Fungsi Homogen Berderajat Sebarang

Diketahui ada kuantitas

f := \sum_{i_{1}, \dots, i_{n} = 1}^{r} M_{i_{1} \dots i_{n}} x_{i_{1}} \dots x_{i_{n}}

$f:=\sum_{i_1,\dots,i_n=1}^r{M}_{i_1\cdots{i}_n}x_{i_1}\cdots{x}_{i_n}$ , sehingga

\frac{\partial f}{\partial x_{i}} = r \sum i_{1}, \dots, i_{n} n \sum k = 1 M_{i_{1} \dots i_{n}} x_{i_{1}} \dots x_{i_{k - 1}} δ_{i i_{k}} x_{i_{k + 1}} \dots x_{i_{n}},

$\frac{\partial{f}}{\partial{x_i}}=\sum_{i_1,\dots,i_n}^r\sum_{k=1}^n{M}_{i_1\cdots{i}_n}{x}_{i_1}\cdots{x}_{i_{k-1}}\delta_{ii_k}{x}_{i_{k+1}}\cdots{x}_{i_n},$

\frac{\partial f}{\partial x_{i}} = n \sum k = 1 r \sum i_{1}, \dots, i_{k - 1}, i_{k + 1}, \dots, i_{n} = 1 M_{i_{1} \dots i_{k - 1} i i_{k + 1} \dots i_{n}} x_{i_{1}} \dots x_{i_{k - 1}} x_{i_{k + 1}} \dots x_{i_{n}},

$\frac{\partial{f}}{\partial{x_i}}=\sum_{k=1}^n\sum_{i_1,\dots,i_{k-1},i_{k+1},\dots,i_n=1}^r{M}_{i_1\cdots{i}_{k-1}ii_{k+1}\cdots{i}_n}{x}_{i_1}\cdots{x}_{i_{k-1}}{x}_{i_{k+1}}\cdots{x}_{i_n},$

$\sum_{i=1}^r{x_i}\frac{\partial{f}}{\partial{x_i}}=\sum_{k=1}^n\sum_{i,i_1,\dots,i_{k-1},i_{k+1},\dots,i_n=1}^r{M}_{i_1\cdots{i}_{k-1}ii_{k+1}\cdots{i}_n}{x}_{i_1}\cdots{x}_{i_{k-1}}x_i{x}_{i_{k+1}}\cdots{x}_{i_n}.$
Karena

${M}_{i_1\cdots{i}_{k-1}ii_{k+1}\cdots{i}_n}x_i=\sum_{i_k=1}^r\delta_{ii_k}{M}_{i_1\cdots{i}_{k-1}i_ki_{k+1}\cdots{i}_n}x_{i_k}$ , maka

$\sum_{i=1}^r{x_i}\frac{\partial{f}}{\partial{x_i}}=\sum_{k=1}^n\sum_{i,i_1,\dots,i_n=1}^r{M}_{i_1\cdots{i}_n}{x}_{i_1}\cdots{x}_{i_n}\delta_{ii_k}.$
Karena

$\sum_{k=1}^n\sum_{i=1}^r\delta_{ii_k}=n$ , maka

$\sum_{i=1}^r{x_i}\frac{\partial{f}}{\partial{x_i}}=nf.$

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Fungsi Homogen Berderajat Sebarang

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