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 $$\sum_{j=1}^{n} x_j \frac{ \partial^{2}f}{ \partial x_j \partial x_i}$ + $\frac{ \partial f}{\partial x_i} \frac{ \partial f x_j}{\partial x_i} $$ ??? ?

  • $\begingroup$ You can write a bigger central equation using the double dollar (just write the formula between two $$ $\endgroup$ Commented Nov 28, 2017 at 22:52

1 Answer 1


$$\begin{align}\frac{\partial}{\partial x_i}\left( \sum_{j=0}^n x_j \frac{\partial f}{\partial x_j}\right) &= \sum_{j=0}^n \frac{\partial}{\partial x_i}\left(x_j \frac{\partial f}{\partial x_j}\right)\\&=\sum_{j=0}^n\left[ \frac{\partial x_j}{\partial x_i}\frac{\partial f}{\partial x_j}+x_j \frac{\partial}{\partial x_i}\left(\frac{\partial f}{\partial x_j}\right) \right]\\&=\sum_{j=0}^n \frac{\partial x_j}{\partial x_i}\frac{\partial f}{\partial x_j} + \sum_{j=0}^n x_j \frac{\partial}{\partial x_i}\left(\frac{\partial f}{\partial x_j}\right)\\&=\frac{\partial f}{\partial x_i} + \sum_{j=0}^n x_j \frac{\partial^2 f}{\partial x_i\partial x_j}\end{align}$$

Since $$\frac{\partial x_j}{\partial x_i} = \begin{cases}1 & i=j\\0& i \ne j\end{cases}$$

  • $\begingroup$ Thank you so much for the explanation, it's definitely lot clearer, can I also ask why we switch from i in the original summation to j? $\endgroup$
    – italy
    Commented Nov 29, 2017 at 12:55
  • 1
    $\begingroup$ Because it is $x_i$ we are differentiating with respect to. In the summation, the index $j$ (or $i$ in your original statement) is a dummy variable. The expression does not depend on a value of $j$. It has the same meaning and same value regardless of what letter is used here. But the partial differentiation is outside the summation. It's index is some given value. It is no longer free to be used for other purposes such as the index of the summation. That would be giving $i$ two different meanings in the same context, which leads to great confusion. $\endgroup$ Commented Nov 29, 2017 at 17:38
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    $\begingroup$ If I had left $i$ as the index of summation, once I moved the differentiation inside the summation, I would have $i$ used in one place to mean the external $i$ given with the problem, and $i$ used in two other places to mean the index of the summation. How would someone reading this expression know which is which? Therefore it was necessary to change the dummy variable used as the index of summation to something else that would not be confused with $i$. $\endgroup$ Commented Nov 29, 2017 at 17:41

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