Convert .csv table to dictionary
Read two column CSV as dict with 1st column as key
To env or not to env Proof of Convergence: Babylonian Method $x_{n+1}=\frac{1}{2}(x_n + \frac{a}{x_n})$ How many combinations can I make? let $n \gt 1$ be an integer, and consider $n$ people; $P_1, P_2,..., P_n$ let $A_n$ be the number of ways these $n$ people can be divided into groups, such that each group have either one or two people  determine $A_1, A_2, A_3$  So I have  $A_1 = 1$ way $A_2 = 4$ way $A_3 = 12$ way  But I am not sure if this is right... Can anyone confirm?  edit:  My way of getting this is  $A_1 = 1 $ because {P1} only has 1 way to sort  $A_2 = 4$ because {P1,P2} = {P1,P2},{P2,P1},{P1P2},{P2P1}  etc. We currently have two rewards checking accounts.  We are nearing the maximums permitted on each account, after which we receive .5% and .25% APY.  It's within our means to perhaps open another one, but at that point we are looking at using our debit cards 50 to 55 times a month.  I feel that running that number of debit transactions is getting out of hands as it's just the two of us.  I realize there are easy ways to game the system, but I want these things to stay around and another person doing just such that could lead to the ends of these accounts.  We are saving towards a new house down payment.  As such we don't require these funds to be fully liquid, but would still like to remain free of as much risk as possible -- however I realize we are unlikely to find anything near our rewards checking accounts (4% APY interest for meeting various rules: direct deposit, a certain number of debit card transactions, and not receiving a paper statement), but would hope we can get at least half of that.  We aren't fully sure when we will be buying, but we are probably a couple of years out.  For that reason I don't think CDs would work.  Even if we ladder them, when we need the money we will need it pretty much instantly.  Do I bite the bullet and just find the best available money market account?  What other options do I have that let me deposit monthly, and not have a timebomb associated to accessing that money that isn't in the initial lump sum? I'm looking for a movie that was set in the US, and all the characters (including the robot) spoke English.  I can't recall the exact plot, but mostly it was about an innocent game robot that shoots small objects for fun and becomes dangerous, because its AI made a wrongful choice about 'winning the game'. The robot starts to arm itself with more dangerous objects (not sure but knives and stuff) and starts to attack people.  It was overall supposed to be fun and adventure movie for kids. Color movie, and probably made in 80~90's.   There is a recurring scene when the robot detects an aggressor, it scans and loudly announces "enemy one, enemy one". The robot was somehow looking similar to Wall-E, and had a saucer-like head part. I read a story once about a man who escapes from his pursuers by doing an uncontrolled jump into hyperspace. If I'm not mixing my stories up he was blind but could see through the eyes of other sighted people/animals nearby. He was able to see through his pet but it dies and he thinks he's finished - he's blind and lost in hyperspace but there's a rat on board which he eventually catches by turning off gravity and then he can see through the rat's eyes. He works out how hyperspace operates and can hop back to where he came from.   Can anyone remember the name of the story and the author? I enjoyed it when I was a kid and would like to read it again. Question:  Let ${a_n}$ be a sequence in $\mathbb{R}$ with $\lim_{n\to\infty}a_n=a$  Prove that  $\lim_{n\to\infty}a_n^2=a^2$  Attempt: Without using any properties of limits (this is a question in a section of the book that came before any list of any limit theorems)  By definition:  $|a_n^2-a^2|&lt;\epsilon$   $|a_n^2-a^2|$ $\leq$ $|(a_n+a)(a_n-a)|$ $\leq$ $|(a_n+a)| |(a_n-a)|$ $\leq$   $(|a_n | + |a|)(|a_n -a|)$ $&lt;$ $\epsilon$  And since I know that $\lim_{n\to\infty}a_n=a$, I'll subsitute that in for $|a_n|^2$ as $a^2$   $\therefore$ $a^2-|a|^2 &lt; \epsilon$ which is true since $a^2-|a|^2$ in my case.  I wonder if this is a possible solution? Sum of $k$-th powers $3^{n+1}$ divides $2^{3^n}+1$ Describe all positive integers,n such that $3^{n+1}$divides $2^{3^n}+1$. I am little confused about what the question asks-if it asks me to find all such positive integers, or if it asks me to prove that for every positive integer n,$3^{n+1}$ divides $2^{3^n}+1$. Kindly clarify this doubt and if it's the former part, please verify my solution-n=1. Finding $\int_0^{\infty}xe^{-\lambda x} \, dx$ How to integrate:  $$\int_0^\infty x \, \lambda e^{-\lambda x} \, dx$$    I tried using integration by parts:  Let $u = x$ and $dv = \lambda e^{-\lambda x} \, dx$  Then $du = dx$  And $v = - \lambda e ^{-\lambda x}$  Correct so far?   Then   $$\begin{aligned} uv - \int v \, du &amp;= -\lambda x e^{-\lambda x} - \int (- \lambda e^{- \lambda x}) dx \\ &amp;= -\lambda x e^{-\lambda x} - \lambda e^{-\lambda x} \end{aligned}$$    The correct answer from lecture notes      UPDATE 2  Let $u=\lambda x$ and $dv = e^{-\lambda x} dx$  Then $du = \lambda \, dx$  $$dv = e^{-\lambda x} \, dx$$  Let $y = -\lambda x$  Then $dy = -\lambda \, dx$  So $dx = -\frac{1}{\lambda} dy$  $$\begin{aligned} v &amp;= \int e^u \cdot - \frac{1}{\lambda} du \\ &amp;= - \frac{1}{\lambda} e^{-\lambda x} \end{aligned}$$  But here I seem to have an extra $- \frac{1}{\lambda}$ in $v$?  If I continue using integration by parts, I get:   $$-x e^{-\lambda x} + \color{red}{\frac{1}{\lambda}} \int e^{-\lambda x} \, dx$$ How to change radius/weight of curve bezier? try scaling, change radius, but its dont work How exactly does Knuth's Up-Arrow notation work? Root Android virtual device with Android 7.1.1 Why can't a union of two intervals be the maximum existence interval of a solution? Assume the IVP $$ x'=x^2, \quad x(0)=1 $$ By separating the variables and using the initial condition $$ x(t)=(1-t)^{-1} $$ Then, my book states that since $x_0=1&gt;0$ then the solution can only be extended to the left with maximum existence interval $\, I=(-\infty,1)$.   My question is, since the solution $x(t)$ we found is defined for every $t\neq1$, why isn't the maximum interval of its existence $I^*=(-\infty,1)\cup(1,+\infty)$? Why does it depend on the sign of $x_0$? Proving $\sum\limits_{k=1}^{n}{\frac{1}{\sqrt{k}}\ge\sqrt{n}}$ with induction I am just starting out learning mathematical induction and I got this homework question to prove with induction but I am not managing.  $$\sum\limits_{k=1}^{n}{\frac{1}{\sqrt{k}}\ge\sqrt{n}}$$  Perhaps someone can help me out I don't understand how to move forward from here: $$\sum\limits_{k=1}^{n+1}{\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{n+1}}\ge \sqrt{n+1}}$$ proof and explanation would greatly be appreciated :) Thanks :)  EDIT sorry meant GE not = fixed :) Why do some people use $+\infty$ instead of $\infty$? Why do some people use $+\infty$ instead of $\infty$? Because we usually denote minus infinity as $-\infty$, I think it is sufficient to denote plus infinity as $\infty$. Are there any reasons for this notation? I saw that many people majoring analysis use like $\int_0^{+\infty}$ or $\sum_{j=1}^{+\infty}$ or $\bigcup_{j=1}^{+\infty}$ which is not related to the unsigned infinity. I've seen some conflicted answers to this question in texts and on the web, in the case of a dipole, for example.   Do magnetic fields do work directly, or is it their induced electric fields that do work? Can a bird, previously at earth potential, get electrocuted by landing on a powerline at high-enough voltage due to the initial "equalization charge"? At the voltage levels of typical overhead transmission lines in the US, a bird can land on one and be just fine (as long as it doesn't do something like spread its wings and touch a tree or something else at lower electric potential).  However, what about a hypothetical powerline at much higher voltage (as in tens of megavolts). Could landing on such a powerline fatally-shock the bird even though it does not complete a circuit for sustained current? (Assume that the distance is long enough that electrical arc'ing is impossible.)  NOTE: My understanding of what happens when a bird flies from an earth object to a powerline (please correct me if I'm wrong) is that - upon contacting the wire - its electric potential changes from earth-potential to the powerline's potential. In order for this to happen, there is an initial transfer of electrical energy (i.e. flow of charge i.e. current) from the powerline to the bird which "equalizes" their electric potential, which happens nearly instantaneously. If this is correct, then my question can be restated more generally as "Can an 'equalization charge' such as this result in a fatal shock, if the potential difference that it's equalizing is high enough?" Is it possible to have two background images?  For instance, I'd like to have one image repeat across the top (repeat-x), and another repeat across the entire page (repeat), where the one across the entire page is behind the one which repeats across the top.  I've found that I can achieve the desired effect for two background images by setting the background of html and body:  html {     background: url(images/bg.png); }  body {     background: url(images/bgtop.png) repeat-x; }   Is this "good" CSS?  Is there a better method?  And what if I wanted three or more background images? What is "function*" in JavaScript?