And that agrees with what I said. If the task is to prove \sum_{k=1}^n \frac{1}{k^2} < 1 - \frac{1}{n} where n \ge 2, then we have a counterexample with the case of n=2.
\begin{align*}\sum_{k=1}^2 \frac{1}{k^2} &< 1 - \frac{1}{2} \\ \frac{1}{1^2} + \frac{1}{2^2} &< \frac{1}{2} \\ 1 +...
Isn't the problem as stated false? \sum_{k=1}^n \frac{1}{k^2} will always be greater than 1 and 1 - \frac{1}{n} will be less than 1. I think what is meant is \sum_{k=2}^n \frac{1}{k^2} instead. How does the condition n \ge 2 effect the start condition of k=1?
A long time ago Mathwonk discussed the School Mathematics Study Group series of books. I managed to find a list of some of the books that were put out by SMSG on a webpage for the utexas archives for the SMSG. I thought it might be useful to have the list here if anyone was trying to track them...
If you are looking to get into the game programming industry then learn C++ or possibly C# as the former will probably remain the standard for a while, and eventually I imagine get overtaken by C#. I'm not in that industry so I can't say that will be the case, but the way MS has been evolving...
You do get 12 hours of electives in the senior year of the CE track, plus it looks like part of the track involves some discrete structures courses, and a algorithms course.
The Comp Sci department doesn't appear to specialize in Software Engineering either. It looks like they do have one...
A good way to learn is to look at other people's code. Lots of good code to look through is available through the GNU license.
http://norvig.com/21-days.html"
Well if you are wanting to do software, Comp Sci would probably be the path to take. It doesn't look like the Engineering school has a software engineering program. Computer Engineering will deal with the design of hardware.
"...if only they'd ask." I think that is the main problem that you will run into with any guy, but probably a little more so with nerdy guys. Nobody likes rejection, so until they learn that it's not the end of the world if a girl says no they will always be hesitant to approach someone they...
Since the task is to simplify you could probably simplify the final expression a bit more:
\frac{x}{\sqrt{x}(x^2-1)} = \frac{x \cdot x^{-1/2}}{x^2-1} = \frac{\sqrt{x}}{x^2-1}