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

It had the following statement:

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.

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).

But I am unable to go further, please help.

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.

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.

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

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?