I'm confused about the 200mm and the 300mm zoom ranges. I am looking for some examples of the same subject, in order to visualize the difference between 200mm and 300mm focal lengths.  I want to compare the results for a very distant subject (like the moon) to a far subject (may be 1km away) and a nearer subject (around 100 meter away).
I have seen many camera review sites illustrate the effect of using different focal lengths on the same photograph using frames. There are also similar illustrations to compare different sensors (). I find this kind of visualization to be very useful in comparing the actual effect of the longer focal length\different sensor size. Of course the ability to do a wider focal length visualization would also be equally useful, but that's not possible.  Is there any ready-made software or plugin that enables this? If not, is there any simple technique to visualize the effect of longer focal lengths on an image (one would of course have to use the zoom information in the image EXIF data to accurately represent this)? All I can think of is to use some trigonometry to do the necessary calculations for the frame sizes for cropping.
Expand the username auto-completion to everyone who can be notified In electromagnetism, why does nature prefer the right-hand rule over the left-hand rule? At school I learnt the  to remember the resulting direction of different phenomena, such as geometrical cross products, mechanical torque, or the direction a screw will move when turning it. This directions are however decided by humans, and we could have easily defined the directions in the opposite way.  However the right hand rule also predicts phenomena which is not arbitrarily set by humans, but is intrinsic to the nature. If you have a coil in which you put some current, and the nature creates a magnetic field in a direction perpendicular to it, the direction depends on the right hand rule.  My question is, why this direction and not the opposite? Nature tends to be symmetric and not have apparent preferences for two equal options, so in a so why is one direction preferred over the other one? Why doesn't a "left hand" rule apply? Consider the following minimal example:  \documentclass{article}  \setlength{\parindent}{0mm}  \usepackage{paralist} \usepackage{tabto}  \begin{document}  \NumTabs{3} \begin{inparaenum} \item text \tab\item text \tab\item text \tab\item text \tab\item text \tab\item text    \end{inparaenum}  \end{document}   Which results in:    It does exactly what I want, but I was wondering whether it was possible to create an environment, say tabbedEnum so that we can eliminate the use of the \tab, in particular, that the first case doesn't need a tab but all others do. Something of the form:   \begin{tabbedEnum}[3] \item text \item text \item text \item text \item text \item text \end{tabbedEnum} Is there a reason why people can't just take the helicopter to mount Everest? Why is it that helicopters can't reach that high? Show that $g\circ f\equiv 0$ implies $\det Df\equiv 0$, where $g(x_1,...,x_n)=x_1^5+...+x_n^5$ Let $f:\mathbb{R}^n \rightarrow \mathbb{R}^n$ be a smooth function and let $g:\mathbb{R}^n \rightarrow \mathbb{R}$ be defined by $g(x_1,...,x_n)=x_1^5+...+x_n^5$. Suppose $g\circ f\equiv 0$. Show that $\det Df\equiv 0$.  I was going to start the solution from $n=2$ and then use induction on $n$.  So, for $n=2$, $f:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ is a smooth function and $g:\mathbb{R}^2 \rightarrow \mathbb{R}$ where $g(x_1,x_2)=x_1^5+x_2^5$. Then $Df=\begin{pmatrix}   f_{x_1x_1} &amp; f_{x_1x_2}  \\   f_{x_2x_1} &amp; f_{x_2x_2}   \end{pmatrix}$  But now I am stuck in proving the result. Am I in the correct track? Can anyone plese give me a hint to continue?  Ok so since $g\circ f$ is $0$, the image of $f$ is the line $x_1+x_2=0$ which has no min, max or saddle.   If det$Df(x_0)&gt;0$ and $f_{x_1x_1}(x_0)&gt;0$ OR det$Df(x_0)&gt;0$ and $f_{x_1x_1}(x_0)&lt;0$, we have contradiction.   But if det$Df(x_0)&gt;0$ and $f_{x_1x_1}(x_0)=0$, then det$Df(x_0)=0$ or, det$Df(x_0)&lt;0$(Since $f_{x_1x_2}=f_{x_2x_1}$). So this is impossible.  If det$Df(x_0)&lt;0$ we have saddle and it is impossible.   So finally we conclude that det$Df(x)=0$ for all $x$. Am I correct? linear dependence proof using subsets Word for something sad and funny at the same time What's a word for conflicting emotions that are felt at the same time? For example, loneliness, sadness, happiness, being content but also feeling as if something isn't right?  To sum up: What's a word for something that is sad and funny at the same time? Word for a modified version of a quote Theoretically if I were able to build some sort of device that let me sit 1 foot away from the surface of the Sun (or any star for that matter) without being vaporized, would a star produce any sort of audible sound at that distance?   From my understanding, sound waves need a medium to propagate through (air, water, etc).   Does this mean that all stars are completely silent (in the audible spectrum of frequencies at least) as they fuse  hydrogen into helium?   I know stars emit radio, xray, infrared, ultraviolet electromagnetic waves, but how about audible (20 Hz to 20Khz) frequencies?   For some reason it seems like this process would produce some sort of sound. Ways to get free and legal access to research papers as a researcher I’ve discovered that website like ResearchGate offer the possibility to ask the paper editor to provide his paper for free. Many are collaborating.  Are there any other ways for researchers to obtain free access to papers? Please give me feedback on my answer to this question. Question: For all $ n\geq1:\binom{2n}{0}+\binom{2n}{2}+\binom{2n}{4}+\cdots+\binom{2n}{2k}+\cdots+\binom{2n}{2n} $   is equal to $ \binom{2n}{1}+\binom{2n}{3}+\binom{2n}{5}+\cdots+\binom{2n}{2k+1}+\cdots+\binom{2n}{2n-1} $. Answer: False, since;  Let $a=1,  b=-1$  and $n=1$  ,$k=0$   , using Binomial expression.  $$(a+b)^n=  \sum   \binom{n}{k}  a^{k}b^{n-k}$$  Then $(1+-1)^{2(1)}=\binom{2}{0}1^{0}(-1)^{2-0}+\binom{2}{1}1^{1}(-1)^{2-1}+\binom{2}{2}1^{2}(-1)^{2-2}=\binom{2}{0}-\binom{2}{1}+\binom{2}{2}$  Thus :For all $n\geq1:  \binom{2n}{0}+\binom{2n}{2}+\binom{2n}{4}+\cdots+\binom{2n}{2k}+\cdots+\binom{2n}{2n}   \ne \binom{2n}{1}+\binom{2n}{3}+\binom{2n}{5}+\cdots+\binom{2n}{2k+1}+\cdots+\binom{2n}{2n-1}$  . Why do you add +1 in counting test questions? What's the process to change a site logo? Who is responsible for defining a site logo? I mean, there are moderators of the site that define it, or stackexchange defines it? How to install Ubuntu on portable external Hard Drive? Assume a p x n data matrix  (e.g. p independent samples of n variables). The n variables are correlated according to a correlation matrix $\rho$. Now if you consider the data matrix as some state t (t=time)  and based on this state I would like to calculate the next timestep (t+1). Therefore the correlation matrix would be allowed to have values &lt; 1 in the diagonal, which means that the diagonal elements would indicate the influence from each variable $x_{p,i}[t]$ on $x_{p,i}[t+1]$. So I would be interested to find a function f like this: $x_{p,i}[t+1] = f(x_{p,i}[t],\rho)$ so that the p x n data matrix  consisting of the $x_{p,i}[t+1]$ replicates the $\rho_{ij}$ (except the diagonal element). f should be something like this: $x_{i,j}[t+1] = \sum_{(k=1\ldots n)}\rho_{i,k} \cdot x_{kj}[t] + e_{i,j}.$ This means specifically, I not only want to calculate a data matrix, that replicates the correlation matrix as such, but also to let each observation be depending on its own state and linear dependent on the previous state of all the intercorrelated variables of its own state. For ease one can assume that the p variables are distributed normal and $e_{ij}$ are independent and distributed with mean$(e_{i,j}) = 0 $. Then for a numerical simulation it would be interesting  to find a formula to calculate the variance/standard deviation of each of the j variables (sd($e_{[.,j]} = ?).$ Thank you for any help! Virtualbox Windows 10 Guest installation is freezing up in 14.04 I've installed VirtualBox using the Software Centre and I'm running a system on Ubuntu 14.04 with 8gb RAM, 64-bit dual core processor.  I want to run Windows 10, I've tried both 64 and 32 bit versions, in virtualbox. However, every time I start up windows 10 it gets to the point where the dots go round in a circle, then it freezes up and crashes back to the virtualbox manager. The only information it gives me is that the virtual machine was aborted.  I've changed the settings around a lot and nothing is working.  Any help would be appreciated.  Oh, and I've run Windows 10 technical preview on this machine before, a while ago  Thanks for any help in advance. Why two inverters cascaded in multiplexer control signal terminals? Why cascaded inverters are used in the multiplexer when it is possible to realize the circuit using a single inverter. Macro for figure position? Why is the following minimal example not working?  I.e., what do I need to do, in order to define the float placement specifier of a figure within a macro? (I know that I should use \newcommand, but in the "full" example the error is part of an involved "superfigure" command. \newcommand is not working too btw.)  \documentclass{scrreprt} \usepackage{blindtext}  \def\PosNotWorking{h} \def\FramboxNotWorking{5cm} \def\CaptionWorking{Caption is working.}  \begin{document}  \Blindtext[1]  \begin{figure}[\PosNotWorking]   \centering     \framebox[5cm]{My special figure. (Pos: \PosNotWorking)}             \caption{\CaptionWorking} \end{figure}  \Blindtext[1]  \end{document}   If I compile this, the figure is at the top (default behaviour). If I enter \figure[h] instead of \figure[\PosNotWorking], the figure is of course between the two paragraphs. What is hard reset and what is factory reset? It is mentioned at  that one should do Factory reset or hard reset. I'm a bit confused about what is hard reset?  Factory reset I suppose is using the option in settings to reset phone to the factory state i.e. when it was unboxed. When applying to a PhD program in the US, how does the admissions process work? If an applicant is weak in a particular area, is it possible to offset that by being strong in a different area?  Note that this question originated from this . Please feel free to edit the question to improve it.