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## #1 2011-08-14 18:48:15

George,Y
Member
Registered: 2006-03-12
Posts: 1,306

### Why Jocabian Determinants?

The 2D case of Jocabian determinants is easy:

If you want to express the area of dxdy by dudv, where (x,y) and (u,v) are different coordinate systems and d stands for differentiate.

dxdy= (dx/du dy/dv - dx/dv dy/du) dudv

which makes sense in geometry. The Jocabian Determinant in the parenthesis is in deed the area of dxdy in coordinate (u,v)

Is there a general proof for transformation of coordinates?

X'(y-Xβ)=0

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## #2 2011-08-16 14:37:18

George,Y
Member
Registered: 2006-03-12
Posts: 1,306

### Re: Why Jocabian Determinants?

I figured it out
Determinant shares the same property with "volume"
x -> ax  V->aV
x -> x+y V->V
x,y -> y,x V->-V
So it is reasonable to define a volumn in Rn by its determinant

X'(y-Xβ)=0

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## #3 2013-11-20 22:33:18

George,Y
Member
Registered: 2006-03-12
Posts: 1,306

anyone?

X'(y-Xβ)=0

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