why did the Doctor choose the face of Lobus Caecilius in the fires of Pompeii the doctor meets Lobus Caecilius     CAECILIUS: Who are you?       DOCTOR: I am Spartacus.       DONNA: And so am I.    CAECILIUS: Mister and Mrs Spartacus.       DOCTOR: Oh no, no, no. We're not, we're not married.       DONNA: We're not together.       CAECILIUS: Oh, then brother and sister? Yes, of course. You look very much alike.    then after saving him from the volcano  at Christmas 2013 the time of the doctor he regenerates into Lobus  face then in series 8 episode 1 Deep Breath he starts wondering why he choose this face     DOCTOR:Er, have you seen this face before?      BARNEY: No.      DOCTOR: Are you sure?      BARNEY: Sir, I have never seen that face.      DOCTOR: It's funny, because I'm sure that I have. You know, I never know where the faces come from. They just pop up. Zap. Faces like this one. Come on, look at it, have a look, come on, look, look, look.    (The Doctor pulls Barney over to look in the mirror on the ground.)       DOCTOR: Look, it's covered in lines. But I didn't do the frowning. Who frowned me this face? Do you ever look in the mirror and think I've seen that face before?      BARNEY: Yes.      DOCTOR: Really? When?      BARNEY: Well, every time I look in the mirror.      DOCTOR: Oh, yes, yes, yes. Fair enough. Good point. My face is fresh on, though.      BARNEY: Er    (Barney moves away from the nutter in the nightshirt.)      DOCTOR: Why this one? Why did I choose this face? It's like I'm trying to tell myself something. Like I'm trying to make a point. But what is so important that I can't just tell myself what I'm thinking?   then in series 9 episode 5 The Girl Who Died he starts to realize why he choose  this face     DOCTOR:I'm not supposed to. Oh. Oh!    (He looks at his reflection in the water and sees what he saw in the mirror back in Deep Breath.)      CLARA: What? What's wrong?       DOCTOR: My face.        DOCTOR [memory]: Who frowned me this face? Why this one? Why did I choose this face?       CLARA: Doctor, what's wrong with your face?       DOCTOR: I think I know why I chose it.       DOCTOR [memory]: It's like I'm trying to tell myself something.       DOCTOR: I think I know what I'm trying to say.       DONNA [memory]: (to Doctor 10) Just someone. Not the whole town. Just save someone.       DOCTOR 10 [memory]: Come with me.    (A memory of Lobus Caecilus cowering as Vesuvius erupts in Fires of Pompeii. Played by Peter Capaldi, if you'd forgotten.)      DOCTOR: I know where I got this face, and I know what it's for.       CLARA: Okay, what's it for?       DOCTOR: To remind me. To hold me to the mark. I'm the Doctor, and I save    people.    (He shouts at the sky - the Time Lords.)      DOCTOR: And if anyone happens to be listening, and you've got any kind of a problem with that, to hell with you!    which made me think why did the Doctor choose the face of Lobus Caecilius? was it like reminder to save people what was it
Have Peter Capaldi's previous acting roles in the Doctor Who universe ever been addressed? Similar to this question posted in August 2014:       Has the fact that  (as  in The Fires of Pompeii, and  in Torchwood’s Children of Earth season) yet been addressed/explained either in, or out of universe?
Prove that for any piecewise smooth curve it is possible to find the parametrisation Show that a ring is commutative if it has the property that ab = ca implies b = c when $a\neq 0$ Show that a ring is commutative if it has the property that ab = ca implies b = c when $a\neq 0$.   This is my proof to show that a ring is commutative if it has the property that ab = ca implies b = c.  We need to show that if x, y ∈ R then xy = yx. Let a = x, b = yx and c = xy. Then ab = x(yx) = (xy)x = ca. Thus, by the hypothesis, b = c, or xy = yx. Thus, for x, y ∈ R we have xy = yx.   How would I show when $a\neq 0$? I'm new to ArcGIS 10 and am trying to do a really simple field calculator equation and it is not working. I'm just trying to find the percentage of a certain demographic trait by doing:  [asian]/[total]*100   It is populating about 10% of the column correctly and then I get an error:     There was a failure during processing, check the Geoprocessing Results   window for details   Obviously my syntax is not incorrect or else it wouldn't start populating the field correctly.   Does anyone have any idea why this is happening? Georeferencing old map in QGIS? I have an old map over a part of Laos and wondering how do I georeference the map with the help of the Grid lines and other Longitude and Latitude information on the scanned map? I have searched for information on the internet and tried many times. My scanned map wont overly correctly on the base map. Any base map would be fine. As long as I can check the result.  Do I add X/East and Y/North values from the grid line crossings as marked and written in some parts of the map? How do I begin georeferencing this map? Do I have to add zeros after the two digits? For example if the Easting is 24 (its in KM right? Do I add like three/four zeros to convert it to meters?  I georeferenced the map with the option "From Map Canvas" in the georeferencer tool and I succeeded but I want to know the coordinate of a point on my old scanned map and write them as X/East and Y/North to georeference the map.   I hope I am not babbling and someone understands what I mean!  We can set the CRS to WGS84, EPSG:4326. How do I check to see if a text file is empty or does not exist in c++? Why do certain sites require more reputation for privileges? So, let's take Skeptics.SE for an example. You can become a trusted user when you acquire 4000 rep., not 20000  I was like, wow, so is it like this because of the lack of users on Skeptics.Se, as in, not so much as Stack Overflow or something?  Why do some sites need less reputation to be a trusted user? Prove that $|\log(1 + x^2) - \log(1 + y^2)| \le |x-y|$ I need to show that $ \forall x,y \in \mathbb R, |\log(1 + x^2) - \log(1 + y^2)| \le |x-y|$  I tried using concavity of log function: $\log(1 + x^2) - \log(1 + y^2)=\log(\frac{1 + x^2}{1 + y^2})=\log(\frac{x^2y^2}{(1 + y^2)y^2}+\frac{1}{1 + y^2}) \ge \frac{2(\log(x)-\log(y))}{1+y^2}$  Also the middle value theorem: $0&lt;\frac{\log(1 + x^2) - \log(1 + y^2)}{x^2-y^2} &lt;1$  But both attempts has not led too far. Breakthrough at the Large Hadron Collider -Leptoquarks skepticism? Undefined control sequence error on \fancy@reset, \f@ncyolh and \f@ncyorh when using newlfm class How to set systemwide "XP style DPI scaling" in Windows 8.1? I run my ThinkPad with a system DPI setting of 140%. In Windows 7 and 8, I enabled the "Use Windows XP style DPI scaling" checkbox in the Custom DPI Setting dialog. When this setting is off, non-DPI-aware applications get DPI virtualization via bitmap stretching, which results in large but fuzzy text. Turning on the checkbox disables this bitmap stretching, so non-DPI-aware apps get smaller text but without the fuzzy stretching. I find most apps to be more readable with the smaller crisper text, so I enable XP style scaling (i.e. disable DPI virtualization).  Windows 8.1 RTM has removed this option completely. After upgrading, all of my non-DPI-aware apps are running with ugly bitmap stretching. I'm able to fix this by going into the Properties for each app and setting the "Disable diplay scaling on high DPI settings" box in the Compatibility tab, but that is a bit of a pain.  Is there any way to restore the systemwide XP-style DPI scaling in Windows 8.1? How to add n hours to a specified time? Which one of these adjectives is correct? I can see that both of them are being used, I'm just not sure which one is grammatically correct. Are there any general rules to follow as to the use of one against the use of the other? WITH ROLLUP WHERE x IS NULL I want to earn Unicoins for free but I can't seem to find the best way mine coins faster. There are also sometimes when I get 20-30 Unicoins at one time, and I think that there is some clue so that we can know which rocks contain lots of Unicoins.. So do you have any idea you can share? Chat related privileges are not site specific I am new to data analysis and now working on a Mixed Effects Logistic Regression Model.  Currently, I have the following data frame (model_data):  Road Id Vehicle_id  entry_time           exit_time          1        1        2017-01-31 00:00:00  2017-01-31 00:00:00 last_veh_time  vehicle_type     status 300            4 wheel          0    Likewise, I have the information (nearly a million rows) of all the vehicles entering a place and exiting the same place.  Status can be either $1$ or $0$ for all of these.  last_veh_time is the last vehicle passed the same place before how much time.    I want to know the probability of a getting status $0$ for a vehicle based on the entry_time and last_veh_time. To do this, I created the model:   model = glmer(status ~last_veh_time + (1 |link ),               data=model_data, REML=FALSE)   Is this the right approach?  Any help is appreciated. Regex match entire words only There are many tutorials that use a pull-up or pull-down resistor in conjunction with a switch to avoid a floating ground, e.g.        Many of these projects use a 10K resistor, merely remarking that it is a good value.  Given a a particular circuit, how do I determine the appropriate value for a pull-down resistor?  Can it be calculated, or is it best determined by experimentation? Does the proof of Bolzano-Weierstrass theorem require axiom of choice? When selecting the terms of subsequence from each bisections, I thought axiom of choice might be required. But I'm not so sure whether or not, so please tell me. [edited] I'm sorry for the lack of explanation. I want to prove this statement: Let $a_1, a_2, \ldots \in \mathbf{R}$, and $(a_n)_{n\in\mathbf{N}}$ is bounded, then $(a_n)$ has some convergent subsequence.  The proof is as follows. Since $(a_n)$ is bounded, for all $n\in\mathbf{N}$,   $a_n \in I = [b, c]$.  Now, let $I_0 = I$ and if $I_n = [b_n, c_n]$,  we define $d_n = (b_n+c_n)/2$ and  if infinite terms of $(a_n)$ is included in $[b_n, d_n]$(resp. $[d_n, c_n]$), we will define  $I_{n+1} = [b_n, d_n]$(resp. $[d_n, c_n]$).If both intervals contain infinite terms, let $I_{n+1}$ be $[d_n, c_n]$. For all $n\in \mathbf{N}$, infinite numbers of $m \in \mathbf{N}$ exist such that $a_m \in I_n$ suffices. We take the sequence of natural numbers $(n(k))_{k\in\mathbf{N}}$which suffices $n(0) &lt; n(1) &lt; \cdots &lt; n(k) &lt; \cdots$ following this procedure: Now we have already selected $a_{n(1)}, \ldots, a_{n(k)}$, there are infinite numbers of $m\in \mathbf{N}$ which suffices $n(k)&lt;m, a_m \in I_{k+1}$, so let's take the minimum m out of it. Applying this process recursively, we obtain a infinite convergent subsequence(?). I think intuitively, by only repeating this process we can't obtain countable infinite terms of subsequence because we have to repeat infinite times. diophantine equation: $x^2 +y^2 = z^n$ Prove that $x^2+y^2=z^n$ has a solution $(x, y, z)$ in $\mathbb{N}$ for all $n\in\mathbb{N}$  I tried to prove this by induction, but couldn't. ( This was probably because the solution for some $n$ isn't necessarily related to the solution for $n+1$)  I can't seem to see any other way other than induction for proving the statement. Any help/hints on solving this problem, or any alternative approach will be highly appreciated :)