Improve Your Algebra Skills with Practice Problems

This seems trivial but I just want to be careful about it. By definition, I would like $\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}P(|X_n-<!--\; -\; -->\mathrm{\mu <!--\; \mu \; -->}|>\mathrm{ϵ<!--\; ϵ\; -->})=0$ for $\mathrm{ϵ<!--\; ϵ\; -->}>0$. But can I just rewrite it as $\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}P(|X_n-<!--\; -\; -->\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{\mathrm{\mu <!--\; \mu \; -->}}_n|>\mathrm{ϵ<!--\; ϵ\; -->})=0$ and bring the limit out from there? That seems really odd to me.

My class notes define a signed measure on a measurable space $(X,\mathcal{R})$ as a $\mathrm{\sigma <!--\; \sigma \; -->}$-additive function $\mathrm{\nu <!--\; \nu \; -->}:\mathcal{R}\to <!--\; \to \; -->\mathbb{R}$. (I take this to mean we're only considering finite measures.) I'm confused on what I'm supposed to interpret $\mathrm{\sigma <!--\; \sigma \; -->}$-additive to mean here. My first guess would be that if $A=\underset{n}{\bigcup <!--\; \bigcup \; -->}{A}_n$ where $A_1,A_2,\dots <!--\; \dots \; -->$ are disjoint, then
$\mathrm{\nu <!--\; \nu \; -->}(A)=\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\mathrm{\nu <!--\; \nu \; -->}(A_n);$
i.e., just the usual definition of $\mathrm{\sigma <!--\; \sigma \; -->}$-additivity. But this seems problematic when the series on the right doesn't converge absolutely, because then its value depends on the order of the $A_i$, which my gut tells me shouldn't be the case. Does $\mathrm{\sigma <!--\; \sigma \; -->}$-additivity here also involve the claim that the series on the right always converges absolutely? The wikipedia page on signed measures does require this, but no other source I found online, or my class notes, explicitly states it.

Let $(X,\Vert <!--\; \Vert \; -->\cdot <!--\; \cdot \; -->\Vert <!--\; \Vert \; -->)$ be a normed space Define
$r=sup\{\Vert <!--\; \Vert \; -->x_1-<!--\; -\; -->x_2\Vert <!--\; \Vert \; -->:x_1,x_2\in <!--\; \in \; -->X,\Vert <!--\; \Vert \; -->x_1+x_2\Vert <!--\; \Vert \; -->=2,\Vert <!--\; \Vert \; -->x_1\Vert <!--\; \Vert \; -->=\Vert <!--\; \Vert \; -->x_2\Vert <!--\; \Vert \; -->=1\}.$
Question: Is $r=0?$

Certain Fraction between Fractions
Is there always a fraction $\frac{r}{s}$ with $\frac{p}{q}<\frac{r}{s}<\frac{p+1}{q}$ and $s<q$ for $0<p<q-<!--\; -\; -->1\in <!--\; \in \; -->\mathbb{Z}$ and $r,s\in <!--\; \in \; -->\mathbb{Z}$?

Algebra Fraction Problem - Variable
I honestly can't figure out how to get this answer, I feel like a complete idiot. I've taken up to Calc 2, so I'm not an idiot just been a while.
I have the problem..
$(n-<!--\; -\; -->1{)}^2-<!--\; -\; -->\frac{(n-<!--\; -\; -->2)(n-<!--\; -\; -->1)}{2}$
The answer is..
$\frac{(n-<!--\; -\; -->1)n}{2}$
How do I get there??
$2\frac{(n-<!--\; -\; -->1)(n-<!--\; -\; -->1)}{2}-<!--\; -\; -->\frac{(n-<!--\; -\; -->2)(n-<!--\; -\; -->1)}{2}\dots <!--\; \dots \; -->$
I'm so lost..

How do you write and solve an inequality represents the values of x for which the area of the rectangle will be at least 35 square feet?

I have an algorithm for calibrating a vector magnetometer. The input is N readings of the x, y, z axes: $(x_1,x_2,\dots <!--\; \dots \; -->,x_N)$, $(y_1,y_2,\dots <!--\; \dots \; -->,y_n)$, and $(z_1,z_2,\dots <!--\; \dots \; -->,z_N)$.
The algorithm fits an ellipsoid to the data by estimating a symmetric $3\times <!--\; \times \; -->3$ matrix A. In order to calibrate the system, it needs to calculate $\sqrt{A}$. I am adapting the algorithm for a microcontroller with very little memory, so cannot load standard matrix manipulation libraries.
Is there an explicit formula for calculating the square root of $3\times <!--\; \times \; -->3$ positive definite matrix?

What is ${\displaystyle 1,600÷4}$ ?

"Canceling out" in division doesn't always work the same way does it?
I've been working on Nested Fractions at the Khan Academy. Recently I was doing a routine problem and came to the correct conclusion but I realized I didn't understand why I wouldn't keep dividing. Here's what I begin with. I'm supposed to reduce it to an equivalent simplified expression. I wish I could link to the problem but I had moved on after posting here. Here's the start:
${\displaystyle \frac{(1+\frac{x}{y})}{\frac{x}{y}}}$
Part way through my problem now looks like:
${\displaystyle \frac{y\cdot <!--\; \cdot \; -->(y+x)}{y(x)}}$
I decide to "cancel out" the y's out side the parentheses. I end up with...
${\displaystyle \frac{y+x}{x}}$
I understood this the be the right answer. But for a moment I was very tempted to cancel the remaining x's and be left with a y. Even though I do this canceling operation all the time I realized then that there is something about it I don't understand. So here I am. Can you explain why it is that I was able to cancel the y earlier but not the x near the end?
Much appreciated.

Given $f_k\in <!--\; \in \; -->L^p(\mathrm{\Omega <!--\; \Omega \; -->})$, I have to prove that ${\Vert \underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{f}_n\Vert }_{L^p(\mathrm{\Omega <!--\; \Omega \; -->})}⩽<!--\; ⩽\; -->\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{\Vert f_n\Vert }_{L^p(\mathrm{\Omega <!--\; \Omega \; -->})}$, where $\Vert f\Vert \doteq <!--\; \doteq \; -->{\left({\displaystyle {\int <!--\; \int \; -->}_{\mathrm{\Omega <!--\; \Omega \; -->}}{\left|f\right|}^p}\right)}^{1/p}$ denotes the p-norm, with $1⩽<!--\; ⩽\; -->p<\mathrm{\infty <!--\; \infty \; -->}$. Since I've already proved the original Minkowski inequality, a simple induction estabilishes the finite case, i.e., ${\Vert \underset{n=1}{\overset{m}{\sum <!--\; \sum \; -->}}{f}_n\Vert }_{L^p(\mathrm{\Omega <!--\; \Omega \; -->})}⩽<!--\; ⩽\; -->\underset{n=1}{\overset{m}{\sum <!--\; \sum \; -->}}{\Vert f_n\Vert }_{L^p(\mathrm{\Omega <!--\; \Omega \; -->})}(\ast <!--\; \ast \; -->)$, $\mathbb{N}\ni <!--\; \ni \; -->m<\mathrm{\infty <!--\; \infty \; -->}$.
So, what I've thought is, since $L^p(\mathrm{\Omega <!--\; \Omega \; -->})$ is a Banach space, to suppose that $\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{\Vert f_n\Vert }_{L^p(\mathrm{\Omega <!--\; \Omega \; -->})}<\mathrm{\infty <!--\; \infty \; -->}$, which would give us that $\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{f}_n$ converges in $L^p(\mathrm{\Omega <!--\; \Omega \; -->})$, by completeness. Hence, the desired inequality follow by making $m⟶<!--\; ⟶\; -->\mathrm{\infty <!--\; \infty \; -->}$ in $(\ast <!--\; \ast \; -->)$, but I'm not completely sure about this.
Any help will be appreciated.

How do you solve ${\displaystyle \frac{a}{5}>10}$ ?

Let $(\mathsf{X},\mathcal{X},\mathrm{\mu <!--\; \mu \; -->})$ be a measure space with $\mathrm{\mu <!--\; \mu \; -->}$ begin $\mathrm{\sigma <!--\; \sigma \; -->}$-finite.
Definition of Random Variable: Let $(\mathsf{Y},\mathcal{Y})$ be a measurable space. A function $\mathrm{\xi <!--\; \xi \; -->}:\mathsf{X}\to <!--\; \to \; -->\mathsf{Y}$ is a random variable if ${\mathrm{\xi <!--\; \xi \; -->}}^{-<!--\; -\; -->1}(A)\in <!--\; \in \; -->\mathcal{X}$ for every $A\in <!--\; \in \; -->\mathcal{Y}$. We say that the random variable is a $\mathcal{X}$-measurable function.
Definition of Radon-Nikodym derivative: Let $\mathrm{\lambda <!--\; \lambda \; -->}$ be $\mathrm{\sigma <!--\; \sigma \; -->}$-finite measure with $\mathrm{\mu <!--\; \mu \; -->}\ll <!--\; \ll \; -->\mathrm{\lambda <!--\; \lambda \; -->}$. Then there exists a $\mathcal{X}$-measurable function $f:\mathsf{X}\to <!--\; \to \; -->[0,+\mathrm{\infty <!--\; \infty \; -->})$ denoted
$f=\frac{d\mathrm{\mu <!--\; \mu \; -->}}{d\mathrm{\lambda <!--\; \lambda \; -->}}$
satisfying
$\mathrm{\mu <!--\; \mu \; -->}(A)={\int <!--\; \int \; -->}_{A}fd\mathrm{\lambda <!--\; \lambda \; -->}\mathrm{\forall <!--\; \forall \; -->}A\in <!--\; \in \; -->\mathcal{X}.$
It seems that the Radon-Nikodym derivative is a $\mathcal{X}$-measurable function so by definition it should be a random variable between $(\mathsf{X},\mathcal{X})$ and $({\mathbb{R}}_{\ge <!--\; \ge \; -->0},\mathcal{B}({\mathbb{R}}_{\ge <!--\; \ge \; -->0}))$. Is this the case that every Radon-Nikodym derivative is a random variable?

How to solve $\log <!--\; \; -->\sqrt[3]{x}=\sqrt{\log <!--\; \; -->x}?$

Let us say that we have some set of independent random variables: $X=\{X_i{\}}_{i=1}^n$ defined over a probability space of $(\mathrm{\Omega <!--\; \Omega \; -->},\mathcal{F},P)$. I want to understand whether the following holds:
If ${\mathcal{F}}_X=\mathrm{\sigma <!--\; \sigma \; -->}\left(\{X_1,X_2,\dots <!--\; \dots \; -->,X_n\}\right)$, then is it, in general, true that: $\left|{\mathcal{F}}_X\right|=\underset{\mathrm{\forall <!--\; \forall \; -->}i\in <!--\; \in \; -->I_n}{\sum <!--\; \sum \; -->}|\mathrm{\sigma <!--\; \sigma \; -->}(\{X_i\})|$?
As we know, as $X$ consists of independent R.V.-s, then we can state that $\mathrm{\forall <!--\; \forall \; -->}i,j\in <!--\; \in \; -->I_n:\mathrm{\sigma <!--\; \sigma \; -->}(\{X_i\})$ and $\mathrm{\sigma <!--\; \sigma \; -->}(\{X_j\})$ are independent. But how to proceed from this fact to the split of the cardinality of ${\mathcal{F}}_X$?
I would appreciate any help, thank you in advance!

Consider a random variable $X$ with $E(X^2)<\mathrm{\infty <!--\; \infty \; -->}$. How do I show that $nP(|X|>\mathrm{\epsilon <!--\; \epsilon \; -->}\sqrt{n})\to <!--\; \to \; -->0$ for each $\mathrm{\epsilon <!--\; \epsilon \; -->}>0$? I don't even know where to start, and would like any hint.

As $X\in <!--\; \in \; -->L^2$, $\{X\}$ is an absolutely continuous collection, wich means that there is an $N\in <!--\; \in \; -->\mathbb{N}$ such that
${\int <!--\; \int \; -->}_{|X|>N}XdP<\mathrm{\epsilon <!--\; \epsilon \; -->}$
Where can I go from there?

What is the difference between a simple "fraction" and a "common fraction"?
I have read about a common fraction in this statement (written in a text book):
Ratio is the simplest form of a common fraction, in which the numerator denotes the antecedent and the denominator denotes the consequent.
The Wordweb dictionary defines a common fraction as
The quotient (defined as "The ratio of two quantities to be divided" / "The number obtained by division") of two integers.
From this definition, a common fraction pretty much seems to be just the same as a fraction. So what is the difference?

MathJax(?): Can't find handler for document MathJax(?): Can't find handler for document Let $Ч$ and $Y$ be absolutely continuous and discrete random variables, respectively. Does random vector $(X,Y)$ have to be absolutely continuous with respect to the product measure on the respective supports of $Ч$ and $Y$?
I think this is true because the following equality holds

so $f_{X,Y}(x,y)$ is the probability density function with respect to the product measure. However, I am not able to prove it formally. I would appreciate some help on this. Thanks.

Are some infinite fractions in one counting system non-infinite in another?
I'm curious whether some infinite periodic fractions in one counting system (e.g. decimal - 10/3 = 3.33333...) turn out to be non-infinite in another system and vice - versa.
Please excuse me if my terminology is not 100% accurate.

PCA formulation: Find the best rank-k subspace of d-dimensional space
Finding the best rank-k subspace of d-dimensional space can be interpreted as the Principle Component Analysis (PCA) problem provided that the goal of optimization is minimizing the error in the sense of variance.
Assume we have access to the data sample $x_1,\dots <!--\; \dots \; -->,x_n\in <!--\; \in \; -->{\mathbb{R}}^n$, prove that the first problem is equivalent to the second one:
$\begin{array}{}\text{(1)}& \stackrel{^<!--\; ^\; -->}{U_n}=\arg <!--\; \; -->min\underset{l=1}{\overset{n}{\sum <!--\; \sum \; -->}}\Vert x_l-<!--\; -\; -->{\mathcal{P}}_U(x_l){\Vert }_2^2\end{array}$
such that
$U\in <!--\; \in \; -->{\mathbb{R}}^{d\times <!--\; \times \; -->k},U^{T}U=I_k$
$\begin{array}{}\text{(2)}& \stackrel{^<!--\; ^\; -->}{U_n}=\arg <!--\; \; -->min\Vert <!--\; \Vert \; -->X_n-<!--\; -\; -->U{U}^T{X}_n{\Vert <!--\; \Vert \; -->}_F^2\end{array}$
such that
$U\in <!--\; \in \; -->{\mathbb{R}}^{d\times <!--\; \times \; -->k},U^{T}U=I_k$
where Ik is Identity matrix, $P_U=U{U}^T$ is a projection matrix, $X_n=\left[\begin{array}{ccc}{x}_1& \cdots <!--\; \cdots \; -->& x_n\end{array}\right]$ is a matrix of data, $\Vert <!--\; \Vert \; -->\cdot <!--\; \cdot \; -->{\Vert <!--\; \Vert \; -->}_F$ is Frobenius norm, i.e. $\Vert <!--\; \Vert \; -->A{\Vert <!--\; \Vert \; -->}_F=\sqrt{\text{Trace}(A^{T}A)}$

Given that measurement along the X, Y and Z axes correspond to the terms "width", "height", and "depth", is there an accepted term for spatial measurement along the W axis when dealing in four dimensions? For that matter, are there such terms for additional dimensional axes?