We've got a mouse in the house. I want to setup traps in the attic, but I don't want to go up there to check them. My thought is to use a simple circuit that uses the trap itself as a switch. I think it needs two circuits: a path for the electricity when the trap is set (circuit A) and a path when the trap has gone off (circuit B).   I think:   Circuit A should be nothing except a resistor. (If it was just metal, the battery would run down real quick... plus... you know... hot). Circuit B should have a resistor and an LED (and a long pair of wires so we can hang the LED out the attic door and it is visible from the hallway.   When the trap is set, circuit A will have less resistance than circuit B, and thus short circuit the light. When the trap goes off, circuit A will break, and B will then be the only path for the electricity thus lighting up the LED.  Questions: How do I calculate out what resistors to use?  In terms of power, how do I decide on 9V? 5V? 1.5V?
I have a simple circuit:    The max rating (current) of the LED is 30mA.  How can I work out what the resistance of the resistor needs to be? By using ohm's law I found it to be \$3V/0.03A = 100 \Omega\$. However using software called , and trial and error I got the minimum possible resistance to be 36&Omega;. However, if I use a 35&Omega; resistor, then the LED breaks. Is the software wrong, or (more likely) am I doing something wrong?
How to calculate what this power series converges against? (double factorials) Show that if $a$ is an odd integer and $b$ is an even integer then $\gcd(a,b)=\gcd(a,b/2)$ Show that if $a$ is an odd en integer and $b$ is an even integer then $\gcd(a,b)=\gcd(a,b/2)$.  I understand that since $a$ is not divisible by $2$ but $b$ is, the gcd of $a$ and $b$ also can't be divisible by $2$ but I'm getting stuck using this to help me. How to prove $\gcd(a^2, b^2) = (\gcd(a, b))^2$?  My attempt:  Let $\gcd(a, b) = d$.  Then $d|a$ and $d|b$ then $d^2|a^2$ and $d^2|b^2$. i.e $d^2$ divides $a^2 ~~\&amp;~~ b^2$. To mean "He is a friend of mine who I love" or "He is a friend of mine that I love", can I say "He is a friend of mine I love"? When I omit "who" or "that" in that sentence, it sounds wrong to me. I think I am not used to that structure. What do you think?  Note: I know (that) relative pronouns can be dropped like in "He is someone (that/who) I love", but in the sentence I gave, I think it is wrong to omit it because "a friend" was already modified by "of mine".  Context: I am talking about a friend, and I am saying I love him. Let $f(z)$ be entire function. Show that if $f(z)$ is real when $|z| = 1$, then $f(z)$ must be a constant function using Maximum Modulus theorem Let $f(z)$ be entire function. Consider the functions $e^{if(z)}$ and $e^{−if(z)}$ and applying the Maximum Modulus Theorem, show that if $f(z)$ is real when $|z| = 1$, then $f(z)$ must be a constant function.  (We take $f(z)=u(z)+iv(z)$)  I am confused as so far I have $|g(z)|=|e^{if(z)}|=|e^{-v(z)}|$ and then since $f(z)$ is real, $f(z)=u(z)$ and $v(z)=0$ so I assumed it would follow that $|g(z)|=|e^{v(z)}|=1$.  Similarly, $|g(z)|=|e^{-if(z)}|=|e^{v(z)}|=1$.  Using Liouville I assumed one could say that both $g(z)$ and $h(z)$ are bounded entire functions, they are constant and so it follows that $v(z)$ is constant, meaning that both its partial derivatives are equal to 0 and, due to Cauchy Riemann, both of the partial derivatives of $u(z)$ are equal to zero.  It would then follow that $f(z)$ is constant.  I don't know how to go about the question using the Maximum Modulus Theorem, also I feel I am overlooking the importance of $|z|=1$ perhaps?  Any help would be much appreciated!! Split comma-separated strings in a column into separate rows Convert IEnumerable to DataTable notation of distribution I have a question:      Does  $$N(0, x)$$ mean a normal distribution with mean $0$ and variance $x$? Or standard deviation $x$?   The notation seems ambiguous sometimes. What Happens When A Gravitational Wave Interacts With Another One? If two gravitational waves came in contact with each other what would happen? In another question entirely, what happens when a higher gravitational field interacts with a weaker one. Let $A$ be a UFD. Assume that $a,b \in A$ are relatively prime, $c \in A$ and $a | bc$. To prove that $a|c$, is the following approach correct (or do you have to use some type of prime factorization argument)?  By the relatively prime assumption, $\gcd(a,b)$ is a unit. Call that unit $u \in A^{\times}$. Then we can write $u = ax + by$ for some $x,y \in A$. But then, $c = c \cdot u = c(ax+by)=cax+cby$. However, it is clear that $a|cax$ and the fact that $a|bc$ implies $a|cby$. Thus, $a|cax+cby=c$. What level is a spell if you cast it without expending a spell slot? enumerate in multicols The following image illustrates what am looking for:    Essentially I want the numbering to occur as shown above but using the multicol and enumitem package. But when I try it by using the following code:  \documentclass{article} \usepackage{amsmath,amssymb} \usepackage{enumitem,multicol} \begin{document} \begin{enumerate} \item Evaluate each of the following if $x$ is a non-zero real number.     \begin{multicols}{2}     \begin{enumerate}[itemsep=1cm]% The item sep is my choice here.     \item $\dfrac{|x|}{x}$     \item $\dfrac{x}{|x|}$     \item $\dfrac{|-x|}{-x}$     \item $|x|-|-x|$     \end{enumerate}      \end{multicols} \end{enumerate} \end{document}   I get:    Essentially, I want the (b) to be where the (c) is. I know this has been done but I can't seem to find it. How to integrate mv command after find command? I am searching for files which name which contain AAA within their path using following command:  find path_A -name "*AAA*"   Given the output showed by the above command, I want to move those files into another path, say path_B. Instead of moving those file one by one, can I  optimize the command by moving those files right after the find command? This thing came up in a combinatorics course I am taking.  Choose a fixed set of primes $p_1,p_2,\dots,p_k$ and let $A_n$ be number of integers in $\{1,2,\dots,n\}$ which are not divisible by any of the $p_i$'s. $A_n$ is given by $ n - \sum_{1\leq i_1 \leq k} \lfloor \frac{n}{p_{i_1}} \rfloor + \sum_{1\leq i_1 &lt; i_2 &lt; k} \lfloor \frac{n}{p_{i_1}p_{i_2}}\rfloor - \dots $. Now, if $n$ is divisible by each of the $p_i$'s then we have the simpler expression : $A_n = n \prod_{i=1}^{k}(1-\frac{1}{p_i})$(added later : note I am not assuming $n = \prod_{i=1}^{k}p_i$ here, this holds true whenever $n = c \prod_{i=1}^{k} p_i $ for some integer c, since each $\lfloor \frac{n}{p_{i_1} \dots p_{i_j}} \rfloor = \frac{n}{p_{i_1} \dots p_{i_j}}\ )$.  Another student pointed out that even if $n$ does not have $\prod_{i=1}^{k} p_i$ as a factor, the approximation $B_n = n \prod_{i=1}^{k}(1-\frac{1}{p_i})$ to $A_n$ is quite close in some specific cases. It is easy to see that $\lim_{n\to\infty}\frac{A_n - B_n}{n}=0$ as $\lim_{n\to\infty}\frac{1}{n}\lfloor \frac{n}{p_{i_1}p_{i_2}\dots p_{i_j}} \rfloor \to \frac{1}{p_{i_1}p_{i_2} \dots p_{i_j}},$ however that is not strong enough to imply that $A_n - B_n$ will be always close. Is there any way to analyze how well this approximation will perform in general? I am interested in the worst case for small to moderately sized $n$.  Just to give a feel of how close $A_n$ and $B_n$ can get, assuming the set of primes is $\{2,3,7\}$ we have (assuming my program is correct):  n     A(n)     B(n) 17    5    4.86   27    8    7.71   37    11      10.57   107   31      30.57   1111  318     317.43   3001  858     857.43   4007  1145    1144.86   5000  1429    1428.57 . The forgetful functor $\mathsf{Groups} \to \mathsf{Sets}$ admits a left adjoint, namely, "forming the free group" functor.  I was wondering if this has a left adjoint.  This seems unlikely, but I don't have a proof.  Is there a functor $\mathsf{Sets} \to \mathsf{Groups}$ that admits a left adjoint at all? I have created several designs for a particular project and they are related, but I don't want to have to keep recreating any similar elements, such as shared materials or textures.  How can I use the existing data in a new Blender file? Why is it called the Millennium Falcon? Let's say I have Windows installed on drive C, however I install all of my programs on the D drive, and I keep my user profile on the D drive as well (things like My Documents). How can I have Windows use those folders instead of the ones on the C drive? How can I prove that all rational numbers are either terminating decimal or repeating decimal numerals? I am trying to figure out how to prove that all rational numbers are either terminating decimal or repeating decimal numerals, but I am having a great difficulty in doing so. Any help will be greatly appreciated. Installing QGIS with ECW support on Mac?