TITLE: Minimization of Floor and Ceiling Functions
QUESTION [0 upvotes]: So the problem at hand is:

Find the minimum value of the following function for $ x> 0 $: $$ \def\lc{\left\lceil}
\def\rc{\right\rceil} \newcommand{\floor}[1]{\lfloor #1 \rfloor} x\lc x \rc + \lc \frac{1}{x} \rc + x  + \floor{x}x + \floor{\frac{1}{x}} $$

My approach was to set $$ f(x) = \def\lc{\left\lceil}
\def\rc{\right\rceil} \newcommand{\floor}[1]{\lfloor #1 \rfloor} x\lc x \rc + \lc \frac{1}{x} \rc + x  + \floor{x}x + \floor{\frac{1}{x}} $$
and then find the minimum using the derivative.  I was reading previous Maths SE posts on the derivative of the floor and ceiling functions:
1) What technique should I apply to find the derivative of a ceiling or floor function e.g d/dx(x*⌈x⌉) and d/dx(x*⌊x⌋)?
2) Derivative of floor function
Using these, I was able to find out the derivative of nearly $50\%$ of $ f(x)$ but I am not able to proceed ahead and find the entire derivative. Could someone shed some light on how to approach this problem?
Any other methods would also be greatly appreciated!

REPLY [0 votes]: Rewrite $f(x)=x\left(\lceil{x}\rceil+\lfloor{x}\rfloor+1\right)+\left\lceil{\tfrac{1}{x}}\right\rceil+\left\lfloor{\tfrac{1}{x}}\right\rfloor$ or shortly $f(x)=x\times m(x)+r(x).$ The functions $m$ and $r$ are constant on each of the intervals $(k,k+1)$ and $\left(\tfrac{1}{k+1},\tfrac{1}{k}\right),\;k\in\mathbb{N}.$
Therefore study the cases

$x=1,$ we have $f(1)=5$
$x\in\mathbb{N},\; x>1$, then $f(x)=2x^2+x+1$ with minimal value $f(2)=11$
$x\in(k,k+1),\;k\in\mathbb{N},$ then $f(x)=x(2k+2)+1.$ Here $f$ doesn't reach a minimal value, but the infimum is $k(2k+2)+1.$ The smallest limit value is $5$ (as $k=1$).
$x=\tfrac{1}{k},\;k\in\mathbb{N},\;k>1,$ compute $f\left(\tfrac{1}{k}\right)=\tfrac{2}{k}+2k=2\left(\tfrac{1}{k}+k\right)$ with minimal value $f\left(\tfrac{1}{2}\right)=5$
$x\in \left(\tfrac{1}{k+1},\tfrac{1}{k}\right),\;k\in\mathbb{N},$ here $f(x)=2x+2k+1.$ Again, $f$ doesn't reach a minimal value, and infimum is $2\times \tfrac{1}{2}+2\times1+1=4$ as $k=1$ and $x\to\tfrac{1}{2}.$

Conclusion the function $f:(0,\infty)\to(0,\infty)$ doesn't have a minimum.


TITLE: Probability density function of a random variable
QUESTION [0 upvotes]: How to find a Probability Density/Mass Function for a random variable without assuming it follows a predetermined distribution, say, Normal or Poisson etc,. Lets say i have two hours of data of vehicle arrives at a specific point and i want to know what kind of distribution it has? Please help me understand this. Thank You, Guys.

REPLY [0 votes]: Usual method to solve this problem is step by step trial and error method. Your data may or may not fit for a known distribution. You may easily calculate mean, median, mood, variance, etc. for your data set. Now try to guess which distribution may be fit for your data. Apply goodness of fit tests to understand if the distribution is fit for your data. If not then try with some other distribution. That's all.


TITLE: Proving $f(x,y)$ is discontinuous at origin
QUESTION [2 upvotes]: Show that the function defined by
$$f(x,y) = \Bigg\{ \frac{x^3 + y^3 }{x-y}, \space when \space x \neq  y$$
and
$$f(x,y) = 0, \space when \space x = y$$
is discontinuous at the origin but possesses partial derivatives $f_x$ and $f_y$ at origin.
I am good with finding $f_x$ and $f_y$ at origin.
I am facing difficulty in proving discontinuous.
I tried to approach continuity as follows
$|f(x,y)-f(0,0)| = \bigg|\frac{x^3 + y^3 }{x-y} - 0 \bigg| \leq \frac{|x|^3 + |y|^3}{|x-y|}$
I couldn't proceed for here on wards, to prove discontinuity.
Please help me to prove discontinuity.

REPLY [0 votes]: In fact this function maps every neighborhood of $(0,0)$ onto $\mathbb R.$ Whenever you see a polynomial denominator whose zero set approaches the origin, be on the lookout for this crazy behavior.