05年回忆题讨论，自己做的，求指正！

frank2363UID: 2199951

2、160
3、1/35
4、y=ln(1+x)
5、C
6、这题有点儿晕，主要是反函数的导数不会求
7、恒负，单调递增
8、不懂
9、q-1
10、5*0.25*0.25^4
11、不懂
12、不懂
13、x=0.5+n且在(0,10)中时
15、极限为0,在[-1,1)收敛
16、若都为正整数，则最小为135，即四个数为1、132、133、134
19、-1/6
22、-f(1)=g'(1)
25、51
26、C
28、1个
30、D
31、A
32、4/3*pi
33、都是错的
34、B
35、0,1
36、有疑问，theta是取0到pi还是0到2pi？
37、解出来是y=2arctan(c*exp(x))
39、有个Eisenstein's criterion 得到2整除k^2
44、答案错了吧
45、导数是1
46、3/8
47、不懂。。。

frank2363UID: 2199951

第6题有点儿明白了，请大家帮我看下对不对？
g(x)=h逆(x)则h(x)=g逆(x)
h'(x)=1/g'(x)=1/(2f(x)f'(x))
所以h'(1)=1/(2*4*5)=1/40

frank2363UID: 2199951

上面这个推导好像不对，h'(x)不等于1/g'(x)。
应该怎么算呢？求助！

frank2363UID: 2199951

又推了一遍，这样对不对？
h(g(x))=x两边对x求导，得
h'(g)*g'(x)=1
所以h'(g0)*g'(x0)=1
其中取g0=1, （由g(2)=[f(2)]^2=1）得对应的x0=2
所以h'(1)=1/g'(2)=1/(2f(2)f'(2))=1/(2*(-1)*3)=-1/6

frank2363UID: 2199951

基本明白了，大家帮确认一下吧。
即：反函数在x0处的导数应该是原函数在y0处的导数的倒数，其中y0为反函数在x0处的值。
普遍的，对于可逆函数g(x)，若其反函数为y=g逆(x)=h(x)，则：
h'(x)=1/g'(y)，其中y=h(x)
取点x=x0，则y=h(x0)=g逆(x0)
在第六题中，x=1, y=g逆(1)=2

zhouxiaofei170UID: 3185081

第三题是否该0.5^6?
第十三题local max是否为arctan(2pi)/pi  + n=0.45+n

treemantanUID: 2770681

zhouxiaofei170 发表于 2012-10-29 11:22
第三题是否该0.5^6?
第十三题local max是否为arctan(2pi)/pi  + n=0.45+n

I consider you are right

treemantanUID: 2770681

目测LZ非数学系的。。。第8题很Classic的题。。q-1OMG

treemantanUID: 2770681

47和8，In general
For N*N invertible matrix on F_q,
the number of these are determined by the following formula,
(q^n(n-1)/2)*(q-1)*(q^2-1)***(q^n-1)
it can be easily followed by the fact that construct the matrix step by step by columns because the previous vector can not span the next.

zhouxiaofei170UID: 3185081

treemantan 发表于 2012-10-29 20:58
目测LZ非数学系的。。。第8题很Classic的题。。q-1OMG

请问高手33题答案是啥？
33  A is a n dimensional matrix. If (A-I)^2=0, which of the following is correct?
     i. A=I
     ii. det(A)=1
     iii. tr(A)=n

LZ说全是错的。
但我怎么觉得ii和iii是对的。至少在n=2的情况下。

谢谢！

treemantanUID: 2770681

zhouxiaofei170 发表于 2012-11-2 16:45
请问高手33题答案是啥？
33  A is a n dimensional matrix. If (A-I)^2=0, which of the following is c ...

2 is right. (A-I)^2=0->A(A-2I)=I->det(A)=1
3 I suppose It's right.
Let B=(A-I )^2=A^2-2A+I
Trace(B)=Trace(A^2)-2Trace(A)+n=0;
Because A is invertible that we can get from (ii) and then A can be diagonalized. Thus,Det(A^2-2A+I)=Det(P^2-2P+I)(which P is the diagonal form of A)
Trace(A^2)=\sigma(\lambda_i)^2 (the dimension of lambda is equal to n)
Trace(A)=\sigma(\lambda_i) Then,
for the i_th row of B, \lambda_i^2-2\lambda+1=0,i.e (\lambda_i-1)^2=0,therefore lambda_i=1;
So, we can get the Trace(A)=\sigma(\lambda_i)=1*n=n.
Probably Wrong..haha

12345drUID: 3394416

我想问一下第3题那道概率题是咋做的？求解答

malandeUID: 3496058

frank2363 发表于 2011-11-9 13:02
基本明白了，大家帮确认一下吧。
即：反函数在x0处的导数应该是原函数在y0处的导数的倒数，其中y0为反函数 ...

YEs

Genius_zyxUID: 3754854

求问46题的3/8怎么出来的啊？

SiuanusUID: 3842712

zhouxiaofei170 发表于 2012-11-2 16:45
请问高手33题答案是啥？
33  A is a n dimensional matrix. If (A-I)^2=0, which of the following is c ...

II and III are both correct.

Let lambda be the eigenvalue of A-I, and v be the corresponding nonzero eigenvector.
Then 0 = (A-I)^2 v = lambda^2 v .
From this we see that lambda =0 , in other words, the only eigenvalue of A-I is 0. So the only eigenvalue of A is 1.

Now consider the Jordan form J of A. Since the eigenvalue of A can only be 1, the diagonal of JNF of A are all 1( we dont care about the size and number of its Jordan blocks ). Since similarity preserves trace and determinant, DetA = DetJ = 1 ; TrA = TrJ = n.