Inverse 3D plot

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Haider al-kanan
Haider al-kanan 2016 年 5 月 6 日
編集済み: Haider al-kanan 2020 年 10 月 24 日
I want to inverse surface z , z=f(x,y).suc


Walter Roberson
Walter Roberson 2016 年 5 月 6 日
This is not well-defined in the way you stated it. z takes two parameters, but inverse takes one parameter. inverse would have to emit two parameters separately.
It would be more consistent if you had two functions, inversex and inversey, and you required
z(inversex(z),inversey(z)) = z
This just might be possible, I guess. But you would not be able to require that
inversex(z(x,y)) == x and inversey(z(x,y)) == y
You would need to allow the inversex and inversey to map to any (x,y) location that would produce the same z.
For example, if z = x^2 + 0*y, then inversex() could return either the positive or negative square root of x, and inversey() could return anything, because z() of that combination together would be 1, but there is no formula that, given (say) 17, can tell you whether that came from x = -sqrt(17) or x = +sqrt(17).

その他の回答 (1 件)

CS Researcher
CS Researcher 2016 年 5 月 6 日

In 2-D you can visualize the transformation as a matrix multiplication. If I is your matrix (x,y), you apply the transformation using a matrix A (it could be rotation, translation, etc.). The transformed matrix becomes

J = AI

If you apply $A^{-1}$ on J you get

 A^{-1}J = A^{-1}AI = I

Since inverse of an invertible matrix (check it out) multiplied by the matrix gives you identity matrix.

Were you looking for something like this?

  1 件のコメント
Haider al-kanan
Haider al-kanan 2016 年 5 月 7 日
Thanks for comment, but I don't think that works, here is my question x=0:0.1:10; y=0:0.1:10; [xx,yy]=meshgrid(x,y); zz=xx.^2-yy.^2; I am not sure if your suggestion works on that, could you please clarify to me how is that works?



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