Exercise 10.7

Question

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\def\dcro{\mbox{]\hspace{-.15em}]}}

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ewcommand{\F}{\mathbb{F}}

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Let $f \in F[x_0,\ldots,x_n]$. One can define the partial derivatives $\partial f/\partial x_0, \ldots,\partial f/ \partial x_n$ in a formal way. Suppose that $f$ is homogeneous of degree $m$. Prove that $\sum_{i=0}^n x_i (\partial f/ \partial x_i) = mf$. This result is due to Euler. (Hint: Do it first for the case that $f$ is a monomial.)

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
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ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof}
For the case that $f = x_1^{a_1}\cdots x_n^{a_n}$ is a monomial, where $a_1+ \ldots + a_n = m = \deg(f)$, then
$$\frac{\partial f}{\partial x_i} = a_i x_1^{a_1}\cdots x_i^{a_i - 1}\cdots x_n^{a_n}, \qquad i=1,\ldots,n.$$
Therefore $x_i \partial f/ \partial x_i = a_i f$, and
$$\sum_{i=1}^n x_i \frac{\partial f}{\partial x_i}  =\left (\sum_{i=1}^n a_i ight ) f = m f.$$
Since the maps $f \mapsto \sum_{i=1}^n x_i \frac{\partial f}{\partial x_i} $ and $f \mapsto mf$ are $FG-$ linear, and since every homogeneous polynomial $f$ is a linear combination of monomial with degree $m$, the relation is true for all such polynomials.

To conclude, every homogeneous polynomial $f \in F[x_0,\ldots,x_n]$ of degree $m$ satisfies
$$\sum_{i=1}^n x_i \frac{\partial f}{\partial x_i}  = m f.$$
\end{proof}

Exercise 10.7

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Exercise 10.7

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