Improve Your Algebra Skills with Practice Problems

How to solve for $x$ in $\sqrt{9+2x}-<!--\; -\; -->\sqrt{2x}=\frac{5}{\sqrt{9+2x}}$?

in my probability theory course, we defined a sequence of random variables $(X_i{)}_{i=1}^{\mathrm{\infty <!--\; \infty \; -->}}$ to be tight if for all $\mathrm{ϵ<!--\; ϵ\; -->}>0$, there is a constant M s.th. $P(|X_n|>M)<\mathrm{ϵ<!--\; ϵ\; -->}$ for all $n\in <!--\; \in \; -->\mathbb{N}$.
I have seen the following criteria for tightness/non-tightness and I was wondering whether they are true or not:
1. $(X_i{)}_{i=1}^{\mathrm{\infty <!--\; \infty \; -->}}$ tight if there exists M s.th. lim${}_{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}P(|X_n|>M)=0$
2. $(X_i{)}_{i=1}^{\mathrm{\infty <!--\; \infty \; -->}}$ not tight if for all M lim${}_{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}P(|X_n|>M)>0$
I am quite sure that the first one is right (we get the condition that the probability is smaller than ϵ for all but finitely many n and can take the maximum of all the remaining M necessary for the finitely many n). For the second one I am not so sure, I was thinking that maybe we need some uniform bound, i.e. limn${}_{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}P(|X_n|>M)\ge <!--\; \ge \; -->\mathrm{ϵ<!--\; ϵ\; -->}>0$. I know that the criterion is right if the limit is equal to 1.

Let $X,Y,Z$ be integrable random variables on the same probability space.
Is it true that $\mathrm{\sigma <!--\; \sigma \; -->}(E[X|Y],E[X|Z])\subseteq <!--\; \subseteq \; -->\mathrm{\sigma <!--\; \sigma \; -->}(E[X|Y,Z])$ ?
My intuition says yes: the approximation of $X$ on $\mathrm{\sigma <!--\; \sigma \; -->}(Y,Z)$ is richer in information than the one on $\mathrm{\sigma <!--\; \sigma \; -->}(Y)$ and the one on $\mathrm{\sigma <!--\; \sigma \; -->}(Z)$. How to prove this?
If it is not true, what if we add $\mathrm{\sigma <!--\; \sigma \; -->}(Y)\subseteq <!--\; \subseteq \; -->\mathrm{\sigma <!--\; \sigma \; -->}(Z)$ ? Does this make the above true?

$\left[\frac{n}{1}\right]+\left[\frac{n}{2}\right]+\cdots <!--\; \cdots \; -->+\left[\frac{n}{n}\right]+\left[\sqrt{n}\right]$ is even
I tried this by taking two different cases, when, $n$ is even and when $n$ is odd. But I could not establish any relation between the two. Please help to solve this problem and thanks a lot in advance.
P.S. Although nothing is mentioned in the question I think $[]$ stands for greatest integer function as otherwise the sum would be in a fraction which is not even as only natural numbers are regarded as even.

Simplifying radical expression $\frac{\sqrt{p-<!--\; -\; -->4q}}{\sqrt{3q^2+4pq+p^2}}\cdot <!--\; \cdot \; -->\frac{\sqrt{p+3q}}{\sqrt{p^2+6pq+8q^2}}÷<!--\; ÷\; -->\frac{1}{\sqrt{p^2+3pq+2q^2}}$
Problem:
${\displaystyle \frac{\sqrt{p-<!--\; -\; -->4q}}{\sqrt{3q^2+4pq+p^2}}}\cdot <!--\; \cdot \; -->{\displaystyle \frac{\sqrt{p+3q}}{\sqrt{p^2+6pq+8q^2}}}÷<!--\; ÷\; -->{\displaystyle \frac{1}{\sqrt{p^2+3pq+2q^2}}}$
My try:
${\displaystyle \frac{\sqrt{p-<!--\; -\; -->4q}}{\sqrt{3p+q}\sqrt{p+q}}}\cdot <!--\; \cdot \; -->{\displaystyle \frac{\sqrt{p+3q}}{\sqrt{p+4q}\sqrt{p+2q}}}\cdot <!--\; \cdot \; -->{\displaystyle \frac{\sqrt{p+2q}\sqrt{p+q}}{1}}=$
${\displaystyle \frac{\sqrt{p-<!--\; -\; -->4q}\sqrt{p+3q}}{\sqrt{3p+q}\sqrt{p+4q}}}=$
${\displaystyle \frac{\sqrt{p^2-<!--\; -\; -->pq-<!--\; -\; -->12q}}{\sqrt{3p^2+13pq+4q^2}}}$
How do I continue?

What does it mean that a function in Sobolev spaces (for example) to be defined only upto a set of measure zero?

How do you write the following in interval notation: ${\displaystyle x>2}$ or ${\displaystyle 5<x}$ ?

Pratt's Lemma is : $\mathrm{\xi <!--\; \xi \; -->},\mathrm{\eta <!--\; \eta \; -->},\mathrm{\zeta <!--\; \zeta \; -->}$ and ${\mathrm{\xi <!--\; \xi \; -->}}_n,{\mathrm{\eta <!--\; \eta \; -->}}_n,{\mathrm{\zeta <!--\; \zeta \; -->}}_n$ such that:
${\mathrm{\xi <!--\; \xi \; -->}}_n\to <!--\; \to \; -->\mathrm{\xi <!--\; \xi \; -->},{\mathrm{\eta <!--\; \eta \; -->}}_n\to <!--\; \to \; -->\mathrm{\eta <!--\; \eta \; -->},{\mathrm{\zeta <!--\; \zeta \; -->}}_n\to <!--\; \to \; -->\mathrm{\zeta <!--\; \zeta \; -->},\text{convergence in probability}$
and ${\mathrm{\eta <!--\; \eta \; -->}}_n\le <!--\; \le \; -->{\mathrm{\xi <!--\; \xi \; -->}}_n\le <!--\; \le \; -->{\mathrm{\zeta <!--\; \zeta \; -->}}_n$, $E{\mathrm{\zeta <!--\; \zeta \; -->}}_n\to <!--\; \to \; -->E\mathrm{\zeta <!--\; \zeta \; -->},E{\mathrm{\eta <!--\; \eta \; -->}}_n\to <!--\; \to \; -->E\mathrm{\eta <!--\; \eta \; -->}$, and $E\mathrm{\zeta <!--\; \zeta \; -->},E\mathrm{\eta <!--\; \eta \; -->},E\mathrm{\xi <!--\; \xi \; -->}$ are finite, prove :
If ${\mathrm{\eta <!--\; \eta \; -->}}_n\le <!--\; \le \; -->0\le <!--\; \le \; -->{\mathrm{\zeta <!--\; \zeta \; -->}}_n$, then $E|{\mathrm{\xi <!--\; \xi \; -->}}_n-<!--\; -\; -->\mathrm{\xi <!--\; \xi \; -->}|\to <!--\; \to \; -->0$.

I know how to prove $E{\mathrm{\xi <!--\; \xi \; -->}}_n\to <!--\; \to \; -->E\mathrm{\xi <!--\; \xi \; -->}$, but $E|{\mathrm{\xi <!--\; \xi \; -->}}_n-<!--\; -\; -->\mathrm{\xi <!--\; \xi \; -->}|\to <!--\; \to \; -->0$ seems not easy proved from it.
My first question is how to prove $E|{\mathrm{\xi <!--\; \xi \; -->}}_n-<!--\; -\; -->\mathrm{\xi <!--\; \xi \; -->}|\to <!--\; \to \; -->0$.
And I don't know why should emphasize the condition "${\mathrm{\eta <!--\; \eta \; -->}}_n\le <!--\; \le \; -->0\le <!--\; \le \; -->{\mathrm{\zeta <!--\; \zeta \; -->}}_n$", so my second question is: Is there any example that $E|{\mathrm{\xi <!--\; \xi \; -->}}_n-<!--\; -\; -->\mathrm{\xi <!--\; \xi \; -->}|\to <!--\; \to \; -->0$ is wrong if the condition is not fullfilled.

Question about an inequality which seems right but not easy to prove
The origin problem is as follows:
let $a,b,c,d$ are positive real numbers,and $a+b+c+d=4$ prove:
$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{d}+\frac{d^2}{a}\ge <!--\; \ge \; -->4+\frac{1}{4}[(a-b{)}^2+(b-c{)}^2+(c-d{)}^2+(d-a{)}^2]$
The solution is easy enough, which is to plug $4=a+b+c+d$ into to $RHS$ and then move it to $LHS$, notice:
$[\frac{(a-<!--\; -\; -->b{)}^2}{b}+\frac{(b-<!--\; -\; -->c{)}^2}{c}+\frac{(c-<!--\; -\; -->d{)}^2}{d}+\frac{(d-<!--\; -\; -->a{)}^2}{a}](a+b+c+d)\ge <!--\; \ge \; -->[(a-b{)}^2+(b-c{)}^2+(c-d{)}^2+(d-a{)}^2]$
which would finally lead to the proof of the problem.
However, when I tried to solve the problem, I applied the Lagrange identical equation to the $RHS$ of the inequality, then I left out some quadratic term which finally leads to the inequality:
$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{d}+\frac{d^2}{a}\ge <!--\; \ge \; -->a^2+b^2+c^2+d^2$
Which seems quite right, but I'm not really sure if it is. However, numerical tests imply that it is right. But I'm still not sure.
That stuff is kind of like Chebyshev inequality but it can't be directly used here.
From another point of view, my question is whether the following strengthening　of the origin inequality is right or not:
let $a,b,c,d$ are positive real numbers,and $a+b+c+d=4$ then:
$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{d}+\frac{d^2}{a}\ge <!--\; \ge \; -->4+\frac{1}{4}[(a-b{)}^2+(b-c{)}^2+(c-d{)}^2+(d-a{)}^2+(a-<!--\; -\; -->c{)}^2+(b-<!--\; -\; -->d{)}^2]$

I`m trying to find an answer, but i have some problems, help.
Let ${\mathbb{P}}_{\mathrm{\theta <!--\; \theta \; -->}}=U[0,\mathrm{\theta <!--\; \theta \; -->}]$.
For $h,{\mathrm{\theta <!--\; \theta \; -->}}_0>0$ and $Z\sim <!--\; \sim \; -->\mathrm{e}\mathrm{x}\mathrm{p}\left(\frac{1}{{\mathrm{\theta <!--\; \theta \; -->}}_0}\right)$ I have to show that:
$\frac{\mathrm{d}{\mathbb{P}}_{{\mathrm{\theta <!--\; \theta \; -->}}_0-<!--\; -\; -->h/n}^n}{\mathrm{d}{\mathbb{P}}_{{\mathrm{\theta <!--\; \theta \; -->}}_0}^n}\stackrel{d,{\mathbb{P}}_{{\mathrm{\theta <!--\; \theta \; -->}}_0}^n}{⟶<!--\; ⟶\; -->}{e}^{\frac{h}{{\mathrm{\theta <!--\; \theta \; -->}}_0}}{\mathbb{1}}_{\{Z\ge <!--\; \ge \; -->h\}}$
I already proved that for $Z_n=n({\mathrm{\theta <!--\; \theta \; -->}}_0-<!--\; -\; -->max\{X_1,\dots <!--\; \dots \; -->,X_n\})$ with $X_1,\dots <!--\; \dots \; -->X_n\sim <!--\; \sim \; -->{\mathbb{P}}_{{\mathrm{\theta <!--\; \theta \; -->}}_0}$ holds $Z_n\stackrel{D}{\to <!--\; \to \; -->}Z$ and this task seems like I have to prove that the pdf is converging too. I'm not sure which technical steps I need to show this and I'm not sure which kind of convergence is meant by $d,{\mathbb{P}}_{{\mathrm{\theta <!--\; \theta \; -->}}_0}^n$.

How do you solve ${\displaystyle 8(1+6n)<392}$ ?

How do you write an inequality for the following statement. 3 is less than or equal to x?

How do you solve ${\displaystyle 11w+99<77}$ ?

MathJax(?): Can't find handler for document MathJax(?): Can't find handler for document Often in physics we integrate by parts

by:

I have a really simple question, how can we assume that $[f(x)\mathrm{\delta <!--\; \delta \; -->}(x-<!--\; -\; -->y){]}_{x_0}^{x_1}=0$?

Intuitively the delta function is zero except for at $x=y$, but what if either $x_0$ or $x_1$ was equal to y?

Is the answer simply 'we must assume separately that $x_0,x_1\ne <!--\; \ne \; -->y$, or is there something obvious that I'm missing, or is there some measure theory reason why we can say it is zero?

Given the step function
$g(x,y)=\{\begin{array}{ll}\frac{1}{x^2}& 0<y<x<1\\ -<!--\; -\; -->\frac{1}{y^2}& 0<x<y<1\\ 0& \text{otherwise}\end{array}$
I want to show that
${\int <!--\; \int \; -->}_{(0,1)}{\int <!--\; \int \; -->}_{(0,1)}g(x,y)d\mathrm{\lambda <!--\; \lambda \; -->}(x)d\mathrm{\lambda <!--\; \lambda \; -->}(y)\ne <!--\; \ne \; -->{\int <!--\; \int \; -->}_{(0,1)}{\int <!--\; \int \; -->}_{(0,1)}g(x,y)d\mathrm{\lambda <!--\; \lambda \; -->}(y)d\mathrm{\lambda <!--\; \lambda \; -->}(x)$
However, when trying to use the relation with Riemann integrability I end up with divergent integrals such as
${\int <!--\; \int \; -->}_0^x{\int <!--\; \int \; -->}_0^1\frac{1}{x^2}dxdy-<!--\; -\; -->{\int <!--\; \int \; -->}_0^1{\int <!--\; \int \; -->}_0^y\frac{1}{y^2}dxdy$
I also considered using limits such as following one, but again I end up with divergence.
$\underset{k\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{\int <!--\; \int \; -->}_{(\frac{1}{k},1)}\underset{k\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{\int <!--\; \int \; -->}_{(\frac{1}{k},1)}g(x,y)dxdy$
I would really appreciate any help!

simplify fractions with exponent
The given fraction is: $(\frac{a}{b}{)}^n\cdot <!--\; \cdot \; -->(\frac{b}{c}{)}^n\cdot <!--\; \cdot \; -->(\frac{c}{a}{)}^{n+1}$
The given solution is: $\frac{c}{a}$
What I have done so far:
$(\frac{a}{b}{)}^n\cdot <!--\; \cdot \; -->(\frac{b}{c}{)}^n\cdot <!--\; \cdot \; -->(\frac{c}{a}{)}^{n+1}$ | multiply $\frac{a}{b}$ and $\frac{b}{c}$ because of same exponent
$(\frac{ab}{bc}{)}^n\ast <!--\; \ast \; -->(\frac{c}{a}{)}^{n+1}$ | get rid of $b$
$(\frac{a}{c}{)}^n\ast <!--\; \ast \; -->(\frac{c}{a}{)}^{n+1}$
Can you please explain how I continue simplifying or what I did wrong? Thanks!

How do I solve this, first I have to factor $\frac{2x}{x-<!--\; -\; -->1}+\frac{3x+1}{x-<!--\; -\; -->1}-<!--\; -\; -->\frac{1+9x+2x^2}{x^2-<!--\; -\; -->1}$
the solution is
$\frac{3x}{x+1}$
The only advance that I have done is factor $x^2-<!--\; -\; -->1$=$(x-<!--\; -\; -->1)$$(x+1)$
do not know how can I factor $1+9x+2x^2$ can someone please guide me in how to solve this exercise.

when the sum of some fractions be $1$
prove if we want that the sum of some fractions be $1$ and the denominators of one of them is $d$ then another denominators should divisible by $d$ or $d$ should be divisible to another denominators.
It seems to be easy I tried to prove it.I first tried some cases.
$1=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}$
Here we can see that $6$ is divisible by $3$. Also here $6$ is divisible by $2$ .But I want to prove one of the denominators but here two of them is possible.After trying a lot I cannot found any proofs.Any hints?
update1: the numerator should be prime

Are ${\displaystyle x\le \frac{2}{3}}$ and ${\displaystyle -3x\le -2}$ equivalent?

Let $f:\mathrm{\Omega <!--\; \Omega \; -->}\to <!--\; \to \; -->B$ be Bochner-measurable, i.e. the point-wise limit of a sequence of simple (i.e. countably-valued measurable) functions $(s_n)$. I know that if
${\int <!--\; \int \; -->}_{\mathrm{\Omega <!--\; \Omega \; -->}}||f(\mathrm{\omega <!--\; \omega \; -->})||d\mathrm{\mu <!--\; \mu \; -->}(\mathrm{\omega <!--\; \omega \; -->})<\mathrm{\infty <!--\; \infty \; -->}(\ast <!--\; \ast \; -->)$
then one can show that f is Bochner-integrable, i.e. there is even a Bochner-integrable sequence of simple functions $({\stackrel{~<!--\; ~\; -->}{s}}_n)$ such that
$\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{\int <!--\; \int \; -->}_{\mathrm{\Omega <!--\; \Omega \; -->}}||f(\mathrm{\omega <!--\; \omega \; -->})-<!--\; -\; -->{\stackrel{~<!--\; ~\; -->}{s}}_n(\mathrm{\omega <!--\; \omega \; -->})||d\mathrm{\mu <!--\; \mu \; -->}(\mathrm{\omega <!--\; \omega \; -->})=0(\ast <!--\; \ast \; -->\ast <!--\; \ast \; -->)$
I am unsure, however, how to derive this sequence when the measure space $\mathrm{\Omega <!--\; \Omega \; -->}$ is not finite. Apparently one is supposed to make use of the fact that due to $(\ast <!--\; \ast \; -->)$, the set
$A:=\{f\ne <!--\; \ne \; -->0\}=\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\bigcup <!--\; \bigcup \; -->}}\underset{=:A_n}{\underset{⏟<!--\; ⏟\; -->}{\{||f||>\frac{1}{n}\}}}$
is $\mathrm{\sigma <!--\; \sigma \; -->}$-finite, as each $A_n$ has finite measure. But how to proceed? Setting ${\stackrel{~<!--\; ~\; -->}{s}}_n=1_A{s}_n$ would not be enough to make the functions Bochner-integrable, as the entire set $A$ might still have infinite measure. Setting ${\stackrel{~<!--\; ~\; -->}{s}}_n=1_{A_n}{s}_n$ would do it and maintain pointwise convergence, but then I don't know how to show the convergence in $(\ast <!--\; \ast \; -->\ast <!--\; \ast \; -->)$ anymore...any tips are much appreciated.