Fields Medal.
Thread posts: 15
Thread images: 4


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