STPM-学校功课討論区

Dexer

你真的好厉害噢。。。
federal study aids的练习好吗？？我都做里面的pass year question...
你读哪里间学校？？

dunwan2tellu

原帖由 Dexer 于 12-8-2006 05:12 PM 发表
你真的好厉害噢。。。
federal study aids的练习好吗？？我都做里面的pass year question...
你读哪里间学校？？

我没有用那本书，所以不清楚。不过题目是蛮不错的。
我来自日新国中。

Dexer

噢。。。希望可以和你拿预考考卷

nikuang04

dunwan2tellu

提示：

1-i = sqrt[2]( cos (-pi/4) + i sin(-pi/4) ) = sqrt[2]*e^(-i*pi/4)

然后用 janjang geometric

nikuang04

这里又有几题不会了！希望各位帮忙！

ss_sky87

原帖由 dunwan2tellu 于 12-8-2006 07:36 AM 发表
d/dx  = -3x^2 * e^(-x^3)

=> integrate  = -1/3 integrate
= -1/3  + C

int(x^2*e^(-x^3), x) => -1/3*e^(-x^3)/ln(e)
   
我也是算到   -1/3 [e^(-x^3)] + C
可是电脑算到的答案为何如此有ln e 的？

dunwan2tellu

15)let z = a+bi , 1/z = 1/(a+bi) = (a-bi)/(a^2+b^2)

z+1/z = (a + a/(a^2+b^2)) + i(b-b/(a^2+b^2))

z+1/z real ==> (b-b/(a^2+b^2)) = 0 ==> b=0 or a^2+b^2 = 1
when b=0 , z = a ( real)
when a^2+b^2 =1 ==> |z| = 1

16)
因为 |z|=1 , 所以可以设 z = cos A + i sin A
那么 2z/(1-z^2) = 2/(1/z - z) = 2/(-2isinA) = i/sinA <-- imaginary
so real part = 0

17)
a)(z^2+z+1)/z = z + 1 + 1/z , to be real z + 1/z must be real .那么这就是 Q15 （之前 prove 过了）

b)如果 z = real <==> z = z*  (这里 z* = conjugate z)
又因为 |x|=|y|=|z|=1 所以 x=1/x* , y=1/y* , z=1/z*

Let A = (x+y)(x+z)(y+z)/(xyz) , A = complex number , 那么

A* = (x*+y*)(x*+z*)(y*+z*)/(x*y*z*)=(1/x+1/y)(1/x+1/z)(1/y+1/z)/(1/xyz)
=(x+y)(x+z)(y+z)/xyz = A

因为 A* = A 所以 A 是 real .

具体例子：

如果 z 是 complex ,i.e z = 3+5i => z* = 3-5i 所以 z =/= z*
如果 z 是 pure imaginary,i.e z = 3i => z* = -3i 所以 z =/= z*
如果 z 是 real , i.e , z = 5 => z* = 5 所以 z = z*

dunwan2tellu

原帖由 ss_sky87 于 15-8-2006 06:19 PM 发表


int(x^2*e^(-x^3), x) => -1/3*e^(-x^3)/ln(e)
   
我也是算到   -1/3  + C
可是电脑算到的答案为何如此有ln e 的？

ln e 不就是 = 1 吗？
可能电脑多此一举吧？

nikuang04

原帖由 dunwan2tellu 于 15-8-2006 19:48 发表
15)let z = a+bi , 1/z = 1/(a+bi) = (a-bi)/(a^2+b^2)

z+1/z = (a + a/(a^2+b^2)) + i(b-b/(a^2+b^2))

z+1/z real ==> (b-b/(a^2+b^2)) = 0 ==> b=0 or a^2+b^2 = 1
when b=0 , z = a ( real)
wh ...

你真的好厉害好厉害啊！
好佩服你
可是我不明白为什么 2/(1/z - z) = 2/(-2isinA)
第16题的
如果不要用sin cos有什么别的方法吗？

dunwan2tellu

Q16 用 z = cos A + isinA 可以说是最快而简单的。（不需要那么多计算）。

由于 z = cos A + isinA ,所以 1/z = 1/(cosA +isinA) = (cosA-isinA)/(cos^2A+sin^2A) = cosA -isinA

因此 1/z - z = (cosA-isinA)-(cosA+isinA) = -2isinA

如果你要用其他方法，最直接就是设 z = a + bi ，和注意到 |z|^2 = a^2+b^2 = 1 ,(warning : 会有不少calculation) ，有耐心的话还是可以用这方法找到答案。

[ 本帖最后由 dunwan2tellu 于 15-8-2006 09:52 PM 编辑 ]

nikuang04

紧急求助！！
Prove (A’ n B n C ) u [ (A’ u (B n C’)]’= (A n B') u ( B n c)

[ 本帖最后由 nikuang04 于 15-8-2006 11:22 PM 编辑 ]

dunwan2tellu

E = semester set

(A’ n B n C ) u [ (A’ u (B n C’)]’
=(A’ n B n C ) u [ A n(B'uC)]
=(A'nBnC)u(AnB')u(AnC)
=(AnB')u {Cn[Au(A'nB)]}
=(AnB')u[Cn(AuB)]
=(AnB')u(AnC)u(BnC)
=(AnB'nE)u[(AnC)n(BuB')]u(BnCnE)    [ 注意这关键的一行，那个 E 和 BuB' 是自己加进去的]
=(AnB'nE)u(AnCnB)u(AnCnB')u(BnCnE)
=[(AnB'nE)u(AnB'nC)]u[(AnBnC)u(BnCnE)]
=[(AnB')n(EuC)]u[(BnC)n(AuE)]
=(AnB')u(BnC)

原因是 EuC=E,AuE=E ，而且任何的A,都有 AnE=A

[ 本帖最后由 dunwan2tellu 于 16-8-2006 03:04 PM 编辑 ]

Dexer

帮忙一下

show
1.(1!)+2.(2!)+3.(3!)+.....(n terms)=(n+1)!-1

hamilan911

原帖由 Dexer 于 17-8-2006 08:57 PM 发表
帮忙一下

show
1.(1!)+2.(2!)+3.(3!)+.....(n terms)=(n+1)!-1

sum k(k!) (k=1 to n) = sum (k+1-1)(k!) (k=1 to n)
                     = sum (k+1)! - k! (k=1 to n)
                     = (k+1)! - k! + k! - (k-1)! ... - 1           (用method of differences)
                     = (k+1)! - 1

邵逸夫

我不是很明白series這個chapter...誰可以教我。。尤其是AP,GP。。好像比中5難很多
這些問題都不是很明白。
1) Tn and Sn are respectively the nth term and the sum of the first n terms of a geometric series.
the first term is non-zero,and the common ratio r is an integer more than 1.
a)show that (Tn+4 - Tn+2)/Tn+1 is an even number which is independent of n.
b) express S2n/Sn in terms of r and n.

2)Sn is the sum of the first n terms of a geometric series with the first term,a,and the common ratio,r. show that Sn= a(1-r^n)/1-r if |r|≤1.and S is the sum to infinity of the series, express S in terms of a and r. Show also S-Sn/S = r^n.
For a geometric series where the third and the 6th terms respectively are 27 and 8, find
i) the value of S
ii) the least value of n for which S-Sn/S < 0.001

3) The sum of the first n terms of a series is given by the expression 6- [2^(n+1)]/[3^(n-1)]. by finding an expression of the nth term of the series, show that the given series is a GP and find the value of the first term and the common ratio.

4) the sum of the first 5 terms of a GP is twice the sum of the terms from the 6th to the 15th inclusive. show that r^5 = 1/2 [(sqrt3) - 1 ]

全部都不會做。。。慘了。。。

[ 本帖最后由 邵逸夫 于 19-8-2006 03:24 PM 编辑 ]

dunwan2tellu

1) Tn and Sn are respectively the nth term and the sum of the first n terms of a geometric series.
the first term is non-zero,and the common ratio r is an integer more than 1.
a)show that (Tn+4 - Tn+2)/Tn+1 is an even number which is independent of n.
b) express S2n/Sn in terms of r and n.

a)
Tn = ar^(n-1) 所以 Tn+4 = ar^(n+3) , Tn+2 = ar^(n+1) , Tn+1=ar^n
=> (Tn+4 - Tn+2)/Tn+1 = (ar^(n+3)-ar^(n+1))/(ar^n) = r^3 - r = r(r+1)(r-1)

当然，我们知道如果连续三个号码相乘(比如 11x12x13)，必定有至少一个是偶数，所以 (r-1)r(r+1) 是 even .

b)用geometric progression 的 sum 的formula .

2)Sn is the sum of the first n terms of a geometric series with the first term,a,and the common ratio,r. show that Sn= a(1-r^n)/1-r if |r|≤1.and S is the sum to infinity of the series, express S in terms of a and r. Show also S-Sn/S = r^n.
For a geometric series where the third and the 6th terms respectively are 27 and 8, find
i) the value of S
ii) the least value of n for which S-Sn/S < 0.001

Sn = a + ar + ar^2 + ... + ar^(n-1) ...(1)
rSn = ar + ar^2 + .. + ar^n ......(2)
(1)-(2) : (1-r)Sn = a - ar^n = a(1-r^n)
必须是 |r|<1 , （没有 = ) ,所以 Sn = a(1-r^n)/(1-r)
Sum to infinity = S = a/(1-r) (因为 n-> oo 时，r^n -> 0 )
应该是要证明 (S-Sn)/S = r^n 吧？带入上面找到的东西进去就可以了。
其他的应该不会难。（运用你所找到的 (S-Sn)/S = r^n 来做(ii)

3) The sum of the first n terms of a series is given by the expression 6- [2^(n+1)]/[3^(n-1)]. by finding an expression of the nth term of the series, show that the given series is a GP and find the value of the first term and the common ratio.

Given Sn = 6- [2^(n+1)]/[3^(n-1)] , 而你的目的是找 Tn (应该知道 Tn的意思吧？不然就看 Q1) 。不过我们知道

Tn = Sn - S_{n-1}  (不难想象因为 Sn = T1+T2+T3 + ... + Tn-1 + Tn = Sn-1 + Tn ==> Tn = Sn - Sn-1)
但你得到 Tn = ? 后，试试和 GP 的pattern 比较看看 (也就是和ar^(n-1)比较）。如果 pattern 一样，那么它就是 GP .

4) the sum of the first 5 terms of a GP is twice the sum of the terms from the 6th to the 15th inclusive. show that r^5 = 1/2 [(sqrt3) - 1 ]

sum of first 5 term = S5
Sum from 6th to 15th term = S15 - S5
已知 S5 = 2(S15 - S5) <==> 3(S5) = 2(S15) ,然后再用 GP sum 的 formula 带入，simplify 后答案出现。

邵逸夫

1)   n
find ∑ ( 2n + 1 - 2r ) in terms of n.
     r=0

可以解釋這題嗎？我不會。。。

[ 本帖最后由 邵逸夫 于 20-8-2006 04:48 PM 编辑 ]

dunwan2tellu

1)   n
find ∑ ( 2n + 1 - 2r ) in terms of n.
     r=0

通常你如果没有头绪的话，最直接就是先写出 first few term 来看到底他是何方神圣。

(2n+1-0) + (2n+1-2) + (2n+1-4) + (2n+1-6) + ... +（2n+1-2(n-1))+( 2n + 1 - 2n)

= (2n+1) + (2n-1) + (2n-3) + (2n-5) + ... + 3 + 1

看到什么吗？

如果看不到，就倒回来看

1 + 3 + ... + (2n-5) + (2n-3) + (2n-1) + (2n+1)

熟悉吗？是的，他就是 1+3+5+7+9 + ... + (2n+1)

表示什么？他不就是一个 AP 的 Sum 吗？

邵逸夫

哦。。他的答案是 n^2 + 2n + 1
但是我就是不懂要怎樣得到。。。