Improve Your Algebra Skills with Practice Problems

Let $(X,\mathcal{A},\mathrm{\nu <!--\; \nu \; -->})$ be a measure space. Suppose that the measure of $X$ is bounded. Show that for disjoint $A_1,A_2,\dots <!--\; \dots \; -->$ $\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\mathrm{\nu <!--\; \nu \; -->}(A_n)=0$.

Since $X$ is bounded from countable additivity we have that
$\mathrm{\nu <!--\; \nu \; -->}\left(\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\bigcup <!--\; \bigcup \; -->}}{A}_n\right)=\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\mathrm{\nu <!--\; \nu \; -->}(A_n)\le <!--\; \le \; -->\mathrm{\nu <!--\; \nu \; -->}(X)<\mathrm{\infty <!--\; \infty \; -->}.$

Thus from Borel-Cantelli
$\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\mathrm{\nu <!--\; \nu \; -->}(A_n)<\mathrm{\infty <!--\; \infty \; -->}⟹<!--\; ⟹\; -->\mathrm{\nu <!--\; \nu \; -->}(\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim sup}{A}_n)=0.$

Now I also have that
$\mathrm{\nu <!--\; \nu \; -->}(\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim sup}{A}_n)\ge <!--\; \ge \; -->\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim sup}\mathrm{\nu <!--\; \nu \; -->}(A_n)$
which implies that
$\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim sup}\mathrm{\nu <!--\; \nu \; -->}(A_n)=0.$

How can I deduce from here that $\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\mathrm{\nu <!--\; \nu \; -->}(A_n)=0$? It seems I would need to use the fact that
$0=\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim sup}\mathrm{\nu <!--\; \nu \; -->}(A_n)\ge <!--\; \ge \; -->\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim inf}\mathrm{\nu <!--\; \nu \; -->}(A_n)\ge <!--\; \ge \; -->0$
so $\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim inf}\mathrm{\nu <!--\; \nu \; -->}(A_n)=0$ also and this would imply that $\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\mathrm{\nu <!--\; \nu \; -->}(A_n)=0$ because of?

find a value of x in a fraction
Can anyone help me with this? I don't know where to start. I assume there is a trick.
Fine the value of x if
$\frac{1}{1+\frac{1}{2+\frac{1}{3+\frac{1}{4+\frac{1}{x}}}}}=\frac{67}{96},$

Let $U\subseteq <!--\; \subseteq \; -->\mathbb{R}$ be open, $f:U\to <!--\; \to \; -->\mathbb{R}$ a convex function and $m_t(x):=\frac{f(x+t)+f(x-<!--\; -\; -->t)-<!--\; -\; -->2f(x)}{2}$.
We may suppose that $U=(a,b)$ wehre $a\in <!--\; \in \; -->\mathbb{R}\cup <!--\; \cup \; -->\{-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}\}$ and $b\in <!--\; \in \; -->\mathbb{R}\cup <!--\; \cup \; -->\{+\mathrm{\infty <!--\; \infty \; -->}\}$. Note that convexity of $f$ implies that $m_t(x)$ is non-negative.
Further, let $\mathbb{P}(A):={\int <!--\; \int \; -->}_A{e}^{-<!--\; -\; -->f(x)}dx$ be a probability measure.
Denote $N_t:=\{x\in <!--\; \in \; -->\mathbb{R}:m_t(x)=0\}$ and suppose that $\mathbb{P}(N_t)>\frac{1}{4}$ for all $t>0$.
Is it true (and is there a simple way to prove) that $f$ is constant on $U$?
It seems to me that this should hold but I have not been able to prove it yet.
Thank you.

Simplify complex fraction
This is a really low level question. I'm trying to simplify
$f(x)=\frac{36x^{-<!--\; -\; -->2}-<!--\; -\; -->3x^{-<!--\; -\; -->1}-<!--\; -\; -->18}{12x^{-<!--\; -\; -->2}-<!--\; -\; -->25x^{-<!--\; -\; -->1}+12}$
After factoring, removing negative exponents, and flipping the second fraction I get
$f(x)=\frac{1}{3(3x+2)(4x-<!--\; -\; -->3)}\frac{(3x-<!--\; -\; -->4)(4x-<!--\; -\; -->3)}{1}$
Then $(4x-<!--\; -\; -->3)$ cancels leaving
$f(x)=-<!--\; -\; -->\frac{3x-<!--\; -\; -->4}{3(3x+2)}$
as my final answer. However the book I have says the correct answer is
$f(x)=-<!--\; -\; -->\frac{3(2x+3)}{4x-<!--\; -\; -->3}$
I've checked my work many times and I don't know how they get this answer. Could someone please help me solve this?

Find the real solution set of the following equations, inequalities, and system of equations. Please provide me a clean, readable, and understandable solution
$\sqrt{2+4x}=3-<!--\; -\; -->\sqrt{3-<!--\; -\; -->4x}$

Simplifying nested/complex fractions with variables
I have the equation
$x=\frac{y+y}{\frac{y}{70}+\frac{y}{90}}$
and I need to solve for x. My calculator has already shown me that it's not necessary to know y to solve this equation, but I can't seem to figure it out. This is how I try to solve it:
$x=\frac{y+y}{\frac{y}{70}+\frac{y}{90}}=2y(\frac{70}{y}+\frac{90}{y})=2y\left(\frac{90+70}{y}\right)=2y\cdot <!--\; \cdot \; -->\frac{160}{y}=\frac{320y}{y}=320$
But according to my calculator, this is not correct. The answer should be 78.75, but I don't know why. Any help would be much appreciated.

Solve the equations and inequalities. Write the answers to the inequalities in interval notation if possible.
$-<!--\; -\; -->|8-<!--\; -\; -->v|\ge <!--\; \ge \; -->-<!--\; -\; -->6$

Hypothesis testing, inference of means
A student in a class randomly selected 25 weekdays and timed how long the 8am bus took to travel from home to school. His data gave an average travel time of 37 minutes and standard deviation of 2.5 minutes.
(a) Use the student's data to construct a 90% confidence interval for the true mean travel time from home to school for the 8am bus on weekdays. Also interpret the confidence interval in the context of this question.
(b) Do the student's data provided sufficient evidence that the true mean travel time is longer than 35 minutes? Carry out an appropriate hypothesis test at the 10% significance level.
I got no problem with (a) its 37+/- 1.711/2. However for (b), while the test statistic using the t-test gives me 4. the answer shows that the alpha value is 0.005 which makes no sense to me as I believe it should be 0.10 since its a one-tailed test.
Hope someone could find the problem here? Am I wrong or the answer is wrong?

I am wondering how the Radon-Nikodym derivative is affected by push-forwarding with a random variable. Formally, let $(\mathrm{\Omega <!--\; \Omega \; -->},\mathcal{F})$ be a measurable space, and $\mathbb{P}$ and $\mathbb{Q}$ be two probability measures on this space such that $\mathbb{Q}\ll <!--\; \ll \; -->\mathbb{P}$. Radon-Nikodym theorem tells us that there exists a $\mathcal{F}$-measurable function $f:\mathrm{\Omega <!--\; \Omega \; -->}\to <!--\; \to \; -->[0,\mathrm{\infty <!--\; \infty \; -->})$ denoted by $\frac{d\mathbb{Q}}{d\mathbb{P}}$, such that
$\mathbb{Q}(E)={\int <!--\; \int \; -->}_{E}f(\mathrm{\omega <!--\; \omega \; -->})d\mathbb{P}(\mathrm{\omega <!--\; \omega \; -->})$
for all $E\in <!--\; \in \; -->\mathcal{F}$.
If somehow we know the expression for $f(\mathrm{\omega <!--\; \omega \; -->})$ already, and $X:\mathrm{\Omega <!--\; \Omega \; -->}\to <!--\; \to \; -->{\mathbb{R}}^d$ is a random vector, is there a way to derive the Radon-Nikodym derivative $\frac{d{\mathbb{Q}}^X}{d{\mathbb{P}}^X}$ for the induced distribution measures ${\mathbb{Q}}^X$ and ${\mathbb{P}}^X$ on $({\mathbb{R}}^d,\mathcal{B}({\mathbb{R}}^d))$? My rough guess was $\frac{d{\mathbb{Q}}^X}{d{\mathbb{P}}^X}(X(\mathrm{\omega <!--\; \omega \; -->}))=f(\mathrm{\omega <!--\; \omega \; -->})$ but am not sure how to prove/disprove it.

I want to implement a Kalman Filter for predicting x/y positions. I have a sensor which gives me the current position (noisy). Now I want to smooth and predict the position. Thus Kalman Filter came to my mind.

How do I have to design the filter taking the time in regards? I want to predict the next state which is ahead of the upcoming measurement. Moreover since i do not have a velocity, I'd like to estimate the velocity by the kalman as well.

State := [xpos,ypos,xvelocity,yvelocity]
Measurement := [xpos,ypos]
ControlInput := predictionTime

I would run the following algorithm, whenever i get a measurement:

1. Measure
2. Update kalman gain
3. Predict
4. Get State estimate

Now, when i use a predictionTime which is newer than the next measurement, the next measurement does not fit to the predicted state.

Is there a strategy to solve this issue? My predicted state and my next measurement do not fit, how could I fix this?

Simplifying this fraction in a different base
Note: I would appreciate a solution that DOES NOT convert back to base 10.
How would one simplify ${\frac{43}{70}}_8$? I assume, like in decimal, I must recognize a common factor and divide by that factor. Keyword is recognize because we are taught for example that $5/10$ has a common factor of 5. Is it the same here? If so, must I ultimately either learn the base (impractical) or convert to base ten?
By the way, this question similar to another question I posted in this forum except that this one simplified the base (possibly directly from the decimal without going through simplification). I would like to mention that I have approximately 12 seconds to do this.\

How do you solve ${\displaystyle 50>\frac{w}{4}}$ ?

Why Is $y^{-<!--\; -\; -->1}$?

If you take the reciprocal in an inequality, would it change the $>/<$ signs?
Example:
$-<!--\; -\; -->16<\frac{1}{x}-<!--\; -\; -->\frac{1}{4}<16$
In the example above, if you take the reciprocal of
$\frac{1}{x}-<!--\; -\; -->\frac{1}{4}=\frac{x}{1}-<!--\; -\; -->\frac{4}{1}$
would that flip the $<$ to $>$ or not?
In another words, if you take the reciprocal of
$-<!--\; -\; -->16<\frac{1}{x}-<!--\; -\; -->\frac{1}{4}<16$
would it be like this:
$\frac{1}{-<!--\; -\; -->16}>\frac{x}{1}-<!--\; -\; -->\frac{4}{1}>\frac{1}{16}$

Showing that ${\int <!--\; \int \; -->}_X\log <!--\; \; -->(f)d\mathrm{\mu <!--\; \mu \; -->}\le <!--\; \le \; -->\mathrm{\mu <!--\; \mu \; -->}(X)\log <!--\; \; -->\frac{1}{\mathrm{\mu <!--\; \mu \; -->}(X)}$ given that ${\int <!--\; \int \; -->}_{X}fd\mathrm{\mu <!--\; \mu \; -->}=1$ for positive $f$ without using Jensen's inequality.
We weren't presented with Jensen's inequality, but with Fubini's Theorem and another theorem which is:
in a finite measure space $f\in <!--\; \in \; -->L_0^{+}(\mathrm{\Sigma <!--\; \Sigma \; -->})$ satisfies ${\int <!--\; \int \; -->}_{X}fd\mathrm{\mu <!--\; \mu \; -->}={\int <!--\; \int \; -->}_{(0,\mathrm{\infty <!--\; \infty \; -->})}\mathrm{\mu <!--\; \mu \; -->}(\{f>t\}dm(t))$
At first I thought I should use $\{|f|<\frac{1}{\mathrm{\mu <!--\; \mu \; -->}(X)}\}$,$\{|f|\ge <!--\; \ge \; -->\frac{1}{\mathrm{\mu <!--\; \mu \; -->}(X)}\}$ and show that if $\int <!--\; \int \; -->\log <!--\; \; -->fd\mathrm{\mu <!--\; \mu \; -->}>\mathrm{\mu <!--\; \mu \; -->}(X)\log <!--\; \; -->(\frac{1}{\mathrm{\mu <!--\; \mu \; -->}(X)})$, then
$\int <!--\; \int \; -->\log <!--\; \; -->fd\mathrm{\mu <!--\; \mu \; -->}\le <!--\; \le \; -->\mathrm{\mu <!--\; \mu \; -->}(\{|f|<\frac{1}{\mathrm{\mu <!--\; \mu \; -->}(X)}\})\log <!--\; \; -->\frac{1}{\mathrm{\mu <!--\; \mu \; -->}(X)}+{\int <!--\; \int \; -->}_{\{|f|\ge <!--\; \ge \; -->\frac{1}{\mathrm{\mu <!--\; \mu \; -->}(X)}\}}\log <!--\; \; -->fd\mathrm{\mu <!--\; \mu \; -->}$ and therefore
$\log <!--\; \; -->\frac{1}{\mathrm{\mu <!--\; \mu \; -->}(X)}(\mathrm{\mu <!--\; \mu \; -->}(X)-<!--\; -\; -->\mathrm{\mu <!--\; \mu \; -->}(\{|f|<\frac{1}{\mathrm{\mu <!--\; \mu \; -->}(X)}\}))=\log <!--\; \; -->\frac{1}{\mathrm{\mu <!--\; \mu \; -->}(X)}(\mathrm{\mu <!--\; \mu \; -->}(\{|f|\ge <!--\; \ge \; -->\frac{1}{\mathrm{\mu <!--\; \mu \; -->}(X)}\}))<{\int <!--\; \int \; -->}_{\mathrm{\mu <!--\; \mu \; -->}(\{|f|\ge <!--\; \ge \; -->\})}\log <!--\; \; -->fd\mathrm{\mu <!--\; \mu \; -->}$
which I can't seem to contradict. Any ideas how this can be shown, without using Jensen's Inequality?

How to simplify equations with a constant times a polynomial to a power
I've started doing integration in class and I often come across parts like this towards the end of my integration:
$\frac{2(6x^2+6x+9{)}^{3/2}}{3}+C$
And I get stuck. How should I go about simplifying these? Do I distribute the two and then try to do something with the exponent, or am I not allowed to because the polynomial is being raised to a power?
All help is appreciated, I am really stuck because almost all of my problems end up in this form at some point or another.

Equivalence of the persistence landscape diagram and the barcode?
I am studying persistent homology for the first time. I was reading Peter Bubenik's paper "Statistical Topological Data Analysis using Persistence Landscapes" from 2015 introducing persistent landscapes. I am quite confused on the approach on finding the values of the persistence landscape function using a barcode/persistence diagram. I feel like I have a naive misunderstanding of this topic as I shall attempt to explain.
Suppose X is a finite set of points in Euclidean space. From my understanding, if we consider the (finite length) persistence vector space given by the simplicial complex homology for a fixed dimension l, $\{H_l(X_k){\}}_{k=1}^n$ with maps $\{{\mathrm{\delta <!--\; \delta \; -->}}_{k,k^{\prime }}\}$, for $1\le <!--\; \le \; -->k,k^{\prime }\le <!--\; \le \; -->n$,$1\le <!--\; \le \; -->k,k^{\prime }\le <!--\; \le \; -->n$, we have
$\{H_l(X_k){\}}_{k=1}^n\cong <!--\; \cong \; -->{\oplus <!--\; \oplus \; -->}_{i=1}^m\mathbb{I}(b_i,d_i)$
(Theorem 4.10 of this paper) for some multiset $\{(b_i,d_i){\}}_{i=1}^m$, where $\mathbb{I}(b,d)$ gives the persistence vector space of length n,
$0\to <!--\; \to \; -->...\to <!--\; \to \; -->0\to <!--\; \to \; -->\mathbb{R}\to <!--\; \to \; -->\mathbb{R}\to <!--\; \to \; -->\mathbb{R}\to <!--\; \to \; -->...\to <!--\; \to \; -->0...\to <!--\; \to \; -->0$
with non-zero vector spaces at values of the specified interval.
This multiset corresponds to the persistence diagram/barcode so that the k-th Betti number can be identified by finding the number of lines of the barcode that intersect the line x=k in ${\mathbb{R}}^2$
Now Bubenik defines the Betti number of the persistence vector space for an interval [a,b] by ${\mathrm{\beta <!--\; \beta \; -->}}^{a,b}=\dim <!--\; \; -->(\text{im}({\mathrm{\delta <!--\; \delta \; -->}}_{a,b}))$, and the persistence landscape functions ${\mathrm{\lambda <!--\; \lambda \; -->}}_k:\mathbb{R}\to <!--\; \to \; -->[-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},\mathrm{\infty <!--\; \infty \; -->}]$, for $k\in <!--\; \in \; -->\mathbb{N}$by
${\mathrm{\lambda <!--\; \lambda \; -->}}_k(t)=sup\{m\ge <!--\; \ge \; -->0\mid <!--\; \mid \; -->{\mathrm{\beta <!--\; \beta \; -->}}^{t-<!--\; -\; -->m,t+m}\ge <!--\; \ge \; -->k\}$
Shouldn't ${\mathrm{\beta <!--\; \beta \; -->}}^{t-<!--\; -\; -->m,t+m}$ then correspond to the number of lines on the barcode that contain the interval $[t-<!--\; -\; -->m,t+m]$], so that ${\mathrm{\lambda <!--\; \lambda \; -->}}_k(t)$ is the largest value of m that has at least k lines of the barcode intersecting $[t-<!--\; -\; -->m,t+m]$?
I am confused on how the triangle construction is equivalent to the persistence landscape function instead. Any help would be much appreciated!

A measure $\mu :B_{\mathbb{R}}\to [0,\infty ]$ is locally finite if $\mu (K)<\infty$ for all K compact. Show that any locally finite measure is $\mathrm{\sigma <!--\; \sigma \; -->}$-finite. Give an example of a $\mathrm{\sigma <!--\; \sigma \; -->}$-finite measure on $\mathbb{R}$, which is not locally finite.
Attempt: Since $\mathbb{R}=\underset{n\in <!--\; \in \; -->\mathbb{Z}}{\bigcup <!--\; \bigcup \; -->}[n,n+1]$ and each $[n,n+1]$ is compact, we have $\mathrm{\mu <!--\; \mu \; -->}([n,n+1])<\mathrm{\infty <!--\; \infty \; -->}$ for each $n$ and $\mathrm{\mu <!--\; \mu \; -->}$ is $\mathrm{\sigma <!--\; \sigma \; -->}$-finite. Is this part of the problem correct?
For the second part, I have no idea. I wanted to take counting measure on a collection of subsets of $\mathbb{R}$, but I know this measure is locally finite on the integers and not so with the usual topology. But I know this measure is not sigma finite on $\mathbb{R}$. Any suggestions?

Where does this fraction come from in this integral?
So you have the integral:
$\int <!--\; \int \; -->\frac{3v}{200-<!--\; -\; -->4v}dv$
I tried to do $u$-substitution at first with $u=200-<!--\; -\; -->4v$, but I could not get the correct answer which is:
$-<!--\; -\; -->\frac{3}{4}v-<!--\; -\; -->\frac{150}{4}ln(200-<!--\; -\; -->4v)+C$
In the worked solution, they did not use a $u$-substitution. The first integral becomes:
$\int <!--\; \int \; -->-<!--\; -\; -->\frac{3}{4}+\frac{150}{200-<!--\; -\; -->4v}dv$
And I cannot see what technique they used to get that. I worked out that if you actually add the 2 fractions you end up back at the first integral, but I do not see how they worked out that is the way it should be re-arranged. I also don't understand why my $u$-substitution didn't work. Should a $u$-substitution have worked? I'm still trying to get my head around this integrating of fractions.

I'm reading this over at the Khan Academy and they use the function f(n)=6n2+100n+300 in their opening discussion of asymptotic boundaries of algorithms. My question is not specific to this example, but general, that is, how are such functions determined from (I'm guessing) raw data? I can see a graph of a parabola and I know from my (bad) school math that a parabola-shaped curve usually means some sort of exponential function.
But in general, by what methodologies are functions derived from data in the real world? So plotting software spits out plots from functions. Is there software that goes the other way and spits out functions from plots or data sets?