Why is this Iris not showing? It worked previously, before I started working on the fur and particle system today and other parts of the model (a monkey model) and at the end I realised that the Iris disappeared. I haven't touched the eyeball, but now the iris doesn't show. I suspect that it has to do with the UVs, but everything is in place, I didn't tweak anything, it should work. I'm attaching a stripped down .blend (only the eyeball) and a screenshot. Can someone have a look into the file, and try to make the iris show?  I should also note, that when I open the previous file, it shows. But when I append that eyeball to the new file, it doesn't show.
My model has a clear plastic (or glass) dome with objects inside.  The objects appear dark.  I've tried the glass shader with color 100% RGB, with very small IOR, and using just a refraction shader.    I know Cycles has problems with caustics, but I just want broad illumination to fall clean through the glass.  The light-to-column path can ignore refraction, but is blocked or tremendously attenuated. The camera-column path needs refraction, and seems to work fine anyway.  This demo image offers all optical paths of interest:   light upon object, seen directly light upon object, seen through glass (looks okay) light through glass onto object, seen directly. This is the shadow on the lit side of the column just below the glass tubing.  I expect full illumination, but there is shadow.  There is some light, but it's diffuse light from the floor. light through glass onto object, seen through glass. This     I did at one point try non-progressive rendering, and find extremely coarse light where there shouldn't be shadow, but this is not a practical way to make nice images. I've even tried cranking up certain sampling settings to many thousands, but still get bad noise though less, at the cost of long render times. I  would like to finish my creations sometime before "Star Trek, the Generation After the Next" becomes real life, if you know what I mean.  The next image is an example. I think Transmission in Sampling was 5000, but not sure.    I'm using Cycles in Blender 2.67 on a quad core Linux machine, 16GB RAM. Machine limits have never been a problem for 3D rendering.  Possibly, I screwed up some setting when I made my startup file, and just need to fix it. OTOH maybe this is just a limitation of path tracing, or of the way Cycles does it. If that's the case, what to real-world Blender artists do when they need to produce images on a budget and schedule?  What is the practical real-world fix?
How do I prove/disprove that $F_n^2 + F_{n\pm 1}^2 = F_{2n\pm 1}$, where $F_n$ is the $n$th Fibonacci number? I was looking at the Fibonacci sequence:  $$1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144,\ldots, F_n : F_n + F_{n - 1} = F_{n + 1} \land F_0 = 0$$  Here I noticed something; $\qquad F_n^2 + F_{n\pm1}^2 = F_{2n\pm1} : n\in \mathbb{N}$  For example; $$\ \ F_3^2 + F_{3 + 1}^2 = F_3^2 + F_4^2 = 2^2 + 3^2 = F_{2\times3 + 1} = F_{6 + 1} = F_7 = 13$$ $$F_3^2 + F_{3 - 1} = F_3^2 + F_2^2 = 2^2 + 1^2 = F_{2\times3 - 1} = F_{6 - 1} = F_5 = 5$$ However, no matter how true this can seem, I cannot seem to prove/disprove it.    My Attempt: $$\because (a + b)^2 = (a + b)(a + b) = a^2 + ab + ba + b^2 = a^2 + b^2 + 2ab : (a,b)\in \mathbb{Z}$$ $$\implies F_n^2 + F_{n\pm 1}^2 = (F_n + F_{n\pm 1})^2 - 2F_nF_{n\pm 1}$$ $$\implies F_n^2 + F_{n - 1}^2 = F_{n + 1}^2 - 2F_nF_{n - 1} : F_n + F_{n - 1} = F_{n + 1}$$ $$\implies F_n^2 + F_{n + 1}^2 = F_{n + 2}^2 - 2F_nF_{n + 1} : F_n + F_{n + 1} = F_{n + 2} \iff F_n + F_{n - 1} = F_{n + 1}$$ $$\implies F_{2n\pm 1} + 2F_nF_{n\pm 1} = F_{n + 2\lor1}^2 : F_n^2 + F_{n\pm 1}^2 = F_{2n\pm 1} \tag{+2 if +, +1 if -}$$ $$\implies F_{2n\pm 1} + 2\sqrt{F_{2n\pm 1} - F_{n\pm 1}^2}\sqrt{F_{2n\pm 1} - F_n^2}= F_{n + 2\lor1}$$ $$\implies F_{2n\pm 1} + 2\sqrt{(F_{2n\pm 1} - F_{n\pm 1}^2)(F_{2n\pm 1} - F_n^2)} = F_{n + 2\lor1}^2$$ $$\implies F_{2n\pm 1} + 2\sqrt{F_{2n\pm 1}(F_{2n\pm 1} - F_n^2 - F_{n\pm 1}^2) - (F_nF_{n\pm 1})^2} = F_{n + 2\lor1}^2$$.  $$\because F_n^2 + F_{n\pm 1}^2 = F_{2n\pm 1}$$ $$\implies F_{2n\pm 1} + 2\sqrt{-(F_nF_{n\pm 1})^2} = F_{n + 2\lor1} \neq F_{2n\pm 1} + 2\sqrt{\big(-(F_nF_{n\pm 1})\big)^2}$$ $$\therefore F_{2n\pm 1} + 2F_nF_{n\pm 1}i = F_{n + 2\lor1} : i = \sqrt{-1}$$  I have looked over every step and cannot see where I am wrong. The only part that must be wrong is that:  $$F_{2n\pm 1} + 2F_nF_{n\pm 1}i = F_{n + 2\lor1} \iff F_n^2 + F_{n\pm1}^2 = F_{2n\pm 1}$$  I can seperate $F_{2n\pm 1}$ as $F_n^2 + F_{n\pm 1}^2$ (if true) so I turn the equation into:  $$F_n^2 + F_{n\pm 1}^2 + 2F_nF_{n\pm 1}i = F_{n + 2\lor1}$$ $$\therefore (F_n + F_{n\pm 1})\bigg(\frac{F_n}{F_{n\pm 1}} + \frac{F_{n\pm 1}}{F_n} + 2i\bigg) = F_{n + 2\lor1}$$ Now I know that the greater the value of $n$, one of the fractions in the above equation (depending on which sign is used in $F_{n\pm 1}$) will get closer to the golden ratio $φ = \frac{1 + \sqrt{5}}{2}$ but $i = \sqrt{-1}$ is not a "real" number, although $F_n$ is.  I then searched for other formulae for finding $F_n$ and came across the one below (being Binet's Formula):  $$F_n = \bigg\{\frac{φ^n - ψ^n}{φ - ψ} : ψ = φ - \frac{\sqrt{20}}{2} = \frac{1 - \sqrt{5}}{2} \ \land \ φ - ψ = \sqrt{5}\bigg\}$$  But I do not know what how to turn that into $F_{n\pm 1}$. Could you please show me where I went wrong (because I must have gone wrong somewhere), and also if you wish, could you please show me how to turn the above fraction into $F_{n\pm 1}$ (if possible).  I only used the $(a + b)^2$ expansion rule because my skill level in mathematics is not particularly high compared to most of the users on this site (especially when logging into the MSE and finding fancy symbols like $\nabla$ and $\partial$ and $\sum$ which all look Greek to me, however the only "fancy" thing I know is that $i = i^5 = \cdots = \sqrt{-1}$) so if you used any other techniques in trying to prove this formula, could you please show me step by step?  Much appreciated. Thank you in advance.    Edit: I believe this question is a duplicate of this other question $\longrightarrow$   and I can go here to serve as a little hint $\longrightarrow$ calculate the mean of trials for each subject in R Which groups have precisely two automorphisms Which groups $G$ have precisely two automorphisms, i.e., precisely one non-trivial automorphism?  Examples: $G= C_3, \mathbf{Z},\ldots$.  I think $G$ has to be abelian. In fact, we have $ \vert G\vert \geq 3$. Therefore, if $G$ is not abelian, we have at least two non-trivial inner automorphisms.  If we can show that $G$ is cyclic the above examples are all of them. PHP: How do I loop through every XML file in a directory? Find the value of $P(1)$ I'm given a inequality as such: $n! &gt; n^3$ Where n > 5,   I've done this so far:  BC: n = 6, 6! > 720 (Works)  IH: let n = k, we have that: $k! &gt; k^3$  IS: try n = k+1, (I'm told to only work from one side)  So I have (k+1)!, but I'm not sure where to go from here.  I've been told that writing out: $(k+1)! &gt; (k+1)^3$ is a fallacy, because I can't sub in k+1 into both sides, but rather prove from one side only.    Any Help on how to continue from here, would be much appreciated. How to assign permanent drive letters to removable USB hard drives I have scripts that expect specific drive letters for hard drives. Windows sometimes assigns the different drive letters (e.g. I expected my Maxtor drive to be E:, but it's now F:, therefore my scripts will fail), depending on the order of how the drives are plugged into my laptop.  In Windows 7, how do I assign a permanent drive letter to a drive? I've looked at the resumes of the type of candidates that get into computer science PhD programs at the top few universities (e.g. Stanford, MIT etc.) and a lot of them seem to have multiple publications (2-4) done during their undergraduate years. Is this representative of the entire pool? Or is this just a self-selected group that puts their resumes online.  I currently have one first-author workshop paper (5 pages), and I wonder how this might affect my chances at entry into these top universities. Most descriptions of SIGTERM and SIGKILL point out that SIGKILL,     In contrast to SIGTERM and SIGINT, [...] cannot be caught or ignored [...] ()   Is this the only difference between SIGTERM and SIGKILL? In particular, if a process does not install a SIGTERM handler, is there any difference at all between sending SIGTERM or SIGKILL to the process? Finding distance between consecutive points using ArcGIS Desktop? Customized prompt does not hold when a child bash is ran I have been learning bash, and while doing so I played around with the .bash_profile file to be able to put greeting messages etc. Later I just removed these changes. But after doing that my default configuration of prompt got lost: originally, the prompt was all in bold, and \w part was blue so that when I was in a deeply nested directory the prompt would not seem so entangled. I think that was the default for Ubuntu (correct me if i'm wrong), however like  I said it got lost and my prompt was non-bold and monochromatic resulting in a mess when I'm in long named directories.  So, last night I tried customizing my prompt in the fashion I mentioned above. It worked well after I added the code to .bash_profile  Code was like this:  export PS1="\[$(tput bold)\]\u@\h:\[$(tput sgr 0)\]\e[1;34m\w$ \e[m"   However if I try to run another bash shell in the current one, the prompt becomes unformatted:    Is that normal? Isn't the whole point of exporting a variable to make it accessible from all the child processes forked from that shell? Why it doesn't that apply to this situation? Mafia 3 no more damage to be dealt barclay mills I did most of the barclay mills missions and brought the damage down to 2550 but now I can find anything else to damage or steal, so the meter is stuck at 2550 and the objective says "find and destroy garbage barges" but when I click it, no waypoint.  Please help. Better HTTPS support for Stack Exchange sites So it looks like Stack Exchange now supports HTTPS (to some extent). Which is awesome! But there are a few problems, the main one being that some content is delivered over the CDN, which is plain HTTP. This causes browsers to complain about "unsecured content": TeX editor with line-spacing preferences A Question On Euler's Proof Of the Basel Problem I've studied the proof that Euler gave for the famous , and it would seem that while it is technically correct, he does not justify all of his steps properly. Namely, he assumes that   $$\frac{\sin(x)}{x}=\left(1-\frac{x}{\pi}\right)\left(1+\frac{x}{\pi}\right)\left(1-\frac{x}{2\pi}\right)\left(1+\frac{x}{2\pi}\right)\dots$$  simply because they have the same roots, which is really not a strong enough condition. How do you really show that the equality holds?  He then notices that if you use the above equality, and consider it against the Taylor expansion for $\frac{\sin(x)}{x}$, then you can equate the coefficients of the two infinite expansions at each order, and the result of the Basel problem follows. But how do you know that if you have two different expansions for a function, then their coefficients at each order must be equal?  I would really appreciate if someone could show me how to make these two intuitive, yet informal, steps rigorous. An infinite series plus a continued fraction by Ramanujan Here is a famous problem posed by Ramanujan     Show that $$\left(1 + \frac{1}{1\cdot 3} + \frac{1}{1\cdot 3\cdot 5} + \cdots\right) + \left(\cfrac{1}{1+}\cfrac{1}{1+}\cfrac{2}{1+}\cfrac{3}{1+}\cfrac{4}{1+\cdots}\right) = \sqrt{\frac{\pi e}{2}}$$   The first series seems vaguely familiar if we consider the function $$f(x) = x + \frac{x^{3}}{1\cdot 3} + \frac{x^{5}}{1\cdot 3\cdot 5} + \cdots$$ and note that $$f'(x) = 1 + xf(x)$$ so that $y = f(x)$ satisfies the differential equation $$\frac{dy}{dx} - xy = 1, y(0) = 0$$ The integrating factor here comes to be $e^{-x^{2}/2}$ so that $$ye^{-x^{2}/2} = \int_{0}^{x}e^{-t^{2}/2}\,dt$$ and hence $$f(x) = e^{x^{2}/2}\int_{0}^{x}e^{-t^{2}/2}\,dt$$ Thus the sum of the first series is $$f(1) = \sqrt{e}\int_{0}^{1}e^{-t^{2}/2}\,dt$$ But I have no idea about the continued fraction and still more I am not able to figure out how it would simplify to $\sqrt{\pi e/2}$ at the end.  Please provide any hints or suggestions.  Update: We have $$\begin{aligned}f(1) &amp;= \sqrt{e}\int_{0}^{1}e^{-t^{2}/2}\,dt = \sqrt{e}\int_{0}^{\infty}e^{-t^{2}/2}\,dt - \sqrt{e}\int_{1}^{\infty}e^{-t^{2}/2}\,dt\\ &amp;= \sqrt{\frac{\pi e}{2}} - \sqrt{e}\int_{1}^{\infty}e^{-t^{2}/2}\,dt\end{aligned}$$ and hence we finally need to establish $$\sqrt{e}\int_{1}^{\infty}e^{-t^{2}/2}\,dt = \cfrac{1}{1+}\cfrac{1}{1+}\cfrac{2}{1+}\cfrac{3}{1+}\cfrac{4}{1+\cdots}$$ On further searching in Ramanujan's Collected Papers I found the following formula $$\int_{0}^{a}e^{-x^{2}}\,dx = \frac{\sqrt{\pi}}{2} - \cfrac{e^{-a^{2}}}{2a+}\cfrac{1}{a+}\cfrac{2}{2a+}\cfrac{3}{a+}\cfrac{4}{2a+\cdots}$$ and it seems helpful here. But unfortunately proving this formula seems to be another big challenge for me. (I'm going to link to this Stackexchange post concerning Nullary Operations: )  In general an  is a function $f:S^n \to S$, where $n$ denotes the arity of the operation, and a  is a function $f:S^0 \to S$.  It is clear that $S^n=S \times S \times \cdots \times S$ denotes the Cartesian product of a set with itself ($n$ times), however $S^0$ makes little sense to me.  One might define $S^0=\emptyset$. At first glance, that seems like a good choice. However then a nullary operation makes no sense, since it must be a function $f: A \to B$ where $A$ and $B$ are two sets representing the domain and codomain of the function. And note a function is defined in terms of relation, where a relation is a subset of the Cartesian product $A \times B$. So when we think of $A=S^0=\emptyset$ and $B=S$, so that we have a nullary operation $f:\emptyset \to S$, what is a a subset of $\emptyset \times S$? What even is $\emptyset \times S$? By definition of the Cartesian product, \begin{equation} \emptyset \times S = \{(a,b):a \in \emptyset\textrm{ and }b \in S\}. \end{equation} But there exists no $a \in \emptyset$! Therefore there exist no ordered pairs in our Cartesian product! Therefore $\emptyset \times S = \emptyset !$. Remember a function is a relation which is a subset of the Cartesian product. Therefore our relation is the empty set, and therefore also $f=\emptyset$.    Edit: We got the definition sorted out in the comments. Thanks especially to Hayden and Christoff for their great help. In summary, we figured out that $S^0=\{()\}$, the set containing only one element, the "empty $n$-tuple". Then if taking $S=\{1,2,...,k\}$, then a nullary operation $f:S^0 \to S$ is a function, i.e. some subset of $S^0 \times S = \{()\} \times S$. Looking at $\{()\} \times S$, this equals $\{((),1),((),2),...,((),k)\}$. Then a function is a special subset of this, so that each element of the domain (and there is only one element, namely $()$) is paired with exactly one element of the codomain. Therefore a function would have the form $f=\{((),a)\}$ where $a$ is one of $1,2,...,k$. Thank you for your help. Why the statement "there exist at least one bound state for negative/attractive potential" doesn't hold for 3D case? Previously I thought this is a universal theorem, for one can prove it in the one dimensional case using variational principal.   However, today I'm doing a homework considering a potential like this:$$V(r)=-V_0\quad(r&lt;a)$$$$ V(r)=0\quad(r&gt;a)$$ and found that there is no bound state when $V_0a^2&lt;\pi^2\hbar^2/8m$.  So what's the condition that we have at least one bound state for 3D and 2D? Getting Started to Xetex for Devangari? Doubt regarding a limit which is related to MVT Let the function $f(x)$ be differentiable  and $f'(x)$ be continuous in $\left(-\infty,\infty \right)$ with $f'(2)=14$ then evaluate the limit  $$\lim_{x\to 0}\frac{f(2+\sin x)-f(2+x\cos x)}{x-\sin x}$$  My attempt:  $\lim_{x\to 0}\frac{f(2+\sin x)-f(2+x\cos x)}{x-\sin x}$  $\lim_{x\to 0}\left(\frac{f(2+\sin x)-f(2+x\cos x)}{\sin x-x\cos x}\right)$ $\frac{\sin x-x\cos x}{x-\sin x}$$=\left(f'(2)\right)(2)=28$  Is the method used here correct