In  there is an answer how to convert to TIFF.  So, I followed that process written on that answer:  ffmpeg -i input.avi -pix_fmt rgb24 output.tiff   Though the TIFF file converted then became lossless compressed TIFF files.  How can I convert it to uncompressed TIFF files?
I would like to use  with a video.  With ffmpeg I can export some (.jpeg) images from a video. Can I convert a .jpeg into a valid .tiff or export directly .tiff images from the video with ffmpeg?
When can an autonomous system be written using a Hamiltonian? Baffling Elevation Profile out of Profile Tool Plugin When using the Profile Tool plugin I get a graph that looks like the 2nd derivative of what I want. That is, it looks like the rate of change of the slope of the elevation profile. I just want the elevation profile! The data is present for good 1' contour lines. Why is the Profile Tool interpreting the data in such a bizarre way? What am I missing? And what are the Min and Max units? Are repository updates secure?  As a bear of little brain from the developer side, I cannot understand why the repository list is http://security.ubuntu.com and the other http(unsecured) sites listed in /etc/apt/sources.list.  Without a certificate chain match this appears as "ask any responder for a list of packages to update" instead of "ask the ubuntu.com site..."  Can any network choose to spoof the update sites, and is this a common practice to provide a locally cached and vetted copy? Play a short sound in iOS If $A$ an integral domain contains a field $K$ and $A$ over $K$ is a finite-dimensional vector space, then $A$ is a field. Prove: If $a\mid m$ and $b\mid m$ and $\gcd(a,b)=1$ then $ab\mid m$ On a series involving harmonic numbers Reading a bit about recent Monica Cellio related posts, I am afraid that I may be using a language that is a lot like English, but might not be the Stack-Exchange-approved version of English. For example, my English has had a certain gender-neutral pronoun in living use for centuries. That pronoun is used throughout the literature written in this version of English. It is used in the English I was taught as a grammar school student and later as a graduate student working on a master's thesis. I believe that this English is good enough to communicate about topics relevant to the various Stack Exchange sites.  The last thing I want from Stack Exchange is to be booted out because I slip up and accidentally use the language that has served me so well for a lifetime.  The topics discussed on Stack Overflow (I almost referred to it as a "sister site;" would that have been allowed? "mother site?" Good thing I didn't?) are often centered on languages which have formal descriptions.  Where is the formal description of the language that is to be used on Stack Exchange?     Strive not with a man without cause, if he have done thee no harm.      He shall die without instruction; and in the greatness of his folly he shall go astray. Prove the identity $\binom{2n+1}{0} + \binom{2n+1}{1} + \cdots + \binom{2n+1}{n} = 4^n$ I've worked out a proof, but I was wondering about alternate, possibly more elegant ways to prove the statement. This is my (hopefully correct) proof:  Starting from the identity $2^m = \sum_{k=0}^m \binom{m}{k}$ (easily derived from the binomial theorem), with $m = 2n$:  $2^{2n} = 4^n = \binom{2n}{0} + \binom{2n}{1} + \cdots + \binom{2n}{2n-1} + \binom{2n}{2n}$  Applying the property $\binom{m}{k} = \binom{m}{m-k}$ to the second half of the list of summands in RHS above:  $4^n = \binom{2n}{0} + \binom{2n}{1} + \cdots + \binom{2n}{n-1} + \binom{2n}{n} + \underbrace{\binom{2n}{n-1} +\cdots \binom{2n}{1} + \binom{2n}{0}}_{\binom{m}{k} = \binom{m}{m-k} \text{ has been applied}}$  Rearranging the above sum by alternately taking terms from the front and end of the summand list in RHS above (and introducing the term $\binom{2n}{-1} = 0$ at the beginning just to make explicit the pattern being developed):  $4^n = (\binom{2n}{-1} + \binom{2n}{0}) + (\binom{2n}{0} + \binom{2n}{1}) + \cdots + (\binom{2n}{n-1} + \binom{2n}{n})$  Finally, using the property $\binom{m}{k} + \binom{m}{k-1} = \binom{m+1}{k}$ on the paired summands, we get the desired result:  $4^n = \binom{2n+1}{0} + \binom{2n+1}{1} + \cdots + \binom{2n+1}{n}$ RX 590 IN Blender and Cycles Render I don't know why, but I cant render anything in CUDA and OptiX but OpenCL work fine. I cant use Cycles render because it's very laggy. I have newest  version of Blender and an RX 590 Red Dragon 8gb. Why does that happen?  How can I turn on my gpu? If $\cos x=\cos y$ and $\sin x=\sin y$ is $x=y$ or $x=2\pi,y=0$ or $x=0,y=2\pi$? $x,y\in[0,2\pi]$ I know that if $\sin x=\sin y$ then $x=n\pi+(-1)^ny$  and $\cos x=\cos y$ implies that $x=y+2k\pi$ or $x=-y+2l\pi$. Since the latter does not fit the form above $x=y+2k\pi$ which implies that $k=0,1,-1$. Hence, we have x=y or $x=2\pi,y=0$ or $x=0,y=2\pi$. I have to show by induction that this function is a multiple of 8. I have tried everything but I can only show that is multiple of 4, some hints? The function is $$5^{n+1}+2\cdot 3^n+1 \hspace{1cm}\forall n\ge 0$$, because it is a multiple of 8, you can say that$$5^{n+1}+2\cdot 3^n+1=8\cdot m \hspace{1cm}\forall m\in\mathbb{N}$$. Proof that every number ≥ $8$ can be represented by a sum of fives and threes. Can you check if my proof is right?  Theorem.  $\forall x\geq8, x$ can be represented by $5a + 3b$ where  $a,b \in \mathbb{N}$.  Base case(s): $x=8 = 3\cdot1 + 5\cdot1 \quad \checkmark\\ x=9 = 3\cdot3 + 5\cdot0 \quad \checkmark\\ x=10 = 3\cdot0 + 5\cdot2 \quad \checkmark$  Inductive step:  $n \in \mathbb{N}\\a_1 = 8, a_n = a_1 + (x-1)\cdot3\\ b_1 = 9, b_n = b_1 + (x-1)\cdot3 = a_1 +1 + (x-1) \cdot 3\\ c_1 = 10, c_n = c_1 + (x-1)\cdot3 = b_1 + 1 + (x-1) \cdot 3 = a_1 + 2 + (x-1) \cdot 3\\ \\ S = \{x\in\mathbb{N}: x \in a_{x} \lor x \in b_{x} \lor x \in c_{x}\}$  Basis stays true, because $8,9,10 \in S$  Lets assume that $x \in S$. That means $x \in a_{n} \lor x \in b_{n} \lor x \in c_{n}$.  If $x \in a_n$ then $x+1 \in b_x$,  If $x \in b_x$ then $x+1 \in c_x$,  If $x \in c_x$ then $x+1 \in a_x$.  I can't prove that but it's obvious. What do you think about this? Is there an official C# guideline for the order of items in terms of class structure?  Does it go:   Public Fields Private Fields Properties Constructors Methods ?   I'm curious if there is a hard and fast rule about the order of items? I'm kind of all over the place. I want to stick with a particular standard so I can do it everywhere.  The real problem is my more complex properties end up looking a lot like methods and they feel out of place at the top before the constructor.  Any tips/suggestions? I am trying to grab the capital letters of a couple of words and wrap them in span tags. I am using  for extract and wrapping purposes, but it's not outputting anything.  preg_replace("/[A-Z]/", "&lt;span class=\"initial\"&gt;$1&lt;/span&gt;", $str) Complete Installation Guide for Android SDK / ADT Bundle on Ubuntu What all needs to be done to install Android SDK / ADT Bundle on Ubuntu?  Each of the steps in the answer was found on some different forum as I encountered obstacles getting started with my Android SDK and posted questions in appropriate forums thereafter. So I thought this would be useful in that it is a one stop solution available, in case one ever needs to re-install Ubuntu and install the Android SDK / ADT Bundle from scratch! Raising a decimal to a power of decimal? The .net framework provides in the Math class a method for powering double. But by precision requirement I need to raise a decimal to a decimal power [ Pow(decimal a, decimal b) ]. Does the framework have such a function? Does anyone know of a library with this kind of function? Android app should send edits to edit queue if necessary rather than rejecting them Comparing Arrays of Objects in JavaScript Verify the identity: $\tan^{-1} x +\tan^{-1} (1/x) = \pi /2$ Verify the identity: $\tan^{-1} x + \tan^{-1} (1/x) = \frac\pi 2, x &gt; 0$  $$\alpha= \tan^{-1} x$$  $$\beta = \tan^{-1} (1/x)$$  $$\tan \alpha = x$$  $$\tan \beta = 1/x$$  $$\tan^{-1}[\tan(\alpha + \beta)]$$  $$\tan^{-1}\left [{\tan\alpha + \tan\beta\over 1 - \tan\alpha \tan\beta} \right]$$  $$\tan^{-1}\left[ {x + 1/x\over 1- x/x }\right]$$  $$\tan^{-1}\left[{x + (1/x)\over 0} \right]$$  I can't find out what I'm doing wrong..