equation solving – Why is Reduce unable to solve this system of inequalities?

I need to solve the following inequalities but the execution just keeps going..

Reduce((1 - 
        Log((delta - beta*(1 - 2 d))/(lamda*delta)))^(rho - 
        1) - (alpha*
       A^rho/d^2) (1 - (1/(lamda*beta*d*(1 - d)))*
         Log((delta - beta (1 - 2 d))/(lamda*delta)))^{alpha*rho - 
        1} - lamda*(delta - 
       beta (1 - 2 d)) (beta*
         A^rho (1 - (1/(lamda*beta*d*(1 - d)))*
             Log((delta - beta (1 - 2 d))/(lamda*delta)))^{alpha*
            rho} + (1 - 
            Log((delta - beta (1 - 2 d))/(lamda*delta)))^
          rho)/(beta*(1 - 2 d)) < 0 && 
  delta - beta*(1 - 2 d) > 
   0 && (delta - beta*(1 - 2 d))/(lamda*delta) < 1 && rho < 1 && 
  A > 0 && 0 < alpha < 1 && 1 > beta > 0 && 1 > lamda > 0 && 
  delta > 0 && 0 < d < 1/2, beta, Reals)

Is it a problem with Reduce ?

inequalities – show this inequality with $frac{d^i}{dx^i}left(frac{x}{ln(1-x)}right)^{1/K} Bigg|_{x=0}>0, ~~~forall iin N^{+}$

Let $K$ be a fixed positive integer,show this $$dfrac{d^i}{dx^i}left(frac{x}{ln(1-x)}right)^{1/K} Bigg|_{x=0}>0, ~~~forall iin N^{+}$$

the problem is from when I solve this:$$left(sum_{i=1}^{n}a_{i}x^iright)^K=dfrac{x}{ln{(1-x)}},show ~that ~a_{i}>0,forall iin N^{+}$$.maybe this $$f(x)=dfrac{x}{ln{(1-x)}}$$ is special function? Is there a background to this conclusion?Thanks

inequalities – “Reversed” Bernstein Inequality

I’m studying harmonic analysis by myself, and I read some online notes that introduce the Bernstein inequality. One of them mention a reversed form of the Bernstein inequality, which is stated below:

Let $mathbb{T} = mathbb{R} / mathbb{Z} = (0,1)$ be the one-dimensional torus. Assume that a function $f in L^{1}(mathbb{T})$ satisfies $hat{f}(j) = 0$ for all $|j| < n$ (vanishing Fourier coefficients). Then for all $1 leq p leq infty$, there exists some constant $C$ independent of $n,p$ and $f$, such that
$$||f’||_{p} geq Cn||f||_{p}$$

It seems that an easier problem can be obtained by replacing $f’$ with $f”$ in the above inequality. The easier problem is addressed in the MO post below:

Does there exist some $C$ independent of $n$ and $f$ such that $ |f”|_p geq Cn^2 | f |_p$, where $1 leq pleq infty$?

However, it seems that the trick of convex Fourier coefficients used in the post above no longer applies to the harder problem (lower bounding the norm of the first derivative). Any suggestions/ideas?

Inequalities between $M_k=sup limits_{mathbb{R}}|f^{(k)}(x)|$

1. Let $fin C^{(n)}left()-1,1(right)$ and $sup limits_{-1<x<1} |f(x)|leq 1$. Let $m_k(I)=inf limits_{xin I} |f^{(k)}(x)|$, where $I$ is an interval (by interval here I mean closed, open or half-open intervals) contained in $)-1,1($. Show that

a) if $I$ is partitioned into three successive intervals $I_1, I_2$ and $I_3$ and $mu$ is the length of $I_2$, then $$m_k(I)leq dfrac{1}{mu}left(m_{k-1}(I_1)+m_{k-1}(I_3)right);$$

b) If $I$ has length $lambda$, then $$m_k(I)leq dfrac{2^{k(k+1)/2}k^k}{lambda^k};$$

2. Let $f$ be twice differentiable on $mathbb{R}$. Let $M_k=sup limits_{xin mathbb{R}}|f^{(k)}(x)|$ for $k=0,1,2$. Then $M_1leq sqrt{2M_0M_2}$.

3. Let $f$ is differentiable $p$ times on $mathbb{R}$ and the quantities $M_0$ and $M_p=sup limits_{xin mathbb{R}}|f^{(p)}(x)|$ are finite, then the quantities $M_k=sup limits_{xin mathbb{R}}|f^{(k)}(x)|$, $1leq kleq p$, are also finite and $$M_kleq 2^{k(p-k)/2}M_0^{1-k/p}M_p^{k/p}$$

Question: I was able to solve problems 1 and 2 but I have issues with problem 3. The hint to this problem says that we need to use problem 1b), 2) and induction.

So I suppose that we need to do induction on $p$ and the base case $p=2$ is just problem 2. I cannot handle the induction step.

I would be very grateful if someone can show the solution. I have spent some time on this problem and also created the post on MSE but it did not help me.

convex geometry – Fourier Motzkin elimination complexity when every eliminated occurs in bounded number of inequalities

Suppose $A(x,y)’leq b$ is a polyhedron and $xinmathbb R^n$ and $yinmathbb R_m$ and we are required to eliminate $y$ and obtain a polyhedron $Bxleq c$ we can resort to Fourier Motzkin elimination.

In my situation every $y_i$ in $y=(y_1,dots,y_m)$ influences $O(1)$ inequalities.

So is it possible to state $B$ has $leq k+poly(m)$ rows where $k$ is number of rows in $A$ and if not is there a way to provide the right bound?

functions – TautologyQ like feature that also works for inequalities

If a=b and b=c, then a=c.

So in mathematica,

TautologyQ((a == b) && (b == c) (Implies) a == c, {a, b, c})

yields ‘True’.

If a<b and b<c, then a<c. (if a,b,c are elements of an ordered set)


TautologyQ((a < b) && (b < c) (Implies) a < c, {a, b, c})

does yield ‘True’.

Instead it yields an error message ‘…is not Boolean valued…’

Because < is not a boolean expression.

Is there a built-in feature in mathematica similar to TautologyQ, which yields true for the latter code ?

My question can be restated as following :
‘extend TautologyQ to include inequality symbol’

If it is possible, I want to go further and further. (first include some basic calcuation symbol like +,-,*,/,mod, then function defining and composition, …)

real analysis – Inequalities regarding odd degree polynomial’s coefficients

Define an odd degree polynomial as a function $f: mathbb{R} to mathbb{R}$ such that


where n is odd.

We all know odd degree polynomials have special characteristics which they don’t share with even degree polynomials, here I propose four of them, which I think are true, but I’m unable to prove. My final goal is to prove something I already know, which is that all odd degree polynomials have at least one real root.

i) $mid x mid gt 1 implies mid frac{a_{n-1}}{x} + … +frac{a_1}{x^{n-1}} + frac{a_0}{x^n} mid lt frac{1}{mid x mid}(mid a_{n-1} mid + mid a_{n-2} mid) + … + mid a_1 mid + mid a_0 mid$

ii) $mid x mid gt max{ 1,2sum_{i=0}^{n-1}mid a_i mid }implies mid frac{a_{n-1}}{x} + … +frac{a_1}{x^{n-1}} + frac{a_0}{x^n} mid lt frac{1}{2}$

iii) $x_1 gt max{ 1,2sum_{i=0}^{n-1}mid a_i mid } implies f(x_1) gt 0$

iv) $x_2 lt -max{ 1,2sum_{i=0}^{n-1}mid a_i mid } implies f(x_2) lt 0$

Any help or proof is appreacieted.


probability – Probabilistic Inequalities with Lower Bounds

Many inequalities (Markov, Chevyshev, Chernoff) give some sort of upper bound to $P(X geq a)$ for some random variable $X$. I am trying to find some results that give a lower bound to this probability.

I have found this result, but unfortunately this only holds for $X geq 0$. Any there any known results that hold more generally?

If it makes any difference, the specific case I’m trying to apply this to has $X$ symmetric and $E(X^n) < infty$ for all $n in mathbb{N}$, but I would be happy to explore any results on this topic.