请教几道math sub的回忆题，多谢！

prettywraithUID: 197049

1, If S is a nonempty finite set with k elements ,then the number of one-to-one functions from S onto S is
答案是 k!
我当时选的是2^k

2，Let f be a real-valued function defined and continuous on the set of real numbers R. Which of the following must be true of the set S= {f(c): 0<c<1} ?
（a）S is a connected subset of R(R表示实数)
（b）S is an open subset of R
（c）S is a bounded subset of R
答案是 （a） 和 （c）正确。c我感觉不一定bounded

3,If A is a subset of the real line R and A contains each rational number, which of the following must be true?
(A)If A is open, then A= R.
(B)If A is closed, then A= R
(C)If A is uncountable, then A=R
(D)If A is uncountable, then A is open
(E)if A is countable, then A is closed

答案选 B， 我觉得是否A也正确，我自己没有找出反例来。


4，Let f be a function with domain [-1, 1] such that the coordinates of each point (x,y) of its graph satisfy x^2 +y^2 =1. The total number of points at which f is necessarily continuous is
答案是 two， 我当时选是无穷多。不清楚该题该如何分析.


5，find a set of orthogonal basis of the column space (1 -1 2, -1 1 -2, -2 3 -5, 3 -2 5)
Column space 基 化简为 （1，0，1）’ 和（0，1，-1）’ 。求所谓的正交的基该如何求？

6,求 k^2/k! 的级数（也就是k从1->无穷的合）

pimp2001UID: 2293623

1. 当然是permutation group了，就差没直接给你答案了。。。

2 note that f is define on the entire R...
Observe f([0,1]) is closed and bounded -> f(0,1) is bounded.

3.(-inf, sqrt(2) ) U (sqrt(2), inf)
4. not sure

5. gram-Schmidt orthogonization process

6. 2e
  Sigma (k^2/k! ) |1-inf
=Sigma (k/(k-1)! )|1-inf
=Sigma((k+1)/k!) |0-inf

(all the following assume sigma from 0 to inf)
But note exp(x)=Sigma (x^k/k!)
->       x*exp(x)=Sigma (x^(k+1)/k!)   
differentiate
->   (x+1)exp(x)=Sigma  ((k+1)*x^k/k!)
x=1
->2e=Sigma((k+1)/k!)

prettywraithUID: 197049

Thanks, buddy!  
It's clear except for "4", now.

vinnieUID: 2104933

3是什么意思啊，能讲解下吗
3# prettywraith

prettywraithUID: 197049

C,D，E清楚吧，和可数，不可数相关的，crack上有。
A，二楼举了一个反例，.(-inf, sqrt(2) ) U (sqrt(2), inf，这个是open的，但！=R
B，当时凭感觉选的，具体不是太清，关键是{所有的有理数}这个不算是闭集这个不太清楚，网上查了一下也没有找到。
   请二楼的大侠有空帮忙解答一下吧！

vinnieUID: 2104933

恩，搞懂了，对了，你那有93 97 的题目吗
我找了一圈没找到瓦
5# prettywraith

pimp2001UID: 2293623

5# prettywraith
the set of rational numbers Q is not closed.
you probably have heard that rational numbers are "dense" in R, meaning that the closure (smallest closed set containing Q) is R.
In other words, every real number can be approximated by a sequence of rationals -- think of the decimals...

prettywraithUID: 197049

明白，非常感谢！

To：vinnie， 我这里只有一套是从精华板下来的97的题目，93的没见过。

soarerzUID: 2721318

4, f must be continuous at -1 and 1. (Take any sequence x_n -> -1 for example,  the arc of the circle must have to go to the extreme point) No restriction at other points

pimp2001UID: 2293623

4, f must be continuous at -1 and 1. (Take any sequence x_n -> -1 for example,  the arc of the circle must have to go to the extreme point) No restriction at other points
soarerz 发表于 2009-11-4 00:41

oh... that makes sense.
thank u very much!

prettywraithUID: 197049

to soarerz:
f must be continuous at -1 and 1. (Take any sequence x_n -> -1 for example,  the arc of the circle must have to go to the extreme point) No restriction at other points
还有有些不太理解， 这个地方“No restriction at other points” ，你的意思是不是说：其他比如Xn->1/2,此时Yn可以等于正的2分之根号3，或者是负的，所以这点可能不收敛？

soarerzUID: 2721318

In your notation, when x_n -> 1/2, y_n may not converge, because it is very easy to construct a function where there exists a subsequence converging to sqrt 3 /2 and another converging to -sqrt 3 /2. Since the question asks for the points where f is NECESSARILY continuous, any point other than -1,1 cannot be included.

pimp2001UID: 2293623

intuitively, for each x in [-1,1], f(x) can either take the upper curve or the lower curve;
for x in (-1,1) you can make f(x) "jump" up and down by taking values alternately;
but when x=-1, 1, it does not matter because the values are all close to 0.

mccrimsonUID: 2636067

3.(-inf, sqrt(2) ) U (sqrt(2), i ...
pimp2001 发表于 2009-11-2 03:59

关于这个题目还是不大明白，你举的这个例子并不包含每个rational number啊....

pimp2001UID: 2293623

关于这个题目还是不大明白，你举的这个例子并不包含每个rational number啊....
mccrimson 发表于 2009-11-5 10:06

square root of 2