Beauty: My Solutions

Well spring break officially starts today, even though I started on Thursday after my managerial economics test. After the exam I went out to a club with a friend and her neighbor. It was the second time that we went to this particular club, luckily this time was more enjoyable than the previous visit. Even though I had a great time, I would not say that this was my best clubbing experience. It had nothing to do the people I went with since they are some of my best friends, it's just that they are not the group of friends that I normally go to clubs with so it was just a little awkward for me and took some getting used to. The next morning I woke up feeling horrible, the only way I know to describe it would be as hung over, but I haven't had any alcohol in about two to three months. In any case it was not fun getting up and going to my 8:30 class on Friday. Even though I felt like crap class was well worth it, since there were very few people there we got to practice our speaking (this is French class by the way).

As of now I am not exactly sure when my ride back home is leaving, all I know is that it will either be on Sunday or Monday. With the free time I have while waiting to leave I should really be doing the homework so that I won't have to do it later. Guess I will start on that once I finish this post. Right now I am just reading some C++ books and working on SPOJ problems. I have solved three more problems the only issue is that I need to find a way to improve my method so that I can fit in their ridiculous time constraints.

Well I am going to post my solutions to some of the problems in the previous post.

1:
Choose $x,y,\in{I}$ , then by the mean value theorem there exist an $a\in(x,y)$ such that $f(x)-f(y)=(x-y)f'(a)$ . From the given it follows that $|f(x)-f(y)|<M(x-y)$ , and $f$ being uniformly continuous trivially follows.

2:
part 1: Since $g$ is continuous at $x=0$ , for every $\epsilon>0$ there exist a $\delta>0$ such that $|g(t)-g(x)|<\epsilon$ when $|t-x|<\delta$ . Now choose an $\epsilon>0$ and its corresponding $\delta$ , then
$\left|\frac{f(t)-f(0)}{t-0}-g(0)\right|=\left|\frac{f(t)}{t}-g(0)\right|=|g(t)-g(0)|<\epsilon$ . Thus $f$ is differentiable at $x=0$ and $f'(0)=g(0)$ .

Part two follows with ease.

6:
Let $g(t)=f(t)-\lambda{t}$ , so $g(t)$ is also differentiable on $[a,b]$ , and $g'(a)<0<g'(b)$ . Thus there exist a $\delta>0$ such that $g(t_1)<g(a)$ and $g(t_2)<g(b)$ when $t_1\in(a,a+\delta)$ and $t_2\in(b-\delta,b)$ . Since $[a,b]$ is compact, we know that $g$ attains a minimum value somewhere on the interval; and because it is differentiable on this interval its derivative at this point is 0. So there exist $c\in(a,b)$ such that $g'(c)=0\Longrightarrow f'(c)=\lambda$ .

Beauty

Saturday, March 08, 2008

My Solutions

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