Showing $f/g$ is identically constant
Show that $f(z) = cg(z)$ is valid for two analytic functions $f$ and $g$ given a certain relationship between them
Mean absolute error OR root mean squared error? Why use Root Mean Squared Error (RMSE) instead of Mean Absolute Error (MAE)??    Hi  I've been investigating the error generated in a calculation - I initially calculated the error as a Root Mean Normalised Squared Error.  Looking a little closer, I see the effects of squaring the error gives more weight to larger errors than smaller ones, skewing the error estimate towards the odd outlier. This is quite obvious in retrospect.   So my question - in what instance would the Root Mean Squared Error be a more appropriate measure of error than the Mean Absolute Error? The latter seems more appropriate to me or am I missing something?  To illustrate this I have attached an example below:                    The scatter plot shows two variables with a good correlation,  the two histograms to the right chart the error between Y(observed ) and Y(predicted) using normalised RMSE (top) and MAE (bottom).     There are no significant outliers in this data and MAE gives a lower error than RMSE. Is there any rational, other than MAE being preferable, for using one measure of error over the other? In the movie, there's one scene that seems very strange to me.  First, the point:  When the team gets in the first level, they find out that Fischer has a militarized subconscious. After they freak out, and argue about it, they also find out that they can't die, otherwise they'd get stuck in limbo.  So, when Eames says that he won't go further, Cobb states that, since the first level sedative will last a week they would be killed. They needed to go on to the next levels as fast as possible, so the combined multi-level kick would make them wake up.  Here's the point: The plan was the multi-level kick would wake them up before the sedative was off     Ariadne: ...is ten years! Who would wanna be stuck in a dream for ten years?   Yusuf: Depends on the dream.    Arthur: So, once we've made the plant, how do we go out? Hope you have something more elegant in mind than shooting me in the head?    Cobb: A kick.    ...   Arthur: Are we going to feel a kick with this kind of sedation?   Yusuf: Well, that's the clever part. I customize the sedative to leave inner ear function unimpaired. That way, however deep the sleep, the sleeper still feels falling, or tipping.     Now, the question:  When they finish the job, they perform the kicks   The van hits the water Arthur detonates the elevator Eames explodes the fortress   But they don't wake up. They are back in the drowned van. So, as far as I could understand, they still had to sit and wait for the first level's week to end.  (We see Fischer talking to his godfather)  Why didn't they wake up with the kicks, as it was planned? And how did they managed not to get killed by Fischer's projections after they got out of the Van? As you can see, the off-topic page isn&#39;t quite right...    Could this (minor) issue be fixed? $g^\frac{p-1}{2} \equiv -1 \ (mod \ p)$ Let $p$ be an odd prime, and let $g$ be a primitive root modulo $p$. Prove that $g^\frac{p-1}{2} \equiv -1 \ (mod \ p)$.  I've seen this solution on the math stack exchange but most of the solutions use quadratic reciprocity. Unfortunately, I skipped that section because it seemed very hard, but the proof that  $g^\frac{p-1}{2} \equiv -1 \ (mod \ p)$ seems easy with it.  Is there another way to solve this This question is proposed in the chapter of Finite Abelian Groups.  DEF:In elementary number theory, an integer $g$ is called a primitive root for the modulus $n$ if $\mathbb{Z}_n^\times$ is a cyclic group and $[g]_n$, is a generator for $\mathbb{Z}_n^\times$. Someone edited my message on StackOverflow, but it really bugs me out.  I'm not sure what's wrong with it:   As you see, the bigger the circle becomes, the more vertices I need for hiding the straight lines. As you see, the bigger the circle becomes, the more vertices I need to hide the straight lines.   The latter (corrected) sentence just bugs me out for some reason. It would sound better if there was a comma after the word "need", perhaps, but I guess that would be even worse in means of grammatical correctness.  Can someone explain why is the first sentence bad and the second sentence correct? And could there even be better solution to fix the first sentence? Blender python custom UI button run custom function? Recently, I accidentally deleted a partition off of my external hard disk. I had a copy of Minitool Power Data Recovery left over when the same drive had become RAW about a year ago. When I selected lost partition recovery, it immediately identified my deleted partition and allowed me to access and recover the files that were there.  My question is this. Is there any way to change the deleted partition back to NTFS without having to have a drive that you need to recover the files to and then move them back after formatting, as I don’t have any space left on either the computer or on my other external hard drive, so that I don’t need to buy a new one. The files on the disk were really important.  I am using Windows 7 Home Premium.  I appreciate any help in solving this crisis. A type of the body is JSON Blob in a form of:  {"key1":"val1";"key2":"val2"}  I wanted to convert it into a Map&lt;String, String&gt;. Do I have to manually deserialize it or there's a class which can handle that? Impossibility of constructing a continuum-size linearly independent set in $\Bbb R$ This is a response to the following exchange at    [Bill constructs a $\aleph_0$ linearly independent set in $\Bbb R$]  Can you extend the idea of this proof to get the right dimension? Currently, you only have a countably infinite independent set, but the dimension is size continuum. –   I believe the answer to this question is no, for the following reason. Suppose we have a spanning set for $\Bbb R$ over $\Bbb Q$, of cardinality $\frak n$. Then by definition we have the cardinality equation ${\frak c}\le^*[{\frak n}]^{&lt;\omega}\cdot \aleph_0$ (where $[{\frak n}]^{&lt;\omega}$ is the cardinality of finite subsets, and $\frak m\le^*n$ means there is a surjection from $\frak n$ to $\frak m$). If $\frak n$ is finite, then we get ${\frak c}\le^*\aleph_0$ which is false, so $\frak n$ is infinite (but not necessarily Dedekind-infinite). If we assume the axiom of choice, then the equation reduces to $\frak c\le n$, hence every Hamel basis is size continuum. But without AC I don't know any way to simplify the original equation to get anything stronger than &quot;$\frak n$ is infinite&quot;. This is at best heuristic evidence for the original question, since it is considering a different problem (spanning sets instead of linearly independent sets). How could one show that there are no ZF-constructible $\Bbb Q$-linearly independent subsets of $\Bbb R$ of cardinality continuum? I would like to count subgroups of each order (2, 3, 4, 6, 8, 12) of $S_4$, and, hopefully, convince others that I counted them correctly.  In order to do this by hand in the term exam, I need a clever way to do this because there can be as many subgroups of a group of order 24 as $2^{23}$.  Do you know how to do this?   (I would be most grateful if you could tell me what part of the answer to the old question answers my question before voting to close.) amsmath align environment: row spacing How can I stretch the spacing between consecutive lines in an align* environment? I have a few lines with \fracs in which both the numerator and the denominator consist of textual phrases, not math symbols, and the result looks too cluttered.  \begin{align*} X &amp;= \frac{\textit{Some text here}}{\textit{And some text here}}\\ Y &amp;= \frac{\textit{Will make everything}}{\textit{Look too close together}} \end{align*} I'm trying to run down a memory leak in a windows forms application.  I'm looking now at a form which contains several embedded forms.  What worries me is that the child forms, in their constructor, take a reference to the parent form, and keep it in a private member field.  So it seems to me that come garbage-collection time:  Parent has a reference to the child form, via the controls collection (child form is embedded there).  Child form is not GC'd.  Child form has a reference to the parent form, via the private member field.  Parent form is not GC'd.   Is this an accurate understanding of how the garbage collector will evaluate the situation?  Any way to 'prove' it for testing purposes? I have been around on Twitter for awhile and I regularly find people using "Please to retweet this" or "Please to help me", etc. Is this proper English? I do not think prefixing the infinitive form with a 'to' is necessary here at all. I want to know how EXACTLY the WPA password gets encrypted  I have been searching through and I was able to catch:   The actual password is converted to some hash Salting is applied to the hash The operation is not reversible The hash comes with the 4-way handshake   So what kind of hash is it converted? like, SHA1? MD5? and how the salting process is done? like, is the SSID appended to the password before the hashing operation?  Or maybe I have some misunderstanding here? This question is pretty simple: I'd like to have the possibility to get the exact size of a picture in LaTeX with, for example, the macro \the.  In fact, I've founded the way \settoheight{} and so on, but it didn't work at all. What I've got doesn't match with what I can see in the image properties (I suppose that's the name in English). The result: 8pt versus 1300 pixels. It's may be a mis-use...  Edit: In the end, my aim was to find a way to fit the picture in LaTeX. If width and height were both less than the dimensions of the body, then it should just be centred. It look like this :     Require:   \usepackage{graphicx} \usepackage{xifthen} \usepackage{calc}   Code (pretty ugly, I agree):   % Centrering+fitting for image, with captation and label (not auto) \newlength\imageheight \newlength\imagewidth \newcommand{\pic}[3]{% \settoheight\imageheight{\includegraphics{pictures/#1.png}}% \settowidth\imagewidth{\includegraphics{pictures/#1.png}}% \ifthenelse{\lengthtest{\imagewidth &gt; \textwidth}}{\ifthenelse{\lengthtest{\imageheight &gt; \imagewidth}}{\par\vspace{0.95em}% \begin{figure}     \centering\includegraphics[height=0.85\textheight]{pictures/#1.png}     \caption{#2}     \label{#3} \end{figure}% \par\vspace{0.95em}}{\par\vspace{0.95em}% \begin{figure}     \centering\includegraphics[width=0.95\textwidth]{pictures/#1.png}     \caption{#2}     \label{#3} \end{figure}% \par\vspace{0.95em}}}{\ifthenelse{\lengthtest{\imageheight &gt; \textheight}}{\par\vspace{0.95em}% \begin{figure}     \centering\includegraphics[height=0.85\textheight]{pictures/#1.png}     \caption{#2}     \label{#3} \end{figure}% \par\vspace{0.95em}}{\par\vspace{0.95em}% \begin{figure}     \centering\includegraphics{pictures/#1.png}     \caption{#2}     \label{#3} \end{figure}% \par\vspace{0.95em}}}} %   To use it : put your png files in a "pictures" directory, then, just simply call the picture like this : \pic{name_whitout_extension}{Captation_if_needed}{Label_if_needed}.    Edit 2 :   Thanks to Martin Scharrer's easyfig package, I've got a more or less 1,6 times faster solution to put my pictures, centred, with the perfect size.  CTAN:   VC:   Version v1.2 – 2012/05/15    This could be useful for some others, I hope. $f,\overline f$ are both analytic in a domain $\Omega$ then $f$ is constant? is it true if $f,\overline f$ are both analytic in a domain $\Omega$ then $f$ is constant? I am not able to  find out what property of holomorphic map I need to apply. please help.Thank you.  $f(z)=u(x,y)+iv(x,y)$, $\bar{f}(z)=u(x,y)-iv(x,y)$ Constructing a circle through a given point, tangent to a given line, and tangent to a given circle While browsing around about problems similar to , I have found references to constructions of all types of circles. For example, not only is it possible to construct a circle tangent to three given circles, but one can construct a circle through any three points, tangent to any three lines, passing through two given points and tangent to a line or circle, passing through a given point and tangent to two given lines or circles, etc. Pretty much any combination of criteria regarding points, lines, and circles, with repetition.  Can we have one of each? I haven't found any resource saying whether it's possible or not to construct a circle through a given point, tangent to a given line, and tangent to a given circle. Is such a construction possible? Thanks. Is there a word for describing people who know many things but superficially? Is there an appropriate word for the antonym of "full-stack", as in programming? Someone who knows a lot of things just as a full-stack programmer, but without mastering them. Reload the page on hitting back button I was working on a tower and painted my own texture for it. I saved halfway though and took a break. When I returned to my little project, I could paint on the image and it shows up in the material view but not the render. I did everything in my power but still couldn't understand why?  I have used:   Nodes, including mapping and insert UV coordinates. Unwrap UVs Used lights Deleted and replace nodes Deleted and replace Mat Removed image and replace it Reload image Painted on mesh   Even though use blender, I am not a programmer. So I find it really helpful if someone could give me an answer.