Will there be a noticable difference in sharpness and color between Nikon and Canon
Is there any significant difference between Nikon and Canon?
$(X,\tau)$ needs to be $T_1$ in order to guarantee that $A'$ is closed? Claim: If $(X,\tau)$ is not $T_1$ then $A'$ is not necessarily closed for any $A$.  Proof: Here is an example. Take $X=\mathbb{N}$ and let $\tau=\lbrace u\subset\mathbb{N}\mid 1\in u\iff 2\in u\rbrace$. Let $A=\lbrace 2,3 \rbrace$. Since any open set containing $1$ also contains $2$ it follows that $1\in A'$. Hence $\overline{A}=\lbrace1,2,3\rbrace$, as it is the smallest closed set containing $A$. But  $X\setminus A'=\mathbb{N}\setminus\lbrace 1\rbrace$ is not open because it contains $2$ and not $1$. The set of accumulation points $A'$ is not closed.  So, in general, trying to prove that $A'$ is closed in a topological space is an effort in futility? I have a shape (rounded rectangle) in PowerPoint 2007 of which I would like to increase the vertical size only (to illustrate de-bottlenecking). So, the width will stay the same but the height will grow (both from the top and bottom, so the centre will remain in the same location).  Custom Animation -> Emphasis only appears to change the whole shape.  Is it possible to increase only the vertical size (stretch the shape)?  This is during animation of the shape. I am trying to burn a shapefile to a raster using GDAL's RasterizeLayer. I pre-create an area of interest raster from a different shapefile, given a specific pixel size. This AOI then serves as a base for all following rasterizations (same number of collumns and rows, same projection and geotransform).  The issue occurs, however, when I go to burn the shapes to their own raster, based on the same pixel size and projections. The link below (don't have enough rep to post the image), shows the original shapefile in tan, and the dark pink where RasterizeLayer has burned data. The light pink is nodata values for the dark pink raster data. The grey is the AOI based on which the shapefile burn was completed.  Given the extents of the shapefile polygons, I would expect to see burn values in the bottom two corners, as well as the two pixels underneath the data that does show. Obviously, however, this is not the case.    As follows is the code that I used to generate these. All of the shapes were created using QGIS, and were all created in the same projection.  (It should be noted that the gridding in the picture shown was just to give an idea of the pixel size I was using.)  from osgeo import ogr from osgeo import gdal  aoi_uri = 'AOI_Raster.tif' aoi_raster = gdal.Open(aoi_uri)  def new_raster_from_base(base, outputURI, format, nodata, datatype):      cols = base.RasterXSize     rows = base.RasterYSize     projection = base.GetProjection()     geotransform = base.GetGeoTransform()     bands = base.RasterCount      driver = gdal.GetDriverByName(format)      new_raster = driver.Create(str(outputURI), cols, rows, bands, datatype)     new_raster.SetProjection(projection)     new_raster.SetGeoTransform(geotransform)      for i in range(bands):         new_raster.GetRasterBand(i + 1).SetNoDataValue(nodata)         new_raster.GetRasterBand(i + 1).Fill(nodata)      return new_raster  shape_uri = 'activity_3.shp' shape_datasource = ogr.Open(shape_uri) shape_layer = shape_datasource.GetLayer()  raster_out = 'new_raster.tif'  raster_dataset = new_raster_from_base(aoi_raster, raster_out, 'GTiff',                                 -1, gdal.GDT_Int32) band = raster_dataset.GetRasterBand(1) nodata = band.GetNoDataValue()  band.Fill(nodata)  gdal.RasterizeLayer(raster_dataset, [1], shape_layer, burn_values=[1])   Is this a bug in GDAL, or is RasterizeLayer burning data based on something other than just the presence or lack of a polygon within a specified pixel area?  The files that I was using can be found . Let $R$ be a commutative ring with unit, $I$ an ideal of $R$ and consider the following three constructions.   The  $R_I$ of $R$ at $I$ (i.e. the localization of $R$ at the multiplicative system $R\setminus I$) gives a morphism $$ f_1:\operatorname{Spec}(R_I)\to \operatorname{Spec}(R)=:X $$ The  $\widehat{R}_I$ of $R$ at $I$ gives a morphism $$ f_2:\operatorname{Spec}(\widehat{R}_I)\to X $$ For the special case $I=(a)$, the localization $R_a$ (i.e. the localization of $R$ at the multiplicative system $\{1,a,a^2,\ldots\}$) gives a morphism $$ f_3:\operatorname{Spec}(R_a)\to X $$   My question is:     What is the geometric meaning of all three constructions and how are they related?   This is what I ''know'' already or describes at least the style of answer I would appreciate. As remarked in the comments, $I$ has to be a prime ideal.   $R\to R_I$ is injective iff $R\setminus I$ contains no zero divisors which is the case if $I$ is a prime ideal. The scheme $\operatorname{Spec}(R_I)$ is the intersection of all the neightbourhoods of $I$ in $X$. This is a little contra-intuitive for me since the last statement sounds like ''$f_1$ is injective'' but the $\operatorname{Spec}$-operator should turn 'injective'' and ''surjective'' around somehow (I know this is literally not true but I only want to get a feeling like ''is contained'', ''is bigger'', ''is smaller'', etc.). $\operatorname{Spec}(R_a)$ is somehow the opposite of $\operatorname{Spec}(R_{(a)})$ (= the intersection of all the neightbourhoods of $(a)$ in $X$) because $\operatorname{Spec}(R_a)$ seems to be something like the union of all the open sets of $X$ not containing the point $(a)$. $R\to\widehat{R}_I$ is injective iff $\cap I^n=(0)$ and this holds very often (e.g. for $R$ noetherian and either an integral domain or a local ring). Hence (this is probably false intuition as remarked before), $f_2$ should be something like a ''projection'' (from something ''big'' into something ''small''). But what is geometrically the difference between localization and completion. I don't have a geometric idea of completion at all.   As remarked by Qiaochu Yuan in the comments below, one should not think of $\operatorname{Spec}$ as injective-surjective switching. Show/Hide hidden files without restarting finder? I came to understand that the following is the way to toggle the visibility of hidden files:  defaults write com.apple.finder AppleShowAllFiles YES # replace YES with NO to hide hidden files killall -HUP Finder /System/Library/CoreServices/Finder.app   Is there a way to show/hide hidden files without killing the Finder? What is the "canary thread" and why is it starving? The following occasionally appears in /var/log/syslog:  rtkit-daemon[1145]: The canary thread is apparently starving. Taking action. rtkit-daemon[1145]: Demoting known real-time threads. rtkit-daemon[1145]: Successfully demoted thread 1431 of process 1368 (n/a). rtkit-daemon[1145]: Successfully demoted thread 1430 of process 1368 (n/a). rtkit-daemon[1145]: Successfully demoted thread 1368 of process 1368 (n/a). rtkit-daemon[1145]: Demoted 3 threads.   What's going on here? How to change the background color only for the current page? How to prove that $\lim\limits_{\epsilon\rightarrow 0}\int_{-\infty}^{\infty}f_\epsilon(x)g(x)dx=g(0)$ (Dirac delta function)) I'm currently studying the Dirac delta function using a textbook which unfortunately provides only partial solutions to its explanations. Why does $\lim\limits_{\epsilon\rightarrow 0}\int_{-\infty}^{\infty}f_\epsilon(x)g(x)dx=g(0)$ if $ f_\epsilon=\frac{1}{2\epsilon}e^{-\frac{|x|}{\epsilon}}$? Any help is very much appreciated. Am I being charged taxes on my grocery purchases at Walmart? Today, I only bought groceries at Walmart and noticed that there is no tax on groceries in California (Google tells me that this is perhaps true for whole of the US?)  So far I have always paid tax at Walmart for my grocery purchases, but that also included items like crackers, shampoo etc  Did Walmart charge me then only for the items that were not groceries, or if your cart has groceries + non-grocery items, you get charged tax on all of them?  Or is this a policy that varies by store, or cashiers' knowledge/discretion, etc?  Would people recommend for me to break down a single mixed purchase into two: one only having groceries and the other with the rest, or is this being paranoid?  My grocery purchases, every two weeks, amount to around $100+.  I pay 8.25% tax, so that is enough to get me free gas if I can figure this out! finding rational roots of polynomials What are the costs of books and materials in Harry Potter? In the first book, Harry sees his Gringotts vault for the first time.  (Quotes thanks to Slytherincess!)     Griphook unlocked the door. A lot of green smoke came billowing out, and as it cleared, Harry gasped. Inside were mounds of gold coins. Columns of silver. Heaps of little bronze Knuts.      "All yours," smiled Hagrid.      All Harry's -- it was incredible. The Dursleys couldn't have known about this or they'd have had it from him faster than blinking. How often had they complained how much Harry cost them to keep? And all the time there had been a small fortune belonging to him, buried deep under London.      Hagrid helped Harry pile some of it into a bag. "The gold ones are Galleons," he explained. "Seventeen silver Sickles to a Galleon and twenty-nine Knuts to a Sickle, it's easy enough. Right, that should be enough fer a couple o' terms, we'll keep the rest safe for yeh."      Harry Potter and the Sorcerer's Stone - pages 58-59 - US Hardcover Edition   Jk Rowling posts in the Harry Potter Lexicon      JKR has stated in an interview (Comic Relief 2001) that she estimates the value of one Galleon to be "about five pounds," which works out to around US$9.75 (the exchange rate at the time of the interview was US$7.33). In the introduction to both QA and FB, US$250-million is stated to be the equivelent of 34 million Galleons. That also works out to a value of £5 to the Galleon, at the exchange rate of the time. The price listed on the back of the books, however, is not correct, since US$3.99 would equal less than half a British pound, or 8 sickles and 15 knuts. The book instead incorrectly lists US$3.99 as being equal to 14 sickles and 3 knuts. (Unfortunately, CNN.com uses this incorrect value for their Knuts-to-dollars converter.)      This gives us approximate values as follows:      1 bronze Knut = £0.01 (US about 2 cents)      1 silver Sickle = £0.29 (about US$0.57)      1 Galleon = £5.00 (more or less US$9.75)      These rates vary as the exchange rate fluctuates - see the Wizarding World Currency Converter for the current rate.   Now looking at the quote      Hagrid helped Harry pile some of it into a bag.   and     Right, that should be enough fer a couple o' terms, we'll keep the rest safe for yeh.   Obviously "some" is unspecified, but presuming the costs of ...     Typical elementary-school textbooks cost more than $100 each   He would be carrying at least 10 galleons for each school book, and going by the list of books just for his first year,  (Thanks Ibid!)     COURSE BOOKS      All students should have a copy of each of the following:         The Standard Book of Spells (Grade 1) by Miranda Goshawk   A History of Magic by Bathilda Bagshot   Magical Theory by Adalbert Waffling   A Beginners’ Guide to Transfiguration by Emeric Switch   One Thousand Magical Herbs and Fungi by Phyllida Spore   Magical Drafts and Potions by Arsenius Jigger   Fantastic Beasts and Where to Find Them by Newt Scamander   The Dark Forces: A Guide to Self-Protection by Quentin Trimble         (Harry Potter and Philospher's Stone, Chapter 5)   That's 80 Galleons just for books. This does not include the list of other expenses Harry has for his school years (wand, potion reagents, etc.)  How many Galleons is "some" or is Diagon Alley selling supplies for third world prices?  Sidenote, is this just a case of bad JK maths? Why does lightning emit light? What exactly is causing the electric discharge coming from the clouds to emit light while traveling through the air. I've read and thought about it a little but with my current knowledge I cant really figure what is happening at a subatomic level. In other words, what causes the emission of photons during lightning. What I want to know is, is there some sort of collisions taking place that produces the photons we see, or is it some other interaction that is causing it?. I don't know too much about particle physics, I've only read a couple of pages in the Introduction to Elementary Particles by Griffihts. So if you could keep the answer simple then that would be great. How to interpret autocorrelation plot in MCMC I am getting familiar with Bayesian statistics by reading the book Doing Bayesian Data Analysis, by John K. Kruschke also known as the "puppy book". In chapter 9, hierarchical models are introduced with this simple example: \begin{align} y_{ji}   &amp;\sim {\rm Bernoulli}(\theta_j)  \\ \theta_j &amp;\sim {\rm Beta}(\mu\kappa, (1-\mu)\kappa)  \\ \mu      &amp;\sim {\rm Beta}(A_\mu, B_\mu)   \\ \kappa   &amp;\sim {\rm Gamma}(S_\kappa, R_\kappa) \end{align} and the Bernoulli observations are 3 coins, each 10 flips. One shows 9 heads, the other 5 heads and the other 1 head.  I have used pymc to infer the hyperparamteres.  with pm.Model() as model: # define the          mu = pm.Beta('mu', 2, 2)     kappa = pm.Gamma('kappa', 1, 0.1)     # define the prior     theta = pm.Beta('theta', mu * kappa, (1 - mu) * kappa, shape=len(N))     # define the likelihood     y = pm.Bernoulli('y', p=theta[coin], observed=y)      # Generate a MCMC chain     step = pm.Metropolis()     trace = pm.sample(5000, step, progressbar=True)     trace = pm.sample(5000, step, progressbar=True)   burnin = 2000  # posterior samples to discard thin = 10  # thinning  pm.autocorrplot(trace[burnin::thin], vars =[mu, kappa])   My question is regarding the autocorrelation. How shall I interpret autocorrelation? Could you please help me to interpret the autocorrelation plot?    It says as the samples gets further from each other the correlation between them reduces. right? Can we use this to plot to find the optimum thinning? Does the thinning affect the posterior samples? after all, what is the use of this plot? How to "pretty" format JSON output in Ruby on Rails I would like my JSON output in Ruby on Rails to be "pretty" or nicely formatted.   Right now, I call to_json and my JSON is all on one line.  At times this can be difficult to see if there is a problem in the JSON output stream.  Is there way to configure to make my JSON "pretty" or nicely formatted in Rails? Recurrence relation for number of ternary strings that contain two consecutive zeros The question is: Find a recurrence relation for number of ternary strings of length n that contain two consecutive zeros.    I know for ternary strings with length one, there are 0. For a length of 2, there is just 1 (00), and for a length of 3, there are 5 (000,001,002,100,200).    I did a similar problem, finding a relation for the number of bit strings of length n with two consecutive zeros: $$a_n = a_{n-1} + a_{n-2} + 2^{n-2}$$    Since you can add "1" to the end of all the $a_{n-1}$ strings, "10" to all the $a_{n-2}$ strings, and "00" any string of size $n-2$.    For the ternary string problem, I'm pretty sure you would replace the $2^{n-2}$ with $3^{n-2}$, but confused about the other terms of the relation. My guess is that it would have the coefficient $2$ in front of the other terms, since you can add either $1$ or $2$ to the end of $a_{n-1}$ and either $01$ or $02$ at the end of $a_{n-1}$.    So I believe the answer for the relation is: $$a_n = 2a_{n-1} + 2a_{n-2} + 3^{n-2}$$    How does that look? Proving a function is Lipschitz continuous Show that the following function is Lipschitz continuous and find a Lipschitz constant  $$y\mapsto f(x,y)\\ f(x,y)=\frac{y}{x}\ln(\frac{y}{x})\text{  , } |x-1|\leq\frac{1}{2}\text{  , } |y-e|\leq\frac{e}{2} $$  I have no clue on where to begin to prove this.   My first question is how I should interpret this function, is $x$ a sort of constant? I was playing terraria today, and my friend came and deleted the my world.  Is there a way for me to recover my world? I've seen in every introductory material on LaTeX and in a lot of code of other people (event experts) that the \newcommand command is used by wrapping the first argument in braces, even if it seems unnecessary to me.   The braces are consistent with the usual way arguments are passed to LaTeX commands (since usually the braces are used even if the argument is made of a single token, for consistency and clarity), but in this single case they appear very ugly to me and in my opinion they lower readability.  But these are opinions, while the question is technical: is there any technical reason why one should write this:  \newcommand{\mycmd}[1]{Hello #1}   instead of this:  \newcommand\mycmd[1]{Hello #1} How to proof the the derivative of $x^2$ is $2x$? I am learning P&amp;S's Introduction of quantum field theory. My teacher said that if the mass of neutrino is exactly 0, then we should not observe any right-handed neutrinos and left-handed anti-neutrinos according to Weyl's theory. It is because $$ \left( i\gamma^\mu \partial_\mu - m \right) \psi = \left( \begin{array}{cc} -m &amp;  i\left(\partial_0 + \vec\sigma \cdot \vec\nabla\right) \\ i\left(\partial_0 - \vec\sigma \cdot \vec\nabla\right) &amp; -m \end{array} \right) \left( \begin{array}{c} \psi_L \\ \psi_R \end{array} \right) =0. $$  If we set m=0, then we will have:  \begin{align} i\left(\partial_0 - \vec\sigma \cdot \vec\nabla\right) \psi_L &amp;= 0; \\ i\left(\partial_0 + \vec\sigma \cdot \vec\nabla\right) \psi_R &amp;= 0. \end{align}  And if these formulas can describe neutrinos we must conclude that there are two types of neutrinos.  My confusions are:   It is known that the mass of neutrino is not 0, therefore there are chances (depending on the mass) to find right-handed neutrinos and left-handed anti-neutrinos. But we don't. If I accept that there are only these two types of neutrinos (right-handed anti-neutrinos and left-handed neutrinos), this property of neutrino causes the parity violation, so parity violation must relate to angular momentum and therefore spinor space. Am I write? Are there any reasons why neutrino is so weird that is has mass but still has helicity + Or - 1. This goes against with what I have learned that only when a particle  moves at the speed of light can its spin aligned with its momentum.