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Let $f(x):\mathbb R^n \to \mathbb R$ be a smooth function, and let $f(0)=0$.

I alway see that someone rewrite the function in the form $f(x)=x_1b_1(x_1)+x_2b_2(\overline {x_2})+...+x_nb_n(\overline x)$,

where the $\overline {x_i}=[x_1\ \ x_2\ \ ...\ \ x_i]^\mathrm{T}$

Can someone provide a proof?

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Find $b_1$ by substituting $x_2 = \cdots = x_n = 0$. Subtract $x_1b_1(x_1)$ and repeat. –  Yuval Filmus Feb 9 '11 at 7:13
    
$f(\overline{x_2})=x_1b_1(x_1)+x_2b_2(x_2)$\\If $x_2=0$,\\ $\rightarrow b_1(x_1)=f(\overline {x_2})/x_1$ is in conflict, because $b_1(x_1)$ have only one variable $x_1$ ?? –  Mroei Feb 9 '11 at 9:18
    
$\overline{x_2}$ depends only on $x_1$ and $x_2$ -- since you've chosen a value for $x_2$, it actually only depends on $x_1$, so $b_1$ does actually only depend on $x_1$. Analogously in the remaining steps. –  joriki Feb 9 '11 at 10:18
    
So the proof:There exist functions $b_i(\cdot)$ such that $f(\overline {x_n})=x_1b_1(x_1)+\cdots$ \\ Step 1:$x_2=x_3=\cdots=x_n=0$\\ $b_1(x_1)=f(x_1)/x_1$\\ Step 2:$x_3=x_4=\cdots=x_n=0$\\ $f(x_2)=x_1b_1(x_1)+x_2b_2(\overline {x_2})$\\ $\Rightarrow b_2(\overline {x_2})=\frac{f(x_2)-x_1b_1(x_1)}{x_2}$\\ (Repeat to step n-1) –  Mroei Feb 10 '11 at 3:07
    
If $x_2\neq 0, x_3\neq 0$, we can't verify $b_1(x_1)$ has variable $x_1$ and $b_2,b_3,\cdots$ ?? –  Mroei Feb 11 '11 at 6:39

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