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chisigma · #15 March 12th, 2009, 04:03 AM

In order to have an idea about how we can solve the problem let’s consider the minimum of this function…

$f(x)= \sum_ {i=1}^{2} |x-2^{i}|$ (1)

… which is represented [in blue] here…

http://digilander.libero.it/luposabatini/MHF1.bmp

The (1) is in fact a ‘straight-line function’ the slope of which changes in $x=2$ and $x=4$ . More exactly, the function starts with negative slope $s=-2$ , in $x=2$ becames ‘flat’ [ $s=0$ , and for $x > 4$ the slope is positive [ $s=2$ ]. The ‘minimum’ of (1) in fact is not rescticted to a single point, but in extended to the interval $2 \le x \le 4$ , where is $f(x)=2$ . In similar way we can ‘attach’ the proposed problem, i.e. finding the minimum of…

$f(x)= \sum_ {i=1}^{16} |x-2^{i}|$ (2)

As the (1), the (2) is also ‘straight-line’ . In $x=0$ is…

$f(0)= \sum_ {i=1}^{16} 2^{i}$ $= 2 \cdot (2^{16}-1)=131070$

The functions starts with negative slope $s=-16$ and each time that $x=2^{i}$ the slope is increased by $+2$ . So we have…

$x=0, s=-16; x=2, s=-14; x=4, s=-12;\dots; x=2^{7}, s=-2; x=2^{8}, s=0;\dots$

So the functions become ‘flat’ in the interval $256 \le x \le 512$ and here exhibits its ‘minimum’ which is [if no mistakes of mine…] …

$f(2^{8})= \sum_{i=1}^{16} |2^{8}-2^{i}|= 2\cdot (2^{8}-1)^ {2}= 130050$

Regards