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In the following notation this is how we determine the partial deriatives of a function using the Implicit Function Theorem, but I haven't seen this notation before, and is a bit unsure of how exactly to read this, specifically the right side of the equation: $\cfrac{\partial \phi_i}{\partial x_j}=\cfrac{\cfrac{\partial (F_{(1)},F_{(2)},\ldots,F_{(n)})}{\partial ( y_1,\ldots,x_j,\ldots,y_n)}}{\cfrac{\partial (F_{(1)},F_{(2)},\ldots,F_{(n)})}{(y_1,\ldots,y_i,\ldots,y_n)}}$

Here is an image providing context for the entire thing, but it is really only the notation on the right side of the previously shown equation that I haven't seen before. Context for entire notation of IFT

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  • $\begingroup$ I recommend looking at the case when $n = 2$. For example take $F(x,y) = x^2 + y^2$ $\endgroup$
    – IAmNoOne
    Commented May 22, 2018 at 8:26
  • $\begingroup$ Ah, so it is the Determinant of the Jacobian. So it'd be a determinant of a $n\times n$ matrix divided by a $n\times n$ matrix? $\endgroup$
    – Johan
    Commented May 22, 2018 at 8:53
  • $\begingroup$ Both should be determinants, see Cramer's rule. Essentially, you have to solve $$\frac{∂F}{∂y}\frac{∂ϕ}{∂x}=-\frac{∂F}{∂x},$$ and the solution can be written down using Cramer. Note that that formula is quite inefficient in practice for dimensions larger 3. The denominator in the numerator should better be written as $∂(y_1,…,y_{i-1},x_j,y_{i+1},…,y_n)$, this is what is hinted at by writing $y_i$ in the denominator of the denominator. $\endgroup$ Commented May 22, 2018 at 9:11
  • $\begingroup$ Could you elaborate on why that is? $\endgroup$
    – Johan
    Commented May 22, 2018 at 9:21

1 Answer 1

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The expression $${\partial(F_1,F_2,\ldots,F_n)\over\partial (y_1,y_2\ldots,x_j,\ldots,y_n)}$$ is meant to be the determinant of the matrix $$\left[\matrix{ {\partial F_1\over\partial y_1}&{\partial F_1\over\partial y_2}&\cdots& {\partial F_1\over\partial x_j}&\cdots &{\partial F_1\over\partial y_n}\cr {\partial F_2\over\partial y_1}&{\partial F_2\over\partial y_2}&\cdots& {\partial F_2\over\partial x_j}&\cdots &{\partial F_2\over\partial y_n}\cr \vdots\cr {\partial F_n\over\partial y_1}&{\partial F_n\over\partial y_2}&\cdots& {\partial F_n\over\partial x_j}&\cdots &{\partial F_n\over\partial y_n}\cr}\right]\ ,$$ and similarly for the expression in the denominator of the quoted formula. Note that a minus sign is missing.

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