TITLE: explain derivative using dot product
QUESTION [1 upvotes]: The question gives me a graph with a function and a vector and asks me to determine the sign of the directional derivative at various points. This part is easy. But then, it asks me to justify my answer using the dot product. How do I do that beyond simply saying "the dot product of the gradient and the vector is zero"?
I know this probably doesn't seem worthy of the holy Stack Exchange, but there are no Stack questions like it, the teacher decided not to release solutions, and he doesn't have office hours until the day before the exam so I'm totally stuck.

REPLY [1 votes]: Note that the gradient, $\nabla f$, points in the direction of maximum change of $f$ and is, therefore, normal to level curves.
The directional derivative, $\frac{df}{du}=\hat u\cdot \nabla f=|\nabla f|\cos(\theta)$, where $\hat u$ is a unit vector along $\vec u$, and $\theta$ is the angle between $\nabla f$ and $\hat u$.
As one example, at the point $Q$, the angle between the $\nabla f$ and $\hat u$ appears to be less than $\pi/2$ and hence the directional derivative should be positive.


TITLE: Bounds on how far away most leaves are from the average height of a binary tree
QUESTION [3 upvotes]: I'm wondering if anyone can help me prove (or disprove) this statement?

Say there is a rooted full binary tree (each non-leaf node has exactly two children) with a height of $h$, an average height of $t$, and $l$ leaves. Then for $k \geq 1$ there are at most $\frac{l}{k}$ leaves above a height of $kt$.

I think the statement is correct and basically I've been trying to assume the contrary and come up with a contradiction using the average height -- but no success.

REPLY [1 votes]: I suppose the terminology of height of a node may be confusing here, so let's make it clear first how I interpret it:
The height of the root is 0; if a node has height $j$, its children has height $j+1$. Being above means having larger height, i.e. further from the root. Thus, one thinks of the root at the bottom and the leaves at the top, not the other way around.
If the tree has $m_j$ leaves at height $j$, where $h$ is the maximal height (i.e. $\sum_j$ means $\sum_{j=0}^h$), then $l=\sum_j m_j$ and $l\cdot t=\sum_j jm_j$. Let me rephrase the problem slightly: for any $a>0$ (where $a=kt$ in the original), there are at most $lt/a$ leaves with height $j\ge a$. This follows from
$$\sum_{j\ge a} m_h
 =\frac{1}{a}\sum_{j\ge a} am_j
 \le\frac{1}{a}\sum_{j=0}^h jm_j
 =\frac{lt}{a}.
$$
The requirement that $k\ge1$ (i.e. $a\ge t$) is not required, only that $a>0$. Of course, the only interesting values of $a$ are integers.
It may also be noted that the result doesn't make use of the fact these are leaves on a binary tree: they are just leaves with values (in this case height) attached to them.


text:
\begin{document}

\setlength{\abovedisplayskip}{3pt}
\setlength{\belowdisplayskip}{3pt}

\title{Matrix Completion and Performance Guarantees for Single Individual Haplotyping}

\author{Somsubhra Barik and Haris Vikalo,~\IEEEmembership{Senior Member,~IEEE}

\thanks{S. Barik and H. Vikalo are with the Department
of Electrical and Computer Engineering, The University of Texas at Austin, Austin,
TX, 78712 USA e-mail: sbarik@utexas.edu.}

}

\markboth{Matrix Completion and Performance Guarantees for Single Individual Haplotyping}
{Matrix Completion and Performance Guarantees for Single Individual Haplotyping}

\maketitle

\begin{abstract}

Single individual haplotyping is an NP-hard problem that emerges when
attempting to reconstruct an organism's inherited genetic variations using data typically
generated by high-throughput DNA sequencing platforms. Genomes of diploid organisms,
including humans, are organized into homologous pairs of chromosomes that differ from
each other in a relatively small number of variant positions. Haplotypes are ordered
sequences of the nucleotides in the variant positions of the chromosomes in a homologous
pair; for diploids, haplotypes associated with a pair of chromosomes may be
conveniently represented by means of complementary binary sequences. In this paper, we
consider a binary matrix factorization formulation of the single individual haplotyping problem
and efficiently solve it by means of alternating minimization. We analyze the convergence
properties of the alternating minimization algorithm and establish theoretical bounds for the
achievable haplotype reconstruction error. The proposed technique is shown to outperform
existing methods when applied to synthetic as well as real-world Fosmid-based HapMap
NA12878 datasets.
\end{abstract}

\begin{IEEEkeywords}
matrix completion, single individual haplotyping, chromosomes, sparsity, alternating minimization.
\end{IEEEkeywords}

\IEEEpeerreviewmaketitle

\input{INTRO}

\input{SYS_MODEL}

\input{ALGO}

\input{RESULTS}

\input{EXP}

\section{Conclusion}
Motivated by the single individual haplotyping problem from
computational biology, we proposed and analyzed a binary-constrained variant of the
alternating minimization algorithm for solving the rank-one matrix factorization problem.
We provided theoretical guarantees on the performance of the algorithm and analyzed
its required sample probability; the latter has important implications on experimental
specifications, namely, sequencing coverage. Performance of haplotype reconstruction
is often expressed in terms of the minimum error correction score; we establish theoretical
guarantees on the achievable MEC score for the proposed binary-constrained alternating
minimization. Experiments with a HapMap sample NA12878 dataset as well as those with
a widely used benchmarking simulated dataset demonstrated efficacy of our algorithm.

\appendices
\input{APPENDIX}

\section*{Acknowledgment}
This work was funded in part by the NSF grant CCF 1320273.

\ifCLASSOPTIONcaptionsoff
  \newpage
\fi

\bibliographystyle{IEEEtran}

\bibliography{ms}

config: arxiv


TITLE: Test Deutsch's algorithm with constant function $f(x) = 1$
QUESTION [1 upvotes]: I am really, really beginner to QP. But for some reason, I am studying about Deutsch-Jozsa algorithm, and I have found this article which really helped me with understanding the pipeline of the algorithm.
There's an example that applies $f(x) = x$ to the algorithm, and I nearly understood it. (Honestly, I cannot sure)
So, I tried to apply $f(x)$ = 1 (constant) to the algorithm, but I have a problem. Here's my (wrong) solution to solve it.

Prepare two qubits ― $(1, 1)$ (I'll notate like this since I'm not used to bra-ket notation)
Operate Hadamard transformation to qubits that I prepared:

$$ (\frac{1}{\sqrt{2}}0 - \frac{1}{\sqrt{2}}1)(\frac{1}{\sqrt{2}}0 - \frac{1}{\sqrt{2}}1) = \frac{1}{2}(00-01-10+11) $$

Apply $U_f$ to each qubit that maps $(x, y)$ to $(x, y + f(x))$. (In this case, as this answer mentions, $U_f$ can be simplified as NOT-gate on second qubit.):
$$ \frac{1}{2}(01-00-11+10) $$
Send output from Step 3 through Hadamard gate again:

$$\scriptsize \frac{1}{2}((\frac{1}{\sqrt{2}}0 + \frac{1}{\sqrt{2}}1)(\frac{1}{\sqrt{2}}0 - \frac{1}{\sqrt{2}}1)-(\frac{1}{\sqrt{2}}0 + \frac{1}{\sqrt{2}}1)(\frac{1}{\sqrt{2}}0 + \frac{1}{\sqrt{2}}1)-(\frac{1}{\sqrt{2}}0 - \frac{1}{\sqrt{2}}1)(\frac{1}{\sqrt{2}}0 - \frac{1}{\sqrt{2}}1)+(\frac{1}{\sqrt{2}}0 - \frac{1}{\sqrt{2}}1)(\frac{1}{\sqrt{2}}0 + \frac{1}{\sqrt{2}}1)) $$
$$ = \frac{1}{2}(\frac{1}{2}(00-01+10-11)-\frac{1}{2}(00+01+10+11)-\frac{1}{2}(00-01-10+11)+\frac{1}{2}(00+01-10-11)) $$
$$ = \frac{1}{4}(00-01+10-11-00-01-10-11-00+01+10-11+00+01-10-11) $$
$$ = \frac{1}{4}(-11-11-11-11) $$
$$ = -11 ? $$
This result is obviously wrong (..or it isn't?), because the final result is negative, and also a bit to measure (a first bit) is one, even if the function $f(x) = 1$ is constant, and it should be zero.
This solution seems to work well to other functions, e.g. the result of $f(x) = 0$ was $(0, 0)$, and $f(x) = \bar{x}$ was $(1, 0)$.
Where did I went wrong?

REPLY [1 votes]: In Deutsch's algorithm you prepare the qubits in the $|01\rangle$ state, not $|11\rangle$. Because the initial condition is incorrect, everything follows will produce a different result. The correct steps should be:

Prepare two qubits $(\color{red}{0}, 1)$
Hadamard: $$ \left( \frac1{\sqrt2} 0 \color{red}+ \frac1{\sqrt2} 1 \right)\left( \frac1{\sqrt2} 0 - \frac1{\sqrt2} 1 \right) = \frac12\left(00 -01 \color{red}+10\color{red}-11\right) $$
Apply $U_f$: $$ \frac12\left(01 -00 \color{red}+11\color{red}-10\right) $$
Hadamard: \begin{align}
& \frac14\bigl( (0+1)(0-1)-(0+1)(0+1) \color{red}+ (0-1)(0-1) \color{red}- (0-1)(0+1) \bigr) \\
&= \frac14 \bigl( 00-01+10-11-00-01-10-11 \color{red}+00 \color{red}-01 \color{red}- 10 \color{red}+ 11 \color{red}- 00\color{red}- 01 \color{red}+ 10  \color{red}+ 11 \bigr) \\
&= \frac14 \bigl( -4\cdot\color{red}01 \bigr) \\
&= -\color{red}01
\end{align}

So the first bit is indeed zero, showing that $f(x)$ is a constant. The negative sign is just a phase which can be ignored.


TITLE: Prime $p$ does not divide $\dbinom{n}{p^e}$
QUESTION [5 upvotes]: I encountered this statement as a lemma in the course of proving the first Sylow theorem.
If $|G|=n=p^em$, $p$ is a prime and $m$ and $p$ are coprime, show that
$p$ does not divide $\dbinom{n}{p^e}$.
I tried expanding to factorials, but it didn’t work.
A solution or directive towards the proof is appreciated.

REPLY [2 votes]: It seems to me that nothing but the definition of the binomial coefficient is
needed for this.
Indeed,
by the definition of the binomial coefficient we have
$$
{n\choose p^e}=
\frac{n(n-1)(n-2)\ldots(n-p^e+1)}{1\cdot2\cdots(p^e-1)\cdot p^e}=
\frac{n}{p^e}\prod_{k=1}^{p^e-1}\frac{n-k}{k}.
$$
It remains to be seen that none of the factors of the last expression is divisible by $p$.


TITLE: Prove through induction that if $f(x) := \frac{2}{1-x^2}$ then $f^{(n)}(x) = n!(\frac{1}{(1-x)^{n+1}}+\frac{(-1)^n}{(1+x)^{n+1}})$
QUESTION [2 upvotes]: The original formula is $f(x) := \frac{2}{1-x^2}$, for $f\colon\mathbb{R}- ({-1,1}) \to \mathbb{R}$
Through Induction prove that $f^{(n)}(x)=n!(\frac{1}{(1-x)^{n+1}}+\frac{(-1)^n}{(1+x)^{n+1}})$.
So I start of with $n=0$ as the base case and see that:
$$f^{(0)}(x)=0!(\frac{1}{(1-x)^{0+1}}+\frac{(-1)^0}{(1+x)^{0+1}})$$
$$f^{(0)}(x)=(\frac{1}{(1-x)}+\frac{1}{(1+x)})$$
$$f^{(0)}(x)=\frac{1(1+x)+1(1-x)}{(1-x)(1+x)}$$
$$f^{(0)}(x)=\frac{2}{(1-x)(1+x)}$$
$$f^{(0)}(x)=\frac{2}{1-x^2}$$
Then I presumed that because it works for $n=0$ then it must also work for all $n∈\mathbb N$. Then I proceeded to state that it must also work for $n=n+1$.
$$f^{(n+1)}(x)=(n+1)!(\frac{1}{(1-x)^{n+1+1}}+\frac{(-1)^{n+1}}{(1+x)^{n+1+1}})$$
$$f^{(n+1)}(x) = f^{(n)'}(x)\\$$
$$=[(n!)((\frac{1}{(1-x)^{n+1}}+\frac{(-1)^n}{(1+x)^{n+1}}))]´$$
$$=(n!)[(\frac{1}{(1-x)^{n+1}}]´ + [\frac{(-1)^n}{(1+x)^{n+1}}]´$$
$$=n!(-(-n-1)(1-x)^{-n-2})+((-1)^n(-n-1)(1+x)^{-n-2})$$
But I'm kinda stuck on how to continue on from there ...

REPLY [1 votes]: you did all right until the last line
$$....=(n!)\left[(\frac{1}{(1-x)^{n+1}}\right]' + \left[\frac{(-1)^n}{(1+x)^{n+1}}\right]'  =(n!)(\frac{n+1}{(1-x)^{n+2}} - \frac{(n+1)(-1)^n}{(1+x)^{n+2}} \\= (n+1)!\left(\frac{1}{(1-x)^{n+2}} + \frac{((-1)^{n+1}}{(1+x)^{n+2}}\right)$$


TITLE: How does: $\sin^3(x) = (\cos^2(x)-1 )(-\sin(x))$
QUESTION [0 upvotes]: So I'm trying to integrate: $\sin^3(x)$ My book says:
$$G(x)=\int \sin^2(x) \times \sin(x) \,\, dx$$
I do not understand the following step:
$$\int -(1-\cos^2(x)) \times -\sin(x) \,\, dx$$

$$\int (\cos^2(x) -1)\,\, d\cos(x)$$
$$\int (u^2 -1) du$$
$$\frac{1}{3}u^3-u+c = \frac{1}{3}\cos^3(x)-\cos(x)+c$$
In other words how do you go from: $$\sin^3(x) = (\cos^2(x)-1) \times -\sin(x)$$
I appreciate it,
Cro-Magnon.

REPLY [1 votes]: $$\left(\cos^2(x) - 1\right)\cdot\left(-\sin(x)\right) =
\left(-\sin^2(x)\right)\left(-\sin(x)\right) = \sin^3(x)$$


TITLE: How many dihedral angles need to be specified to uniquely specify a triangulated polyhedron?
QUESTION [4 upvotes]: Suppose you are given a simplicial complex $K$ homeomorphic to the sphere and for each each edge of the complex a label specifying a length of that edge (this gives us a polyhedral metric on $K$). In other words, I'm specifying the local geometry and combinatorics of a polyhedron, but not its realization.
We'll call a realization of $K$ a polyhedron in $\mathbb{R}^3$ whose vertex, edge, and face structure is given by $K$ and whose edges are equivalent to the given edge length labeling. Suppose also that I know that at least one realization exists (side question: is it NP-hard to determine this?).
I know that thanks to the familiar examples, like the "house" made of a cube (with triangulated sides for consistency) and a pyramid on top (and its alternative the house with an inverted roof), the edge labeling is not enough to uniquely specify the realization (barring something like requiring convexity and then invoking Cauchy's theorem). On the other hand, clearly if I specify all the dihedral angles as well I've uniquely specified the realization, but this feels like overkill. My question is: beyond knowing the edge lengths, how much extra information is needed to uniquely specify the polyhedron? It seems pretty clear that if I take a spanning tree of the dual graph, and specify the dihedral angles represented by each dual edge then that should be enough, but even this feels like it's too much information. Can I always get away with less?
Any references to work on this or related problem is appreciated.

REPLY [4 votes]: It is an open problem to determine if the combinatorics plus the dihedrals
determine a convex polyhedron.
More precisely Stoker's conjecture from 1968—still unsolved—says that the face lattice of a convex polyhedron and its dihedral angles determine the face angles:

Stoker, James J. "Geometrical problems concerning polyhedra in the large." Communications on Pure and Applied Mathematics 21, no. 2 (1968): 119-168.

This doesn't address the OP's "beyond knowing the edge lengths,...",
but rather what is needed beyond knowing the dihedral angles (for convex polyhedra).
There has been relatively recent progress on Stoker's conjecture:

Mazzeo, Rafe, and Grégoire Montcouquiol. "Infinitesimal rigidity of cone-manifolds and the Stoker problem for hyperbolic and Euclidean polyhedra." Journal of Differential Geometry 87, no. 3 (2011): 525-576.
  (arXiv abs.)
"We prove here an infinitesimal rigidity result valid for cone angles less than $2\pi$, stating that infinitesimal deformations which leave the dihedral angles fixed are trivial in the hyperbolic case, and reduce to some simple deformations in the Euclidean case."


TITLE: Problem on normal distribution probability
QUESTION [0 upvotes]: An investment portfolio contains stocks of a large number of companies. Over the past\ year the rates of return on these corporate stocks followed a normal distribution with-As mean 12.5% and standard deviation 7.4%. Assuming that the rates of return for different companies are independent of each other, select the appropriate statement below.

A. The probability of a rate of return above 15% for these companies
is 0.4801, a level of return of 82.59% was exceeded by the top 10% of
stocks, and the average rate of return over any 4 selected stocks in
this portfolio has a distribution that is normally distributed with a
mean of 12.5% and a standard deviation of 3.7%.
B. The probability of a rate of return above 15% for these companies
is 0.3669, a level of return of 21.97% was exceeded by the top 10% of
stocks, and the average rate of return over any 4 selected stocks in
this portfolio has a distribution that is normally distributed with
a. mean of 50% and a standard deviation of 14.8%.
C. The probability of a rate of return above 15% for these companies
is 0.3669, a level of return of 21.97% was exceeded by the top 10% of
stocks, and the average rate of return over any 4 selected stocks in
this portfolio has a distribution that has an unknown form but with a
mean of 12.5% and a standard deviation of 3.7.4.
D. The probability of a rate of return above 15% for these companies
is 0.2709, a level of return of 23.4% was exceeded by the top 10% of
stocks, and the average rate of return over any 4 selected stocks in
this portfolio has a distribution that is normally distributed with a
mean of 12.5% and a standard deviation of 3.7%.

E. The probability of a rate of return above 15% for these companies
is 0.3669, a level of return of 21.97% was exceeded by the top 10%
of stocks, and the average rate of return over any 4 selected stocks
in this portfolio has a distribution that is normally distributed
with a mean of 12.5% and a standard deviation of 3.7%.

REPLY [0 votes]: I don´t write the percentage sign in my calculations.
$\textrm{... a level of return of x% was exceeded by the top 10% of stocks,}$
The equation is $P(X\geq x)=1-P(X\leq x)=0.1$
$$1-\Phi\left(\frac{x-12.5}{7.4} \right)=0.1$$
$$0.9=\Phi\left(\frac{x-12.5}{7.4} \right)$$
$$\Phi\left(0.9\right)^{-1}=\frac{x-12.5}{7.4} $$
$$1.282=\frac{x-12.5}{7.4} $$
$$x=21.9868$$
So the answer is approximately $21.97\%$. The differences in decimal digits are not so important. You can exclude the options $A$ and $D$.
$\textrm{ and the average rate of return over any 4 selected stocks in this portfolio}$ $\textrm{ has a distribution that is normally distributed with a mean of y% and a standard deviation of z%}$
Let $X_1, X_2, ..., X_n$ be n independent, identically and normally distributed random variables as $X_i\sim \mathcal N(\mu, \sigma^2)$.
And let $\overline X_n=\frac{X_1+ X_2+ ...+ X_n}{n}$. Then the distribution is $\overline X_n\sim \mathcal N\left(\mu, \frac{\sigma^2}n\right)$. That means in your case $$\overline X_4\sim \mathcal N\left(12.5, \frac{ 7.4^2}4\right)$$
Therefore the standard deviation of $\overline X_4$ is $\sqrt{\frac{ 7.4^2}4}=\frac{7.4}{2}=3.7$


TITLE: A question on a linear system on a complex surface.
QUESTION [3 upvotes]: For a complex surface $X$ with a line bundle $L$, the base locus Bs$|L|$ consists of $0$-dimensional and $1$-dimensional components. The fixed part of $|L|$ is the $1$-dimensional locus and of Bs$|L|$ and we denote it by $F$.

Why is $h^0(X,O(F))=1$?
How should I think of the map $L(-F)\hookrightarrow L$, which induces an isomorphism $H^0(X,L(-F))\cong H^0(X,L)$?

As to the second question, I have trouble about how to understand the line bundle $L(-F)=L\otimes O(-F)$.
If I understand global sections of $L(-F)$ are global sections of $L$ which are zeros on $F$, the last isomorphism $H^0(X,L(-F))\cong H^0(X,L)$ is reasonable; since any global section of $L$ has zeros on $F$, they are global section of $L(-F)$.

REPLY [4 votes]: Let's illustrate with an example.  Say $X$ is the blow-up of $\mathbb P^2$ at a point, with exceptional divisor $E$.  Think about the line bundle $\mathcal O_X(H+E)$, where $H$ is a line not through the blown-up point.  I think it's easier to think about this in the language of divisors and linear systems, but you can translate it back to line bundles.
What does the linear system $|H+E|$ look like in our case?  You can check that a divisor in this linear equivalence class consists of $E$ together with some line $L$ (or $2E+R$, where $R$ is the strict transform of a line through the blown-up point). In particular, $E$ is in the base locus, and the fixed part $F$ is just $E$.
What's $dim H^0(X,\mathcal O_X(F))$?  There's no way it can be greater than 1 -- if it were, it would mean that $F$ moves in a linear system of dimension at least $1$.  But that means that at least one component of $F$ isn't in the base locus of $F$ at all (it's not fixed), and there's no way such a component could be in the base locus of $L$.  So it shouldn't have been included in $F$ in the first place.
At the level of linear series, the map $L(-F) \to L$ just says to take an element of $L(-F)$, add on $F$, and you get something in the linear series $|L|$.  In our case, something $|L(-E)|$ will just be a line, and you get something in $L$ by throwing in $E$ too.
Your interpretation at the end seems correct to me.  Note that there's always a map $H^0(X,L(-F)) \to H^0(X,L)$ whenever $F$ is effective, but it's only an isomorphism if $F$ is the fixed part.


text:
{\bf Problem.} The pattern of Pascal's triangle is illustrated in the diagram  shown. What is the fourth element in Row 15 of Pascal's triangle? $$
\begin{array}{ccccccccccccc}\vspace{0.1in}
\textrm{Row 0}: & \qquad & & & & & 1 & & & & & & \\ \vspace{0.1in}
\textrm{Row 1}: & \qquad & & & & 1 & & 1 & & & & &\\ \vspace{0.1in}
\textrm{Row 2}: & \qquad & & & 1 & & 2 & & 1 & & & &\\ \vspace{0.1in}
\textrm{Row 3}: & \qquad & & 1 && 3 && 3 && 1&& \\ \vspace{0.1in}
\textrm{Row 4}: & \qquad & 1&& 4 && 6 && 4 && 1
\end{array}
$$
{\bf Level.} Level 3
{\bf Type.} Counting & Probability
{\bf Solution.} In Pascal's triangle, the $k^\text{th}$ element in the row $n$ has the value $\binom{n}{k-1}.$ Row $15$ starts with $\binom{15}{0},$ $\binom{15}{1},$ $\binom{15}{2},$ $\binom{15}{3},$ so the fourth element is $$\binom{15}{3}=\frac{15!}{3!(15-3)!}=\frac{15\cdot14\cdot13}{3\cdot2\cdot1}=5\cdot7\cdot13=\boxed{455}.$$

set_name: MATH


text:
\DOC MK_CONJ

\TYPE {MK_CONJ : thm -> thm -> thm}

\SYNOPSIS
Conjoin both sides of two equational theorems.

\DESCRIBE
Given two theorems, each with a Boolean equation as conclusion, {MK_CONJ}
returns the equation resulting from conjoining their respective sides:
{
         A |- p <=> p'   B |- q <=> q'
      ----------------------------------- MK_CONJ
           A u B |- p /\ q <=> p' /\ q'
}

\FAILURE
Fails unless both input theorems are Boolean equations (iff).

\EXAMPLE
{
  # let th1 = ARITH_RULE `0 < n <=> ~(n = 0)`
    and th2 = ARITH_RULE `1 <= n <=> ~(n = 0)`;;
  val th1 : thm = |- 0 < n <=> ~(n = 0)
  val th2 : thm = |- 1 <= n <=> ~(n = 0)

  # MK_CONJ th1 th2;;
  val it : thm = |- 0 < n /\ 1 <= n <=> ~(n = 0) /\ ~(n = 0)
}

\SEEALSO
AP_TERM, AP_THM, MK_BINOP, MK_COMB, MK_DISJ, MK_EXISTS, MK_FORALL.

\ENDDOC

subset_name: curated


TITLE: What is this theorem about measurable functions saying?
QUESTION [1 upvotes]: Theorem:
Let $(\Omega,\mathcal{F})$ be a measurable space and let $f:\Omega \rightarrow Y$ be a given function. Let $\mathcal{A}$ be a collection of subsets of $Y$.
If $f^{-1}(A) \in \mathcal{F}$ for every $A \in \mathcal{A}$, then $f^{-1}(A) \in \mathcal{F}$ for every $A \in \sigma(\mathcal{A})$.
Questions:
I understand that $\Omega$ is the underlying set and $\mathcal{F}$ is a class of subsets of $\Omega$ that is closed under finite unions and complementation (i.e., $\mathcal{F}$ is a field). But what is $Y$? And what is the purpose of defining $\mathcal{A}$ as a collection of subsets of $Y$, with $A$ an element in that collection? Moreover, what is the message of the theorem?
The notation is throwing me off.

REPLY [1 votes]: $Y$ is defined in $f:\Omega\to Y$ (it is a space, containing the image of $f$).
$\mathcal{A}$ is a family of subsets of $Y$, not necessarily a $\sigma$-algebra. The point of the theorem is exactly to say that it is sufficient to check the measurability of $f$ for family of sets generating the $\sigma$-algebra on $Y$, instead of the whole $\sigma$-algebra on $Y$ (e.g. if $\sigma(\mathcal{A})$ is the $\sigma$-algebra on $Y$, then it is sufficient to check the measurability for all elements $A\in \mathcal{A}$ instead of all elements $A\in\sigma(\mathcal{A})$).


TITLE: Partial derivative of two variable function: $f(x,y)=\frac{2x^3-3y^2}{x^2+y^2}$
QUESTION [1 upvotes]: What is the partial derivatives  of this function at the point $(0,0)$?
$$f(x,y)=
  \begin{cases}
    \frac{2x^3-3y^2}{x^2+y^2} & \text{if }(x,y)\ne(0,0) \\
    0 & \text{if }(x,y)=(0,0)
  \end{cases}
$$
Thanks.

REPLY [0 votes]: $$ \begin{cases}
f_x(0,0): \lim_{h\to0} f(h,0)=\lim_{h\to0}\frac{2h^3}{h^3}=2 \implies f_x(0,0)=2\\f_y(0,0): \lim_{h\to0} f(0,h)=\lim_{h\to0}\frac{-3h^2}{h^3}=\pm\infty \implies f_y(0,0) \quad N.D.
\end{cases}$$


TITLE: Find a point given two other points and an angle
QUESTION [0 upvotes]: I have the 3D coordinates for three points $A, B, C$. I can find the angle formed between the vectors $\vec {AB}$ and $\vec {BC}$ by using a dot product.
However, I want to move point $B$ to $B^{\prime}$ such that the new vectors $\vec {AB^{\prime}}$ and $\vec {B^{\prime}C}$ form an angle $\theta$. $B^{\prime}$ is constrained to lie in the same plane as $A, B, C$. In addition, the distance between $B^{\prime}$ and $C$ must remain the same as the distance between $B$ and $C$.
Is there a way for me to compute $B^{\prime}$ given $\theta$?

REPLY [0 votes]: Let $\vec n = \vec{BA} \times \vec{BC}$. Let $b = \|\vec{CB}\|$, and let $\vec a = \vec{CA}$. $\vec n$ is normal to the plane of $A, B, C$.
We want to define a vector $\vec v$ such that $B' = C + b\vec v$. Your first constraint requires $\|\vec v\| = 1$. The coplanar constraint requires $\vec n \cdot \vec v = 0$, and finally the angle condition requires
$$(A - (C + b\vec v))\cdot(-b\vec v) = \|A -(C + b\vec v)\|b\cos \theta$$
That is:
$$(\vec a - b\vec v)\cdot(-b\vec v) = \|\vec a - b\vec v\|b\cos\theta$$.
This gives you three equations in the three unknown coordinates of $v$. Solve them to find the two possible values of $\vec v$, then calculate $B' = C + b\vec v$.


TITLE: Prove absolute convergence of $\sum_{n=1}^{\infty} \frac{2^nn!}{(z+1)(z+3)...(z+2n+1)}$
QUESTION [1 upvotes]: I need to prove that
$$\sum_{n=1}^{\infty} \frac{2^nn!}{(z+1)(z+3)...(z+2n+1)}, \operatorname{Re}z \gt \frac{1}{2}$$
is absolutely convergent.
I've tried to use d'Alambert ratio test, but I got $$\lim\limits_{n\to \infty}\vert{\frac{a_{n+1}}{a_n}}\vert = \lim\limits_{n\to \infty}\vert\frac{2^{n+1}(n+1)!(z+1)(z+3)...(z+2n+1)}{2^nn!(z+1)(z+3)...(z+2n+1)(z+2n+3)}\vert= 1$$ and so it's inconclusive.
Also I've tried to rewrite denominator as $\frac{(z+2n+1)!!}{(z-1)!!}$. Then I think I can use the fact that $(2k+1)!! = \frac{(2k+1)!}{2^kk!}$, the property of gamma function that $Γ(z+1)=zΓ(z)$ for $\forall z \in \mathbb{C}$ and the Stirling formula for gamma function, but I don't know how to apply all of this correctly there. How should I do it?
Or can you give any other ideas how to prove this?
Thanks for your help.

REPLY [0 votes]: For any $z$ such that $\text{Re}(z)>1$ the given series equals
$$ \frac{1}{z+1}\sum_{n\geq 1}\frac{n!}{\left(\frac{z+1}{2}+1\right)\cdots\left(\frac{z+1}{2}+n\right)}=\frac{1}{2}\sum_{n\geq 1}\color{blue}{\frac{\Gamma(n+1)\,\Gamma\left(\tfrac{z+1}{2}\right)}{\Gamma\left(\frac{z+3}{2}+n\right)}}$$
or
$$ \frac{1}{2}\sum_{n\geq 1} B\left(n+1,\frac{z+1}{2}\right) = \frac{1}{2}\int_{0}^{1}(1-u)^{\frac{z-1}{2}}\sum_{n\geq 1}u^n\,du $$
or
$$ \frac{1}{2}\int_{0}^{1}u (1-u)^{\frac{z-3}{2}}\,du=\frac{1}{2}B\left(2,\frac{z-1}{2}\right)=\color{red}{\frac{2}{z^2-1}} $$
and the absolute convergence follows by applying Gautschi's inequality to the blue term.
On the other hand we cannot have absolute convergence for any $z$ with real part in $\left(\frac{1}{2},1\right]$ because $\frac{2}{z^2-1}$ has a pole at $z=1$!


TITLE: How to come up with formula to get minimum sell price to profit on commodity exchange
QUESTION [2 upvotes]: I'm stuck trying to come up with an equation that will calculate the minimum sale price of a commodity (such as Bitcoin) in order to at least break even. That in it of itself is no problem, sell for same price it was purchased for at the minimum.
The issue I'm running into is, based on my several searches, one variable is dependent on another variable. To be specific, if I purchase something for $100, with .5% of the purchase being a fee, how do I figure out what my sale price of that same quantity will need to be in order to break even? If I'm thinking about this correctly, my independent variable is sale price, while the dependent variable is the sale fee. I've tried different things, but I can't seem to get the result I'm looking for. I'll use the following variables:
a =  purchase price
b =  purchase fee (constant, 0.5% or 0.005)
c =  sale price
d =  sale fee (constant, 0.5% or 0.005)

The first equation I come up with, and with the help of online formula solver to simplify:
$$c = (a * (1 + b)) / (1 + d).$$
But the fee is not supposed to be additive, this comes out of the purchase and sale prices
$$(c - (c * .005)) - (a - (a * .005)).$$
But I don't even know what c is. So if I use the online formula solver to solve for c, I get:$$ c = a.$$
I can trial and error in a spreadsheet until I get my profit to show $0.00 so I know what my minimum sale price is. But trying to convert that into a C# program is proving to test my limited knowledge in math. Can anybody point me in the right direction on how to logic this out to the point I can program it?

REPLY [1 votes]: I find the math a bit easier to work with starting with the purchase price and the sales price as your variables. So,  let's say you buy at a purchase price of $a=100$,  which means $b=.50$ and you have spent $100.50$ and have $100$ nominal value.   If you sell at price $c$ you have to pay $.005c$ as a fee, so you only earn back $c-.005c=.995c$ in return.   So if you want to make $100.50$,  we can set $100.50=.995c$ and solve for $c$ getting $c=\frac {100.50} {.995}\approx 101.005025$
Now we can replace the $100.50$ in this example with a generic purchase price of $a$ and a sales fee of $.005$ meaning you have spent $1.05a$  Setting that equal to $.995c$ we get
$$1.05a=.995c$$
$$c=\frac {1.05} {.995}a\approx 1.010050251a$$
So the price has to go up by a little over $1\%$ for you to break even, which makes sense, you have to cover the two $.5\%$ costs plus the little bit extra where the fees effectivly overlap
Generalizing this for other fees $b$ and $d$ written in decimal, we get
$$c=\frac {1+b} {1-d}a$$


TITLE: Determine limits of the following sequences and prove
QUESTION [1 upvotes]: The two sequences are $a_n = \frac{n}{n^2 + 1}$
and $s_n = \frac{1}{n} \sin(n)$.
I sort of know what do here
Obviously $\lim_{n\to\infty} a_n=0$, and you can do some sidework and say
$$\left|\frac{n}{n^2 + 1} - 0 \right| = \frac{|n|}{|n^2 + 1|} < \epsilon$$
Dont really know where to go from here.
I am wondering if n has to be a natural number.

REPLY [2 votes]: We want to find
$$\lim_{n\to\infty}\frac{n}{n^2+1}.$$
The limit is obviously $0$, and it seems to me that saying any more is a waste of time.
However, if we want to practice $\epsilon$-$N$ language, which can be useful elsewhere, we want to show that for ant $\epsilon\gt 0$ there is an $N$ such that if $n\get N$ then
$$\left|\frac{n}{n^2+1}-0\right|\lt \epsilon.$$
Note that $0\lt \frac{n}{n^2+1}\lt \frac{1}{n}$. It follows that if $n$ is any integer greater than $\frac{1}{\epsilon}$, then $\frac{n}{n^2+1}\lt \epsilon$.
So we can take $N=\left\lceil \frac{1}{\epsilon}\right\rceil$, where $\lceil x\rceil$ is the ceiling function.
The second problem is essentially the same. The limit is clearly $0$, since $\sin(n)$ stays between $-1$ and $1$, and the $n$ at the bottom crushes it.  Formally, we want to show that for every $\epsilon\gt 0$ there is an $N$ such that if $n\gt N$ then
$$\left|\frac{\sin(n)}{n}-0\right|\lt \epsilon.$$
We observe that $\left|\frac{\sin(n)}{n}\right|\le \frac{1}{n}$ and finish like before.

REPLY [1 votes]: For the first sequence $a_n$, note that
$$a_n=\frac{n}{n^2+1}=\frac{n/n}{(n^2+1)/n}=\frac{1}{n+1/n}.$$
So as $n\to\infty$, since $\frac{1}{n}\to0$, then $n+\frac{1}{n}\to\infty$, and thus
$$\lim_{n\to\infty} a_n=\lim_{n\to\infty} \frac{1}{n+1/n}=0.$$
For the second sequence $b_n$, note that $0\leq|\sin(n)|\leq1$. As $n\to\infty$, $n$ is eventually positive, so:
$$0\leq \left|\frac{1}{n}\sin(n)\right|\leq \frac{1}{n}$$
Using the squeeze theorem, because $\frac{1}{n}\to0$ then $\left|\frac{1}{n}\sin(n)\right|\to0$ also, therefore $\lim_{n\to\infty}\frac{1}{n}\sin(n)=0$.


TITLE: Mouse trapped in a box with 3 exits - DTMC problem
QUESTION [2 upvotes]: A mouse is trapped in a box, with 3 exits. Exit 1 leads back to the box in 3 steps,
Exit 2 leads back to the box in 1 step and Exit 3 never leads back to the box. The mouse chooses the exit at random. Define a Markov chain that describes the problem and derive its transition matrix.

Attempt. We need to define $\{X_n\}$ such that $X_{n+1}$ is determined by only
$X_n$ and not its predessecors. $X_n$ being equal to $1$ if the mouse is in the box and $0$ otherwise is not suitable for our problem.

Thank you in advance.

REPLY [1 votes]: Note that you need to keep track of the location of the mouse, because every chosen exit has different consequences. If we would have that Exit 1 would lead back to the box in one step, the following diagram could be used:

In that case you have four states: $0$ (box), $1$ (exit 1), $2$ (exit 2) and $3$ (exit 3). From state $0$ you go to the three other states with equal probability. If you enter state $3$, you do not return. If you enter state $1$ or $2$, you return back to the box (state $0$).
The problem states that you only return from exit 1 to the box in 3 steps. Therefore, if the mouse chooses exit 1, you have to keep track of the number of steps already taken, to make the chain Markovian. This is not possible with the current states, forcing us to extend the state space with states that keep track of the number of steps if the mouse has chosen exit 1. One way to add a step is via the following scheme, which can be extended to include more steps:


TITLE: Find the coordinates of a point on a circle
QUESTION [40 upvotes]: I have a circle like so
$r$, with angle $\theta$ to the $y$-axis">
Given a rotation θ and a radius r, how do I find the coordinate (x,y)?  Keep in mind, this rotation could be anywhere between 0 and 360 degrees.
For example, I have a radius of 12 and a rotation θ of 115 degrees. How would you find the point (x,y)?

REPLY [2 votes]: The standard circle is drawn with the 0 degree starting point at the intersection of the circle and the x-axis with a positive angle going in the counter-clockwise direction.
Thus, the standard textbook parameterization is:
x=cos t
y=sin t
In your drawing you have a different scenario.  The way it is drawn, the starting point is at the top and increasing degrees is in the clockwise direction.  Thus, the standard parameterization must be modified to accommodate your situation.
Take a look at what the value of x is in your picture at the starting point and then what happens as t increases.  x starts at 0 and then increases to a maximum of 1 and then returns to 0 when t = Pi.
Now you want to compare that behavior to a standard graph of sin and cos to decide which one matches that need.  X=sin t behaves that way, so now you have the parameterization of x.  Note, it is not x=cos t as a standard math book teaches you because in trigonometry class they typically have 0 degrees at the intersection of the x-axis and the unit circle.
Now, y in your drawing starts out at 1 and then decreases until you hit 0 and then -1 at PI.  A look at the a graph of either sin or cos shows that cos behaves that way.
So... Y=cos t.
I'm posting this because it can aid someone that knows about sin and cos, but has a problem in which the 0 degree starting point is in a non-standard position and the direction of positive degrees is in the clockwise direction, not the ccw direction.


TITLE: Find all smooth solutions of $\frac{\partial f}{\partial \bar{z}}=g$ for various $g$
QUESTION [1 upvotes]: This question comes from Complex Variables: An Introduction (pg 226, question 5):

Find all solutions $f\in\mathscr{E}(\mathbb{C})=\mathcal{C}^{\infty}(\mathbb{C})$ to the equation $\dfrac{\partial f}{\partial \bar{z}}=g$ for:
(a): $g(z)=z$
(b): $g(z)=\bar{z}$
(e): $g\in\mathscr{H}(\mathbb{C})$

(I skipped some of the questions)
I'm honestly at a bit of a loss as to how to approach this problem. For (a) and (e), we know such a solution exists by Theorem 3.2.1 (which is the solution to the inhomogeneous Cauchy-Riemann equation, and says that a smooth solution will exist if $g$ is smooth) but how would you classify all (smooth) solutions?
For (b), we can apply $\dfrac{\partial }{\partial {z}}$ to the equation to conclude that $f$ is harmonic, but I'm still at a bit of a loss to reduce the number of solutions here.

REPLY [2 votes]: All given $g$ don't involve $\bar z$ very much: it appears to the first power, at most. So it makes sense to look for $f$ of similar form: $$f(z) =  \bar z f_1(z)+\bar z^2 f_2(z)\tag1$$ where $ f_1,f_2$ are holomorphic. (More generally, one could consider $\sum_{n=0}^N \bar z^n f_n(z)$, etc.)
Applying the $\partial/\partial\bar z$ operator to $(1)$ and equating coefficients with the given right hand side
will give you a solution.
Once you have one solution, all other solutions are obtained by adding an arbitrary holomorphic function $f_0$. Indeed, for any two solutions of the inhomogeneous Cauchy-Riemann equation, their difference solve the homogeneous equation, i.e., it is holomorphic.


TITLE: Criterion that $R[\sqrt{D}]$ be a Dedekind ring.
QUESTION [3 upvotes]: I'm trying to work out a criterion that $R[\sqrt{D}]$ be a Dedekind ring.
Suppose the $R$ is integrally closed in its quotient field $F$, $2$ is a unit in $R$, and let $F(\sqrt{D})$ be a quadratic extension for some $D\in R$ nonzero.
I have been able to show that $x+y\sqrt{D}\in F(\sqrt{D})$ is integral over $R$ iff $x,Dy^2\in R$.

Now suppose $R$ is Dedekind, $2$ is a unit in $R$, and $(D)$ is not divisible by the square of a prime ideal in $R$. Apparently it follows in this case that $R[\sqrt{D}]$ is Dedekind.

Why? My hunch is that $R[\sqrt{D}]$ is the integral closure of $R$ in $F(\sqrt{D})$, in which case $R[\sqrt{D}]$ would be a Dedeking ring by a well known theorem.
By the above criterion, its clear that $R[\sqrt{D}]$ is integral over $R$. On the other hand, suppose $x+y\sqrt{D}\in F(\sqrt{D})$ is integral over $R$. Again by the above criterion, $x\in R$, and $Dy^2\in R$. It is enough to show $y\in R$.
The fact that $Dy^2\in R$ and $(D)$ is not divisible by the square of a prime ideal seem like they should be related to show $y\in R$, but I'm not sure of the link here. Since $D\mid Dy^2$ and $y^2\mid Dy^2$ in $R$, and $R$ is Dedekind, there exist ideals $I$ and $J$ such that $(Dy^2)=(d)I=(y^2)J$. I could further factor these into prime ideals, but that's it.

REPLY [2 votes]: If $y$ is not in $R$, then since $R$ is Dedekind, there is a nonzero prime ideal $\mathfrak{p}$ of $R$ such that $v_{\mathfrak{p}}(y)$ -- the exponent of $\mathfrak{p}$ in the prime-power factorization of the fractional ideal $(y)$ -- is negative.  You want to play that off against the condition that $D$ is not divisible by the square of any prime ideal.
If you don't see how to do this, try it in $\mathbb{Z}$ (in which $2$ is not a unit, but never mind that) first: if $y \in \mathbb{Q}$, $D$ is a squarefree integer, and $Dy^2 \in \mathbb{Z}$, then show that $y \in \mathbb{Z}$.  The case of a general Dedekind domain is very similar.


TITLE: A functional "composition" of homotopies is also a homotopy.
QUESTION [6 upvotes]: Let $h,h':X\to Y$ be homotopic and $k,k':Y\to Z$ be homotopic. The idea is to show that $k\circ h$ and $k'\circ h'$ are homotopic.
Let $F$ be a homotopy between $h(x)$ and $h'(x)$ and $G$ a homotopy between $k(x)$ and $k'(x)$. Define $H:X\times[0,1]\to Z$ by $G(F(x,t),t)$. Then $H$ is continuous as a composition of continuous maps and  for $x\in X$ we have
\begin{align}
H(x,0)&=G(F(x,0),0) = G(h(x),0) = k\circ h(x)\\ H(x,1) &= G(F(x,1),1) = G(h(x),1) = k'\circ h'(x),
\end{align}
so $H$ is a homotopy between $k\circ h$ and $k'\circ h'$.
My question is this: We cannot exactly compose $F$ and $G$, since both are functions of two inputs, and one output, but instead we compose $G\circ (F, t\mapsto t)$. It turns out that this is correct, but how did we know a priori to choose $t\mapsto t$? Why not, e.g. $t\mapsto t^2$? This turns out to be a homotopy as well, since
\begin{align}
G(F(x,0),0^2) = G(F(x,0),0) = k\circ h(x)\\
G(F(x,1),1^2) = G(F(x,1),1) = k'\circ h'(x).
\end{align}
But clearly $t\mapsto 1-t$ would not work. Is it the case that we can choose any continuous $\alpha : [0,1]\to[0,1]$ with $\alpha(0)=0$ and $\alpha(1)=1$ and $G\circ(F,\alpha)$ would be a homotopy?

REPLY [6 votes]: Given a homotopy $H : A \times I \to B$ and $t \in I = [0,1]$ define $H_t : A  \to B, H_t(a) = H(a,t)$.
Now let us denote your homotopy by $G * F$. This is certainly the most obvious solution because $(G * F)_t = G_t \circ F_t$, i.e. the homotopy $G * F$ is obtained by composing $G$ and $F$ levelwise.
Now let $\alpha, \beta : I \to I$ be continuous maps such that $\alpha(0) = \beta(0) = 0$ and $\alpha(1) = \beta(1) = 1$. Then $F^\alpha = F \circ (id_X \times \alpha)$ and $G^\beta = G \circ (id_Y \times \beta)$ are also homotopies between $h,h'$ and $k,k'$, respectively. Therefore also $G^\beta * F^\alpha$ is a homotopy as desired. But it definitely has no benefit to invoke $\alpha, \beta$ as additional ingredients.


TITLE: Calculate $\sum_\limits{k=1}^{100}\left \lfloor \sqrt{k}+\frac{1}{k} \right \rfloor$
QUESTION [5 upvotes]: Determine: $\sum_\limits{k=1}^{100}\left \lfloor \sqrt{k}+\frac{1}{k} \right \rfloor$.
It's well-known that $x-1<\lfloor x \rfloor \leq x, \forall x \in \mathbb{R}$, but this didn't help me at all.
I computed it and the sum equals 627, but I found no useful properties.

REPLY [4 votes]: Note that for $k\ge5$,
$$
\sqrt{k+1}-\sqrt{k}=\frac1{\sqrt{k+1}+\sqrt{k}}\gt\frac1k\tag{1}
$$
so that
$$
\sqrt{k}\lt\sqrt{k}+\frac1k\lt\sqrt{k+1}\tag{2}
$$
Taking the floor of $(2)$ says that
$$
\left\lfloor\sqrt{k}\right\rfloor\le\left\lfloor\sqrt{k}+\frac1k\right\rfloor\le\left\lfloor\sqrt{k+1}\right\rfloor\tag{3}
$$
The only time that $\left\lfloor\sqrt{k}\right\rfloor\lt\left\lfloor\sqrt{k+1}\right\rfloor$ is when $k+1$ is a perfect square, which, in light of $(2)$ means that $\left\lfloor\sqrt{k}+\frac1k\right\rfloor\lt\left\lfloor\sqrt{k+1}\right\rfloor$. Thus, we get that for $k\ge5$
$$
\left\lfloor\sqrt{k}+\frac1k\right\rfloor=\left\lfloor\sqrt{k}\right\rfloor\tag{4}
$$
We can also verify $(4)$ for the case $k=4$.
Thus,
$$
\begin{align}
\sum_{k=1}^{100}\left\lfloor\sqrt{k}+\frac1k\right\rfloor
&=\overbrace{2+1+2+10}^{k=1,2,3,100}+\sum_{k=4}^{99}\left\lfloor\sqrt{k}\right\rfloor\\
&=15+\sum_{j=2}^9j\left((j+1)^2-j^2\right)\quad\text{floor times number of terms with that floor}\\
&=15+\sum_{j=2}^9\left(2j^2+j\right)\\
&=15+\sum_{j=2}^9\left(4\binom{j}{2}+3\binom{j}{1}\right)\\
&=15+\left[4\binom{j+1}{3}+3\binom{j+1}{2}\right]_1^9\\[3pt]
&=15+4\binom{10}{3}+3\binom{10}{2}-4\binom{2}{3}-3\binom{2}{2}\\[12pt]
&=15+480+135-0-3\\[18pt]
&=627
\end{align}
$$


text:
\begin{definition}[Definition:Kuratowski Closure Operator/Definition 2]
Let $S$ be a [[Definition:Set|set]].
Let $\cl: \powerset S \to \powerset S$ be a [[Definition:Mapping|mapping]] from the [[Definition:Power Set|power set]] of $S$ to itself.
Then $\cl$ is a '''Kuratowski closure operator''' {{iff}} it satisfies the following axioms for all $A, B \subseteq X$:
{{begin-axiom}}
{{axiom | n = 1
        | t = $\cl$ is a [[Definition:Closure Operator|closure operator]]
}}
{{axiom | n = 2
        | t = $\map \cl {A \cup B} = \map \cl A \cup \map \cl B$
        | rc= $\cl$ preserves [[Definition:Binary Operation|binary]] [[Definition:Set Union|unions]]
}}
{{axiom | n = 3
        | t = $\map \cl \O = \O$
}}
{{end-axiom}}
\end{definition}

config: wiki


TITLE: Proof that for all positive reals $a, b$ , if $a <b$, then $n^a$ is little of oh $n^b$.
QUESTION [0 upvotes]: For all positive real numbers $a$ and $b$, and $a < b$, then $n^a$ is little oh of $n^b$.
The tricky part of this proof is that for all positive constant $c$, $c(n^b) \geq n^a$ for all $n > n_0$. What would $n_0$ and $c$ be in this case to prove the question? Thanks

REPLY [0 votes]: Use the definition $f(x)=o(g(x)) \iff \lim_{x\to\infty}\frac{f(x)}{g(x)}=0.$
So if $a<b$ we have $$\lim_{n\to\infty}\frac{n^a}{n^b}=\lim_{n\to\infty}\frac{1}{n^{b-a}}=0.$$ Thus $n^a=o(n^b)$ whenever $a<b$.
Since $n^b$ is nonzero for all $b>0$ and all $n>0$, my definition is equivalent. Observe that if for all $\varepsilon>0$ there exists $N$ such that when $n\ge N$ we have $$n^a<\varepsilon n^b\iff \forall\varepsilon>0~\exists N~s.t.~\forall n\ge N, ~\frac{n^a}{n^b}<\varepsilon \iff \lim_{n\to\infty}\frac{n^a}{n^b}=0.$$


TITLE: Problem with Deriving Curl in Spherical Co-ordinates.
QUESTION [1 upvotes]: The formula for the curl of a vector field $\vec{A}$ in spherical co-ordinates is given as:
$$ \begin{align}\nabla \times \vec{A}=
  \frac{1}{r\sin\theta} \left(
    \frac{\partial}{\partial \theta} \left(A_\varphi\sin\theta \right)
  - \frac{\partial A_\theta}{\partial \varphi}
  \right) &\hat{\mathbf r} \\
{}+ \frac{1}{r} \left(
    \frac{1}{\sin\theta} \frac{\partial A_r}{\partial \varphi}
  - \frac{\partial}{\partial r} \left( r A_\varphi \right)
  \right) &\hat{\boldsymbol \theta}  \\
{}+ \frac{1}{r} \left(
    \frac{\partial}{\partial r} \left( r A_{\theta} \right)
  - \frac{\partial A_r}{\partial \theta}
  \right) &\hat{\boldsymbol \varphi}.
\end{align}$$
Del in spherical co-ordinates is given as (slide 20):
$$\nabla = \frac{\partial}{\partial r} \hat{r}+ \frac1r \frac{\partial}{\partial \theta}\hat{\theta} + \frac1{r\sin{\theta}}\frac{\partial}{\partial \varphi} \hat{\varphi}.$$
Given the above definition of $\nabla$ in spherical co-ordinates, evaluating the curl manually using the determinant gives:
$$ \begin{array}{rcl} \nabla \times \vec{A} & = &
\begin{vmatrix} \hat{r} & \hat{\theta} & \hat{\varphi} \\
 \frac{\partial}{\partial r} & \frac1r \frac{\partial}{\partial \theta} & \frac1{r\sin{\theta}}\frac{\partial}{\partial \varphi} \\
 A_r & A_{\theta} & A_{\varphi} \end{vmatrix} \\
 &  &
\begin{align}=\left(\frac1r \frac{\partial}{\partial \theta} A_{\varphi} - \frac1{r\sin{\theta}}\frac{\partial}{\partial \varphi} A_{\theta} \right)\hat{r} \\
+\left(\frac1{r\sin{\theta}} \frac{\partial}{\partial \varphi} A_r - \frac{\partial}{\partial r} A_{\varphi} \right)\hat{\theta} \\
+\left(\frac{\partial}{\partial r} A_{\theta} - \frac1r \frac{\partial}{\partial \theta} A_r \right)\hat{\varphi}. \end{align}
\end{array} $$
The two formulae are very similar, however in the top one, various terms have been 'factored' out from inside the partial derivatives (by multiplying the term inside correspondingly), for example the very first term:
$$ \frac1{r \sin{\theta}} \frac{\partial}{\partial \theta} A_{\varphi} \sin{\theta} $$
which has been obtained from:
$$\frac1r \frac{\partial}{\partial \theta} A_{\varphi}.$$
Is there a mistake in my working, or is there some other reason it is ok to move terms such as $\sin{\theta}$ in and out of the partials freely? (I know that curl is defined fundamentally in terms of integrals etc however I am trying to avoid this explanation.)

REPLY [0 votes]: According to my sources, the curl should be written as
$$
 \nabla \times \mathbf{A} =
\frac{1}{r^2\sin\theta}\begin{vmatrix} \hat{r} & r\hat{\theta} & r\sin\theta\hat{\varphi} \\
 \frac{\partial}{\partial r} & \frac{\partial}{\partial \theta} & \frac{\partial}{\partial \varphi} \\
 A_r & rA_{\theta} & r\sin\theta A_{\varphi} \end{vmatrix} \\
$$
That would resolve your problems.


TITLE: Comparing Lebesgue integral and Riemann Stieltjes integral
QUESTION [4 upvotes]: Please how can I justify that these two definitions can be the same thing or let me say how can I compare the following two definitions.

If $g$ is a right continuous, increasing step function on $\mathbb{R}$ with  countable discontinuous points $x_1,x_2,\cdots,x_n$. Then for any function $f$, we define the integral $$\int_{\mathbb{R}
}f dg =\sum_{k=1}^{n} f(x_k)\Delta g(x_k)$$where $\Delta g(x_n) = g(x_n^+)-g(x_n^-)> 0$.
Let $(X, \mathcal{H}, \mu)$ be a measure space and let $f: X \rightarrow [0, \infty]$ be a  measurable function. We define the Lebesgue integral of $f$ over $X$ as $$\int_{X}f d \mu= \sup \left\{ \int_{X}g d \mu: 0 \le g \le f, g \,\, \text{is simple} \right\}.$$

I would be glad if a proof or correspondence can be given.
Thanks for your help.

REPLY [2 votes]: The second definition contains the first:
Let $g:\mathbb{R}\rightarrow\mathbb{R}$ as in 1.. With the classical extensions theorems, one can construct a measure $\mu_{g}$ over the borelians such that $\mu_{g}(]a,b])=g(b)-g(a)$. It can be show that $\mu_{g}(A)=\sum_{x_{n}\in A}\Delta g(x_{n})$.

If $f=\sum_{i=1}^{m}a_{i}1_{A_{i}}$ is measurable (and the $A_{i}$ disjoint) then $$\int fd\mu_{g}=\sum_{i=1}^{m}a_{i}\mu_{g}(A_{i})=\sum_{i=1}^{m}a_{i}\sum_{x_{n}\in A_{i}}\Delta g(x_{n})=\sum_{i=1}^{m}\sum_{x_{n}\in A_{i}}f(x_{n})\Delta g(x_{n})=\sum_{x_n}f(x_{n})\Delta g(x_{n})$$
If$f$ is positive and measurable, then there exist simple functions $f_{k}$ such that $f_{k}(x)\rightarrow f(x)$ monotonously. We deduce, by the monotone convergence theorem and the previous result, that $\int fd\mu_{g}=\sum_{n}f(x_{n})\Delta g(x_{n})$


TITLE: Using the chain rule on $P \left( Y = y \mid X = x \right)$, with $y \in \mathbb{R}^2$
QUESTION [0 upvotes]: I have a probability of the form $P \left( Y = y \mid X = x \right)$, with $y \in \mathbb{R}^2$, and want to find a way to express it in terms of $y_1$ and $y_2$ separately. Would the following be correct?
$$\begin{align*}
P \left( Y = y \mid X = x \right) & = P \left( Y_1 = y_1, Y_2 = y_2 \mid X = x \right) \\
& = P \left( Y_2 = y_2 \mid Y_1 = y_1, X = x \right) \cdot P \left( Y_1 = y_1 \mid X = x \right)
\end{align*}$$

REPLY [1 votes]: If $\def\eq{\hspace{0.25ex}{=}\hspace{0.25ex}}\vec Y\eq\langle Y_1,Y_2\rangle$ and $\vec y\eq\langle y_1,y_2\rangle$, then indeed:-
$$\begin{align}\mathsf P(\vec Y\eq\vec y\mid X\eq x)&=\mathsf P(Y_1\eq y_1, Y_2\eq y_2\mid X\eq x)\\[1ex]&=\mathsf P(Y_1\eq y_1\mid X\eq x)\,\mathsf P(Y_2\eq y_2\mid Y_1\eq y_1, X\eq x)\end{align}$$
This is just an application of the Chain Rule for Conditional Probability.


TITLE: What conditions must be present in a word problem to set it up and solve as a system of linear equations (2 variables)?
QUESTION [0 upvotes]: I was solving various word problems such as mixing solutions, differing interest rates, boat speeds, etc., and I thought what's the common underlying factor among these problems which allow us to represent them as a system of linear equations?
I'm having trouble identifying the common elements in them because the scenarios are so vastly different. One problem I'm solving an investment problem involving two different interest rates, the next I'm solving the speed of a boat going down river versus its speed going upriver, and another problem I'm trying to find the number of nickels and dimes in a purse that adds up to $1.10.
Bonus question: How much information is too little? Or rather, what's the minimum amount of information needed to set up a problem as a linear system of equations?
EDIT:
I thought it might be helpful to provide some examples, verbatim, from the textbook I'm using:
1

A boat's crew rowed 16 kilometers downstream, with the current, in 2 hours. The return trip upstream, against the current, covered the same distance, but took 4 hours. Find the crew's rowing rate in still water and the rate of the current.

2

You invested 7000 dollars in two accounts paying 6% and 8% annual interest. If the total interest earned for the year was 520 dollars, how much was invested at each rate?

3

A wine company needs to blend a California wine with a 5% alcohol content and a French wine with a 9% alcohol content to obtain 200 gallons of wine with a 7% alcohol content. How many gallons of each kind of wine must be used?

4

A student has two test scores. The difference between the scores is 12 and the mean, or average, of the scores is 80. What are the two test scores?

5

A coin purse contains a mixture of 15 coins in nickels and dimes. The coins have a total value of $1.10. Determine the number of nickels and the number of dimes in the purse.

6

A new restaurant is to contain two-seat tables and four-seat tables. Fire codes limit the restaurant's maximum occupancy to 56 customers. If the owners have hired enough servers to handle 17 tables of customers, how many of each kind of table should they purchase?

Please note, I know how and have already solved these, so please no answers explaining how to solve them. Thank you! :)

REPLY [0 votes]: An old engineering professor once said, "N equations, N unknowns."
When I look at these examples, the common theme is: there are two things you are looking for answers to; there are two things you know, which may be relationships.  Let's pick one of them apart:

A student has two test scores. The difference between the scores is 12 and the mean, or average, of the scores is 80. What are the two test scores?

It's asking for two scores, meaning the two unknowns.  You have two things you know: the difference between the scores, and the average of the scores, which are the two equations.
The challenge is assessing whether you have enough information to solve the problem (specifically, N equations, and up to N unknowns), that the information is not inherently in conflict, and if are there any statements are irrelevant to the problem.


TITLE: dendrogram node height
QUESTION [0 upvotes]: In my lectures, we are told that the height of the node connecting two clusters in a dendrogram should be the distance between the clusters. However, on Wikipedia, it says that the height of the node should be half of the distance between the clusters. This is according to the Wikipedia article because of the ultrametric constraint $d(x,z) \leq max(d(x,y), d(y,z))$. I can't find any information online clarifying this.

REPLY [0 votes]: It is traditional to make the height in a dendrogram equal to the (generalized) distance between clusters, but one can scale this height arbitrarily so long as the vertical axis is properly labeled.


TITLE: Approach to the integral $\int d^3u \exp(-\alpha|\mathbf{w}-\mathbf{u} |^2)\delta(\mathbf{k}\cdot\mathbf{u})$
QUESTION [1 upvotes]: I am trying to evaluate the integral
$\int d^3u \exp(-\alpha|\mathbf{w}-\mathbf{u} |^2)\delta(\mathbf{k}\cdot\mathbf{u})$
where I have 3 vectors w,u, and k, a constant alpha, and the integral is taken over all u (3 spatial coordinates). Usually for an integral of this form of the exponential I would just use a coordinate to make the exponential symmetric about the orgin. The issue here is that the delta function prevents (It makes the problem seemingly worse) me from effectively making this shift. I have played around with using the gaussian delta function sequence to integrate and then do the limit afterwards but this did not go so well.
I was thinking that if I align $ \mathbf{u}$ along $\mathbf{w}$ and $\mathbf{k}$ along $\mathbf{w}$ so that
$\mathbf{k}=k_\perp cos(\theta)\hat{x}+k_\perp sin(\theta)\hat{y}+k_\parallel \hat{w}$
$\mathbf{u}=u_\perp cos(\theta)\hat{x}+u_\perp sin(\theta)\hat{y}+u_\parallel \hat{w}$
then I'd have $\mathbf{k}\cdot\mathbf{u}=k_\perp u_\perp+k_\parallel u_\parallel$
and I could attempt the integral in cylindrical coordinates.
Any suggestions on an approach to deal with the delta function would be great.
Also If anyone has encountered a similar but slightly different integral Id be curious to know what approach you've used.

REPLY [1 votes]: If $k=(0,0,1)$, then the integral $I$ is actually two-dimensional and over the space $\{u_3=0\}$ and the integrand is $\exp(-\alpha(w_1-u_1)^2)\cdot\exp(-\alpha(w_2-u_2)^2)$. The change of variable
$$(u'_1,u'_2)=(u_1-w_1,u_2-w_2)
$$
shows that $I$ is the square of
$$
\int_\mathbb R\mathrm e^{-\alpha v^2}\mathrm dv=\sqrt{\frac{\alpha}\pi},
$$
hence
$$
I=\frac{\alpha}\pi.
$$
The general case follows, by a homothety and a rotation.


TITLE: Finding perimeter knowing area of right angle triangle and ratio of sides
QUESTION [0 upvotes]: Hi I am stuck on this question (GCSE extended maths) I can sort of get to the correct answer but I am not clear why. There is no clear explanation (or at all) how to do this in any texts I have looked at. The question looks like a curve ball to me, but I am sure I am missing something:

The area of a right-angled triangle is $51.2cm^{2}.$
If the two perpendicular sides are in the ratio $1:1.6,$ find the lengths of all three sides of the triangle.

REPLY [0 votes]: The ratio of sides is 1:1.6
or lengths $x$ and $1.6x$
Area $=  \frac 12 bh\\ 0.8 x^2 = 51.2\\x^2=64\\x = 8$
Two of the sides then are $8$ and $12.8$
The 3rd side you can find using the Pythagorean theorem.


TITLE: Matrix times eigenspace associated to its eigenvalue
QUESTION [2 upvotes]: Let $A$ be a square matrix of dimension $n$ and let $λ ∈ C$ be an element of the spectrum of $A$. $X_λ$ denotes the eigenspace of $A$ associated to $λ$;In which case do we have $AX_λ = \{0\}$?
The solution is $\lambda=0$. However it is not clear to me what the product between the matrix A and the eigenspace of $\lambda$ actually means, given that $X_λ$ is actually a set of vectors. What is the product between a matrix and a set of vectors?

REPLY [3 votes]: In this case you should think of $AX_\lambda$ as $\{A \mathbf x : \mathbf x \in X_\lambda\}$, ie the set of vectors obtained by multiplying every vector in $X_\lambda$ with $A$.
You see this kind of shorthand used quite often in maths - where $f$ is some function defined on a set $S$, $f(S)$ can denote the image of $f$. For example, $2\mathbb Z$ is sometimes used to mean the even integers.


TITLE: re-scaling a ratio to reflect a change in value
QUESTION [1 upvotes]: Please assume you are replying to somebody with limited knowledge in mathematics. I am drawing a blank.
I made up a formula which quantifies how similar two graphs are G and G' (Conceptual Graphs to be precise), scaled between 0 and 1, with 0 being completely dissimilar, and 1 being identical, where nodes can belong to either a set of concepts C or a set of relations R (but not both):
\begin{equation}
C_{G} \sim C'_{G'} = \frac{C(G \cap G')}{ \mid C_{G} \mid + \mid C_{G'} \mid }.
\end{equation}
\begin{equation}
R_{G} \sim R'_{G'} = \frac{R(G \cap G')}{ \mid R_{G} \mid + \mid R_{G'} \mid }.
\end{equation}
\begin{equation}
E_{G} \sim E'_{G'} = \frac{E(G \cap G')}{ \mid E_{G} \mid + \mid E_{G'} \mid }.
\end{equation}
By obtaining the ratio of identical concepts, relations and Edges,
I can calculate the normalised average graph similarity as shown below:
\begin{equation}
G \sim G' = \frac{ (C_{G} \sim C'_{G'}) + (R_{G} \sim R'_{G'}) + (E_{G} \sim E'_{G'})}{3}.
\end{equation}
This works, but it is biased towards nodes (concepts C and relations R),
since the ratio of nodes to edges is 2:1.
Now assume I have some graph metric, e.g., 0.79 (or 79%) and I want to adjust it, so that the graph ratio for nodes and edges is 1:1, so that the change in the ratio reflects the change in the metric.
I know that since the ratio is 2:1, thus 0.33 of 0.79 is attributed to edges, hence edges provide 0.26 of the total.

How can I adjust the formula so that the ratio of Edges and Nodes is 1:1?
Is it possible to adjust the actual value (e.g., 0.79) to reflect the ratio change, while keeping it normalised between 0 and 1?

REPLY [1 votes]: I don't know what exactly you're doing, but it sounds like you want to use some kind of weighted average. Given three quantities $x,y,z$, their weighted average with respective weights $a,b,c$ would be $\dfrac{ax+by+zc}{a+b+c}$. Often it is convenient (and always possible) to choose weights whose sum is $1$, so that the denominator vanishes, but that's not important. In your case, if you want to give $x$ twice as much weight as $y$, and say for example give $y$ and $z$ the same weight, then it would be $\dfrac{2x+y+z}{4}$.


TITLE: Important examples of measures which are not $\sigma$-finite
QUESTION [14 upvotes]: I think a measure which is not $\sigma$-finite is pain in the ass.
I wish I could safely assume all the measures are $\sigma$-finite.
I wonder if my wish is reasonable.
Here's my question:
what are important examples of measures which are not $\sigma$-finite?
Of course, "important" is a subjective word. So this is a soft question.
This is also a big-list question.

REPLY [9 votes]: I believe the most important class of non-$\sigma$-finite measures is provided by the Hausdorff measures.
$\mathcal{H}_{\alpha}$ on $\mathbb{R}^d$ where $d > \alpha$ is not $\sigma$-finite. (Because any subset of finite $\mathcal{H}_{\alpha} $ measure is of null Lebesgue measure)


TITLE: Integration of $x^2 \sin(x)$ by parts
QUESTION [3 upvotes]: How would I integrate the following?
$$\int_0^{\pi/2} x^2\sin(x)\,dx$$
I did $u=x^2$ and $dv=\sin(x)$
I got
$x^2-\cos(x)+2\int x\cos(x)\,dx.\quad$ I then used $u=x$ and $dv=\cos(x).$
I got
$$x^2-\cos(x)+2[x-\sin(x)-\int\sin(x)]$$
then
$x^2-\cos(x)+-2 \sin(x)(x)-\cos(x)\Big|_0^{\pi/2} =\dfrac{\pi^2}{4}-0-2$

REPLY [3 votes]: $u=x^2$ then $du=2xdx$ and $dv=\sin(x)dx$ then $v=-\cos(x)$. So integral becomes
$=-(x^2\cos(x))+2\int x\cos(x)dx$
integration by parts for $$\int x\cos(x)dx$$
$x=u, dx=du$ and $dv=\cos(x)dx, v=\sin(x)$
then
$$=-(x^2\cos(x))-2\int \sin(x)dx+2x\sin(x)$$
the integral of $\sin x$ is $-\cos x$ so we get
$$=2\cos(x)-x^2\cos(x)+2x\sin(x)$$


TITLE: Probability on selecting balls
QUESTION [1 upvotes]: If I have B black balls and W white balls in a bag, what is the probability that the last one I select is white?

How shall I solve this problem?
I am not sure how to make a start, is it correct to first assume $B \lt W$?

REPLY [0 votes]: Consider the permutation of the ball, by symmetry the probability of choosing white ball as the last ball is the same as choosing the white ball as the first ball.
Thus P=$W\over {W+B}$


TITLE: Integral and its limit
QUESTION [6 upvotes]: Evaluate: $$ \int_{0}^{\pi/2}n \left(1-\sqrt[n]{\cos x}\right) \mathrm{d}x$$
Rewriting this as $$I(n)= \int_{0}^{\pi/2}n \left(1-\sqrt[n]{\cos x}\right) \mathrm{d}x$$ and then Differentiating under the Integral Sign with respect to $n$. I was unable to go much further with this.
$$$$ Original problem: $$ \large\lim_{n \to \infty} \int_{0}^{\pi/2}n \left(1-\sqrt[n]{\cos x}\right) \mathrm{d}x$$

REPLY [3 votes]: I think you should solve the actual problem by solving the limit first.
The limit can be taken into the integral, so now the problem becomes
$$\int_{0}^{\pi /2}\lim_{n\rightarrow \infty }n(1-\sqrt[n]{\cos x})dx$$
By L'Hôpital's rule,
$$\lim_{n \to \infty }n(1-\sqrt[n]{\cos x})=\lim_{n \to \infty }\frac{-\cos^{\frac 1n}x\cdot \ln \cos x\cdot \frac {-1}{n^2}}{-\frac{1}{n^2}}=-\ln \cos x$$
Hence we have
$$\lim_{n \to \infty }\int_{0}^{\pi /2}n(1-\sqrt[n]{\cos x})dx=-\int_{0}^{\pi /2}\ln \cos x dx$$
Now the integral on the right side is $-\frac{\pi }{2}\ln2$. You can refer to here.
Thus
$$\lim_{n \to \infty }\int_{0}^{\pi /2}n(1-\sqrt[n]{\cos x})dx=\frac{\pi }{2}\ln2$$


TITLE: How to solve this fractional PDE?
QUESTION [0 upvotes]: $F_x = F*F_y = F^2*Fz$,
where $F = F(x,y,z)$.
So far I could solve the reduced problem considering either of the equalities:
$F_x = F*F_y$
Which yields me:
$F(x,y) = \frac{-y}{x+c}$

REPLY [1 votes]: $$F_x = F*F_y = F^2*Fz$$
If I well understand, you have two dependent PDEs.
FIRST PDE :$\quad F_x - F*F_y=0$
With the method of characteristics, the system of characteristic equations is :
$$\frac{dx}{1}=\frac{dy}{-F}=\frac{dF}{0}$$
A first family of characteristic curves comes from $\quad dF=0 \quad\to\quad F=c_1$
A second family of characteristic curves comes from $\quad \frac{dx}{1}=\frac{dy}{-c_1} \quad\to\quad y+c_1x=c_2\quad\text{or}\quad y+xF=c_2$
The general solution of the PDE is expressed on the form of the implicit equation :
$$F=\Phi(y+xF)$$
where $\Phi(X)$ is an arbitrary differentiable function.
The particular solution that you found $\quad F=\frac{-y}{x+c}\quad$ corresponds to the case of the particular function $\quad \Phi(X)=-X/c$ .
Since $F$ includes an independent variable $z$, the correct writing is :
$$F(x,y,z)=\Phi\big(z\:,\:y+xF(x,y,z)\big)$$
where $\Phi$ is an arbitrary differentiable function of two variables.
SECOND PDE :$\quad F_y - F*F_z=0$
The same method leads to the general solution on implicit form :
$$F(x,y,z)=\Psi\big(x\:,\:z+yF(x,y,z)\big)$$
where $\Phi$ is an arbitrary differentiable function of two variables.
NOTE :
Alternatively we could solve the PDE : $\quad F_x = F^2*Fz\quad$ which leads to
$$F(x,y,z)=\Theta\big(y\:,\:z+xF^2(x,y,z)\big)$$
where $\Theta$ is an arbitrary differentiable function of two variables.
But the functions $\Phi$ , $\Psi$ and $\Theta$ are not independent. So only two of them are sufficient to solve the problem.
RESULT :
The general solution of the associated PDEs $\quad F_x = F*F_y = F^2*Fz \quad$ is :
$$F(x,y,z)=\Phi\big(z\:,\:y+xF(x,y,z)\big)=\Psi\big(x\:,\:z+yF(x,y,z)\big)$$
where $\Phi$ and $\Psi$ are two independent arbitrary differentiable functions, each one of two variables.
Usually the arbitrary functions are determined according to some boundary conditions. Since there is no boundary condition specified in the wording of the question, one cannot go further.


TITLE: Set the variables $a$ and $b$ to a random value from 0-9. How many different lines can $ax+by=0$ form?
QUESTION [0 upvotes]: I recently had a Math Olympiad test, and I memorized a question that I couldn't find an effective solution for.

Set the variables $a$ and $b$ to a random value from 0-9. How many different lines can $ax+by=0$ form?

The way I tried in the test (which was very, very ineffective, and I didn't even have enough time to solve the question) was to list how many duplicates of a fraction there were, and then subtract the amounts from $10^2$.
This is obviously not a plausible solution, and I still don't know what the answer is.
NOTE: I don't remember the exact words that the question used. If you need me to clarify anything, please comment.

REPLY [0 votes]: You have to find unique lines of the form $ax+by=0$ where $a$ and $b$ are integers between $0$ and $9$.
I would simply count them -
i) When either $a$ or $b$ is zero : $ \ 2$ solutions
ii) when $a$ and $b$ are both non-zero
Start with $a = 1$ and that gives $9$ solutions $(1 \leq b \leq 9)$. Now note that $a \geq 2$ do not give new solutions if both $a, b$ have a common gcd other than $1$ (including $a, b$). So,
$a = 2, 4, 8$ will give $5$ more solutions each.
$a = 3, 9$ will give $6$ more solutions each.
$a = 5, 7$ will give $8$ more solutions each.
$a = 6$ will give $3$ more solutions.
That is total of $57$ different lines.


TITLE: Finding partitions $\{A,B\}$ of the set $\mathbb N_n:=\{1,...,n\}$ such that $\Big|\prod_{i \in A}p_i-\prod_{i\in B}p_i\Big|=1$
QUESTION [2 upvotes]: For a fixed $n$ , for what  partitions $\{A,B\}$ of the set $\mathbb N_n:=\{1,...,n\}$ do we have
$\Big|\prod_{i \in A}p_i-\prod_{i\in B}p_i\Big|=1$ ? , where $p_m$ denotes the $m$th prime for example $p_1=2,p_2=3...$ etc.
When the quantity is not $1$ , I can conclude that $p_{n+1} \le \Big|\prod_{i \in A}p_i-\prod_{i\in B}p_i\Big| $ this is why I am
interested in it . Also I would like to ask , for every integer $n>1$ does there exist a partition $\{A,B\}$
of  $\mathbb N_n:=\{1,...,n\}$ such that $\Big|\prod_{i \in A}p_i-\prod_{i\in B}p_i\Big|=1$ ? If not then for which $n$ there does not
exist such a partition ?

REPLY [1 votes]: There is some information about this problem in Guy, Lacampagne, and Selfridge, Primes at a glance, Mathematics of Computation 48, #177 (January 1987) 183-202, available here. See in particular Table 4, also the reference to D H Lehmer, On a problem of Stormer, Illinois J Math 8 (1964) 59-79, which should be available here.
EDIT: Carl Pomerance, Ruth-Aaron numbers revisited, in the book, Paul Erdos and his Mathematics, 567-579, writes that it is conjectured that 714, 715 is the last pair of consecutive numbers whose product is the product of the first $k$ primes for some $k$. The paper is here.


TITLE: Linear Transformation and Linear Independence Question
QUESTION [0 upvotes]: I'm wondering if $T$ is a linear transformation from $V$ into $W$ and $T(\alpha_1),\ldots,T(\alpha_n)$ is linearly independent would $\alpha_1,\ldots, \alpha_n$be linearly independent? I understand that that if $T$ is non-singular and $\alpha_1,\ldots, \alpha_n$ is linearly independent then $T(\alpha_1),\ldots,T(\alpha_n)$ is linearly independent but does the converse also apply?

REPLY [1 votes]: Suppose that that $T:V\to W$ is a linear transformation and that there are $\alpha_1,...,\alpha_n\in V$ such that $T(\alpha_1),...,T(\alpha_n)$ are linearly independent in $W$. Assume that there are $\lambda_1,...,\lambda_n\in k$ (where $k$ is the base-field, it is probably $\mathbb{R}$ in your case), such that:
$$\lambda_1\alpha_1+\lambda_2\alpha_2+...+\lambda_n\alpha_n=0 .$$
Applying the linear transformation $T$ to both sides of the above equation we get:
\begin{align}
0=T(0)&=T(\lambda_1\alpha_1+\lambda_2\alpha_2+...+\lambda_n\alpha_n)\\
      &=\lambda_1T(\alpha_1)+\lambda_2 T(\alpha_2)+...+\lambda_nT(\alpha_n).
\end{align}
Since $\{T(\alpha_1),...,T(\alpha_n)\}$ is a linearly independent set, we must have:
$$\lambda_1=\lambda_2=...=\lambda_n=0.$$
Thus, by definition $\alpha_1,...,\alpha_n$ are linearly independent.


TITLE: Are finitely generated objects in a locally Noetherian category also finitely presented?
QUESTION [2 upvotes]: Let $\mathcal C$ be a locally Noetherian category. Then, we know that every finitely generated object is Noetherian.
I cannot seem to be able to prove that Noetherian objects are finitely presented. A locally Noetherian category is locally coherent and so we know that finitely presented objects in it are coherent.
But are Noetherian objects automatically coherent?
I would be grateful for a proof or a precise reference.

REPLY [1 votes]: If your locally noetherian abelian category is also a Grothendieck category, it is true that every finitely generated object of it is finitely presented. Indeed, in a locally finitely generated Grothendieck category, a finitely generated object X is finitely presented if and only if for any short exact sequence 0 -> T-> Y -> X -> 0 with Y finitely generated, T is finitely generated. As finitely generated=noetherian is stable under subobjects in a locally noetherian category, this condition is free. Alternatively, you can use the classical equivalence between a locally noetherian Grothendieck category and the category of left-exact functors from a small artinian abelian category to abelian groups. If the locally noetherian category does not satisfy (AB5), what is your definition of "finitely generated"?