Find all functions $f\colon \mathbb R \to \mathbb R$ such that $$\large f(f(x)(y + 1)) = x[f(y) + 1], \forall x, y \in \mathbb R$$   Let $P(x, y)$ be the assertion $f(f(x)(y + 1)) = x[f(y) + 1]$.  For $P(0, -1)$ and $P(x, 0)$, we respectively have that $f(0) = 0$ and $f^2(x) = x, \forall x \in \mathbb R$.  This implies that $f(x)$ is a bijective function, (although I am not sure why).  For $P(x, x - 1)$, we have that $f(xf(x)) = x[f(x - 1) + 1], \forall x \in \mathbb R$  Then... I don't know what to do next.
Determine all $f: \mathbb{R} \rightarrow \mathbb{R}$ such that, for all $x, y \in \mathbb{R}$, $$f((x + 1) f(y)) = y (f(x) + 1).$$    Source: Vuong Lam Huy, on a Facebook group.
Profitable strategy in coin tossing? There is a coin with a probability $p$ of heads, and $1-p$ of tails. Tosses are independent of each other. When you bet an amount of money $x$, you receive $2x$ if it lands heads, and you lose what you bet if it lands tails.  We are going to use the following strategy:  You start betting an initial amount $x_0$. Every time you lose, you bet twice what you bet in the previous throw. You keep doing this, until you win, at which point you retire, or start again, betting $x_0$.  The interesting case of course is when $p &lt; 0.5$.  The question is: How profitable is this strategy? Intuitively I feel that eventually I can always win an arbitrarily large amount of money. But that can't be entirely right ("the house always wins!"). So what am I missing here? What's the expected win in each run?  Also, does this sort of strategy have a name in the literature?  P.S.: Found this later, For which Natural $n\ge2: \phi(n)=n/2$ Docker ppa package for Ubuntu 19.10 Does someone knows if it exists some ppa packages source for the newest Ubuntu version 19.10 i had problem with that when i upgraded from Ubuntu 19.04 to 19.10 i didn't work? What metadata does a LibreOffice file contain and how to remove it? There is a command l available on my machine which appears to do nothing.  which l also produces no output. Is this a real command, and does it actually do anything? I need scale up video from one DOS game using . But with  on left side and  filter on the other one.  Input (320x200)    Output (1280x800)    Here is a command, where's xBR processing whole screen...  ffmpeg     -i input.avi     -sws_flags neighbor     -vcodec libx264 -strict -2     -preset veryslow -qp 0     -filter:v "xbr=4"     output.mp4 Is it possible to dynamically set Views' pager settings? Say for instance, you have a view that displays when viewing a certain node type, but you want the pager settings to be different for each individual node. I was thinking of making a CCK field like "pager setting" and then give it an integer option of x-y. But I don't know if it's possible to somehow dynamically plug that field into the Views' settings. Or is there another way to do this maybe? How to Delete Elements from List1 appearing in List2? Something special about energy eigenstates when it comes to time evolution? A particle is subject to an infinite square well potential with  $$V(x)= \begin{cases}  0  &amp;   −a \lt x \lt a\\   \infty &amp;  \,\,\,\,\text{otherwise} \end{cases}$$  At a time $t=0$ its wavefunction is given by $$\psi(x, t=0)=\frac{1}{\sqrt{5a}}\cos\left(\frac{\pi x}{2a}\right)+\frac{2}{\sqrt{5a}}\sin\left(\frac{\pi x}{a}\right)$$       (a) What are the possible results of an energy measurement at $t = 0$, and with what probabilities?     Re-writing the terms in $\psi(x,t=0)$ as $$u_1(x)=a^{-1/2}\cos\left(\frac{\pi x}{2a}\right)\qquad \text{&amp;} \qquad u_2(x)=a^{-1/2}\sin\left(\frac{\pi x}{a}\right)$$ such that $$\psi(x,t=0)= \frac{1}{\sqrt{5}}u_1(x)+\frac{2}{\sqrt{5}}u_2(x)$$ with energies $$E_1=\frac{\hbar^2 \pi^2}{8ma^2}\qquad \text{&amp;} \qquad E_2=\frac{\hbar^2 \pi^2}{2ma^2}$$ and probabilities $$P(E_1)=\frac15 \qquad \text{&amp;} \qquad P(E_2)=\frac45$$       (b) If no measurement is performed, what is the wavefunction $\psi(x, t)$ at all times $t$? What are the possible energies and their probabilities if a measurement is first performed at time $t$?     At a time $t$ the wavefunction is given by $$\psi(x, t=0)=\frac{1}{\sqrt{5a}}\cos\left(\frac{\pi x}{2a}\right)\exp\left(\frac{-i E_1 t}{\hbar}\right)+\frac{2}{\sqrt{5a}}\sin\left(\frac{\pi x}{a}\right)\exp\left(\frac{-i E_2 t}{\hbar}\right)$$    The problem is that I cannot answer the second part of the question (b).    The answer states that:      Because the two terms are energy eigenstates, the relative probabilities do not change and are as in the previous part.     I have some questions about the above statement: why does an "energy eigenstate" mean that the probablities don't change with time? Or, put in another way, does a "momentum eigenstate" (for example) have constant relative probabilities when measured at any time?   The answer seems to be implying there is something special about energy eigenstates when it comes to time evolution. The eigenvalues of energy measurements don't seem to change regardless of how many times you measure them, and no matter how long you wait until you measure them. I'm not sure if this is the case, but could someone please explain why all quantum operators (except Hamiltonian) do not exhibit this behaviour. How to get Scilab 6.0.1 working on Ubuntu 18.04 LTS? I have installed clean Ubuntu MATE 18.04 LTS and installed Scilab from repository here with    sudo apt-get update sudo apt-get install scilab   After installation I tried to launch it from GUI (Applications → Other → Scilab) - it does not start. Then I tried to launch it from terminal with scilab command. Terminal output is as follows:  $ scilab /usr/bin/scilab-bin: error while loading shared libraries: libjava.so: cannot open shared object file: No such file or directory   so it can't find Java. I reported  about this problem.  I checked existence of libjava.so:  $ dpkg -S libjava.so   openjdk-11-jre-headless:amd64: /usr/lib/jvm/java-11-openjdk-amd64/lib/libjava.so   so it is in place.  What should I do to get Scilab 6.0.1 working on Ubuntu 18.04 LTS? Layover in Europe to the US - allowed? Qt: Best practice for a single instance app protection Pre-generate Review audits to avoid the lag? Classic ASP SQL Injection Protection What is a strong way to protect against sql injection for a classic asp app?  FYI I am using it with an access DB.  (I didnt write the app) Homology Groups of $S^n$ Proof of Russian Peasant Multiplication I'm studying a combinatorics text () in preparation for taking discrete math this upcoming semester. I've taken a fair amount of math in college, through Calculus II, but this is my first 500+ level math course.    In chapter 2 of the  book, the author introduces an algorithm for multiplication called Russian Peasant Multiplication. I want to prove that this algorithm will always result in the product of two numbers.  I've been chewing on the proof for a while now, and just can't seem to get any traction. Thanks in advance for any help.  Description of Algorithm  I'll try to explain the algorithm clearly. The goal of the algorithm is to find the product of two numbers $m$ and $n$. The below table is filled out for $m$ and $n$:  $$\begin{array}{c} m&amp;&amp;n&amp;\\ \hline \bigl\lfloor \frac{m}{2} \bigr\rfloor&amp;&amp; 2n \\ \bigl\lfloor \frac{m}{4} \bigr\rfloor&amp;&amp; 4n \\ \bigl\lfloor \frac{m}{8} \bigr\rfloor&amp;&amp; 8n \\ \bigl\lfloor \frac{m}{16} \bigr\rfloor&amp;&amp; 16n \\ \bigl\lfloor \frac{m}{32} \bigr\rfloor&amp;&amp; 32n \\ \end{array}$$  The sequence of numbers in both columns is the sequence $a_n = 2a_{n-1}\text{ where }a_1=2$. The columns are extended down until the value in the left column is equal to the number one. After filling out the table, the product is found by adding the values of the right column where the left column is an odd number.  For example, to multiply 18 and 37 with the algorithm:  $$\begin{array}{c} m&amp;&amp;n&amp;\\ \hline 18 &amp;&amp; 37 \\ 9 &amp;&amp; 74 \\ 4 &amp;&amp; 148 \\ 2 &amp;&amp; 296 \\ 1 &amp;&amp; 592 \\ \end{array}$$  To find the product, cross out values of the left column where the value of the right column is even:  $$\begin{array}{c} m&amp;&amp;n&amp;\\ \hline \require{enclose} \enclose{horizontalstrike}{18} &amp;&amp; \enclose{horizontalstrike}{37} \\ 9 &amp;&amp; 74 \\ \enclose{horizontalstrike}{4} &amp;&amp; \enclose{horizontalstrike}{148} \\ \enclose{horizontalstrike}{2} &amp;&amp; \enclose{horizontalstrike}{296} \\ 1 &amp;&amp; 592 \\ \end{array}$$  And add the remaining numbers in the right column to get the result: $18\times37=74+592=666$  Alternate Description of Algorithm  As an aside, here is a shorter way to describe the algorithm, repeat the below steps until the left column == 1:  • Halve the last number in the first column • Write the result below (discard the remainder) • Double the last number in the first column. • Write the result below.   Now, for every even number in the first column, delete the adjacent number in the second column. Add the remaining numbers in the second column.  What I've done so Far  I haven't been able to get much of anywhere on the proof. One thing I did was create a bit of a different way of arriving at the same outcome:  $$\begin{array}{c} m&amp;&amp;n&amp;&amp;\text{index}&amp;&amp;\text{remainder}&amp;&amp;\text{if remainder == 1, }n\times2^{Index}&amp;\\ \hline 18 &amp;&amp; 37 &amp;&amp; 0 &amp;&amp; 0 \\ 9 &amp;&amp; 74 &amp;&amp; 1 &amp;&amp; 1 &amp;&amp; 74\\ 4 &amp;&amp; 148 &amp;&amp; 2 &amp;&amp; 0 \\ 2 &amp;&amp; 296 &amp;&amp; 3 &amp;&amp; 0 \\ 1 &amp;&amp; 592 &amp;&amp; 4 &amp;&amp; 1 &amp;&amp; 592\\ \end{array}$$  Which brings to light the relationship between this algorithm and binary numbers (the fourth column in the above table is the product in binary, read bottom to top).   I also wrote some Python functions to help me explore the problem, Convergence of the sequence $\sqrt{1+2\sqrt{1}},\sqrt{1+2\sqrt{1+3\sqrt{1}}},\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1}}}},\cdots$ I recently came across this problem     Q1 Show that $\lim\limits_{n \rightarrow \infty} \underbrace{{\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots+n\sqrt{1}}}}}}}_{n \textrm{ times  }} = 3$   After trying it I looked at the solution from that book which was very ingenious but it was incomplete because it assumed that the limit already exists.  So my question is     Q2 Prove that the sequence$$\sqrt{1+2\sqrt{1}},\sqrt{1+2\sqrt{1+3\sqrt{1}}},\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1}}}},\cdots,\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots+n\sqrt{1}}}}}$$ converges.   Though I only need solution for Q2, if you happen to know any complete solution for Q1 it would be a great help .  If the solution from that book is required I can post it but it is not complete as I mentioned.   Edit: I see that a similar question was asked before on this site but it was not proved that limit should exist. Find all the values of $w$ ∈ C that satisfy the equation. What causes a "Java Hotspot 64-Bit Server VM Warning"? I get this warning:     Java HotSpot(TM) 64-Bit Server VM warning: Using incremental CMS is deprecated and will likely be removed in a future release   What does it mean? How can I bring back the old menu style on Chrome 28 onwards, now that the --disable-new-menu-style switch has been removed?