I have seen this type of circuit design a few times, but am not sure why it is needed. Considering the schematic below, what is the purpose of resistor R3?
Here in the Figure R5=50K. This resistor value is not in the equations to obtain output voltage, and in many tutorials non-inverting input is directly grounded without a resistor. What might be the reason to use it here?
How to prove that the only subgroup of the symmetric group $S_n$ of order $n!/2$ is $A_n$?  Why isn't there other possibility?  Thanks :) Why is "letter" not plural in "two letter words"? Why is "letter" not plural in "two letter words"?  For me it's very strange as the equivalent in French would be plural but my English friend finds it totally normal. 3-4 old computers = general purpose cluster? Recently when skimming through my physics-text I encountered an interesting definition of Force $$F(x) = -\frac{\mathrm dU(x)}{\mathrm dx}$$  We were taught that some functions are continuous but not differentiable. So for the force to exist $U(x)$ has to be differentiable. So how can we prove that $U(x)$ is differentiable everywhere? I'm wondering how to approach duct work, piping, and wiring? In the past, I have used the spin tool to make corners, but I find it very cumbersome. How do you do duct work and piping? I have seen people use curves, but I find it's hard to get straight 90 degree angles or just uniform angling across the whole model. It's also hard connecting piping together if they started from different locations. Here's some reference: Prove the following equality using mathematical induction: Prove the following equality using mathematical induction: For any integer $n \ge 1$  $$\sum_{i=1}^{n} \frac{1}{i(i+1)} = \frac{n}{n+1}$$  I understand for the base base I need to have $n=1$. If I substituted $n$ for 1 and $i$ for 1 I would get   \begin{align} \sum_{i=1}^{1} \frac{1}{1(1+1)} &amp;= \frac{1}{1+1} \\ \sum_{i=1}^{1} \frac{1}{2} &amp;= \frac{1}{2} \end{align}  As for the inductive step, I inputted $n +1$ but it did not work for me because honestly I did not know what to do with the $i$ and what I should do next.  How would I go about doing the next step and an explanation for $i$ would be helpful. $Q_8$ is isomorphic to a subgroup of $S_8$, but not isomorphic to a subgroup of $S_n$ for $n\leq 7$. Question is to prove that :     $Q_8$ is isomorphic to a subgroup of $S_8$, but not isomorphic to a subgroup of $S_n$ for $n\leq 7$.   I see that $Q_8$ is isomorphic to subgroup of $S_8$ by left multiplication action.  Hint given was to prove that stabilizer of any point contains $\{\pm 1\}$.  To prove Cayley's theorem, stating any group is isomorphic to a subgroup of $S_n$ we take action of given group on a set $A$ having same cardinality.  with that motivation I want to check if there is a Isomorphism then there is a map from $G\times A \rightarrow A$. i.e., $G$ gives a permutation group $S_A$.  I tried in same manner. Suppose $Q_8$ is isomorphic to subgroup of $S_n$ with $n\leq 7$  then it should come from a group action of $Q_8$ on a set of cardinality atmost 7.   Suppose $Q_8$ acts on a set $A$ with possible cardinality at most 7.  $stab(a)=\{g\in Q_8 : g.a=g \forall a\in A \}$  $cl(a)=\{g.a : g\in Q_8\}$  I know number of elements in class of $a$ equals to index of stabilizer.  As $cl(a)=\{g.a : g\in Q_8\}\in A$ i.e., $cl(a)\subseteq A$ and as $|A|\leq 7$ i see that $|cl(a)|\leq 7$.  But, $|cl(a)|=|Q_8:stab(a)|$ for any element $a\in A$.  So, $|Q_8:stab(a)|=|cl(a)|\leq 7$ for all $a\in A$.   So, $stab(a)$ should be non trivial subgroup of $Q_8$ if not then $|Q_8:stab(a)|=8$  non trivial subgroup (proper) of $Q_8$ contains $\{\pm1\}$.  So, In the worst case, $\{\pm 1\}\subseteq stab(a)$ for all $a\in A$.  As $Ker(\eta)=\cap_{a\in A}stab(a)$ (where $\eta$ is the action of $Q_8$ on $A$)   we see that $\{\pm 1\}\subseteq Ket(\eta)$ which means that $Ker(\eta)$ is non trivial.  thus there is no isomorphism (coming from $\eta$) between $Q_8$ and any subgroup of $S_7$.  I would be thankful if someone can check whether my approach is correct or if there is any other simple possible way.  P.S : Usually what i do to see whether two groups are isomorphic or not is to check for cardinality, abelian property, no of elements with same order and so on. But I was having no idea when i fail in all these ways. With this Group actions i could see possibility for getting a precise conclusion on Isomorphisms.I would like to Thank Mr. Jyrki Lahtonen (a user of Math.SE) who made me to get used to Group actions.  P.S $2$: If any thing is wrong in my idea, it is entirely my fault, and if anything is correct in this whole credit should go to Mr. Jyrki Lahtonen Is JavaScript a pass-by-reference or pass-by-value language? float to int unexpected behaviour Are The Lazarus Effect and Dune in the same universe? There are similar aspects between Dune, and the technology in The Lazarus Effect.   Noticeable similarities:   Lasguns - this weapon exists in both of the books, and has similar architecture in both books.   Merman and Ixians - are both technologically sophisticated in similar ways. Rely on mechanical technology as opposed to organics or similar.   Is this merely Herbert sticking to his original style? Area of an ellipse. (Calculus) the composition was  defined as follow:  (a,b) \in (R;S) &lt;=&gt; there is c | (a,c) \in R  and (c,b) \in S .   If our two relations R and S are two convex polygon   Is there a geometric interpretation of the composition of two convex polygon ? for example :   the green polygon was the result of composition. How can i describe geometrically the result !!! "At the time" versus "at that time" Terminal returns not found for most commands Mac OSX I'm really new to Mac OSX and UNIX based systems. I wanted to run a few Windows games on my Mac Mini so I started installing Wine and MacPorts. I think the installs are correct, but when I type sudo port install wine I see a not found error in my terminal. I tried other basic commands like say and clear and all of them return the same not found error. My research shows that my PATH might be incorrectly set, but it lacks steps I can implement. Specifically, I ran this command: echo $PATH and it returned:  /opt/local/bin:/opt/local/sbin:/opt/local/bin:/opt/local/sbin:/opt/local/bin  I'm a complete newbie to mac and have no idea where .bash_profile or any of those files are. My skills let me install Xcode and the Command Line tools for Xcode and I sense I need to level up on path management. At this point I need help managing my dot files and changing my path to fix these specific errors on Mountain Lion. What are my next steps? Which browsers are officially supported by Stack Exchange?  Also, what else is needed in order to use every feature of the site? Is there a Unix command to get the absolute (and canonicalized) path from a relative path which may contain symbolic links? Consider the question :    The large number of down-votes displays as 199 rather than a score of -199.  This is clearly unintentional (and unfortunate for questions that possess such voting!), but can this be&nbsp;corrected? hotel occupancy counting problem The multiple-choice question:  A hotel has four vacant rooms. Each room can accommodate a maximum of four people. In how many different ways can six people be accommodated in the four rooms? (A) 4020 (B) 4068 (C) 4080 (D) 4096.  My attempt:  If there were no 'four person' constraint, the six people could be accommodated in $4^6=4096$ ways. however, this number includes combinations with 5 or 6 people in one room, which should be excluded.  If one room has 5 occupants, one other room has 1 occupant. The number of '1,5' combinations is ${{4} \choose {2}}$=12. If one room has 6 occupants, the other three rooms have no occupants, and there are four such combinations.  I believe the answer should be $4096-12-4=4080$, but am not confident. I'd be grateful for any comments on my attempt. How do I add a condition to a query? How to make a char string from a C macro's value? For example, how to avoid writing the 'func_name' twice?  #ifndef TEST_FUN #  define TEST_FUN func_name #  define TEST_FUN_NAME "func_name" #endif   I'd like to follow the  rule.  Version of C preprocessor:  $ cpp --version cpp (GCC) 4.1.2 20070626 (Red Hat 4.1.2-14)