Comparative form of shy I've been looking for this answer a lot and people say diferent things are true. Is it shyer and the shyest, shier and shiest or more shy and the most shy? According to Merriam-Webster dictionary it's shyer and the shyest, shier and shiest. () The same goes to Dictionary.com (). So does this mean both are right?
"shyer" or "shier" My Longman dictionary states that the comparative of 'shy' is 'shyer'. However, at least two online dictionaries also give the form 'shier' as being acceptable:  and . On an English language  I came across a reference to British (shyer) vs. American (shier) spelling. But an Ngram chart shows that even in American English 'shyer' is much more used.  My problem is that I've been told that it is definitely wrong, but if it's in dictionaries then... has there been a change to what is wrong?   P.S.:
I came across a  where a comprehensive example would be very useful. So I wrote an example program showing one way to solve their problem.  My problem is that when I was done creating it, and testing to make sure it worked, it was over 400 lines in 18 files. My example included source code, header files, and scripts for building the example in a variety of circumstances.   It's a working and tested, so it's of huge value to illustrate the tricky solution, but because it's all there, my response was huge! Is there a better way of posting my code dump?  Edit:  I guess there isn't a current "best way" to include a large chunk of code, so I made this a feature request and tried to add some ideas.  Perhaps code entries that are longer then a certain number of lines, or ones that have been specifically marked could initially be displayed as "collapsed", and someone could expand them if they want to see them?  Or maybe, like attaching images, users with enough reputation could attach an archive of files. They could be displayed as "attached code" (initially collapsed also).  Some problems have simple answers that unfortunately touch disproportionately large parts of the program, or can be solved in slightly different ways depending on the environment. I feel that stackoverflow needs a way of capturing these large solutions. Sometimes, they're large chunks of code. In my case, the code was small, but there were many files.  Using github or a similar code repository is great, but we risk linkrot (even for big guys like github) and loose the community editing, commenting, and voting abilities.  I understand the problems that could come from with people dumping whole "I can't get this to work" packages, but surely we already have a solution for that in moderation.  Edit again:  I'm starting to see that this may have been already addressed in posts like    , and    If you feel this a dupe, I'm open to having it "closed as already addressed" How to animate noise texture "seed"? It's possible to animate a change noise texture by animating the texture coordinates with a mapping node or by animating the texture space, however this only works on one axis at a time (on the other axes the texture will appear to be sliding across the object's surface).  For example, the top face has the desired effect but the other faces appear to be moving/sliding upwards:    Even with all three axes location and rotation animated it still appears to be sliding rather than statically changing.    Is there some way to make each face appear like the top face in the first gif but still be seamlessly blended with the other faces? This should work for other shapes as well.  In other words, I'm looking for a way to animate the "seed" of the texture. How do I calculate the date six months from the current date using the datetime Python module? I am using the datetime Python module.  I am looking to calculate the date 6 months from the current date. Could someone give me a little help doing this?  The reason I want to generate a date 6 months from the current date is to produce a review date.  If the user enters data into the system it will have a review date of 6 months from the date they entered the data. Prime spectrums bijection but not homeomorphism Equivalent of shell 'cd' command to change the working directory? cd is the shell command to change the working directory. How do I change the current working directory in Python? What makes a field not groupable I'm trying to use an aggregate query, but one of the fields I want to group by is not groupable (saw it in the soql error and then verified groupable=false in the metadata).  It's just a number field that holds an integer (number 2,0 in salesforce field spec-speak).  Is there a trick to make it groupable?  Any way around this problem? I don't want to write hundreds of lines of code just to average a list by this field. Let $$ f(x) =    \begin{cases}\begin{align}     1\quad&amp;\text{ if }\; 0\leq x \leq 1 \\     2 \quad&amp;\text{ if }\; 1&lt;x \leq 2 \\   \end{align}\end{cases}$$  Then $\int^2_0 f(x)dx=3$ and an anti derivative of f(x) is $$F(x)=\begin{cases}\begin{align}     x\quad&amp;\text{ if }\; 0\leq x \leq 1 \\     2x \quad&amp;\text{ if }\; 1&lt;x \leq 2 \\   \end{align}\end{cases}$$ But, $F(2)-F(0)=4-0=4 \neq 3.$ Why has this happened?  Find an anti-derivative $G(x)$ of $f(x)$ such that $G(2)-G(0)=3$, the correct answer.  My attempt: $F(2)-F(0)=4 \neq 3$ because we need to break it up into two parts like this: $\int^2_1 2x dx + \int^1_0 x dx = x^2|^2_0 + \frac{1}{2} x^2|^1_0 =4-1+\frac{1}{2}-0=3.5$ but this is not right either. :(  For the "Find an anti-derivative $G(x)$..."I'm completely lost but perhaps if we solve the first part then I'll understand what it's asking for there.  Any help would be appreciated. Are the contents of a loot box determined at the time it is awarded or when it is opened?  For example, if you collect multiple loot boxes before a special event but opened them during the special event, would they have the potential of spawning rewards from that event? Evaluating $\int P(\sin x, \cos x) \text{d}x$ Suppose $\displaystyle P(x,y)$ a polynomial in the variables $x,y$.   For example, $\displaystyle x^4$ or $\displaystyle x^3y^2 + 3xy + 1$.  Is there a general method which allows us to evaluate the indefinite integral     $$ \int P(\sin x, \cos x) \text{d} x$$   What about the case when $\displaystyle P(x,y)$ is a rational function (i.e. a ratio of two polynomials)?   Example of a rational function: $\displaystyle \frac{x^2y + y^3}{x+y}$.    This is being asked in an effort to cut down on duplicates, see here:   and here: G acts on X transitively, then there exists some element that does not have any fixed points Let $X$ be a transitive $G$-set. ($G$ acts on  $X$ transitively.) If $X$ is finite and has at least two elements, show that there is some element $g$ $\in$ G which does not have any fixed points; that is, such that $g$$($$x$$)$ $\ne$ $x$ for all $x$ $\in$ $X$  I am trying to use contradiction, but it is not very clear to me why this is true. MacBook is displaying the message "no hardware installed" when I click on Wi-Fi I have a MacBook Pro 2009 running MacOS El Capitan. Recently the Wi-Fi has stopped working.   When I click on the Wi-Fi icon it displays the message "No hardware installed", but Ethernet still works.   Is there any way of fixing this by using a Wi-Fi dongle or something of the sort, if so do I need a specific dongle and how do I install it? Group presentation for discrete Heisenberg group Proving by induction $4^n  0$ Consider:  I need to do this. I need do this.  My English grammar knowledge tells me that &quot;need&quot; doesn't have the same status as the modal verbs &quot;may&quot;, &quot;can&quot;, &quot;should&quot; and what not. Hence the second usage where two verbs appear consecutively is incorrect. But yesterday, my native English speaking friends (Americans) told me that &quot;I need compute this.&quot; is a perfectly grammatical sentence, and one is simply omitting the &quot;to&quot;. How can this be? Is it a colloquial usage but grammatically incorrect, or is it grammatically correct? If it is grammatically correct, is it because &quot;need&quot; is a semi-modal verb? EDIT: in particular, is it okay to use &quot;need compute&quot; in a scientific paper? EDIT2: the exact phrase that raised the question was &quot;The advantage of this representation is that we need only compute sums and products&quot; How do I integrate the following? $\int{\frac{(1+x^{2})\mathrm dx}{(1-x^{2})\sqrt{1+x^{4}}}}$ When estimating variance, why do unbiased estimators divide by n-1 yet maximum likelihood estimates divide by n? I am totally confused: On the one hand you can read all kinds of explanations why you have to divide by n-1 to get an unbiased estimator for the (unknown) population variance (degrees of freedom, not defined for sample size 1 etc.) - see e.g.  or .  On the other hand when it comes to variance estimation of a supposed normal distribution all of this doesn't seem to be true anymore. There it is said that the maximum likelihood estimator for variance includes only a division by n - see e.g. .  Now, can anyone please enlighten me why it is true here but not there? I mean normality is what most models boil down to (not least due to the ). So is the choice "division by n" yet the best choice for finding the best estimation for the true population variance after all?!? What does $\dot\sim$ notation (dot over tilde) mean, in the context like $x \mathrel{\dot\sim} \mathcal N(0,1)$?  Turns out it is easier to find how to typeset it correctly:  that one should type \mathrel{\dot\sim} instead of simply \dot\sim to fix the spacing issue -- than to find what it actually means. It has only been used  on CV until now; is it standard? What is the meaning of "this" in Java? If $|A| > \frac{|G|}{2} $ then $AA = G $ I'v found this proposition.  If $G$ is a finite  group , $ A \subset G $ a subset and $|A| &gt; \frac{|G|}{2} $ then $AA = G $.  Why this is true ? The cube of any number not a multiple of $7$, will equal one more or one less than a multiple of $7$ Yeah so I'm kind of stuck on this problem, and I have two questions.   Is there a way to define a number mathematically so that it cannot be a multiple of $7$? I know $7k+1,\ 7k+2,\ 7k+3,\ \cdots$, but that will take ages to prove for each case.  Is this a proof? $$(7k + a)^3 \equiv (0\cdot k + a)^3 \equiv a^3 \bmod 7$$    Thank you.