## 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 -
beta)*(-(1/lamda)*
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 -
beta)*(-(1/lamda)*
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?

## co.combinatorics – Number of linear inequalities describing a polyhedron with prescribed number of vertices

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

1. Let $$fin C^{(n)}left()-1,1(right)$$ and $$sup limits_{-1. Let $$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.

## Are there any relationships (inequalities) between \$|x’Hy|\$ and \$|x’y|\$?

where x,y are $$n times 1$$ vectors, H is an $$ntimes n$$ hatmatrix (i.e. $$H = Z(Z’Z)^{-1}Z’$$).

## 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)

But

``````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

$$f(x)=x^n+a_{n-1}x^{n-1}+…+a_1x+a_0$$

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.

Thanks

## 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.