By differentiating the equation in Euler's theorem with respect to x_i and summing over i, prove that if f(x) is homogenous of degree m then
$$ m(m-1)f(x) = x^{T}H(f(x))x $$
Eulers Theorem gives $ mf(x) = \sum_{i=0}^{n} x_i \frac{\partial f}{\partial x_{i}}$
I am not sure how to differentiate this with respect to xi can someone please help me this part? Thank you
I don't follow the answer in my book, $$ m\frac{ \partial f }{\partial x_i} = \sum_{j=1}^{n} x_j \frac{ \partial^{2}f}{ \partial x_j \partial x_i} + \frac{ \partial f}{\partial x_i} $$
why is it not $$\sum_{j=1}^{n} x_j \frac{ \partial^{2}f}{ \partial x_j \partial x_i}$ + $\frac{ \partial f}{\partial x_i} \frac{ \partial f x_j}{\partial x_i} $$ ??? ?
$$\begin{align}\frac{\partial}{\partial x_i}\left( \sum_{j=0}^n x_j \frac{\partial f}{\partial x_j}\right) &= \sum_{j=0}^n \frac{\partial}{\partial x_i}\left(x_j \frac{\partial f}{\partial x_j}\right)\\&=\sum_{j=0}^n\left[ \frac{\partial x_j}{\partial x_i}\frac{\partial f}{\partial x_j}+x_j \frac{\partial}{\partial x_i}\left(\frac{\partial f}{\partial x_j}\right) \right]\\&=\sum_{j=0}^n \frac{\partial x_j}{\partial x_i}\frac{\partial f}{\partial x_j} + \sum_{j=0}^n x_j \frac{\partial}{\partial x_i}\left(\frac{\partial f}{\partial x_j}\right)\\&=\frac{\partial f}{\partial x_i} + \sum_{j=0}^n x_j \frac{\partial^2 f}{\partial x_i\partial x_j}\end{align}$$
Since $$\frac{\partial x_j}{\partial x_i} = \begin{cases}1 & i=j\\0& i \ne j\end{cases}$$