Minimum of Ʃ[itex]^{n}_{i=1}[/itex][itex]\left|x_{i}-z\right|[/itex]

[no_alone]

Homework Statement

Hello,
I have a question what is
What is the minimum of Ʃ$^{n}_{i=1}$$\left|x_{i}-z\right|$
x$_{i}$ are constants

Homework Equations

Not Much

The Attempt at a Solution

I split the sum to 3 sums
Ʃ$^{}_{x_{i}>z}$$(x_{i}-z)$ + Ʃ$^{}_{x_{i}<z}$$(-x_{i}+z)$+Ʃ$^{}_{x_{i}=z}$$(x_{i}-z)$
then I split the sums again:
Ʃ$^{}_{x_{i}>z}$$(x_{i})$+Ʃ$^{}_{x_{i}>z}$$(-z)$ + Ʃ$^{}_{x_{i}<z}$$(-x_{i})$ + Ʃ$^{}_{x_{i}<z}$$(z)$+Ʃ$^{}_{x_{i}=z}$$(x_{i}-z)$
This is Equall to
Ʃ$^{}_{x_{i}>z}$$(x_{i})$-$n^{}_{x_{i}>z}$$*z$ + Ʃ$^{}_{x_{i}<z}$$(-x_{i})$ + $n^{}_{x_{i}<z}$$*z$+Ʃ$^{}_{x_{i}=z}$$x_{i}-z$-$n^{}_{x_{i}=z}$$z$

Then I derivative by z and =0 to find maximum
I get:
-$n^{}_{x_{i}>z}$+ $n^{}_{x_{i}<z}$-$n^{}_{x_{i}=z}$=0

If x_{i}≠z for every i then
-$n^{}_{x_{i}>z}$+ $n^{}_{x_{i}<z}$=0

$n^{}_{x_{i}<z}$=$n^{}_{x_{i}>z}$

If there is i that x_{i}=z
$n^{}_{x_{i}<z}$ = $n^{}_{x_{i}>z}$+$n^{}_{x_{i}=z}$

What we get is that the minimus is when z is the median of x_{i}

But I have a problem, when I try for example to derivate:
$n^{}_{x_{i}<z}$$*z$
n is dependent on z soo... I cannot do it! I don't know how to derivate it...

Thank's for the help

 

[MathematicalPhysicist]

This is a nonnegative function, so you want to find for which value of z it gets the minimal value.

As always we have the arithmetic-geometrical means inequality:

$$\sum_{i=1}^{n}\frac{|z-x_i|}{n} \geq \prod_{i=1}^{n} (|z-x_i|)^{\frac{1}{n}}$$

So the minimum value of this function is achieved when the above LHS equals the RHS, and this happens when:

$$|z-x_i|=|z-x_j|$$ for all $$i\neq j$$.

 

[Ray Vickson]

Homework Statement

Hello,
I have a question what is
What is the minimum of Ʃ$^{n}_{i=1}$$\left|x_{i}-z\right|$
x$_{i}$ are constants

Homework Equations

Not Much

The Attempt at a Solution

I split the sum to 3 sums
Ʃ$^{}_{x_{i}>z}$$(x_{i}-z)$ + Ʃ$^{}_{x_{i}<z}$$(-x_{i}+z)$+Ʃ$^{}_{x_{i}=z}$$(x_{i}-z)$
then I split the sums again:
Ʃ$^{}_{x_{i}>z}$$(x_{i})$+Ʃ$^{}_{x_{i}>z}$$(-z)$ + Ʃ$^{}_{x_{i}<z}$$(-x_{i})$ + Ʃ$^{}_{x_{i}<z}$$(z)$+Ʃ$^{}_{x_{i}=z}$$(x_{i}-z)$
This is Equall to
Ʃ$^{}_{x_{i}>z}$$(x_{i})$-$n^{}_{x_{i}>z}$$*z$ + Ʃ$^{}_{x_{i}<z}$$(-x_{i})$ + $n^{}_{x_{i}<z}$$*z$+Ʃ$^{}_{x_{i}=z}$$x_{i}-z$-$n^{}_{x_{i}=z}$$z$

Then I derivative by z and =0 to find maximum
I get:
-$n^{}_{x_{i}>z}$+ $n^{}_{x_{i}<z}$-$n^{}_{x_{i}=z}$=0

If x_{i}≠z for every i then
-$n^{}_{x_{i}>z}$+ $n^{}_{x_{i}<z}$=0

$n^{}_{x_{i}<z}$=$n^{}_{x_{i}>z}$

If there is i that x_{i}=z
$n^{}_{x_{i}<z}$ = $n^{}_{x_{i}>z}$+$n^{}_{x_{i}=z}$

What we get is that the minimus is when z is the median of x_{i}

But I have a problem, when I try for example to derivate:
$n^{}_{x_{i}<z}$$*z$
n is dependent on z soo... I cannot do it! I don't know how to derivate it...

Thank's for the help

Setting the derivative to zero is a mistake: the function $f(z) =\sum_i |z - x_i|$ is piecewise differentiable only, and may not have a derivative at all at the minimizing point. For example, draw the graph of $f(z) = |z-1|+|z-2|+|z-3|.$

You are correct that the minimum is at the median, but there is an easier way to get this: imagine that the x_i are sorted in increasing order, so $x_1 < x_2 < \cdots < x_n$; I'll let you worry about the modifications needed in the argument if some of the x_i are equal. So, to the left of x_1 the slope of the graph of f(z) is -n, because
$$f(z) = \sum_{i=1}^n x_i - nz$$ when z < x_1 . As z increases through x_1 the slope increases by 2, to become -n+2. Every time z passes through the next x_i the slope increases by 2.

 

[Michael Redei]

If you're looking for a minimum of the form $\sum|x|$, you can look at the function $\sum x^2$ instead, since |x| and x² are extreme at the same places.

If you do that for this case, you'll be led to the arithmetic mean of the x_i and not their median.

 

[Ray Vickson]

If you're looking for a minimum of the form $\sum|x|$, you can look at the function $\sum x^2$ instead, since |x| and x² are extreme at the same places.

If you do that for this case, you'll be led to the arithmetic mean of the x_i and not their median.

In this problem the minimum is at the median, not at the mean. The two forms |x-z| and (x-z)^2 give different things when there is more than one such term.