All Questions
30
questions
1
vote
1
answer
38
views
Bound of an Integrable function (Analysis)
For non-negative Riemann integrable function f in [a,b], and dissection $\mathcal D= {x_0,x_1,...,x_n } $, if $p(f,\mathcal D) $ is defined as $$p(f,\mathcal D)=\prod_{k=1}^n [1+(x_k-x_{k-1}) \inf_{x\...
- 165
2
votes
1
answer
81
views
Prove or find a counterexample $\forall x>0: f(2x)-f(x)<g(3x)-g(2x)$, given information about $f, g$.
Let $f,g:(0, \infty) \rightarrow \mathbb{R}$ be two functions that satisfy the following for all $x>0$: $g'(x)>f'(x)$ and $f''(x)>0$.
Prove or find a counterexample:
$$
\forall x>0: f(2x)-...
- 2,510
3
votes
1
answer
91
views
An inequality involving a quasiconvex function with binomial and power terms
The question is as follows.
Let us consider a positive integer number $x \in \{1,2,3,...\}$ and a positive real number $q \in [1,x]$. Show that
\begin{equation}
\sum_{m=0}^{x-1} \left( \frac{x+2-q}{m+...
- 109
8
votes
2
answers
171
views
Proving that $\frac{f(b)-f(a)}{b-a}+ \left(\frac{g(b)-g(a)}{b-a}\right)^2\le \max_{t\in [a,b]}\{f'(t)+(g'(t))^2\}$
Let $f,g\in C^1([a,b])$ with $a<b$ then prove that
$$\frac{f(b)-f(a)}{b-a}+ \left(\frac{g(b)-g(a)}{b-a}\right)^2\le \max_{t\in [a,b]}\{f'(t)+(g'(t))^2\}$$
It smells like there is some mean ...
- 24.2k
2
votes
2
answers
155
views
Is there an easy way to decide the sign of $\sqrt x \ln (x + 1) - \sqrt {x + 1} \ln x$?
Is there an easy way to decide the sign of $\sqrt x \ln (x + 1) - \sqrt {x + 1} \ln x$?
The original problem is to decide when ${(n + 1)^{\sqrt n }}$ is greater than ${n^{\sqrt {n + 1} }}$.
It seems ...
- 1,519
2
votes
1
answer
552
views
Inequalities about arctan and tanh.
Prove that for all real numbers $x\geq 0$ and $y \geq 0$ the following inequalities are true:
$$\arctan(x+y)\leq\arctan(x)+\arctan(y) \qquad \tanh(x+y)\leq \tanh(x)+\tanh(y)$$
I tried to use both ...
- 2,158
4
votes
4
answers
1k
views
Prove that $\forall x > 0, x - 1 \ge \ln(x)$
Prove that $\forall x > 0, x - 1 \ge \ln(x)$ .
Here is my proof:
We prove the inequality on two intervals, $(0,1]$ and $[1,+\infty)$.
First the easier one, $[1,+\infty)$.
Notice that at $x=1$, ...
- 1,527
0
votes
1
answer
32
views
If $\frac{d}{d t} (f(t)^2) \le 2f(t)g(t)$ then $\frac{d}{d t} \sqrt{K+f(t)^2} \le g(t)?$
Let $f(t), g(t)$ be a real-valued functions of $t\in(0,\infty)$ and $f(t), g(t)\ge 0$.
If $$\frac{d}{d t} (f(t)^2) \le 2f(t)g(t)$$
then how can I conclude that
$$
\frac{d}{d t} \sqrt{K+f(t)^2} \le g(t)...
0
votes
1
answer
370
views
Exponential type of $\sin z$
An entire function $f$ is of exponential type if $\,\lvert\, f(z)\rvert\le C\mathrm{e}^{\tau\lvert z\rvert},\,$ for all sufficiently large values of $\lvert z\rvert$.
The exponential type of $f$ is ...
- 680
3
votes
0
answers
222
views
Two-leg games in Elo rating for football teams
Do you know Elo rating for association football? It is a numerical estimation of strength of football clubs using simple mathematical formula based past results allowing predictions for the future.
...
- 777
4
votes
1
answer
300
views
Hypergeometric functions inequality
Let $_2F_1(a,b;c,z)$ be the (Gauss) hypergeometric function, and $m$ and $n$ positive integers.
From a simple plot it looks like
$_2F_1(m+n,1,m+1,\frac{m}{m+n})>\frac{m}{n} \,_2F_1(m+n,1,n+1,\...
- 594
1
vote
1
answer
140
views
Inequality holds?
Can anyone prove that
$$
\frac{\sum\limits_{i=1}^{k*} a_i i (x+\epsilon)^{(i-1)}}{\sum\limits_{i=1}^{k*} a_i (x+\epsilon)^{i}+k^*-1}>\frac{\sum\limits_{i=1}^{k*} a_i i x^{(i-1)}}{\sum\limits_{i=1}^{...
- 7,892
1
vote
4
answers
123
views
Does this inequality hold
Let $f(x)$ be a positive function on $[0,\infty)$ such that $f(x) \leq 100 x^2$. I want to bound $f(x) - f(x-1)$ from above. Of course, we have $$f(x) - f(x-1) \leq f(x) \leq 100 x^2.$$ This is not ...
- 95
0
votes
1
answer
141
views
Is the hypergeometric function $F(5/4,3/4; 2, z)$ bounded on $(0,1]$
Consider the classical hypergeometric function $F(5/4,3/4; 2, z)$ for $z\in (0,1]$. Is this bounded by some real number (independent of $z$)?
I'm aware of Euler's formula:
$$F(5/4,3/4; 2, z) = \frac{...
- 526
1
vote
0
answers
78
views
Lower bounds for holomorphic functions on annuli with explicit bounds on their power series
Let $f$ be a holomorphic function on $\mathbf{C}$ and consider its restriction to the annulus $X=B(0,1) - B(0,3/4)$ in the complex plane. (Here $B(0,r)$ is the open disc with radius $r$ around $0$.) ...
- 221