MHB What is the minimum value of y in the Absolute Value Function?

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The discussion revolves around finding the minimum value of the function y defined as y=|x-1|+|2x-1|+|3x-1|+...+|nx-1|, where n is a natural number and both x and y are real numbers. Participants agree that the goal is to minimize y with respect to x, considering it as a function of n. Clarifications are made regarding the nature of the variables involved. The focus remains on the mathematical approach to achieve the minimization of y. The conversation emphasizes the importance of understanding the absolute value function in this context.
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Determine the minimum of $y$ where $y=|x-1|+|2x-1|+|3x-1|+\cdots+|nx-1|$.
 
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The minimum of $y$ is zero and is achieved when $n = 1$ and $x = 1$.

Seriously though, should we minimize $y$ over $x$ as a function of $n$? Also, are $x, y$ real? (I assume $n$ is a natural number).
 
Bacterius said:
The minimum of $y$ is zero and is achieved when $n = 1$ and $x = 1$.

Seriously though, should we minimize $y$ over $x$ as a function of $n$? Also, are $x, y$ real? (I assume $n$ is a natural number).

Ops...my problem isn't very clear, and yes, Bacterius, your instinct are all right, $n$ is a natural number and both $x$ and $y$ are real, and we should minimize $y$ over $x$ as a function of $n$ instead.:o
 
My solution:

Note that if we are to graph of $y=|x-1|+|2x-1|+|3x-1|+\cdots+|nx-1|$, we need to first determine the function for each piece from different interval, and each part in its interval is necessarily piecewise linear and hence the graph of $y$ will always looks like "elbow(\/)".

Therefore the minimum of $y=|x-1|+|2x-1|+|3x-1|+\cdots+|nx-1|$ occurs at $kx-1=0$ by the time we see the changes of the gradient of the linear function turns from negative to positive.

I will illustrate this with an example.

If we have $y=|x-1|+|2x-1|+|3x-1|+|4x-1|$ and we plot its graph, we see that we need to break the absolute function into 5 intervals:

[TABLE="class: grid, width: 820"]
[TR]
[TD]For $x>1$:

$y=x-1+2x-1+3x-1+4x-1=10x-4$[/TD]
[TD][/TD]
[/TR]
[TR]
[TD]For $\dfrac{1}{2}<x<1$:

$\begin{align*}y&=-(x-1)+2x-1+3x-1+4x-1\\&=-x+2x+3x+4x+1-1-1-1\\&=8x-2\end{align*}$[/TD]
[TD]For $\dfrac{1}{3}<x<\dfrac{1}{2}$:

$\begin{align*}y&=-(x-1)-(2x-1)+3x-1+4x-1\\&=-x-2x+3x+4x+1+1-1-1\\&=4x\end{align*}$[/TD]
[/TR]
[TR]
[TD]For $\dfrac{1}{4}<x<\dfrac{1}{3}$:

$\begin{align*}y&=-(x-1)-(2x-1)-(3x-1)+4x-1\\&=-x-2x-3x+4x+1+1+1-1\\&=-2x+2\end{align*}$[/TD]
[TD]For $x<\dfrac{1}{4}$:

$\begin{align*}y&=-(x-1)-(2x-1)-(3x-1)-(4x-1)\\&=-x-2x-3x-4x+1+1+1+1\\&=-10x+4\end{align*}$[/TD]
[/TR]
[/TABLE]
View attachment 4309From the graph, we can see that minimum of $y$ occurs at $3x-1=0$, i.e. $x=\dfrac{1}{3}$. This happens when the gradient of the piecewise linear function changes from $4$ ($y=4x$) to $-2$ ($y=-2x+2$). We can evaluate the minimum value by evaluating the function $-x-2x-3x+4x+1+1+1-1=-2x+2$ at $x=\dfrac{1}{3}$.

Now, back to the original problem to find the minimum of $y=|x-1|+|2x-1|+|3x-1|+\cdots+|nx-1|$. I know I need to focus on finding which $k$ value (from $kx-1$) does the gradient of the piecewise function changes from negative to positive. Since the only thing that matters is the gradient of the linear function, I will consider only all the coefficients of $x$ terms and discard the constants in finding for $k$. Here, $k$ is natural number.

$(1,\,2,\,3,\,\cdots,\,k-1),\,\,(k,\,k+1,\,\cdots,\,n)$

I know I need the sum from $(1,\,2,\,3,\,\cdots,\,k-1)$ is less than the sum from $(k,\,k+1,\,\cdots,\,n)$.

Therefore I get

$k\left(\dfrac{k-1}{2}\right)<(n+k)\left(\dfrac{n+1-k}{2}\right)$

I then solve the inequality above for natural number $k$, and obtain

$2k^2-2k-(n^2+n)<0$

$k=\left\lfloor\dfrac{1}{2}+\sqrt{\dfrac{1}{4}+\dfrac{n^2+n}{2}}\right\rfloor$

Now, to find for its minimum value, we subtract the sum of $kx-1+(k+1)x-1+\cdots+nx-1$ from the sum of $x-1+2x-1+\cdots+(k-1)x-1$ and remember to replace $x=\dfrac{1}{k}$ and get:

$\begin{align*}y_{\text{min}}&=(kx-1+nx-1)\left(\dfrac{n+1-k}{2}\right)-(x-1+(k-1)x-1)\left(\dfrac{k-1}{2}\right)\\&=(kx+nx-2)\left(\dfrac{n+1-k}{2}\right)-(kx-2)\left(\dfrac{k-1}{2}\right)\\&=\left(\dfrac{n}{k}-1\right)\left(\dfrac{n+1-k}{2}\right)+\left(\dfrac{k-1}{2}\right)\end{align*}$

where $k=\left\lfloor\dfrac{1}{2}+\sqrt{\dfrac{1}{4}+\dfrac{n^2+n}{2}}\right\rfloor$.
 

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