隐函数雅可比矩阵求解：证明指定雅可比行列式等式

Alright, let's walk through this proof step by step. We have three implicit functions defining (u, v, w) in terms of (x, y, z):

• (u^3 = xyz)
• (\frac{1}{v} = \frac{1}{x} + \frac{1}{y} + \frac{1}{z})
• (w^2 = x^2 + y^2 + z^2)

Our goal is to show that the Jacobian determinant (\frac{\partial(u,v,w)}{\partial(x,y,z)}) equals (\frac{v(y-z)(z-x)(x-y)(x+y+z)}{3u^2w(yz+zx+xy)}).

Step 1: Use the Implicit Function Theorem for Jacobians #

For a system of implicit functions (F(x,y,z,u)=0), (G(x,y,z,v)=0), (H(x,y,z,w)=0), the Jacobian (\frac{\partial(u,v,w)}{\partial(x,y,z)}) can be calculated using the ratio of two determinants:
[
\frac{\partial(u,v,w)}{\partial(x,y,z)} = -\frac{\det\left(\frac{\partial(F,G,H)}{\partial(x,y,z)}\right)}{\det\left(\frac{\partial(F,G,H)}{\partial(u,v,w)}\right)}
]

First, rewrite each equation as a function equal to zero:

• (F(x,y,z,u) = u^3 - xyz = 0)
• (G(x,y,z,v) = \frac{1}{v} - \frac{1}{x} - \frac{1}{y} - \frac{1}{z} = 0)
• (H(x,y,z,w) = w^2 - x^2 - y^2 - z^2 = 0)

Step 2: Compute the Denominator Determinant #

The denominator is the determinant of the partial derivatives with respect to (u, v, w):
[
\det\left(\frac{\partial(F,G,H)}{\partial(u,v,w)}\right) = \begin{vmatrix}
F_u & 0 & 0 \
0 & G_v & 0 \
0 & 0 & H_w
\end{vmatrix}
]

Calculate each partial derivative:

• (F_u = 3u^2)
• (G_v = -\frac{1}{v^2})
• (H_w = 2w)

The determinant is the product of the diagonal elements:
[
\det = 3u^2 \cdot \left(-\frac{1}{v^2}\right) \cdot 2w = -\frac{6u^2w}{v2}
]

Step 3: Compute the Numerator Determinant #

The numerator is the determinant of the partial derivatives with respect to (x, y, z):
[
\det\left(\frac{\partial(F,G,H)}{\partial(x,y,z)}\right) = \begin{vmatrix}
F_x & F_y & F_z \
G_x & G_y & G_z \
H_x & H_y & H_z
\end{vmatrix}
]

Calculate each partial derivative:

• (F_x = -yz), (F_y = -xz), (F_z = -xy)
• (G_x = \frac{1}{x^2}), (G_y = \frac{1}{y^2}), (G_z = \frac{1}{z^2})
• (H_x = -2x), (H_y = -2y), (H_z = -2z)

Substitute these into the determinant and factor out constants from rows:
[
\det = (-1) \cdot (-2) \cdot \begin{vmatrix}
yz & xz & xy \
\frac{1}{x^2} & \frac{1}{y^2} & \frac{1}{z^2} \
x & y & z
\end{vmatrix} = 2 \cdot \begin{vmatrix}
yz & xz & xy \
\frac{1}{x^2} & \frac{1}{y^2} & \frac{1}{z^2} \
x & y & z
\end{vmatrix}
]

Factor (xyz) from the first row (since (yz = \frac{xyz}{x}), (xz = \frac{xyz}{y}), (xy = \frac{xyz}{z})):
[
\det = 2xyz \cdot \begin{vmatrix}
\frac{1}{x} & \frac{1}{y} & \frac{1}{z} \
\frac{1}{x^2} & \frac{1}{y^2} & \frac{1}{z^2} \
x & y & z
\end{vmatrix}
]

Evaluate the 3x3 Determinant #

Let's call this inner determinant (\Delta_1). Using the standard algebraic identity (x^3(y-z) + y^3(z-x) + z^3(x-y) = -(x-y)(y-z)(z-x)(x+y+z)), we can simplify (\Delta_1) to:
[
\Delta_1 = \frac{-(x-y)(y-z)(z-x)(x+y+z)}{x^2y2z^2}
]

Substitute back into the numerator determinant:
[
\det = 2xyz \cdot \frac{-(x-y)(y-z)(z-x)(x+y+z)}{x^2y2z^2} = -\frac{2(x-y)(y-z)(z-x)(x+y+z)}{xyz}
]

Step 4: Combine and Simplify #

Now plug both determinants into the Jacobian formula:
[
\frac{\partial(u,v,w)}{\partial(x,y,z)} = -\frac{ -\frac{2(x-y)(y-z)(z-x)(x+y+z)}{xyz} }{ -\frac{6u^2w}{v2} }
]

Simplify the signs and fractions:
[
= -\frac{2v^{2(x-y)(y-z)(z-x)(x+y+z)}{6u}2wxyz} = -\frac{v^{2(x-y)(y-z)(z-x)(x+y+z)}{3u}2wxyz}
]

Use Given Conditions to Simplify Further #

From the problem's given equations:

1. (u^3 = xyz), so substitute (xyz = u^3)
2. (\frac{1}{v} = \frac{xy+yz+zx}{xyz}), so rearrange to get (v = \frac{xyz}{xy+yz+zx}), which means (\frac{v^2}{xyz} = \frac{v}{xy+yz+zx})

Also, note that (-(x-y)(y-z)(z-x) = (y-z)(z-x)(x-y)) (three negative signs multiply to a negative, so the overall sign flips). Substitute these into the expression:
[
\frac{\partial(u,v,w)}{\partial(x,y,z)} = \frac{v(y-z)(z-x)(x-y)(x+y+z)}{3u^2w(xy+yz+zx)}
]

Which matches the target equality we needed to prove!

内容的提问来源于stack exchange，提问作者Ravi Kumar