I want to have an option to capture wide-angle shots. Since the wide-angle lens is expensive, I want to give a try to 0.4x wide-angle converter lenses. I found one such lens here on .  QUESTION: I found that all the lens converters have a specification that they can be mounted on the lens with  filter diameter 52 mm, 58 mm and 67 mm. But I checked that my 18-55mm kit lens has a specification Ø55.      Is it possible for me to use the above mentioned wide-angle lens   converter with my kit lens?   My Camera: Nikon D5600
I'm a beginner at photography and own a Canon 20D with two lens; a 50mm prime and a 17mm-85mm zoom.  I've purchases all my filters at 67mm to fit my zoom lens and use a  to use them with my prime lens.  I'm looking at buying a . My question is, will the same principal work? Can I buy a 67mm wide angle lens and use it on my 52mm prime without any trouble.  One of the reasons for the purchase of a wide angle is my cropped sensor makes the 50mm prime rather close - I'm hoping the wide angle will make it easier to capture more.  Many thanks for any help in advance. Pete
Where is the .Trash folder? I no longer need an answer to a question; what should I do to the question? nginx httpMapModule regex variables The  states that:     Since 1.0.4, case insensitive regular expressions can be used by prefixing the pattern with ~*.   And gives the example:  map $uri $myvalue {     /aa                   /mapped_aa;     ~^/aa/(?&lt;suffix&gt;.*)$  $suffix; }   I updated to nginx/1.2.7 to make use of this, and naive regexes work okay, but variables don't. When I try the above syntax I get the error:  nginx: [emerg] pcre_compile() failed: unrecognized character after (?&lt; in "^/aa/(?&lt;suffix&gt;.*)$" at "suffix&gt;.*)$"   I also tried the following syntaxes, but none of them worked:  ~^(.*)$  $1; ~^(.*)$  \1;   but neither of them worked either.  Does anyone know how to get variables to work in regular expressions in HttpMapModule? In what occasion could the word “precious” be taken disapprovingly, or sarcastically? I understand that the word “precious” in the following quote of Maureen Dowd’s article “Heart of Darkness” (NYT, March 20) literally means “2. valuable or important and not to be wasted” as defined in OALD:     Congressman Jones read an e-mail from a former boss of General   Allen’s, giving the congressman this unvarnished assessment:   “Attempting to find a true military and political answer to the   problems in Afghanistan would take decades. Would drain our nation of   precious resources, with the most precious being our sons and daughters. Simply put, the United States cannot solve the Afghan   problem, no matter how brave and determined our troops are.   But I didn’t know “precious” has another meaning as defined in OALD: 5. (disapproving, people and their behavior) very formal, exaggerated and not natural, and Readers English Dictionary: 2.(colloquial) terrible, almost worthless, until I was told by my respected forum mate a few days ago.   Being encouraged by his suggestion for me to “ask another question if I’m not sure of what he means by describing his input as “precious” is actually rather “precious” phrasing, I would like to ask:   What is an example of “precious” being used as “disapproving” or sarcastic way (as neither OALD nor CALD provides examples)? Is there an easy way to discern the instance “precious” being used as “disapproving” from the case being used in the sense of genuine “great value” as used in the above quote other than judging from the context? I look today on a few fresh question, and the view count was 0. Even after I refreshed the page a few time. This is definitely a bug, since the question isn't mine.  Also I don't think the all page was cached, since that the DateTime mark was change every reload.  But maybe the SQL result is cached. How do you keep user.config settings across different assembly versions in .net? Strange black shading. Cannot fully light mesh. (Cycles) Is there a constructive way to exhibit a basis for $\mathbb{R}^\mathbb{N}$? UTF-8 -> ASCII in C language How to prove that $\lim \limits_{n\rightarrow \infty} \frac{F_{n+1}}{F_n}=\frac{\sqrt{5}+1}{2}$ How would one prove that $$\lim_{n\rightarrow \infty} \frac{F_{n+1}}{F_n}=\frac{\sqrt{5}+1}{2}=\varphi$$  where $F_n$ is the nth Fibonacci number and $\varphi$ is the Golden Ratio? Cube of an integer $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=k$ and $x, y, z, k$ are integers.  Prove that $xyz$ is cube of some integer number.    I was wondering about giving a parametrization for the rational points on the elliptic curve in $\mathbb{P}^2(\mathbb{R})$ $$ x^2 y + y^2 z + z^2 x = k xyz$$ but I am not too confident in the topic, so I am asking you to shed some light.  Thank you for your help. About $\lim \left(1+\frac {x}{n}\right)^n$ When to use "Well" or "Good" The general advise is that you should not call GC.Collect from your code, but what are the exceptions to this rule?  I can only think of a few very specific cases where it may make sense to force a garbage collection.   One example that springs to mind is a service, that wakes up at intervals, performs some task, and then sleeps for a long time. In this case, it may be a good idea to force a collect to prevent the soon-to-be-idle process from holding on to more memory than needed.   Are there any other cases where it is acceptable to call GC.Collect? Quick question about curly braces not showing up Ii am writing a short code, so I doubt anything else in the code is causing the problem. This is the problematic line of code:  \[ M_R := \max {|f(z)|: |z - z_0| = R }\le M\]   It doesn't show the curly braces. I've tried using \left{ and \right} but to no avail.  What should I do?  Thank you integral of $\int_{0}^{1}\frac{\ln(x^{2}+1)}{x+1}dx$ using contour integration? I have an interest in contour integration. I am not that good at it, but I enjoy learning what I can about it.  Here is a version of a rational log integral rarely encountered.  $\displaystyle \int_{0}^{1}\frac{\ln(x^{2}+1)}{x+1}dx=\frac{3}{4}\ln^{2}(2)-\frac{{\pi}^{2}}{48}$  I can do this using real methods (via double integral and substitution). I can post my workings if anyone would be interested.  My question is, can this be evaluated using contour integration due to the limits being $[0,1]$ instead of $[0,\infty)$?.  Contours may not be the most efficient way to go about it, but what is the course of action when the limits are 0 to 1 instead of 0 to infinity?. I'm working on an exercise from functional analysis.  Let $E$ be a vector space and $\|\cdot\|_1$ and $\|\cdot\|_2$ be two complete norms on $E$. Now suppose that $E$ satisfies the following property:  $\bullet$ if $(x_n)$ is a sequence in $E$ and $x,y\in E$ such that $\|x_n-x\|_1\to 0$ and $\|x_n-y\|_2\to 0$, then $x=y$.  Now we want to show that the norms $\|\cdot\|_1$ and $\|\cdot\|_2$ are equivalent.  My idea is as follows: If for any $n&gt;0$, there is an element $x_n\in E$ such that $\|x_n\|_1&gt;n\|x_n\|_2$. Then consider $(\frac{x_n}{\|x_n\|_1})_{n\geq 1}$. Clearly, $(\frac{x_n}{\|x_n\|_1})_{n\geq 1}$ converges to $0$. However, I cann't get a contradicition from this. Maybe my idea is wrong.   In fact, I even don't konw how to show that a Cauchy sequence in norm $\|\cdot\|_1$ is also a Cauchy sequence in norm $\|\cdot\|_2$.   Anyone can give me some hints or a counter example? Thank you very much. came in Low Quality Posts review which was  as "This is commentary on another post, not an answer" by *. But the comment was added 5 times automatically on behalf of that user.        First four comments' time is 2013-07-15 10:23:18Z while the last one's time is 2013-07-15 10:23:19Z. Those comments are still exist. 10K+ users can see that.  Is this a bug or something else?  * Because the comment is added on behalf of that user. Why do you need to know who I know on Google? I just created a SE account using my Google account. It said that it needs to "Know who you are on Google" (I guess my connections). As far as I know, SE is not a social network. I can't even follow other people (seeing when they post questions and answers).  What is the reason you need to know my Google connections?  This was on code review. Actually I have an account there, but for some reason it asked me to reconnect. This is the same on MSO; one sees this explained as "This app is requesting permission to associate you with your public Google profile": The boundary of union is the union of boundaries when the sets have disjoint closures