I am stuck with the following problem that says :
If $u_r=\frac{x_r}{\sqrt{1-x_1^2-x_2^2-x_3^2 \cdot \cdot \cdot-x_n^2}}$ where $r=1,2,3,\cdot \cdot \cdot ,n$, then prove that the jacobian of $u_1,u_2,\cdot \cdot, u_n$ with respect to $x_1,x_2,\cdot \cdot, x_n$ is $(1-x_1^2-x_2^2-\cdot \cdot \cdot -x_n^2)^{-\frac12}$
My try: Now, I have to calculate the value of
\begin{vmatrix}
\frac{\delta u_1}{\delta x_1} & \frac{\delta u_1}{\delta x_2} & \frac{\delta u_1}{\delta x_3} & \cdots & \frac{\delta u_1}{\delta x_n} \\
\frac{\delta u_2}{\delta x_1} & \frac{\delta u_2}{\delta x_2} & \frac{\delta u_2}{\delta x_3} & \cdots & \frac{\delta u_2}{\delta x_n} \\
\vdots & \vdots & \vdots & \ddots &\vdots \\
\frac{\delta u_n}{\delta x_1} & \frac{\delta u_n}{\delta x_2} & \frac{\delta u_n}{\delta x_3} & \cdots & \frac{\delta u_n}{\delta x_n} \\
\end{vmatrix}
Now,the value of $\frac{\delta u_1}{\delta x_1}=\frac{1-2x_1^2}{\{1-x_1^2\}^\frac32}$ , $\frac{\delta u_1}{\delta x_2}=\cdot \cdot =\frac{\delta u_1}{\delta x_n}=0$..
So, things are getting complicated. Can someone show me the right direction?
Best Answer
As the partial derivatives read: $$ \frac{\partial u_i}{\partial x_j}= \frac{\delta_{ij}}{\left(1-\sum_{k=1}^nx_k^2\right)^{1/2}} +\frac{x_ix_j}{\left(1-\sum_{k=1}^nx_k^2\right)^{3/2}}, $$ the jacobian matrix is of a form: $$ {\cal J}=c(I+ v^Tv), $$ with $c=\frac1{\left(1-\sum_{k=1}^{n}x_k^2\right)^{1/2}}$ and $v=\frac{(x_1,x_2,\dots, x_n)}{\left(1-\sum_{k=1}^{n}x_k^2\right)^{1/2}}$.
Therefore by the Matrix determinant lemma: $$\begin{array}{} \det{\cal J}&=c^n\det(I+ v^Tv)=c^n(1+vv^T)\\ &\displaystyle=\frac{1}{\left(1-\sum_{k=1}^{n}x_k^2\right)^{\frac n2}} \left[1+\frac{\sum_{k=1}^{n}x_k^2}{1-\sum_{k=1}^{n}x_k^2}\right]\\ &\displaystyle=\frac{1}{\left(1-\sum_{k=1}^{n}x_k^2\right)^{\frac n2+1}}. \end{array} $$
As stated already in a comment the correct result deviates from that claimed.