TITLE: Change this integral $\frac{1}{a+b} \int_{a}^{b} x \left[ f(x) + f(x+1) \right] dx$.
QUESTION [3 upvotes]: It is given that $f(a+b+1 -x) = f(x)$ where $a$ and $b$ are positive real numbers then $\frac{1}{a+b} \int_{a}^{b} x \left[ f(x) + f(x+1) \right] dx$ is equal to

$\int_{a-1}^{b-1} f(x) dx$

$\int_{a+1}^{b+1} f(x+1) dx$

$\int_{a-1}^{b-1} f(x+1) dx$

$\int_{a+1}^{b+1} f(x) dx$

I tried using integration by parts along with this property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$ and the one given in the question but I'm unable to get anything. It's that extra $x$ and the fraction $\frac{1}{a+b}$ which is making everything quite uneasy.

REPLY [1 votes]: $$I=\frac{1}{a+b} \int_{a}^{b} x \left[ f(x) + f(x+1) \right] dx=I_1+I_2$$
$$I_1=\frac{1}{a+b}\int_{a}^{b}xf(x) dx$$
$$I_2=\frac{1}{a+b} \int_{a}^{b} x f(x+1) dx = \frac{1}{a+b} \int_{a+1}^{b+1} (u-1)f(u) du$$
N0w apply the property:
$$\int_{p}^{q} g(x) dx=\int_{p}^{q} g(p+q-x) dx$$
Then $$I_2=\frac{1}{a+b} \int_{a+1}^{b+1} (a+1+b+1-u-1)f(a+b+2-u) du$$
Use $f(a+b+2-u)=f(a+b+1-(u-1))= f(u-1)$. then
$$I_2=\frac{1}{a+b} \int_{a+1}^{b+1} (a+b+1-u) f(u-1) du.$$
Let $u-1=w$, then $$I_2=\frac{1}{a+b} \int_{a}^{b} (a+b-w)f(w) dw$$
Finally $$I=I_1+I_2=\int_{a}^{b} f(w) dw =\int_{a-1}^{b-1} f(x+1)dx$$


TITLE: Induction for recurrence
QUESTION [1 upvotes]: I'm trying to understand an induction proof that aims to prove some function is in $O(n\log{ n})$.
It's on page 5 of this PDF:
https://courses.engr.illinois.edu/cs573/fa2010/notes/99-recurrences.pdf
The function is
$$T(n)=\sqrt{n}\, T(\sqrt{n}) +n$$
We want to prove that
$$T(n) \leq an \log{ n}$$ for sufficiently large $n$ and some constant $a$ to be determined later.
The author uses induction as follows:
\begin{align*}
T(n) &= \sqrt{n}\,T(\sqrt{n}) +n \\
     &\le \sqrt{n}\,a\sqrt{n}\, \log{\sqrt{n}} +n \qquad \text{[induction hypothesis]}  \\
     &=  (a/2)n \log{ n} +n   \\
     &\le an \log{n }
\end{align*}
First, this is an induction proof, so there should be a base case, but i understand that one needn't prove the base case first. That can always be done later, as the author acknowledges further down the page.
Second, and this is what i don't understand, how is this induction if the proof isn't going from some $n$ to an $n+1$? Is the jump from $\sqrt{n}$ to $n$ equivalent to the jump from $n$ to $n+1$? Or is something else going on.
btw, i asked this question at maths stackexchange, and i got no good answer. That's why I am reposing here.

REPLY [0 votes]: It is not clear exactly what the author had in mind. I think there are a couple of
possible interpretations.
One possibility is that the author was implicitly using an inductive step $n \rightarrow n^2$ rather than the usual $n \rightarrow n + 1.$
This works for numbers that are of the form $b^{2^k}$ for $k \in \mathbb N$ and a suitable base case $b.$  That is, the base case is really $k = 0$ (for which the problem size is $b^{2^0}$) and we do a simple induction on $k$ in $b^{2^k}.$
The disadvantage of this interpretation is that it doesn't say what happens with problems whose size isn't exactly of the form $b^{2^k}.$ If the answer is that we have to "pad out" the problem to the next larger such size, I don't think we get $n \lg \lg n$ time in general when $n$ can be any positive integer.
Another (more likely?) possibility is that "$\sqrt n$" (in the author's notation) is really
$\lceil \sqrt n \rceil,$ which could happen if the approach to a problem of size $k^2 + 1$ is to pad it out to a problem of size $(k + 1)^2.$
Then we can use strong induction to reason about $T(n)$ for any  positive integer $n.$
I suspect this is how such an algorithm would actually play out.
Since $\lceil \sqrt n \rceil < \sqrt n + 1$ for any positive integer $n,$
I think the inductive step would still hold. (I haven't checked.)
What we can't do (as far as I can see) is to prove case $T(n)$ for an arbitrary (but sufficiently large) integer $n$ in terms of $T(\sqrt n)$ if $\sqrt n$ means literally what it says, because in general  $\sqrt n$ is not an integer and therefore $\sqrt n \notin \{ 1, 2, 3, \ldots, n - 1 \}.$


TITLE: Find a continuous bounded function $f:(0,1]\to \mathbb{R}$ that is not uniformly continuous.
QUESTION [2 upvotes]: Find a continuous bounded function $f:(0,1]\to \mathbb{R}$ that is not uniformly continuous. Extend $f$ with continuity in such a way that $f(0)=0$ and find the oscillation $\omega _f(0)$ of $f$ at $0$.
I'm not sure if I have to extend the function with continuity, but I think so. I'm struggling to find a function, I've tried $\sin(1/x)$ but can't extend it with continuity. Any hint?

REPLY [3 votes]: How about $f(x) = sin(\frac{1}{x})$? Clearly $sin$ is bounded and continuous. As $x$ goes to $0^+$ , $sin(\frac{1}{x})$ will oscillate between -1 and +1.


TITLE: Does this principle have a name? $A\subset B\subset A \implies A=B$
QUESTION [2 upvotes]: I have a proof about some particular sets, which starts with something like $A \subset B$, and after a sequence of steps I show finally that $B \subset A$.
The proof from a high level is something like $A \subset B = C = D = E = F \subset A$, therefore $A=B$
But we know that if $A\subset B$ and $B \subset A$ then $A=B$.
QUESTION: Does this principle have a name?
I don't think it's the definition of set equality per se, but rather something which is equivalent to it.  Neither do I think that this is the axiom of extensionality, yet again it is closely related to it.

REPLY [1 votes]: Do you mean $\subseteq$ instead of $\subset$, proper subset?
Such theorems are called squeeze theorems.
For example, a <= b and b <= a implies a = b.


TITLE: Upper bound on $\sum_{n=1}^N \frac{1}{n}$
QUESTION [1 upvotes]: are there any analytical upper bounds on
\begin{align}
\sum_{n=1}^N \frac{1}{n}
\end{align}
Clearly, one upper bound is $N$. But there has to be a better one. There is an approach with Harmonic number but it's not "analytic".
Thank you

REPLY [5 votes]: $\sum_{n=1}^N {1 \over n} \le 1+ \ln N$.
This follows from $\int_{n-1}^n {1 \over t} dt = \ln n - \ln (n-1) > {1 \over n}$.

REPLY [1 votes]: I think the standard answer is $$\ln N+\gamma+\frac{1}{2N}$$ But I believe that with some effort that can be improved to $$\ln N+\gamma+\frac{1}{2N+\frac{1}{3}}$$


TITLE: Origin of 2π normalization factor in chemical $J$-coupling compared to the Heisenberg model
QUESTION [1 upvotes]: In chemistry, particularly the field of NMR spectroscopy, the interaction between two (nucleic) spins (or so I guess?) is governed by the Hamiltonian:
\begin{align}
\mathcal{H}=2\pi\cdot J_{ij}{\vec {S_{i}}}\cdot {\vec {S_{j}}}
\label{eq1}
\end{align}
https://en.wikipedia.org/wiki/J-coupling
This interaction is called $J$-coupling and it is usually heuristically explained by interaction between two electronic spins which is governed by the Pauli principle and the electron-nucleus interaction which in the end favors antiparallel alignment of the nucleic spins over parallel and thereby leads to a splitting of energy levels. Note that mainly chemists cover this topic and their approach is usually very hands-on in the sense that the values of $J_{ij}$ are usually measured on a spectrometer.
Now the above equation looks very similar to the Heisenberg model, governing the general relationship between two spins:
\begin{align}
 H_{\text{Heis}}=-J\sum _{\langle i,j\rangle }{\vec {S_{i}}}\cdot {\vec {S_{j}}}\qquad {\text{with }}i,j\,{\text{next neighbors}}
\label{eq2}
\end{align}
https://en.wikipedia.org/wiki/Heisenberg_model_(quantum)
This surely looks similar, here $J$ is the value of the exchange integral, which I would assume is also what defines the scalar coupling tensor $J_{ij}$ in the first equation. However, I am by no means sure those two $J$s are the same and if they are, where does the normalization factor of $2\pi$ coming from? Is it just common among chemists to divide their exchange integrals by $2\pi$? I would be very grateful if someone could shed a little light on that.

REPLY [0 votes]: Yes, chemists like J couplings in Hz. The NMR J or scalar coupling and the Heisenberg model occur in very different contexts. In NMR, the scalar coupling is typically more than 1000 times smaller than the Zeeman term in the Hamiltonian. Moreover, the temperature is more than 1000 times larger than the Zeeman term. The NMR J coupling is a residual magnetic dipole interaction, which remains after the main magnetic dipole interaction has been averaged away by motion. It can be of either sign. All of this is very different from the usual situation and concerns in applications of the Heidelberg model.


TITLE: Specific Integral calculus
QUESTION [2 upvotes]: I was surprised by the value of an integral I saw in Wolfram Alpha so I wanted to prove it on my own. This is the following integral :
$$ \displaystyle I=\int_{0}^{+\infty}\frac{\sin^2\left(t\right)}{1+t^2}\text{d}t=\frac{\pi}{2e}\text{sinh}\left(1\right)$$
Is there an easy way to show this equality ?
In order to prove it, my idea was to introduce its twin integral $ \displaystyle J=\int_{0}^{+\infty}\frac{\cos^2\left(t\right)}{1+t^2}\text{d}t$.
Then we have two results :
$$I+J=\frac{\pi}{2}$$
$$I-J=\int_{0}^{+\infty}\frac{\cos\left(2t\right)}{1+t^2}\text{d}t$$
I would like to show that for $w>0$
$$\int_{0}^{+\infty}\frac{\cos\left(wt\right)}{1+t^2}\text{d}t=\frac{\pi}{2}e^{-t}$$
I've searched for a differential equation with its derivative $\displaystyle -\int_{0}^{+\infty}\frac{t\sin\left(wt\right)}{1+t^2}\text{d}t$ but it doesnt seem that nice even when I integrate by parts... Any ideas ?

REPLY [5 votes]: Too long for a comment. Let $\omega\geq 0$ and define:
\begin{align}
f(\omega)= \int^\infty_0 \frac{\cos(\omega t)}{t^2+1}\,dt
\end{align}
The Laplce transform of it:
\begin{align}
\mathcal{L}(f)(s)=\int^\infty_0 \int^\infty_0 \frac{\cos(\omega t)}{t^2+1}e^{-s\omega}\,dt\,d\omega &= \int^\infty_0 \int^\infty_0  \frac{\cos(\omega t)}{t^2+1}e^{-s\omega}\,d\omega\,dt\\
&= \int^\infty_0 \frac{s}{(t^2+1)(s^2+t^2)}\,dt\\
&=\frac{s}{s^2-1}\int^\infty_0 \frac{1}{1+t^2}-\frac{1}{t^2+s^2}\,dt\\
&=\frac{s}{s^2-1} \left( \frac{\pi}{2} -\frac{\pi}{2s}\right)\\
&=\frac{\pi}{2}\left( \frac{s}{s^2-1}-\frac{1}{s^2-1}\right)\\
&=\frac{\pi}{2}\frac{1}{s+1}
\end{align}
assuming $s>0$.
Now inverting that yields:
\begin{align}
f(\omega) = \frac{\pi}{2}e^{-\omega}
\end{align}
For this method you must know the Laplace Transfrom and inverting it (this one was relatively easy though). Finally notice that some steps need to be filled which I leave it to you.


TITLE: A bag contains $3$ red and $2$ black balls . One ball is drawn at random . Its color is noted and put in the bag. A second draw is made and the same
QUESTION [0 upvotes]: A bag contains $3$ red and $2$ black balls . One ball is drawn at random . Its color is noted and put in the bag. A second draw is made and the same procedure is repeated . Find the probability of drawing two red balls.
Well, here goes my solution:

The number of ways to select two red balls is $\binom{3}{2}$ . The total number of ways to choose two balls is $\binom{5}{2}$. The required probability is $3/10$.

However if we do this in the following way:

We consider two events $A$: drawing a red ball at first draw and $B$: drawing a red ball at second draw. We know that, $P(A)P(B)=P(A\cap B)$ when $A$ and $B$ are independent events. So, $P(A)=3/5=P(B)$ . So, $P(A\cap B)=9/25$.

The answer given as the solution is $9/25$. However , if we solve this taking into consideration that these two events are independent to each other , then we get a different answer than solving it in the "normal way"(i.e the 1st method). Which method is valid? Why is the other one not valid? Why is the problem occuring?

REPLY [3 votes]: It isn't a question of one way or the other being "normal"
There are two distinct problems.
$\dfrac{\binom32}{\binom52}$ is for drawing without replacement
$\frac35\cdot\frac35$ is for drawing with replacement


TITLE: Prove that $\sqrt n \log n = O (n)$
QUESTION [1 upvotes]: Can someone please help me prove this problem? I don't know the right direction to take this problem in.

REPLY [1 votes]: By the Cauchy-Schwarz inequality, for any $n\geq 1$
$$ \color{red}{\log n} = \int_{1}^{n}\frac{dx}{x}\color{red}{\leq} \sqrt{\int_{1}^{n}1\,dx \int_{1}^{n}\frac{dx}{x^2}} =\sqrt{n+\frac{1}{n}-2}=\color{red}{\sqrt{n}-\frac{1}{\sqrt{n}}}$$
hence $\log(n)=O(\sqrt{n})$ for $n\to +\infty$, that is enough to prove your claim.


TITLE: On estimating the number of iid samples
QUESTION [5 upvotes]: Suppose we have iid samples $X_1,\cdots,X_n$, with the number of samples $n$ unknown, but I can sample from their sum $m=\sum_{i=1}^n X_i$. Further suppose $\mathbb{E}[X_i]=\mu$ and $Var[X_t]=\sigma^2$, with both $\mu$ and $\sigma$ known.
If I want to estimate the number of samples $n$, intuitively, one would find the nearest integer from $\frac{m}{\mu}$ (or is there any better way to estimate $n$?) If I want the estimate to be 95% trust-worthy, I guess there should be some requirements on the variance $\sigma^2$ and the true sample number $n$.
My attempt:

Suppose $n$ is huge and according to central limit theorem, the distribution of $\frac{m}{n}$ is approximately $\mathcal{N}(\mu,\frac{\sigma^2}{n})$. But I have a trouble handling the "rounding function". And probably central limit theorem is probably not proper for this circumstance, since it says what would happen for $n$ goes to infinity, but what we are trying to do here is exactly estimating $n$.

I tried to use Hoeffding's inequality, but since $n$ is stochastic here, I am not sure Hoeffding's inequality is proper for this circumstance.

REPLY [0 votes]: This may not be the most useful answer, since there aren't any answers yet, I figured it is better than nothing.
If the $X_{i}$ are $N(\mu, \sigma^{2})$, the sum $m$ is distributed as $N(n\mu, n\sigma^{2})$, so the problem of estimating $n$ is equivalent to the problem of estimating $\theta$ for $Y \sim N(\theta, k\theta)$ for known proportionality $k$.
As pointed out in the comments, one approach is the maximum likelihood estimator, which is rather efficient in general. The log-likelihood is
\begin{align*}
    L(\theta | y)
    &= -\log(k\theta) - \frac{1}{2}\left( \frac{y - \theta}{\sqrt{k\theta}}\right)^{2} + \mathrm{const} \\
    &= -\log(\theta) - \frac{1}{2}\left(\frac{y^{2}}{k\theta} - 2 \frac{y}{k} + \frac{\theta}{k} \right) + \mathrm{const}
\end{align*}
The derivative with respect the $\theta$ is
$$ L'(\theta|y) = -\theta^{-1} + \frac{y^{2}}{k} - \frac{1}{2k},$$
so the maximum likelihood estimator is
$$ \hat{\theta} = \frac{2y^{2} - 1}{2k} = \frac{y^{2}}{k} - \frac{1}{2k}.$$
Since $k=\sigma^{2}/\mu$ and $n = \theta/\mu$,
$$ \hat{n} = \frac{m^{2}}{\sigma^{2}} - \frac{1}{2\sigma^{2}}. $$
Interestingly, this is basically a method of moments estimator like you suggested, but using the second moment instead of the first.
If this were unbiased, it would be a minimum-variance unbiased estimator (since maximum likelihood estimators satisfy the Cramer-Rao bound), but unfortunately it is not since
$$ \mathrm{E}[y^{2}] = \theta^{2} + k\theta$$
implies
$$ \mathbb{E}[\hat{\theta}] - \theta = \frac{\theta^{2} - 1/2}{k}.$$
However, it is probably still reasonably efficient.
The confidence interval is a bit tougher, since you cannot use the exact distribution of $\hat{n}$, which is non-central $\chi^{2}$ with degrees of freedom $n$. However, if $n$ is large, a normal approximation may be accurate.


TITLE: Probability that maximum and minimum of two segments is less than a given value
QUESTION [0 upvotes]: I'm working on probability theory and recently I've got stuck with the following task:
Random point $A$ divides the segment $[0, 1]$ on two segments. Let $M$ be the size of biggest segment and $m$ be the size of least one. Calculate the probabilites $P(M \le x)$ and $P(m \le x)$ for every $x \in [0; 1]$.
This is my solution:
Suppose that $X$ is the length of the second segment(from $0$ to $A$). And $Y$ be the length of the second(from $A$ to $1$).
The first slippery moment:
$P(X \le x) = x$ and $P(Y \le x) = 1-x$
By symmetry $P(X \le Y) = P(Y \le X)$
Than $P(m \le x) = P(X \le Y)P(X \le x)+ P(Y \le X)P(Y \le x) = \frac{1}{2}x+ \frac{1}{2}(1-x)=\frac{1}{2}$
$P(M \le x) = P(X \ge Y)P(X \le x)+ P(Y \ge X)P(Y \le x) = \frac{1}{2}x+ \frac{1}{2}(1-x)=\frac{1}{2}$
I'm very unsure about this solution. Am I right?

REPLY [1 votes]: It appears to me that you've made an error:
$$(m≤x)=P(X≤Y)P(X≤x)+P(Y≤X)P(Y≤x)$$ This is incorrect. Intuitively, why? First, it is correct that this probability is the summation of the two equiprobable cases $Y>=X$ and $Y<=X$ . But, when X<=Y, then we must multiply it by the CONDITIONAL probability $P(X<=x)$ given that $X<=0.5$. This information makes a difference. If $x=0.6$, then $P(X<=x)=1$ given that$ X<=Y $(or that X<=0.5). Your error is from the use of the absolute probability $P(X<=x)$ rather than the conditional one, which is $P(X<=x)_{|(X<=Y)}=x$ if $x<=0.5$, and 1 otherwise.
My solution would be that $M$ is uniformly distributed over $[0.5,1]$ and $m$ over $[0,0.5]$, and similar to the analysis above, I solve for $P(m<=x)$:
For a randomly selected $x$ in $[0,1]$, it is equivalent to being distributed over $[0,0.5]$ with a 0.5 chance, and over $[0.5,1]$ with a 0.5 chance. In the first case, $x$ has the same distribution as $m$, and hence $P(m<=x)=0.5$ for that case. In the second case, $x>=m$ with certainty, so $P(m<=x)=1$. As mentioned, these two cases are equiprobable, so we sum the conditional probabilities multiplied by 0.5 each: $$P(m<=x)=0.5*0.5+0.5*1=0.75$$
$P(M<=x)$ follows using the same technique. The critical difference between my approach and yours is that I used conditional probabilities whereas you neglected them in your definition, rendering your result invalid.
A hint that points to the invalidity of your result is that $m<M$ in all cases, so it doesn't make sense that they are have the same probability of being greater than x.


TITLE: Give a context-free grammar that generates the language
QUESTION [0 upvotes]: Give a context-free grammar that generates the language:
$\{a^i b^j c^k d^h \mid i, j, h \geq 0, k>0 \text{ and } i+j \leq h\}$
This is what I've done so far:
$S \rightarrow aSb \mid bSc \mid cSd \mid D$
$D \rightarrow dD \mid d$

REPLY [1 votes]: Your construction fails to make sure that the $a$'s stay in front of the $b$'s, note that it allows $S \to bSc \to baSbc \to baDbc \to badbc$.  This example also shows that you need to make sure there are at least as many $d$'s as $a$'s and $b$'s combined.
The following rules should work:
$$
S \to aSd \mid T \\
T \to bTd \mid U \\
U \to Ud \mid cU \mid c
$$
Note how the rules $S \to aSd, T\to bTd$ ensure that $i+j \le h$, and the addition of the intermediate nonterminal $T$ makes sure the $a$'s are all in front of the $b$'s.


TITLE: Stirling's formula for derivatives of the Gamma function
QUESTION [1 upvotes]: Let $s = \sigma + it$ be a complex number such that $|\arg s| \leq \pi-\varepsilon$. Fix $\sigma > 0.$ For $t \neq 0$, Stirling's formula gives
$$\Gamma(\sigma+it) = \sqrt{2\pi}(it)^{\sigma-\tfrac{1}{2}}e^{-\frac{\pi}{2}|t|}\left(\frac{|t|}{e}\right)^{it}\left\{1+O\left(\frac{1}{|t|}\right)\right\}.$$
Let $k$ be a large positive integer. Is it possible to prove a similar expression for the derivatives $\Gamma^{(k)}(\sigma + it)$, where $\Gamma^{(k)}(\sigma+it)$ denotes the $k$-th derivative with respect to $t$. In particular, can one show that $\Gamma^{(k)}(\sigma + it)$ has exponential decay as $t \to \infty$?
Edit: A previous version of the question defined $\Gamma^{(k)}(z)$ to mean the complex derivative. What I am really after is derivative with respect to $t$.

REPLY [0 votes]: Rewrite it in the form $$\Gamma(s)=F(s)(1+O(1/|t|))=F(s)G(s)=F(s)+H(s)$$
with $F(s)$ analytic non-vanishing in the region and $G(s)=\Gamma(s)/F(s)=O(1/|t|)$.
The differentiation
$$\Gamma^{(k)}(s)=\sum_{m=0}^k {k\choose m}F^{m}(s)G^{(k-m)}(s)= F^{(k)}(s)+O(1/|t|)\sum_{m=0}^{k-1} {k\choose m}F^{m}(s)$$ $$ = F^{(k)}+O(1/|t|) F(s+kr)$$
works because the Cauchy integral formula gives that for a fixed $r$
$G^{(m)}(s)=O(\sup_{|z-s|=r} |G(z)|)= O(1/|t|)$ and $F^{(m)}(s)=O(\sup_{|z-s|=r} |F^{(m-1)}(z)|)=O(F^{(m-1)}(s+r))$, where the $O$ constants depend on $r$.


text:
\begin{document}

\pagenumbering{roman}

\thispagestyle{empty}

\begin{center}
  {\large\textbf{\thesistitle}}
  \vspace{.7in}

  by
  \vspace{.7in}

  \thesisauthor
  \vfill

\begin{doublespace}
  A dissertation submitted in partial fulfillment\\
  of the requirements for the degree of\\
  Doctor of Philosophy\\
  Department of Mathematics\\
  New York University\\
  \graddate
\end{doublespace}
\end{center}
\vfill

\noindent\makebox[\textwidth]{\hfill\makebox[2.5in]{\hrulefill}}\\
\makebox[\textwidth]{\hfill\makebox[2.5in]{\hfill\thesiscoadvisor\hfill}}\\

\vspace{20pt}\\
\noindent\makebox[\textwidth]{\hfill\makebox[2.5in]{\hrulefill}}\\
\makebox[\textwidth]{\hfill\makebox[2.5in]{\hfill\thesisadvisor\hfill}}
\newpage

\thispagestyle{empty}
\begin{center}
  \copyright ~Arjun Krishnan. \\
  All rights reserved, 2014.
\end{center}
\newpage

\thispagestyle{empty}
\vspace*{0in}
\newpage

\section*{Dedication}\addcontentsline{toc}{section}{Dedication}
\vspace*{\fill}
\begin{center}
  \thesisdedication
\end{center}
\vfill
\newpage

\section*{Acknowledgements}\addcontentsline{toc}{section}{Acknowledgements}
\input{acknowledge}
\newpage

\section*{Abstract}\addcontentsline{toc}{section}{Abstract}
\input{abstract}
\newpage

\tableofcontents

\listoffigures\addcontentsline{toc}{section}{List of Figures}
\newpage

\pagenumbering{arabic}

\part{Homogenization theorem and variational formula\label{part:one}}
\input{intro}
\input{preliminaries}
\input{outline_of_proof}

\input{proof_discrete_homogenization}
\input{proof_variational_formula}
\part{Applications and Discussion\label{part:two}}

\input{recap}
\input{abstract_algorithm}
\input{comparing_two_distributions}
\bookmarksetup{startatroot}
\addtocontents{toc}{\smallskip}
\input{future_work}

\appendix
\input{miscellaneous_proofs}

\bibliographystyle{chicago}

\begin{onehalfspace}
  \bibliography{My_Zotero_Library.bib}
\end{onehalfspace}
\addcontentsline{toc}{chapter}{Bibliography}

\end{document}

config: arxiv


TITLE: $\frac{2-a}{a+\sqrt{bc+abc}}+\frac{2-b}{b+\sqrt{ca+abc}}+\frac{2-c}{c+\sqrt{ab+abc}}\ge1$
QUESTION [3 upvotes]: For $a,b,c\ge0: ab+bc+ca+abc=4$ then: $$\frac{2-a}{a+\sqrt{bc+abc}}+\frac{2-b}{b+\sqrt{ca+abc}}+\frac{2-c}{c+\sqrt{ab+abc}}\ge1$$

I used the condition and get: $a+\sqrt{bc+abc}=a+\sqrt{4-a(b+c)}\le a+2$
So we need to prove that: $$\frac{2-a}{a+2}+\frac{2-b}{b+2}+\frac{2-c}{c+2}\ge1$$
I tried to full expand but the rest seems complicated for me.
Can anyone help me full my idea? Every thinking is welcomed, thanks!

REPLY [0 votes]: Futhermore, I tried C-S inequality and saw something "tricky":
It is obvious to get equivalent inequality: $$\frac{2+\sqrt{bc+abc}}{a+\sqrt{bc+abc}}+\frac{2+\sqrt{bc+abc}}{a+\sqrt{bc+abc}}+\frac{2+\sqrt{ca+abc}}{b+\sqrt{ac+abc}}\ge4$$
By C-S: $$2(a+b+c)=\sqrt{\left(a(b+c)+bc+abc\right)\left((b+c-a)^2+4a(b+c)\right)}\ge 2a(b+c)+\sqrt{bc+abc}(b+c-a)$$
Or: $$\frac{2+\sqrt{bc+abc}}{a+\sqrt{bc+abc}}\ge \frac{(a+b+c)\sqrt{bc+abc}+2a(b+c)+\sqrt{bc+abc}(b+c-a)}{(a+b+c)(a+\sqrt{bc+abc})}=\frac{2(b+c)}{a+b+c}$$
The rest is similar to above solution.


TITLE: calculus area problem
QUESTION [0 upvotes]: Consider the function $f(x)=\frac1{2x-1}$.
For every $p>1/2$, there is a rectangle with vertices $(0,0), (p,0), (p,f(p)),$ and $(0,f(p))$ with area $A(p)=p/(2p-1)$. For certain values of $a$ and $b$, $A(p)=a+b(f(p))$. Find $a$ and $b$.
I found $f(p)$ and tried plugging in, but failed. Any help would be greatly appreciated and will cut down on my Nexium and Xanax bills.

REPLY [1 votes]: You have
$$
a+bf(p)={2ap+b-a\over2p-1}
$$
Compare this with
$$
A(p)={p\over2p-1}
$$
What values of $a$ and $b$ make the right hand sides equal?


TITLE: Complex numbers involving roots of unity
QUESTION [7 upvotes]: Let $z\in \mathbb{C}$ and $n\in \mathbb{N}, n \geq 1$. Solve the following equation: $$z+z^2+\dots+z^n=n|z|^n$$
Obviously, $(z,n)=(0,n)$ and $(1,n)$ are solutions, $\forall n \geq 1$. Considering $z \neq 0,1$ I tried to use $$z+z^2+\dots+z^n=z(1+z+\dots+z^{n-1})=z(z-\epsilon)(z-\epsilon^2)\dots(z-\epsilon^{n-1})$$
where $\epsilon=\cos \frac{2\pi}{n}+i\sin \frac{2\pi}{n}$, such that $$z(z-\epsilon)(z-\epsilon^2)\dots(z-\epsilon^{n-1})=n|z|^n$$
I then tried to take the norm on both sides and obtain some sort of inequalities, but didn't manage to get somewhere.

REPLY [2 votes]: Expanding on the work of @WB-man by looking at the solutions graphically.

$n=2$
$$
f(z) = z+ z^2 -2 \left| z\right|^{2}
$$
Find the roots of the real and imaginary components. Solve
$$
\begin{align}
\text{Re } f &= x - x^2 - 3 y^2 = 0, \\
\text{Im } f &= y + 2 x y2 = 0, \\
\end{align}
$$
The solution for the real component is plot in $\color{blue}{blue}$, imaginary component in $\color{red}{red}$.
The solutions are the intersection of the $\color{blue}{blue}$ and $\color{red}{red}$.

$n=3$
$$
f(z) = z + z^2 + z^3 - 3 \left| z\right|^{3}
$$
Solve
$$
\begin{align}
\text{Re } f &= -3 x^2 \sqrt{x^2+y^2}-3 y^2 \sqrt{x^2+y^2}+2 x^2+x-2 y^2 = 0, \\
\text{Im } f &= y + 4 x y = 0, \\
\end{align}
$$

$n=4$
$$
f(z) = z + z^2 + z^3 + z^{4} - 4 \left| z\right|^{4}
$$
Solve
$$
\begin{align}
\text{Re } f &= x + 2 x^2 - 3 x^4 - 2 y^2 - 14 x^2 y^2 - 3 y^4 = 0, \\
\text{Im } f &= y + 4 x y + 4 x^3 y - 4 x y^3 = 0, \\
\end{align}
$$

$n=5$
$$
f(z) = z + z^2 + z^3 + z^{4} + z^{5} - 5 \left| z\right|^{5}
$$
Solve
$$
\begin{align}
\text{Re } f &= x^7+x^6-21 x^5 y^2+x^5-15 x^4 y^2+x^4+35 x^3 y^4-10 x^3 y^2+15 x^2 y^4-6 x^2 y^2-7 y^6 \sqrt{x^2+y^2}-21 x^2 y^4 \sqrt{x^2+y^2}+2 x^2-7 x^6 \sqrt{x^2+y^2}-21 x^4 y^2 \sqrt{x^2+y^2}-7 x y^6+5 x y^4+x-y^6+y^4-2 y^2 = 0, \\
%
\text{Im } f &= 5 x^4 y+4 x^3 y-10 x^2 y^3-4 x y^3+4 x y+y^5+y = 0, \\
\end{align}
$$


TITLE: prove equation equals $-1$
QUESTION [4 upvotes]: I was wondering if it was possible to prove that
$\left( \frac{( a^2 - c^2 + (\frac{d}{c}a)^2-d^2)}{ \sqrt{(a-c)^2+(\frac{d}{c}a)-d)^2} \sqrt{(a+c)^2+(\frac{d}{c}a)+d)^2}} \right) = -1$ when $a,d,c \in [-1,1]$ and $|a|<c$, and $|b|<d$?
Attempt: By Asdrul, I know that
$$\left[(a-c)^{2}+\bigg(\frac{d}{c}a-d\bigg)^{2}\right]\cdot\left[(a+c)^{2}+\bigg(\frac{d}{c}a+d\bigg)^{2}\right]=\frac{1}{c^4}( c^{2}+d^{2}) ^{2}\left( a^2-c^2\right) ^{2}$$
So applying this, I have that the above equals
$\frac{c^2(a^2-c^2+(\frac{da}{c})^2-d^2)}{(c^2+d^2)|a^2-c^2|}$

REPLY [2 votes]: Hint: note that
$$\left[(a-c)^{2}+\bigg(\frac{d}{c}a-d\bigg)^{2}\right]\cdot\left[(a+c)^{2}+\bigg(\frac{d}{c}a+d\bigg)^{2}\right]=\frac{1}{c^4}( c^{2}+d^{2}) ^{2}\left( a^2-c^2\right) ^{2}$$