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.