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  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -

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#1 2011-08-15 16:48:15

George,Y
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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)

But how about n-dimensional case?

Is there a general proof for transformation of coordinates?
yikes


X'(y-Xβ)=0

#2 2011-08-17 12:37:18

George,Y
Super Member

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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

#3 2013-11-21 21:33:18

George,Y
Super Member

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Re: Why Jocabian Determinants?

anyone?


X'(y-Xβ)=0

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