# number of vectors in the decomposition of vector

Let $x \in R^n$ be vector with entries $x_1\geq\cdots \geq x_n$.

I would like to decompose this vector into vectors $x^1=(x_1,\ldots,x_m,0,\ldots0), x^2=(0,\ldots,0,x_{m+1},\ldots,x_1,0,\ldots,0)$..., where $2x_m>{x_1}$; $2x_{m+1}<{x_1}$; $x_{m+1}<2x_l;x_{l+1}>2x_{l+1}$ and so on.

How many of those vectors $x^i$? (My filing it should be $log$ terms, but I would like to see the proof of it. Any good sours will be also helpful).

Thank you.

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It will depend on $x$; you may end up using anywhere from $1$ to $n$ vectors. For example, if $x = (4,4,4,3,3,3,3)$ then you will only have one vector in the decomposition. But if $x = (10000,4000,1000,400,100,40,10)$ then you will need $n = 7$ vectors.