## Proof of inequality $\lceil x \rceil \le x+1$

I went through the Master Theorum extension for floors and ceiling section 4.6.2 in the book Introduction to Algorithms

Using the inequality $$\lceil x \rceil \le x+1$$

But I haven’t seen the inequality anywhere and could not understand the verifiability of inequality.

Instead the Chapter Floors and ceilings defined floors and ceilings as:

$$x-1 \lt \lfloor x \rfloor \le x \le \lceil x \rceil \lt x+1$$

Please clear my doubt over this.

On how to use this identity and which identity to be considered when because both of them define completely different inequalities.

Thank you.

Posted on Categories proxies

## A tricky mutual information inequality

Let $$X_0, X_1, X_2$$ be three independently distributed bits, let $$B$$ be a random variable such that $$I(X_0:B)=0$$, $$I(X_1:B)=0$$, and $$I(X_2:B)=0$$, I need to prove that $$I(X_0, X_1, X_2:B)\leq 1$$

(where $$I(M:N)$$ is Shannon’s mutual information).

I can demonstrate that $$I(X_0, X_1, X_2:B)\leq 2$$, by using chain rule of mutual information $$I(X_0, X_1, X_2 : B)= I(X_0:B)+I(X_1,X_2:B|X_0) = H(X_1,X_2|X_0) – H(X_1,X_2|B,X_0) = 2 – H(X_1,X_2|B,X_0) \leq 2$$.

(where $$H(.)$$ is Shannon’s binary entropy).

Posted on Categories proxies

## Positiveness of two variable inequality

How to show function

g(a, n) =-4a^2+8a^3-4a^4+16an-24a^2n+8a^3n-12n^2-4an^2+12a^2n^2+12n^3-8an^3  

of two variables is positive in a\in (2,n-2) 2 \leq n < infinity. See photo Posted on Categories cheapest proxies

## Confused in how to insert a slack variable in a constrain inequality

According to my understanding , we should put a slack variable to equate an inequality constrain by inserting the slack varaible in the side that is less than the other side , for example if we have 4x+2<2 this will be 4x+2+slack_variable=2 .But here in Wikipedia :
https://en.wikipedia.org/wiki/Slack_variable In the example section , it says the following “By introducing the slack variable y>=0, the inequality Ax<=b can be converted to the equation y-Ax+b = 0” Which means that the slack variable is inserted in the bigger side ! Please some one explain this confusion.

Posted on Categories proxies

## How to plot a graphic with solution of some inequality?

I defined the function $$z(n)$$ by

z[n_] := Catch[Do[i; If[Mod[Fibonacci[i], n] == 0, Throw[i]], {i, 100000}]]

and now, I would like to plot the graphic (n,z(n)) for points such that $$z(n)/n<\epsilon$$, for some given value of $$\epsilon>0$$. For instance, for $$\epsilon=0.05$$, I used the command

Catch[Do[n; If[z[n] < 0.05*n, Print[n]], {n,1, 3000}]]

which provides me many values of $$n$$, but I do not know how to make a graphic with them (they appear in some collumn). Thanks in advance.

Posted on Categories cheapest proxies

## Numerical results for Reduce on an inequality

I am trying to determine the regions parameters (Te, ymin) must belong to for a function to be negative for any argument x greater than zero. I have tried using Resolve

       Resolve[         ForAll[x, (a/x^2)*(Log[x - b] + 1) - (a/x)*(1/(x - b)) - (1/         x^2)*(-b^3 + b) < 0] && x > 0, Reals] 

but no expression could be provided. This is not surprising, as I suppose such expression would involve the solution of trascendental equations.

On the other hand, is there a way to get a numerical output from Resolve? For example, a set of (a, b) parameters values on the boundary of the region for which the function is positive?

I thought maybe Nsolve could take an inequality as an expression, but could not find any examples and could not make it work.

A “brute force” alternative I thought about is to create a grid of (a,b) values, and for each couple of values verify if the function is negative along all the positive x-axis (the last step does not seem obvious neither). Is there a maybe more clever way, fully exploiting Mathematica’s capabilities?

I got intrigued by the answer this post, Reduce a complex inequality but I could not make it work for my case, for example

RegionPlot[ ForAll[x, (a/x^2)*(Log[x - b] + 1) - (a/x)*(1/(x - b)) - (1/    x^2)*(-b^3 + b) < 0], {a, 0.01, 0.5}, {b, 0.001,  0.01}] 

does not seem to be correct syntax.

Thanks a lot

## triangle inequality TSP is NP-complete?

I have been reading online source and it mentioned triangle inequality TSP is NP-complete but without proof. In general, the reduction from HAM-cycle problem to TSP works for asymmetric and symmetric TSP problem. So, I wonder how to start a basic approach to prove triangle inequality TSP is NP-complete.

Posted on Categories proxies

## Changing ContourStyle based on an inequality

I would like to create a bifurcation diagram with ContourPlot. Suppose I want to plot the contour 1-2*a*Q+3*Q^3==0 with a the x-axis, Q the y-axis. I also want linestyle/thickness of the contour plot to change depending on the expression expr = 9*Q^2-2*a. If expr>0, I want it to be dashed; if expr<0, I want it to be a thick line. How to accomplish this?

Some previous posts about similar topic suggested putting an If statement inside the plot, but when I try something like

ContourPlot[If[expr>0, 1-2*a*Q+3*Q^3 == 0], {a,-2,2}, {Q,-2,2}] 

it gives error.

## inequality of distances in a graph

Certainly it’s obvious but I can’t catch the reason behind it.

Why do we have :

Let $$D= (V,A)$$ be a directed graph, $$w:A \to \mathbb R$$ be arc weights and $$s \in V$$. Denote with $$d(s,v)$$ the length of the shortest path from $$s$$ to $$v$$ in $$D$$, subject to $$w$$.

If there are no negative cycles in $$D$$, then we have $$\forall (u,v) \in A : d(s,v) \leq d(s,u) + w(u,v) \iff d(s,v)- d(s,u) \leq w(u,v).$$

Why?

## Inequality for integro-differential system

Reading a paper about homogenization of stochastic coefficients [A. Gloria et al., Invent. math. (2015)], I found the following lemma that gives an estimate of the solution for a given ODE system. The lemma is:

Let $$1\le p,\gamma<\infty$$ and $$a(t),b(t)\ge 0$$. Suppose that there exists $$C_1 <\infty$$ such that for all $$t\ge0$$, $$\begin{gather} \tag{1}\label{eq1} a(t) \le C_1 \left( (t+1)^{-\gamma} + \int_{0}^{t} (t-s+1)^{-\gamma} b(s) \,ds \right), \ b(t)^p \le C_1\left(-\frac{d}{dt}a(t)^p\right). \end{gather}$$ Then there exists $$C_2<\infty$$ depending only on $$C_1,p$$ and $$\gamma$$ such that $$\begin{equation} \tag{2}\label{eq2} a(t) \le C_2 (t+1)^{-\gamma}. \end{equation}$$

My question is: Is it possible to prove a similar result replacing \eqref{eq1} by $$\begin{equation*} a(t) \le C_1 \left( t^{-\gamma} + \int_{0}^{t} (t-s)^{-\gamma} b(s) \,ds \right)? \end{equation*}$$ How would \eqref{eq2} change?