Question

For any $ x\in R $ , minimum value of $ \left| x-1 \right|+\left| 2x-1 \right|+\left| 3x-1 \right|+...+\left| 119x-1 \right| $ is _____. \[\]

Answer 1

Hint: We recall that the definition of absolute value function $ \left| x \right| $ where minimum value occurs at breaking point $ x=0 $ . We see that the given function will have a minimum at one of the breakpoints $ x=\dfrac{1}{m} $ where $ m\in \left\{ 1,2,3,...,119 \right\} $ . We take the sum of the slopes for $ x=\dfrac{1}{m} $ and minimize the sum to get $ m $ . \[\]

Complete step by step answer:
Let us denote the given function as
\[f\left( x \right)=\left| x-1 \right|+\left| 2x-1 \right|+\left| 3x-1 \right|+...+\left| 119x-1 \right|\]
We know that the absolute value function is not differentiable at break points and the break point of $ \left| ax-b \right| $ is $ x=\dfrac{b}{a} $ since $ \left| ax-b \right|=ax-b $ if $ x\ge \dfrac{b}{a} $ and $ \left| ax-b \right|=-\left( ax-b \right) $ if $ x < \dfrac{b}{a} $ .We see in the given function there are 119 absolute value functions $ \left| x-1 \right|,\left| 2x-1 \right|,\left| 3x-1 \right|,...,\left| 119x-1 \right| $ whose break points are $ x=1,x=\dfrac{1}{2},x=\dfrac{1}{3},...,x=\dfrac{1}{119} $ . We see that at $ x=\dfrac{1}{n},n=1,2,3,...118 $ the slope of the function $ f\left( x \right) $ increases by $ 2n $ .
 So we have
\[\begin{align}
  & f\left( x \right)=x-1+2x-1+...+119x-1\text{ if }x\ge 1 \\
 & f\left( x \right)=-\left( x-1 \right)-\left( 2x-1 \right)-...-\left( 119x-1\text{ } \right)\text{ if }x < \dfrac{1}{119} \\
\end{align}\]
So $ f\left( x \right) $ will have negative slope beginning from $ x=\dfrac{1}{119} $ and then will continue decreasing until a point $ x=\dfrac{1}{m} $ where minimum will occur with zero slope until $ x=\dfrac{1}{m+1} $ and then slope of $ f\left( x \right) $ will increase after $ x=\dfrac{1}{m+1} $ . So the function can be defined as
\[f\left( x \right)=1-x+1-2x+....+1-\left( m-1 \right)x+\left( m\cdot \dfrac{1}{m}-1 \right)+\left( m+1 \right)x-1+...+119x-1\]
We see that the slopes of the absolute value terms up to $ x=\dfrac{1}{m} $ are $ 1,2,3,...,m $ and after $ x=\dfrac{1}{m} $ are $ m+1,m+2,...119 $ . So the sum of the slopes of $ f\left( x \right) $ can be given as
\[\begin{align}
  & -\sum\limits_{i=1}^{m}{i}+\sum\limits_{i=m+1}^{119}{i} \\
 & \Rightarrow -\sum\limits_{i=1}^{m}{i}-\sum\limits_{i=1}^{m}{i}+\sum\limits_{i=1}^{m}{i}+\sum\limits_{i=m+1}^{119}{i} \\
 & \Rightarrow -2\sum\limits_{i=1}^{m}{i}+\sum\limits_{i=1}^{119}{i} \\
 & \Rightarrow -2\cdot \dfrac{m\left( m+1 \right)}{2}+\dfrac{119\times 120}{2} \\
 & \Rightarrow -{{m}^{2}}-m+7140 \\
\end{align}\]
We need to minimize the above sum of slopes because larger is the sum larger is the value of the function $ f\left( x \right) $ . The minimum of the above quadratic function will occur at zeros which are
\[\begin{align}
  & -{{m}^{2}}-m+7140=0 \\
 & \Rightarrow {{m}^{2}}+m-7140=0 \\
 & \Rightarrow {{m}^{2}}+85m-84m-7140=0 \\
 & \Rightarrow \left( m+85 \right)\left( m-84 \right)=0 \\
 & \Rightarrow m=84,-85 \\
\end{align}\]
So the slopes $ f\left( x \right) $ will remain the same in the interval $ \left[ \dfrac{1}{84},\dfrac{1}{85} \right] $ . So the value of the function in the interval $ \left[ \dfrac{1}{84},\dfrac{1}{85} \right] $ will be minimum which is given by

\[\begin{align}
  & f\left( \dfrac{1}{84} \right)=1-\dfrac{1}{84}+1-2\cdot \dfrac{1}{84}+...+1-83\cdot \dfrac{1}{84}+0+85\cdot \dfrac{1}{84}-1+...+119\cdot \dfrac{1}{84}-1 \\
 & \Rightarrow f\left( \dfrac{1}{84} \right)=\left( \dfrac{83}{84}+\dfrac{82}{84}+...+\dfrac{1}{84} \right)+\left( \dfrac{1}{84}+\dfrac{2}{84}+...+\dfrac{35}{84} \right) \\
 & \Rightarrow f\left( \dfrac{1}{84} \right)=\dfrac{1}{84}\left( \dfrac{83\times 84}{2} \right)+\dfrac{1}{84}\left( \dfrac{35\times 36}{2} \right) \\
 & \Rightarrow f\left( \dfrac{1}{84} \right)=\dfrac{83}{2}+\dfrac{1}{12\times 7}\times \dfrac{7\times 5\times 36}{2} \\
 & \Rightarrow f\left( \dfrac{1}{84} \right)=41.5+7.5=49 \\
\end{align}\]
So the minimum value is 49. \[\]

Note:
 We note that we have frequently used here formula for sum of first $ n $ terms $ 1+2+3...+n=\dfrac{n\left( n+1 \right)}{2} $ . We can also put $ x=85 $ to get the minimum 49. We can alternatively solve using the triangle inequality property $ \left| x-a \right|+\left| x-b \right|\ge \left| x-a+x-b \right| $ to have
\[\left| x-1 \right|+\left| 2x-1 \right|+\left| 3x-1 \right|+...+\left| 119x-1 \right|\ge \left| x-1+2x-1+3x-1+...+119x-1 \right|\]. Since minimum exists all $ x $ ’s have to be cancelled out so we need such $ n $ such that $ 1+2+3...n=\left( n+1 \right)+\left( n+2 \right)...+119 $ . We solve for $ n $ to get $ n=84 $ .