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Find Upper Bound for $T(n)=T(n-<!--\; -\; -->1)+T(\frac{n}{2})+n$ with recursive tree method.

Prove that the height of a 2-3 tree is between ${\log }_3<!--\; \; -->N$ and $\lg <!--\; \; -->N$

Find the tight upper and lower bounds for the following recurrence:
$T(n)=2T\left(\frac{n}{2}\right)+6t\left(\frac{n}{3}\right)+n^2,\text{if}n\ge <!--\; \ge \; -->3=1\text{if}n\le <!--\; \le \; -->2$

Solve a laser's upper level $\frac{dy}{dx}=\frac{1-<!--\; -\; -->y}{A}-<!--\; -\; -->\frac{y\times <!--\; \times \; -->f(x)}{B}$ where y = y(x) and A, B are constants.

Is there an algorithm which for a formula P of PA outputs ${\mathrm{\Sigma <!--\; \Sigma \; -->}}_m^0/{\mathrm{\Pi <!--\; \Pi \; -->}}_m^0$ such that the output is the level P belongs to in Arithmetical hierarchy? If that's not computable is there an algorithm which outputs an upper bound on the level?

I've been given the recursive function:
$$T(n)=n^{\frac{2}{3}}T(n^{\frac{1}{3}})+C$$
$$T(n<=2)=O(1)$$
With $n=2^{3^p}$ where p is a positive integer.

a is a factor of 10 and b is a factor of 15.
$\begin{array}{|cc|}\hline \text{Column A}& \text{Column B}\\ \text{The smallest value that ab must be a factor of}& 30\\ \hline\end{array}$
Is Column A greater than/ smaller than / equal to Column B?

I have an upper and lower bound number:
upper: 21
lower: 3
I then have a second number that can be anywhere between this range, I would like the second number to increment faster when it is closer to the lower bound and slower when it is closer to the upper bound.
How can I achieve this using mathematics?

Сheck the repeatability of $T(n)=T(n/2)+n$ by replacing

Let $f:K\subset <!--\; \subset \; -->{\mathbb{R}}^n\to <!--\; \to \; -->\mathbb{R}$ be a function on a compact K. I have encountered the following statement: ${\int <!--\; \int \; -->}_{K}f=\mathrm{\infty <!--\; \infty \; -->}$ if $\mathrm{v}\mathrm{o}\mathrm{l}(\{x\in <!--\; \in \; -->K\mid <!--\; \mid \; -->f(x)\ge <!--\; \ge \; -->r\})=\mathrm{\Omega <!--\; \Omega \; -->}(r^{-<!--\; -\; -->1})$ as $r\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}$ . It is trivial for $n=1$ but I am at a loss how to approach it for $n>1$ .

I am having some trouble wrapping my head around quasiconcavity. I have a couple of definitions given the function: $f(𝑥,𝑦)={𝑥}^2+{𝑦}^2$
1) A function f defined on a convex set S is quasiconcave if every upper level set of f is convex. That is, $P_a=x\in S:f(x)\ge a$ is convex for every value of a.
2) A function $f:S\to <!--\; \to \; -->\mathbb{R}$ defined on a convex subset S of a real vector space is quasiconcave if $f(\mathrm{\lambda <!--\; \lambda \; -->}x+(1-<!--\; -\; -->\mathrm{\lambda <!--\; \lambda \; -->})y)\ge <!--\; \ge \; -->min\{f(x),f(y)\}$ .

${\int <!--\; \int \; -->}_0^{2\mathrm{\pi <!--\; \pi \; -->}}\frac{1}{{\sin }^4<!--\; \; -->x+{\cos }^4<!--\; \; -->x}$
1) I've tried multiplying by ${\sec }^4<!--\; \; -->x$ and doing u substitution, but It used crazy amounts of upper level maths
2) I've tried double/half angle identities to simplify down to ${\displaystyle \frac{1}{1-<!--\; -\; -->\frac{1}{2}{\sin }^2<!--\; \; -->2x}}$
3) I've mixed strategies and came up with ${\displaystyle \frac{4\cos <!--\; \; -->2x}{4\cos <!--\; \; -->2x-<!--\; -\; -->\cos <!--\; \; -->3x\cos <!--\; \; -->x}}$

I am trying to solve the following recurrence $T(n)=T\left(\frac{n}{3}\right)+T\left(\frac{2n}{3}\right)+O(n)$ by either substitution or recursion tree method. I have seen examples online that solve this recursion for
$\mathrm{\Omega <!--\; \Omega \; -->}$
lower bound but not tight upper bound. How can this be solved for tight upper bound?

There are x+y different books on the shelf, x books have blue binding, and y books have white bindings.
a) how many ways can the books be arranged so that the books with blue bindings occupy the first x positions?
b) how many ways can we arrange the books so the books with blue bindings are together?

The exact $(1-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->})$ level confidence interval lower limit is given by
$$\underset{k=y}{\overset{n}{\sum <!--\; \sum \; -->}}(\genfrac{}{}{0ex}{}{n}{k}){p_L}^k(1-<!--\; -\; -->p_L{)}^{n-<!--\; -\; -->k}=\mathrm{\alpha <!--\; \alpha \; -->}/2$$
and the upper limit analogously. Why does the resulting C.I. has greater coverage than the nominal $(1-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->})$ ? I thought that the above equation has a valid solution for $p_L$ at any $\mathrm{\alpha <!--\; \alpha \; -->}$ and so should provide the correct coverage.

Give evidence: Let X and Y be random variables with moment generating functions $M_X$ and $M_Y$, respectively, existing in open intervals about 0. Then $F_X(z)=F_Y(z)$ for all $z\in <!--\; \in \; -->\mathbb{R}$ if and only if $M_X(t)=M_Y(t)$ for all $t\in <!--\; \in \; -->(-<!--\; -\; -->h,h)$ for some $h>0$ .

I'm struggling with the intuition behind why Chernoff bounds differ in the upper and lower tails. That is, for the lower tail we have:
$Pr(X\le <!--\; \le \; -->(1-<!--\; -\; -->\mathrm{\delta <!--\; \delta \; -->})\mathrm{\mu <!--\; \mu \; -->})\le <!--\; \le \; -->e^{-<!--\; -\; -->\frac{\mathrm{\mu <!--\; \mu \; -->}{\mathrm{\delta <!--\; \delta \; -->}}^2}{2}}$
Whereas for the upper tail:
$Pr(X\ge <!--\; \ge \; -->(1+\mathrm{\delta <!--\; \delta \; -->})\mathrm{\mu <!--\; \mu \; -->})\le <!--\; \le \; -->e^{-<!--\; -\; -->\frac{\mathrm{\mu <!--\; \mu \; -->}{\mathrm{\delta <!--\; \delta \; -->}}^2}{3}}$
Both bounds are found in roughly the same manner, so at a high level, why would they differ in value when we don't know that the distribution is necessarily asymmetric?

W denotes a Wiener process in filtration $\mathcal{F}$ , where X is an ${\mathcal{F}}_0$ -measurable
a, Exponential
b, Cauchy
distributed random variable. Let us define
$$\mathrm{\tau <!--\; \tau \; -->}=inf\{t\ge <!--\; \ge \; -->0:\left|W_t\right|\ge <!--\; \ge \; -->X\}.$$ .
What is the probability, that $\mathrm{\tau <!--\; \tau \; -->}$ is finite, so $\mathbf{P}(\mathrm{\tau <!--\; \tau \; -->}<\mathrm{\infty <!--\; \infty \; -->})=?$ And what is $\mathbb{E}\left(\mathrm{\tau <!--\; \tau \; -->}\right)=?$ and $\mathbb{E}(W_{\mathrm{\tau <!--\; \tau \; -->}}\cdot <!--\; \cdot \; -->{\mathrm{\chi <!--\; \chi \; -->}}_{\{\mathrm{\tau <!--\; \tau \; -->}<\mathrm{\infty <!--\; \infty \; -->}\}})=?$ , where ${\mathrm{\chi <!--\; \chi \; -->}}_{\{\mathrm{\tau <!--\; \tau \; -->}<\mathrm{\infty <!--\; \infty \; -->}\}}$ is the indicator variable of the $\{\mathrm{\tau <!--\; \tau \; -->}<\mathrm{\infty <!--\; \infty \; -->}\}$ event?

If we have $f(r)=r^n$ , where $r^n=(x^2+y^2+z^2{)}^{n/2}$ , and one wishes to examine the level curves of f(r), namely $f(r)=k$ for some real number k, then one might proceed as follows: there would be two cases - (i) take $n>0$ and (ii) $n<0$ .
Now, for case (i), taking $n=1$ , we'll have the upper part of a sphere in 3-space of radius $\sqrt{k}$ . Taking $n=2$ , we'll obtain a whole sphere of radius $\sqrt{k}$ . But what about taking $n>2$ ? That's where I'm getting lost. What kind of a surface are we going to get?
For case (ii), it becomes even more complicated. For example, for $n=-<!--\; -\; -->2$ , we'll have something like a "reversed sphere".
The question is that we can actually view these surfaces in two ways:
1) For example, for $n=5$ , we can arrive at $x^2+y^2+z^2=k^{2/5}$ . For $n=-<!--\; -\; -->2$ , we can arrive at $x^2+y^2+z^2=\frac{1}{k}$ , and both are just spheres in 3-space.
2) We can expand (for $n=5$ ) the expression $(x^2+y^2+z^2{)}^{5/2}=k$ and get a very long expression with powers of 4 and less.
Which approach is "more correct" - (1) or (2)?

Creating an application with a level system
$\text{Level}=1,XP=100-<!--\; -\; -->299$ , Equation:
$\frac{\text{user's XP}}{(\text{Level}+2)\cdot <!--\; \cdot \; -->50}$
If the user has an XP of 205 the answer is 1.36666. Less than 2, so they stay at level 1. This works all the way up to 299, and when they hit 300, they will level up to 2.
Problem 1: anything below 150, and the answer is less than 1.
Problem 2: you can lose XP as well as gain it, so I have to find an equation that also takes into account XP that decreases.