TITLE: Evaluate $\lim_{x \to 0} ( \frac{1}{x-\sin x}\int_{0}^{x}t\sin t \text{d}t)$
QUESTION [1 upvotes]: Evaluate
$$\lim_{x \to 0}  \frac{1}{(x-\sin x)}\displaystyle \int_{0}^{x}t\sin t  \text{d}t$$
I know about limits and I also know about integrals but combining them in this form is confusing
the only thing I can say about this expression is that as x approaches zero , the upper bound of the integral will become zero and then the whole expression will be zero !

REPLY [1 votes]: here is the limit without l'hospitals. i will use $$\sin x = x - x^3/6 + \cdots$$
then we have $$x - \sin x = x^3/6 + \cdots, \int_0^x t\sin t \, dt = x^3/3+\cdots .$$ therefore the $$\lim_{x \to 0}\frac 1 {x - \sin x}  \int_0^x t\sin t \, dt = 2.$$


TITLE: Can we show that $\sum_{k=n+1}^{\infty}\frac{1}{k^2} \leq \frac{1}{n}$
QUESTION [0 upvotes]: Can we show the above or looking at one of the proofs of $\sum_{n\geq 1}\frac{1}{n^2}$, say using Fourier series, and try to adjust the proof to terms greater than $n$ is the only way? Thanks.

REPLY [2 votes]: You can do this directly.
Hint: $\frac{1}{k^2} \leq \frac{1}{k(k-1)} = \frac{1}{k-1} - \frac{1}{k}$.
Sum up all the terms.

If you want a proof via Fourier series, be specific in the question.

In fact, we have
$\frac{1}{k^2} \leq \frac{1}{k^2 - \frac{1}{4}} = \frac{1}{k-\frac{1}{2} } - \frac{1}{k + \frac{1}{2}}.$
So we can strengthen the bound to
$$\sum_{k=n+1}^\infty \frac{1}{k^2} \leq \frac{1}{n+\frac{1}{2}}.$$


TITLE: Is flow speed between two pressures dependent on absolute or relative pressure difference?
QUESTION [0 upvotes]: Imagine you have a pressure chamber which is divided by a removable wall in two equal parts.
Scenario A) Air pressure on the left side is 1 bar, on the right side it is 2 bar.
Scenario B) Air pressure on the left side is 10 bar, on the right side it is 11 bar.
The absolute pressure difference between the two sides is the same (1 bar in both scenarios). The relative pressure difference is greater in scenario A than in scenario B.
When the wall in the middle is removed air flows from the right side to the left side. Now, is the flow speed of the air the same for both scenarios or is it greater in scenario A? Intuitively, I assume the flow speed to be greater in scenario A (because of a greater relative pressure difference), but I can't back it up with a solid argument.
What is happening?

REPLY [2 votes]: Well, let's look at the equations for a flow initially at rest:
$$ \frac{\partial \rho u}{\partial t} = -\frac{\partial p}{\partial x}$$
So, if we assume density is constant in the flow, you get:
$$ \frac{\partial u}{\partial t} = -\frac{1}{\rho}\frac{\partial p}{\partial x}$$
and so you can see that in both your cases, the gradient in pressure is the same size regardless of the magnitude of pressure. So, if your case B is hotter such that the two densities between case A and B are the same, the acceleration should be the same.
On the other hand, if your temperatures are the same, then your acceleration will be higher in case A because density is lower.
Since your system is closed and transient, it doesn't make sense to look at the total flow velocity because it is always changing. But the acceleration will change depending on how you changed your pressures.
Another way to look at it. If we assume isentropic flow, the total pressure is conserved. This means that:
$$ P_0 = P_s + \frac{1}{2}\rho u^2 \rightarrow \Delta P = \frac{1}{2}\rho u^2$$
where $\Delta P = P_0 - P_s$ somewhere along the streamline. So again we can see that the velocity depends on the change in pressure and the density and we're back at the same argument we had above.


text:
\section{More details about Algorithm \ref{alg:gd}}

\subsection{The initialization scheme\label{sec:Initialization-scheme}}

For self-completeness, we record in this section the detailed initialization
procedure employed in the two-stage nonconvex algorithm proposed in
\cite{cai2019nonconvex} (namely, Algorithm~\ref{alg:gd}). This
is summarized in Algorithm~\ref{alg:init}, with auxiliary procedures
detailed in Algorithm~\ref{alg:localization}. As a high-level interpretation,
Algorithm~\ref{alg:init} estimates the subspace spanned by the tensor
factor $\{\bm{u}_{l}^{\star}\}_{1\leq l\leq r}$ via a spectral method (similar to PCA-type methods \cite{montanari2018spectral,zhang2018heteroskedastic,cai2019subspace}),
whereas Algorithm~\ref{alg:localization} attempts to retrieve estimates
for individual tensor factors from this subspace estimate $\bm{U}_{\mathsf{space}}$.
Here and throughout, we denote $\bm{T}^{\mathsf{obs}}:=[T_{i,j,k}^{\mathsf{obs}}]_{1\leq i,j,k\leq d}$,
where we set $T_{i,j,k}^{\mathsf{obs}}=0$ for any $(i,j,k)\notin\Omega$.

\begin{algorithm}[H]
\caption{Spectral initialization for nonconvex tensor completion}
\label{alg:init} \begin{algorithmic}[1] \State Let $\bm{U}_{\mathsf{space}}\bm{\Lambda}\bm{U}_{\mathsf{space}}^{\top}$
be the rank-$r$ eigen-decomposition of
\begin{align}
\bm{B}:=\mathcal{P}_{\mathsf{off}\text{-}\mathsf{diag}}(\bm{A}\bm{A}^{\top}),\label{B_alg}
\end{align}
where $\bm{A}=\mathsf{unfold}\big(p^{-1}\bm{T}^{\mathsf{obs}}\big)$
is the mode-1 matricization of $p^{-1}\bm{T}^{\mathsf{obs}}$, and
$\mathcal{P}_{\mathsf{off}\text{-}\mathsf{diag}}(\bm{Z})$ extracts
out the off-diagonal entries of $\bm{Z}$. \State \textbf{Output:}
an initial estimate $\bm{U}^{0}\in\mathbb{R}^{d\times r}$ on the
basis of $\bm{U}_{\mathsf{space}}\in\mathbb{R}^{d\times r}$ using
Algorithm~\ref{alg:localization}. \end{algorithmic}
\end{algorithm}

\begin{algorithm}[H]
\caption{Retrieval of low-rank tensor factors from a given subspace estimate.}
\label{alg:localization} \begin{algorithmic}[1] \State \textbf{Input:}
number of restarts $L$, pruning threshold $\epsilon_{\mathsf{th}}$,
subspace estimate $\bm{U}_{\mathsf{space}}\in\mathbb{R}^{d\times r}$
given by Algorithm~\ref{alg:init}. \For{$\tau=1,\dots,L$} \State
Generate an independent Gaussian vector $\bm{g}^{\tau}\sim\mathcal{N}(0,\bm{I}_{d})$.
\State $\big(\bm{\nu}^{\tau},\lambda_{\tau},\mathsf{spec}\text{-}\mathsf{gap}_{\tau}\big)\gets\Call{Retrieve-one-tensor-factor}{\bm{T}^{\mathsf{obs}},p,\bm{U}_{\mathsf{space}},\bm{g}^{\tau}}$.
\EndFor \State Generate tensor factor estimates $\big\{(\bm{w}^{1},\lambda_{1}),\dots,(\bm{w}^{r},\lambda_{r})\big\}\gets\Call{Prune}{\big\{\big(\bm{\nu}^{\tau},\lambda_{\tau},\mathsf{spec}\text{-}\mathsf{gap}_{\tau}\big)\big\}_{\tau=1}^{L},\epsilon_{\mathsf{th}}}$.
\State \textbf{Output:} initial estimate $\bm{U}^{0}=\big[\lambda_{1}^{1/3}\bm{w}^{1},\dots,\lambda_{r}^{1/3}\bm{w}^{r}\big].$
\end{algorithmic}
\end{algorithm}

\begin{algorithm}[htb]
\label{alg:one-factor} \begin{algorithmic}[1] \Function{Retrieve-one-tensor-factor}{$\bm{T},p,\bm{U}_{\mathsf{space}},\bm{g}$}
\State Compute \begin{subequations}
\begin{align}
\bm{\theta} & =\bm{U}_{\mathsf{space}}\bm{U}_{\mathsf{space}}^{\top}\bm{g}=:\mathcal{P}_{\bm{U}_{\mathsf{space}}}(\bm{g}),\\
\bm{M} & =p^{-1}\bm{T}^{\mathsf{obs}}\times_{3}\bm{\theta},\label{eq:defn-Mtau}
\end{align}
\end{subequations} where $\times_{3}$ is defined in Section~\ref{subsec:Notations}.
\State Let $\bm{\nu}$ be the leading singular vector of $\bm{M}$
obeying $\langle\bm{T}^{\mathsf{obs}},\bm{\nu}^{\otimes3}\rangle\geq0$,
and set $\lambda=\langle p^{-1}\bm{T}^{\mathsf{obs}},\bm{\nu}^{\otimes3}\rangle$.
\State \Return $\big(\bm{\nu},\lambda,\sigma_{1}(\bm{M})-\sigma_{2}(\bm{M})\big)$.
\EndFunction \end{algorithmic}
\end{algorithm}

\begin{algorithm}[htb]
\label{alg:prune} \begin{algorithmic}[1] \Function{Prune}{$\big\{\big(\bm{\nu}^{\tau},\lambda_{\tau},\mathsf{spec}\text{-}\mathsf{gap}_{\tau}\big)\big\}_{\tau=1}^{L},\epsilon_{\mathsf{th}}$}
\State Set $\Theta=\big\{\big(\bm{\nu}^{\tau},\lambda_{\tau},\mathsf{spec}\text{-}\mathsf{gap}_{\tau}\big)\big\}_{\tau=1}^{L}.$
\For{$i=1,\dots,r$} \State Choose $(\bm{\nu}^{\tau},\lambda_{\tau},\mathsf{spec}\text{-}\mathsf{gap}_{\tau})$
from $\Theta$ with the largest $\mathsf{spec}\text{-}\mathsf{gap}_{\tau}$;
set $\bm{w}^{i}=\bm{\nu}^{\tau}$ and $\lambda_{i}=\lambda_{\tau}$.
\State Update $\Theta\gets\Theta\setminus\left\{ \big(\bm{\nu}^{\tau},\lambda_{\tau},\mathsf{spec}\text{-}\mathsf{gap}_{\tau}\big)\in\Theta:|\langle\bm{\nu}^{\tau},\bm{w}^{i}\rangle|>1-\epsilon_{\mathsf{th}}\right\} $.
\EndFor \State \Return $\big\{(\bm{w}^{1},\lambda_{1}),\dots,(\bm{w}^{r},\lambda_{r})\big\}.$
\EndFunction \end{algorithmic}
\end{algorithm}

\subsection{Choices of algorithmic parameters\label{subsec:Choices-of-algorithmic-pars}}

To guarantee fast convergence of Algorithm~\ref{alg:gd}, there are
a couple of algorithmic parameters --- namely, the number of restart
attempts $L$, the pruning threshold $\epsilon_{\mathsf{th}}$ in
Algorithm~\ref{alg:localization}, as well as the learning rates
$\eta_{t}$ --- that need to be properly chosen. Unless otherwise
noted, this paper adopts the following choices suggested by \cite{cai2019nonconvex}:
\begin{equation}
L=c_{4}r^{2\kappa^{2}}\log^{3/2}d,\quad \eta_{t}\equiv\frac{c_{5}\lambda_{\min}^{\star4/3}}{p\lambda_{\max}^{\star8/3}} \quad\text{and}\quad \epsilon_{\mathsf{th}}=c_{6}\bigg(\frac{\mu r\log d}{d\sqrt{p}}+\frac{\sigma_{\min}}{\lambda_{\min}^{\star}}\sqrt{\frac{rd\log^{2}d}{p}}+\sqrt{\frac{\mu r\log d}{d}}\bigg),\label{eq:choice-L-epsilon}
\end{equation}
where $c_{4}>0$ is some sufficiently large constant, and $c_{5},c_{6}>0$
are some sufficiently small constants. The interested reader is referred
to \cite{cai2019nonconvex} for justification.

config: arxiv


TITLE: Find $f:f(xf(x)+f(y))=f(x)^2+y$
QUESTION [3 upvotes]: Find $f:f(xf(x)+f(y))=f(x)^2+y$

Domain and co-domain is real numbers
I did the following:

Let $s=f(0)$

Then $f(f(y))=s^2+y$ so $f$ is surjective
Also, $f(x)=f(y)\implies f(xf(x)+f(y))=f(xf(x)+f(x))\implies x=y$ so $f$ is injective
So, $f$ is bijective. Letting $f(x)=0, y=0$

we get $f(f(0))=0 \implies s=f(0)=0$

In fact $f(f(x))=x$ for ALL $x$
Letting $x=1,y=0$ we get

$f(f(1))=f(1)^2\implies 1=f(1)^2 \implies f(1)=1$ or  $f(1)=-1$
If $f(1)=1$ we let $x=1$ to get

$f(y+1)=f(y)+1$

By induction this leads to $f(x)=x$ for all integers
If $f(1)=-1$ we let $x=1$ to get

$f(y-1)=f(y)+1$

By induction we get $f(x)=-x$ for all integers
How to extend the domain over real numbers I don't know, any help would be appreciated

REPLY [1 votes]: Let $x=f(u)$, by using $f(x)=f(f(u))=u$ one gets now $f(u f(u)+f(y))=u^2+y$. By replacing $u$ with $x$, it follows that for all $x$, $f(x)^2 = x^2$, hence for all $x$, $f(x) = \pm x$. It remains to see if the sign depends on $x$.


text:
{\bf Problem.} These two spinners are divided into thirds and quarters, respectively. If each of these spinners is spun once, what is the probability that the product of the results of the two spins will be an even number? Express your answer as a common fraction.

[asy]

size(5cm,5cm);

draw(Circle((0,0),1));

draw(Circle((3,0),1));

draw((0,0)--(0,1));

draw((0,0)--(-0.9,-0.47));

draw((0,0)--(0.9,-0.47));

draw((2,0)--(4,0));

draw((3,1)--(3,-1));

label("$3$",(-0.5,0.3));

label("$4$",(0.5,0.3));

label("$5$",(0,-0.5));

label("$5$",(2.6,-0.4));

label("$6$",(2.6,0.4));

label("$7$",(3.4,0.4));

label("$8$",(3.4,-0.4));

draw((0,0)--(0.2,0.8),Arrow);

draw((3,0)--(3.2,0.8),Arrow);

[/asy]
{\bf Level.} Level 2
{\bf Type.} Counting & Probability
{\bf Solution.} We will subtract the probability that the product is odd from 1 to get the probability that the product is even. In order for the product to be odd, we must have both numbers be odd. There are $2\cdot2=4$ possibilities for this (a 3 or 5 is spun on the left spinner and a 5 or 7 on the right) out of a total of $3\cdot4=12$ possibilities, so the probability that the product is odd is $4/12=1/3$. The probability that the product is even is $1-1/3=\boxed{\frac{2}{3}}$.

set_name: MATH


TITLE: A closed path is made up of 11 line segments. Can one line, not containing a vertex of the path, intersect each of its segments?
QUESTION [5 upvotes]: This  problem is taken from the book Mathematical Circles by Dmitri Fomin, et al., translated by Mark Saul and published by the American Mathematical Society. Can anyone describe what the question actually means?

REPLY [4 votes]: Alas, I didn't read the question too well.  The OP was not asking for a proof, but a proof is here in the hidden portion.

 The answer is no.  Suppose contrary that a line $L$ exists with the required property.  This line cuts the plane into two open half plane.  Note that two consecutive vertices cannot belong in the same half plane.  Do you see a contradiction now?


TITLE: If falling from height into water, at what height would a buoyancy aid be more likely to cause injury?
QUESTION [1 upvotes]: There are strict requirements for lifejacket / buoyancy aids in the offshore oil & gas industry, requiring them to be worn when working above the water. Some places have been noted to be using a buoyancy aids rather than an auto-inflatable device. I know that theoretically this would cause them to a more sudden stop as they hit the water, increasing the danger, but I can't find any source of information that gives a clear criteria for what height should be considered dangerous.
I know there's a lot of subjectivity, since a slim person with a buoyancy aid could equate to the same force on impact as a larger person, but if we consider like-sized average adult males, one with a 150N rated buoyancy device, another with 100N, 50N and then nothing. What would the difference be?

REPLY [1 votes]: It shouldn't change the impact much.
The breaking force due to surface tension is the most damaging, and we can probably suppose this force would be the same in all three cases (that wouldn't be true if the flotation device were large, but that force is anyway highly variable, depending on the position with which the person hits the water surface).
The forces the person is subject once in the water are weight $F_g$, buoyancy $E$, and viscous forces $F_v$. Without a flotation device, $E$ and $F_g$ approximately cancel out. The only force left, $F_v$, is again strongly dependent on the person's position (how hydrodynamic, or "streamlined" it is) so we can consider it too to be independent of the presence of a buoyancy device.
Thus the buoyancy rate of each device (be it 50N, 100N, or 150N) is the precisely extra force the person will be subjected to, once submerged. These forces don't seem large and are fully present only after the initial impact with the water surface, so they shouldn't offset the advantages of wearing them.


TITLE: The intention behind the sub-mulipcative condition of matrix norms
QUESTION [0 upvotes]: I noticed that the definition of 'matrix norm' includes an extra sub-multiplicative condition (apart from the definition of 'norm'), $$\lVert AB \lVert \leq \lVert A \lVert\lVert B \lVert,~~~~~~~~(*)$$ to make sense the continuity of matrix mulitiplication relative to the topological vector space induced by the same. But the definition of operator norm $\lVert T \lVert$, where $T$ is a linear operator from a normed linear space $(X,\lVert \cdot \lVert_X)$ to the normed linear space $(Y,\lVert \cdot \lVert_Y)$, via $$\lVert T \lVert=\inf\{\alpha>0~:~\lVert Tx \lVert_Y \leq \alpha \lVert x \lVert_X,~\text{for all}~x \in X\},$$
immediately allows the result $\lVert Tx \lVert_Y \leq  \lVert T \lVert\lVert x \lVert_X$, for all $x \in X$. Consequently, for the composition $T_1T_2$, we have $$\lVert T_1T_2\lVert \leq \lVert T_1\lVert\lVert T_2\lVert,~~~~~~~~~(**)$$ since $\lVert T_1T_2(x) \lVert_Z \leq \lVert T_1 \lVert\lVert T_2x \lVert_Y \leq \lVert T_1 \lVert\lVert T_2 \lVert\lVert x \lVert_X$ for the linear maps $T_1:(Y,\lVert \cdot \lVert_Y) \to (Z,\lVert \cdot \lVert_Z)$ and $T_2:(X,\lVert \cdot \lVert_X) \to (Y,\lVert \cdot \lVert_Y)$ with $T_2(X)=Y$.
My confusion:
However, the definition of operator norm itself imply the inequality $(**)$, what would be the need for such a sub-multiplicative condition $(*)$? Any conceptual mislead here?

REPLY [1 votes]: I believe the problem here is that the definition you are given is that of a normed  algebra, and the two ingredients are not clearly separated.
In principle, a norm does not have to satisfy such condition. Actually, a norm only requires an underlying vector space structure and the usual axioms of norm.
If your space has an additional multiplication operation, then it makes sense to require its continuity with respect to the factors. This is what you call a normed algebra.
In the space of matrices you have both a vector space structure and a multiplication, so it makes sense to require the extra continuity property. But in principle you can have norms not satisfying that property.
It turns out that if you consider the operator norm, the extra condition is automatically satisfied.
I am not aware that any norm in the space of matrices corresponds to an operator norm for some underlying norm in the vector space. If that is the case, the extra requirement would be redundant of course.


TITLE: Show that if $r>1$ and $a>0$ then $(a + r^n)^{1/n}$ converges.
QUESTION [1 upvotes]: Show that if $r>1$  and $a>0$ then $(a + r^n)^{1/n}$ converges.
This answer must rely principally on the idea that the limit of a sum is the sum of the limits, and unfortunately, cannot employ consequences of this theorem.

REPLY [0 votes]: If allowed, this is a routine application of L'Hopital's rule after writing
$(a + r^n)^{1/n}$ as $e^{(ln (a + r^n))/n}$.


TITLE: What is Validity and Satisfiability in a propositional statement?
QUESTION [17 upvotes]: I tend to see these words a lot in Discrete Mathematics. I assumed these were just simple words until I bumped into a question.
Is the following proposition Satisfiable? Is it Valid?
$(P \rightarrow Q) \Leftrightarrow (Q \rightarrow R ) $
Then I searched in the net but in vain. So I'm asking here. What do you mean by Satisfiable and Valid? Please explain.

REPLY [0 votes]: Valid: always true regardless of the values of its variables, e.g., P OR NOT(P)..

Slightly more formally: A valid formula is one which is always true, no matter what truth values its variables may have.

Satisfiable: Can be solved, i.e., true and false values can be assigned to its variables, in a way that the final outcome is true.

Slightly more formally: A satisfiable formula is one which can sometimes be true, i.e., there is some assignment of truth values to its variables that makes it true.
An example of an unsatisfiable formula is P AND NOT(P)..

Why are satisfiability and validity treated together?
Because they are related:
To check that G is valid,
we can check that NOT(G) is not satisfiable.

If you want to read more about satisfiability and validity in an accesible way, check the chapter 3 of Mathematics for Computer Science by Lehman, Thomson and Meyer.


TITLE: Equations of significance probabilities
QUESTION [1 upvotes]: Consider a population of independent light bulbs with an exponential
lifetime distribution with mean $\mu$. It is claimed that their expected
lifetime is 1000 hours. A definition of a 100(1−$\alpha$)% confidence interval
obtained from an observation t0 is the set of all $\mu_0$ which are not rejected
in a test of a null hypothesis  $\mu$= $\mu_0$ against an alternative hypothesis $\mu$$\neq$$\mu_0$.

One particular light bulb fails after 622 hours. Solve the equations of
the two significance probabilities Pr(T ≥ 622 |$\mu_0$) = 0.05 (for a test
of  $\mu$= $\mu_0$ versus $\mu$> $\mu_0$) and Pr(T ≤ 622 |$\mu_0$) = 0.05 (for a test of
$\mu$= $\mu_0$ versus $\mu$< $\mu_0$) for $\mu$. Determine the range of values of $\mu$  such
that both of the probabilities Pr(T ≥ 622 | $\mu$) and Pr(T ≤ 622 |$\mu$) are
at least 0.05. (This range gives an equi-tailed 90% confidence interval
for $\mu$.)

I don't seem to understand what they mean by 'solve the equations'. Do I have to find a specific value for T or compute Pr(T ≥ 622 |$\mu_0$), Pr(T ≤ 622 |$\mu_0$)  and compare with 0.05? I believe I will get the second part after I understand this bit.

REPLY [1 votes]: I also found the wording somewhat difficult.
For the first problem, you need to find the value of $\mu_0$ such that an exponentially distributed random variable $T$ with mean $\mu_0$ has probability $0.05$ of being $\ge 622$. You probably know what the preceding sentence means. But to make sure, we sketch the calculation.
An exponentially distributed rv with mean $\mu_0$ has density function $\frac{1}{\mu}e^{-t/\mu_0}$ (for $t\ge 0$), and therefore cdf $1-e^{-t/\mu_0}$. So the probability that $T\ge 622$ is $e^{-622/\mu_0}$.
Set this equal to $0.05$ and solve for $\mu_0$.
For the second problem, we want to find $\mu_0$ such that an exponential $T$ with mean $\mu_0$ is $\le 622$ with probability $0.05$.
The range for $\mu_0$ that you will get from the two calculations turns out to be very large. Basically that says you can't learn much about the mean lifetime of lightbulbs by observing a single one.


TITLE: Why the sign in the definition of the discriminant?
QUESTION [3 upvotes]: Consider the split monic $f=\prod_{i=1}^n(x-x_i)\in \mathbb Z[x_1 ,\dots ,x_n,x]$. Its discriminant is usually defined as $$(-1)^{n(n-1)/2}\prod_{i=1}^nf^\prime(x_i)=\prod_{1\leq i<j\leq n}(x_i-x_j)^2.$$
What is the reason for taking this definition as opposed to $\prod_{i=1}^nf^\prime(x_i)$? The product of the derivatives at the roots "feels" to me more canonical than the product on the RHS.

REPLY [1 votes]: The definition of the resultant $\text{res}(f,g)$ of two (monic) polynomials $f,g\in {\mathbb Z}[x]$ as the determinant of the corresponding Sylvester matrix, readily implies
that if $\alpha_1,\ldots,\alpha_n$ are the roots of $f(x)$ (say, in ${\mathbb C}$) then
$$\text{res}(f,g)=\prod_{i=1}^ng(\alpha_i).$$
Now, if you define the discriminant of the (monic) polynomial $f(x)$ as
$$\Delta(f(x))=\text{res}(f(x),f'(x))=\prod_{i=1}^nf'(\alpha_i),$$
writing $f(x)=(x-\alpha_1)\cdots (x-\alpha_n)$ it follows that
$$f'(\alpha_j)=\prod_{i\neq j}^n(\alpha_j-\alpha_i).$$
In this equality, once you order the indices, for each pair $i<j$ there are two factors $\pm(\alpha_i-\alpha_j)$ and thus
\begin{align*}
\Delta(f(x))&=\prod_{i=1}^nf'(\alpha_i)=\prod_{i\neq j}^n(\alpha_j-\alpha_i)\\
&=(-1)^{n(n-1)/2}\prod_{i< j}^n(\alpha_j-\alpha_i)^2
\end{align*}
where the sign is because there are precisely $n(n-1)/2$ pairs $\pm(\alpha_i-\alpha_j)$.


TITLE: Which is probability that the student enter at least one university?
QUESTION [0 upvotes]: A student give exams to 3 universities. The chance to enter the first is: 80%, to the second is 70% and in the third 60%.

Which is probability that the student enter at least one university?

REPLY [1 votes]: The probability he doesn't get into any is (1-.8)(1-.7)(1-.6) = .2*.3*.4 = 3/125
The probability you're looking for is the compliment of the above probability. So the probability he gets into at least 1 university is 1- 3/125= 122/125.


TITLE: Nilpotent ring and Nilpotent groups.
QUESTION [4 upvotes]: Let $R$ be a ring (associative and with unity) and $B$ be a subring with the property that $B^n = 0$ i.e.
  $$ \forall\; x_1, x_2, \dots, x_n \in B: \; x_1 \cdot x_2 \cdots x_n = 0$$
  My aim is to prove that the set $G = \{1+x \mid x\in B\}$ is a nilpotent group.

It is easy to see that $G$ is a group. For $n=3$, I found
$$[1+x,1+y]_G = 1+[x,y]_R,$$
where $$[1+x,1+y]_G = (1-x+x^2)(1-y+y^2)(1+x)(1+y)$$ is the group commutator
and $[x,y]_R = xy-yx$ is the ring commutator. And $G$ is a nilpotent group of degree no more than $3$.
But if $n=4$, I got
$$[1+x, 1+y]_G = 1 + x^2y+xy+xyx+y^2x-yx-yxy,$$ and I don't know what to do.

REPLY [3 votes]: First, note that since every element of $B$ is nilpotent, we have the following explicit formula for the inverse of $1-x$:
$$(1-x)^{-1}=1+x+x^2+...+x^{n-1}$$
So let's define our descending series of normal subgroups as:
$$G_k=1+B_k$$ Where $B_k$ is the additive subgroup of $B$ generated by elements of the form $x_1 x_2 ...x_m$ for $x_i\in B$, $m\geq k$.
One may check that $G_k$ is a normal subgroup in $G$, and we have $G=G_1$ and $G_n=1$.
So to show $G$ is nilpotent, it suffices to show that $[g,G_k]\subset G_{k+1}$ for all $g\in G$, where $[,]$ is the group commutator.
Doing this in the case of the element of $G_k$ being $1-x_1 x_2 ....x_m$:
$(1-y)(1-x_1 x_2 .... x_m)(1+y+y^2...+y^{n-1})(1+x_1 x_2 ..x_m + (x_1 x_2...x_m)^2+...)$=
$\big(1-(1-y)x_1 x_2 ...x_m(1+y+y^2+...y^{n-1})\big)(1+x_1 x_2 ..x_m + (x_1 x_2...x_m)^2+...)$=
$\big(1-x_1 x_2 ...x_m(1+y+y^2+...y^{n-1})+y(x_1 x_2 ...x_m(1+y+y^2+...y^{n-1})\big)(1+x_1 x_2 ..x_m + (x_1 x_2...x_m)^2+...)$
$\alpha=-x_1 x_2 ..x_m(y+y^2+...y^{n-1})$
$\beta= y x_1 x_2 ...x_m(1+y+y^2+...y^{n-1})$
=$(1-x_1 x_2 ..x_m +\alpha +\beta)(1+x_1 x_2 ..x_m + (x_1 x_2...x_m)^2+...)$
=$1-(x_1...x_m)^2+(\alpha+\beta)\big((x_1 x_2...x_m)^2+...(x_1...x_m)^{n-1}\big)+(1-x_1 x_2 ..x_m)(x_1 x_2...x_m)^2+...)+(\alpha+\beta)(1+x_1 ...x_m)$
Which is in $G_{k+1}$, since all the terms after the $1$ in this expression are of length at least $m+1$.
The general element case is the exact same, its just a bit more of a mess.


TITLE: Find all functions $f:[0;1]\rightarrow[0;1]$ which satisfying the following conditions: $f(0)=0, f(1)=1$ $f(x+y)-f(x)=f(x)-f(x-y)$
QUESTION [2 upvotes]: Find all functions $f:[0;1]\rightarrow[0;1]$  which satisfying the following conditions: $f(0)=0, f(1)=1$
$$f(x+y)-f(x)=f(x)-f(x-y)$$
for any $x,y\ge0$ such that $(x-y), (x+y) \in [0;1]$.
My work so far:
Let $x=y$. Then $$f(2x)=2f(x)$$ for $0\le x \le \frac12$
Let $y=2x, y=3x,...$. Then $$f(nx)=nf(x)$$ for $0\le x \le \frac1n$
Edit: This is an IMO1986 Longlisted problem (FIN 1)

REPLY [4 votes]: Picking up essential points from @wythagoras: For $x,y,x+y\in [0,1]$ the functional equation gives
$$ 2f \left( \frac{x+y}{2} \right) = f(x+y) + f(0) = f(x)+f(y) $$
and $f(0)=0$ shows that
$$ f(x+y)=f(x)+f(y)$$
$f\geq 0$ implies that $f$ is monotone increasing. Iterating we get $1=f(1)=n f(1/n)$ and then $f(k/n)=k/n$ for $0\leq k\leq n$, $n\geq 1$. Then for any real $x\in (0,1)$ (and rational q's): $$ x=\sup\{q<x\} \leq f(x) \leq \inf\{q>x\} =x$$
The important point is positivity (or boundedness) of $f$. Without this it is possible to construct other (non-measurable) functions using a Hamel basis of the reals over ${\Bbb Q}$. See:
Cauchy's Functional equation


TITLE: Can you self-teach yourself Mathematics?
QUESTION [3 upvotes]: If you are not a natural genius, can you teach yourself Mathematics (through textbooks) enough to make a successful mathematical career?

REPLY [3 votes]: The answer to this question has a simple answer: Maybe. The ability to learn mathematics independently from textbooks depends on (1) your willingness do so, and (2) how well you can read mathematics. Many students who take introductory college math courses (i.e., College Algebra, and Calculus) learn new concepts primarily by attending lectures, and usually avoid reading any kind of textbook. And if students accompany their lectures with a ton of practice problems, they usually get good test scores, and end up never reading a math textbook (unless they're majoring in math). The issue that arises is with this that these students probably don't fully grasp the concepts behind the problems they are working. In order to obtain a true understanding of a mathematical concept, read a textbook about it. But know that reading mathematics is not like reading literature. You have to  read slowly and ensure that you fully understand each statement. Work examples as you read them to make sure you understand each step. It's also important to know that Mathematical statements are very compact. Most of the most important and influential theorems in mathematics can be stated in one or two sentences. So if you feel like it takes you "too long" to read math textbooks, you're wrong. So to summarize: Yes, it is possible for anyone to learn mathematics on their own through textbooks. But it is a very time consuming, and frustrating endeavor. And if you don't have the passion and dedication to learn mathematics, you probably won't learn it. Especially not on your own. If you really want to become good at mathematics on your own, read some math books, and do a lot of practice problems.

As to making a successful career in mathematics, I don't think many will hire you without a degree.


TITLE: Prove this limit of ratios of divergent integrals
QUESTION [2 upvotes]: Define
$$ m_{n}(a)\equiv\int_{-\infty}^{\infty}x^{n}e^{x^{2}-ax^{4}}dx $$
I would like to prove that
$$ \lim_{a\to 0^+} \frac{m_{2}^{2}(a)}{m_{0}(a)m_{4}(a)} = 1. $$
This limit bears out numerically, but I can't figure out how to prove it analytically. All of the integrals individually diverge. $a$ is real and positive.
OP Edit:
I don't know if this helps, but since both limits of the integrand are zero when $a>0$, we can deduce that:
$$\int_{-\infty}^{\infty}\frac{d}{dx}\left(x^{n}e^{x^{2}-ax^{4}}\right)dx=0$$
$$\Rightarrow2m_{n+1}(a)-4am_{n+3}(a)+nm_{n-1}(a)=0$$
$$\Rightarrow2m_{2}(a)-4am_{4}(a)+m_{0}(a)=0$$
This relation felt like progress.

REPLY [2 votes]: Let $n = 2k$ be even and we first make a simple change of variables $x \mapsto \sqrt{x}$ to write
$$ m_{2k}(a) = \int_{0}^{\infty} x^{k-\frac{1}{2}} e^{-(ax^2 - x)} \, dx. $$
Now set $u^2 = 4a(ax^2 - x) + 1 = (2ax - 1)^2$. When $u \geq 0$, this relation gives two inverses
$$ x_{+}(u) = \frac{1 + u}{2a} \quad \text{for } u \in [0,\infty), \qquad x_{-}(u) = \frac{1 - u}{2a} \quad \text{for } u \in [0,1]. $$
Now splitting the integral into two parts, one on $[0, \frac{1}{2a}]$ and the other on $[\frac{1}{2a}, \infty)$, and applying the substitution $x = x_{\pm}(u)$ appropriately,
\begin{align*}
m_{2k}(a)
&= \int_{0}^{\frac{1}{2a}} x^{k-\frac{1}{2}} e^{-(ax^2 - x)} \, dx + \int_{\frac{1}{2a}}^{\infty} x^{k-\frac{1}{2}} e^{-(ax^2 - x)} \, dx \\
&= \int_{0}^{1} \left(\frac{1-u}{2a}\right)^{k-\frac{1}{2}} e^{-\frac{u^2-1}{4a}} \, \frac{du}{2a} + \int_{0}^{\infty} \left(\frac{1+u}{2a}\right)^{k-\frac{1}{2}} e^{-\frac{u^2-1}{4a}} \, \frac{du}{2a} \\
&= \frac{e^{1/4a}}{(2a)^{k+\frac{1}{2}}} \int_{0}^{\infty} \left( (1-u)_{+}^{k-\frac{1}{2}} + (1+u)^{k-\frac{1}{2}} \right) e^{-\frac{u^2}{4a}} \, du.
\end{align*}
Here $x_+ = \max\{0, x\}$ denotes the positive part of $x$. Symmetrizing the last integral and rewriting a little bit, we finally obtain the expression
$$ m_{2k}(a) = \frac{\sqrt{2\pi} e^{1/4a}}{2^{k+1} a^k} \int_{-\infty}^{\infty} \left( (1-|u|)_{+}^{k-\frac{1}{2}} + (1+|u|)^{k-\frac{1}{2}} \right) \frac{1}{\sqrt{4\pi a}} e^{-\frac{u^2}{4a}} \, du. $$
Notice that the last integral involves the Gaussian kernel which 'converges' to $\delta_0(x)$ as $a \to 0^+$. This tells that
$$ m_{2k}(a) = \frac{\sqrt{2\pi} e^{1/4a}}{(2a)^k} (1 + o(1)) $$
as $a \to 0^+$. Plugging this back, all the term cancels out to prove that the limit is indeed $1$.
I know little about physics theory, but somehow the exponent reminds me of the $\phi^4$-theory. If it is indeed related to the theory, I guess they might have a fancy way of explaining the limit.