I am trying to prove the following statement:  If $f(z)$ is entire, show that the family formed by all the functions $f(kz)$ with constant k is normal in the annulus $r_1 &lt; |z|&lt;r_2$ If and only if $f$ Is a polynomial.  I have been trying to write, unsuccessfully, a concise proof of this fact just using Arzela-Ascoli, Montel’s theorem, Marty’s theorem or results in this line.  Thank you for your help.
Suppose that $f : \mathbb{C} \to \mathbb{C}$ is entire. Let $\Omega = \{ z : \frac{1}{2} &lt; |z| &lt; 2\}$. Furthermore, suppose that $ \mathcal{F} = \{f(kz) : k \in \mathbb{C}\}$ is a normal family of analytic functions defined on $\Omega$; the family $\mathcal{F}$ is normal on $\Omega$ iff each sequence from $\mathcal{F}$ has a subsequence converging uniformly on compact subsets of $\Omega$. Prove that $f$ is a polynomial.  I'm studying for my qualifying exam in complex analysis, and this question appears on an old exam. Usually, when I want to show that some entire $f$ is a polynomial, I just try to show that there exists some $N$ so that $f^{(k)}(0) = 0$, all $k \ge N$, but I'm having trouble getting that strategy to work here. I've been looking at the sequence $f(\frac{z}{n})$ trying to make something work, but have not had any success.   I would prefer to receive a hint or two, and then I'll try to post my own solution. Thanks in advance!
Are there any official specifications regarding the torque for camera mounts? I've been wondering why I've never seen any official specifications of the maximum allowed torque that can be exerted upon lens mounts. It seems to me like a very useful specification, but I can't seem to find it from any manufacturer.  indicates that there might be no specifications for Nikon mounts. Are there really no manufacturer that publish these specification of their mounts? declare property as object? Consider a complete metric space $(X,d)$ and $T\colon X\to X$. Suppose there exists $n\in\mathbb{N}$ such that the n-th power of $T$ is $q$-contractive. Show that then $T$ has exactly one fixed point $\overline{x}\in X$.      The n-th power is defined inductively: $$ T^{n+1}(x):=T(T^{n}(x)), n\in\mathbb{N}. $$ And for a function $T\colon X\to X$, $q$-contractive means  $$ \exists 0\leq q&lt;1~\forall~x,y\in X: d(T(x),T(y))\leq q d(x,y). $$    Now to the proof.  I think I have to show, that $T$ is q-contractive, because then it follows with Banach, that $T$ does have exactly one fixed point. So I have to show, that there exists a $0\leq q&lt;1$ so that for all $x,y\in X$ it is $$ d(T(x),T(y))\leq q\cdot d(x,y). $$ And most likely I have to use the q-contractivity of $T^n$, i.e. that there exists a $0\leq q &lt;1$ so that for all $x,y\in X$ it is $$ d(T^n(x),T^n(y))\leq q\cdot d(x,y). $$  Can you help me? Possible Duplicate:          Consider the field extension $\mathbb{Q}(\sqrt2)$. I want to show that $\sqrt5 \notin \mathbb{Q}(\sqrt2)$. If this were not the case, then we could write $\sqrt5 = a + b\sqrt2$ for $a,b\in\mathbb{Q}$. However, I do not see the contradiction here. Is there a better/easier way to prove this? R script line numbers at error? How is arccos derived? Stupid question but I need to understand:  If $ \cos = \dfrac{\text{adjacent}}{\text{hypotenuse}} $  What is arcosine? $ \text{adjacent} \cdot \text{hypotenuse}$? Is this the same for arcsine and arctangent? I read this quite a while back, part of my parents library (they're in their 50s now) I think the book was probably published in the 70s or 80s, my copy was paperback. I don't recall the cover art, but I have vague recollections of a satellite on the cover and maybe blue with grid-lines below it?  I've made concerted efforts to find it again but to no avail. My parents don't remember it either, but mum's an avowed book-aholic and buys a lot of books, reads them once and forgets about them, so that's not so surprising.  The key parts I remember:   Both the Soviets and Americans were working on psychic techniques and technologies, with real and tangible results including Remote Viewing. MK-ULTRA was explicitly namedropped in relation to the American psychic program. The Soviets and Americans both had orbiting superweapons, One was a microwave laser (or something very similar) and the other used slivers of the brains of psychics to blow people's brains up in a wide area. A super-psychic (male) took offence to the state of the cold-war and took control over the orbital weapons by manipulating the onboard computers psychically. There's a gruesome description of a psychic weapon being tested on monkeys in a lab, the POV character throws up afterwards in horror. I think there might have been more than one POV character. There's an extended description of the psychic orbital weapon as experienced by people in its area of effect. Similar to the monkeys, it builds up, starting with alzheimer-esque hand-shaking and building up to seizures and screaming before their brain liquifies fully and either comes out the nose and ears or outright blows the skull open. (this is one of the more vivid recollections I have of the book) I remember that the super-psychic was working wholly alone and did most of his psychic stuff from motels halfway around the world from events. I believe the super-psychic also killed other psychics explicitly. The weapon satellites had code-names Both weapons were used on cities in the course of the story I vaguely recall an office building getting targeted specifically at one point. Is `if (condition = value)` the correct syntax for comparison? Subgroup of $\mathbb{R}$ either dense or has a least positive element? Windows Media Player 12 shows wrong cover art when playing How do JavaScript closures work? How would you explain JavaScript closures to someone with a knowledge of the concepts they consist of (for example functions, variables and the like), but does not understand closures themselves?  I have seen  given on Wikipedia, but unfortunately it did not help. Why use a prime number in hashCode? Live USB for installation does not boot, black error screen followed by visual artifacts on a purple screen I'm trying to install Ubuntu 18.04 desktop with Rufus on a computer with the following specs:  CPU: AMD Ryzen 5 2600 MOBO: ASUS Prime B450M-A/CSM RAM: G-Skill Ripjaws DDR4 F4-2400C15D-16GVB GPU: Nvidia GeForce GTX750 PSU: Apexgaming AG-650M SSHD: Seagate Firecuda Compute mATX: Rosewill SCM-01  Basically it boots from the USB (FAT32) and brings up a black error screen with words and numbers. Then it transitions to the purple splash screen with &quot;ubuntu&quot; and 5 dots, however the screen shows various visual glitches depending on which USB port I plug into. When trying to boot the computer, it boots to a black screen with the following error message: ACPI Error: Needed [Integer/String/Buffer], found [Region]    that shows what I mean. Overstaying Schengen Visa for 1 day I was in Germany last September, my visa said I had 14 days to stay in the Schengen area.  When I was leaving back for Egypt the passport control officer noticed that I was leaving on the 15th day. I didn't even realise that as I had written previously in my visa application that I was staying for 15 days and the embassy in Cairo had my complete airline ticket with specific dates.  Anyway the officer just filed a report on the incident but didn't stamp my passport with my exit date from the Schengen area.  Do any of you guys know if that would make it harder for me to be granted a visa in the future? Proof that any simple connected graph has at least 2 non-cut vertices. I'm trying to prove that any simple connected graph with at least $3$ vertices ($|V| \ge 3$) has at least $2$ vertices whose removal will not lead to the increment of number of components. In other words, there are at least $2$ vertices that are non-cut vertices.  Attempt at solution:   I tried to prove it using induction by the number of vertices: $n - t \ge 2$, where $n$ is the number of vertices our graph has, and $t$ is a number of cut vertices.   Base  Let $n =3$, then the graph will look like this:  $О-О-О$  Clearly, it has one cut vertex (the middle one). So, $3-1 \ge 2$ holds.  Or it can be a full graph, then $3 - 0 \ge 2$ also holds. Assumption  Let $n = k$, and the formula holds, $k - t &gt;= 2$ Step  Prove for $n = k + 1$ If we add one vertex to the graph, it could either be a cut vertex or non-cut vertex. In case it is a cut vertex, we have $$\begin{align*}&amp;(k+1) - (t+1) \ge 2\\ &amp;k + 1 - t - t\ge 2\\ &amp;k\ge 2\end{align*}$$   In case it is not a cut vertex...and I'm stuck here, because anything can happen. The number of cut vertices might increase, might decrease, or stay the same.  I'm sure there is better and shorter proof to this, and I really hope someone could help me. Why are 50mm Lenses Cheaper? Pressing Ctrl Alt NumPad 0 causes the camera to teleport over and align itself to the current view in the 3D viewport. However, it usually takes up only a small portion of the view, meaning that you have to move the camera back to fit the same view you had before. Is there a way to set the camera view to fill more of the 3D viewport when using this feature? Can't convert string into float/double How to get the filename without the extension from a path in Python? How to uncompress separated tgz files? I am receiving this error message when I'm trying to uncompress a .tgz archive:  $ tar -zxvf OVM_EL5U3_X86_EBIZ12.1.1_DB_VIS_PVM.tgz.1 gzip: stdin: not in gzip format tar: Child returned status 1 tar: Error is not recoverable: exiting now   As you can see, the file name ends with .tgz.1.  When the file name is .tgz.0 it will be uncompressed with no error by the tar -zxvf command.