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lsdup Description:Oct. 23, 2009 — It’s been about five years since The Twilight Saga: New Moon hit theaters. But despite the franchise’s success, it’s not clear that Stephenie Meyer’s books will ever make enough money to pay the bills.
The books are popular but not wildly successful. They’re hard to translate on the big screen. There are no signs of a Twilight movie series. And some fans have left the Twilight crowd because of the explicit scenes in the books.
Which is a shame, because that may be a part of the story too.
Meyer told The Associated Press in a recent interview that her first book, Twilight, sold about 3.8 million copies. But that number was only a fraction of the total books sold. The movie didn’t make money either, reportedly losing $90 million.
So, where do all those copies of the books go? Why haven’t movie theaters been able to make money on the Twilight franchise?
And what does that say about the books? Do they have staying power? Do they really have an audience that would buy several of them?
We asked Dr. Kurt Beals, a popular science writer, blogger and science educator, who also happens to be a Twilight fan to get his take.
Can you explain why a franchise, in this case the Twilight series, is a hard sell?
Beals: Stephenie Meyer’s books, especially the Twilight series, are fascinating. They’re really well-written, and they are engaging. They really are. They have a lot of interesting details and interesting ideas.
But there are a lot of books out there that are like that. People have these engaging ideas, but they are just stories. They’re not really well-written, not particularly compelling.
We have to think about our books, our books on its own, in the context of what they are: A book that is about the idea of love. It’s a book that has a really compelling idea.
There are a lot of books, especially science fiction and fantasy, that give us compelling ideas, but the idea is not
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lsdup Screenshots:X_f(w)+{{\varepsilon}}\eta\bigl(w,V(X_f(w))\bigr)\,dw,$$ where ${{\varepsilon}}$ is as in Theorem \[mt\].
\[T6\] Let $X$ be a topologically mixing system with the factor map $f:X\to X$ such that there exists a finite set $Y\subset X$ whose factor $\bar f:Y\to Y$ is transitive. Let ${{\mathcal A}}$ be a transitive subshifts with the associated Markov map $V:\,\mathcal A\to{{\mathcal A}}$ and the factor map $f:X\to{{\mathcal A}}$. Then, for every ${{\varepsilon}}>0$, there exist a finite set $Y\subset{{\mathcal A}}$, a subset $Y_0\subset Y$, and a finite set $W\subset X$ such that the set $S=Y_0\setminus W$ satisfies the following. For any $y_1,y_2\in Y_0$ and for any $x_1,x_2\in W$, $$d_{T(S)}\bigl(f(x_1),f(x_2)\bigr)0$ as in the proof of Theorem \[mt\]. We may assume that $Y\subset{{\mathcal A}}$ is $\rho$-dense in ${{\mathcal A}}$. For $n\geq 0$, we define $$Y_n:=\{y\in Y\,;\, V^ny\cap B_n(0)
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lsdup is a handy and reliable command line application designed to scan folders for files with duplicate content.
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Description:
Summary:
A simple utility for copying multiple files into one archive.
Name:
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Version:
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Date:
2012-04-30
Filesize:
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Files:
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Author:
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KeyDown method inside a class
I need to create a method that will handle KeyDown events. The problem is I don’t have a reference to the form or form1 because it is a class. What would be the best way to do this?
Thank you
A:
You should add an event handler to your form’s class in the constructor.
Assuming your form has a single public constructor,
public Form()
{
// Instantiates your control, or the component the user added to the form
// Here you handle the KeyDown event.
}
If you have several forms you’d want to handle the event on, you’d do something like this:
public Form1()
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}
public Form2()
{
InitializeComponent();
this.KeyDown += MyForm_KeyDown;
}
private void MyForm_KeyDown(object sender, KeyEventArgs e)
{
// Your logic goes here
}
~~~
evanpw
From the post: “Once you do find a closed form (i.e. exactly calculable in
terms of elementary functions), the method of finding the convergence of the
iterative series is pretty straightforward.”
The challenge (it seems) is not finding the converging series, but instead
finding a closed form for the iterative series.
~~~
jcfrei
Yep, not only the “closed form” but also the convergence. I’m surprised that
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