数学Paper 1讨论专区

Lov瑜瑜4ever

尝试~
To prove (z1/z2)*=z1*/z2*,
Consider
(z1/z2)*x(z2)*
={(z1/z2)x(z2)}*the conjugate of a ...
HanQing 发表于 17-12-2010 03:55 PM

我也使觉得有点奇怪。。。看不懂。。。

HanQing

回复 2461# Lov瑜瑜4ever

To prove (z1/z2)*=z1*/z2*,
Consider
(z1/z2)*x(z2)* （1）
={(z1/z2)x(z2)}*the conjugate of a product is the product of the conjugates （2）
=(z1)* ↔ (z1/z2)*=(z1*/z2*) （3）
这问题限制我们以(z1/z2)*x(z2)*为开头，并利用the conjugate of a product is the product of the conjugates来证明the conjugate of a quotient is the product of the quotient，所以我才写下此步骤（2），至于（3）[ 以英文来写吧==华语不好]if and only if (z1/z2)*=(z1*/z2*) , then {(z1/z2)x(z2)}*=(z1)*, not sure about that but thats the way how i argue out the part, but it seems vague =.= would be better for other to try this =X just want to share my point of view heh

peaceboy

回复 2456# Lov瑜瑜4ever


    the conjugate of a product is the product of the conjugates  
means z1z2 = z1*z2*

(z1/z2)*x(z2)*= [(z1/z2)x z2]*
                      =(z1)*

(z1/z2)*x(z2)*/z2* = z1*/z2*


你有没有答案？

Lov瑜瑜4ever

回复  Lov瑜瑜4ever


    the conjugate of a product is the product of the conjugates  
means z1 ...
peaceboy 发表于 18-12-2010 06:35 PM

没有哦。。。不过看起来你这个比较容易明白。。。

hongji

the probaility that a student is late for school on any day is 1/6.  Find the probaility that the student is

(a)late for school at least once in any two consecutive days


(b) never late for school in any two consecutive days.

peaceboy

回复 2465# hongji


    the probaility that a student is late for school on any day is 1/6.  Find the probaility that the student is

(a)late for school at least once in any two consecutive days

1/6 * 5/6 + 5/6*1/6 + 1/6*1/6 =11/36


(b) never late for school in any two consecutive days.

5/6*5/6 = 25/36

hongji

我想问关于probability distributions的normal distrbution

given Z is the standard normal distribution variable .
find the values of the following.

a)P(Z<-0.6)        <<这个怎样算到0.2743?

b)P(-1<Z<=0.2)

Log

a) P(Z<-0.6)=P(Z>0.6)=0.2743  ---因为Z-graph is symmetrical about the f(Z) axis..so, use calculator/table to calculate
b.) P(-1<Z<=0.2)=1-P(Z<-1)-P(Z≥0.2)
                          =1-P（Z>1)-P(Z≥0.2)
                           = 1-0.15866-0.42074=0.4206
做这类的问题最好skecth出来the graph。。。

Minute

integral 1/(e^x+e^-x) dx
by using the substitution u=e^x


ans:tan^-1 e^x + c

peaceboy

回复 2469# Minute


    integral 1/(e^x+e^-x) dx
by using the substitution u=e^x

u=e^x ---@
du/dx = e^x
         =u
dx=du/u

int 1/(e^x+e^-x) dx = int 1/(u+1/u) du/u
                             = int 1/[(u^2+1)/u] du/u
                             = int 1/(u^2+1) du

let u=tan a---*
du/da = sec^2 a
du = sec^2a da

int 1/(u^2+1) du = int [1/(tan^2 a+1)] (sec^2 a) da
since tan^2 a +1 = sec^2 a
therefore ,
int [1/(tan^2 a+1)] (sec^2 a) da = int [1/(sec^2 a)] (sec^2 a) da
                                               =int 1 da
                                               =a+c
                 
from * , u = tan a
            a = tan^-1 u
from @ , u= e^x
a= tan^-1  e^x

therefore , integral 1/(e^x+e^-x) dx=tan^-1 e^x + c

Minute

回复 2470# peaceboy


   谢谢你！

797

我想问你们这几题哦~~chapter 8 integration

differentiate x^2/2x-1 with respect to x and hence evaluate ∫x(x-1)/(2x-1)^2 dx.

我做的
dy/dx= 2x^2-1 . -(2x-1)^-1-1 (2)
        =2x（-2）/ (2x-1)^2

可是答案是=2x（x-1）/ (2x-1)^2

请问我哪里做错了？

2. given that y=(x+3)&#11905;(x^2+1),show that dy/dx=3x/ㄏ(2x-3)

   dy/dx=(x+3)1/2(2x-3)^-1/2 + (2x-3)^1/2 （1）
            = ？

peaceboy

回复 2472# 797


   
differentiate x^2/2x-1 with respect to x and hence evaluate ∫x(x-1)/(2x-1)^2 dx.

y= x^2/2x-1
dy/dx = [(2x-1) (2x) - (x^2) (2)] /(2x-1)^2
         = (4x^2-2x-2x^2) / (2x-1)^2
         = (2x^2-2x) / (2x-1)^2
         = 2x(x-1) /  (2x-1)^2


2. given that y=(x+3)&#11905;(x^2+1),show that dy/dx=3x/ㄏ(2x-3)

(x+3)[(x^2+1)^(1/2)] = (x+3) (1/2)(2x)(x^2+1)^(-1/2) + (x^2+1)^(1/2)
                               = [(x)(x+3)]/(x^2+1)^(1/2)  + (x^2+1) / (x^2+1)^(1/2)
                                 =(x^2+3x+x^2+1) /  (x^2+1)^(1/2)
                                 = 2x^2+3x+1/  (x^2+1)^(1/2)
你问题有没有给错？

whyyie

回复 2472# 797

跟peaceboy兄找到的答案一样. 题目应该是y=(x+3)&#11905;(2x-3)吧.

Minute

integration~~
evaluate ∫sin3XsinXdx

whyyie

回复 2475# Minute

用integration by part?

whyyie

integration~~
evaluate ∫sin3XsinXdx
Minute 发表于 5-1-2011 07:26 PM

我知道问题出在哪里了. Integration by parts 的答案应该不是参考书给的答案吧?

用cos A- cos B = -2 sin ((A+B)/2) sin ((A-B)/2)
=>sin 3x sin x = (cos 2x - cos 4x)/2

∫sin3x sinx dx = sin^3 x cos x

peaceboy

回复 2477# whyyie


    不稀饭trigo
这问题应该有part A吧

whyyie

回复 2478# peaceboy

嗯, 说得也是.
你最近有在忙些什么吗?

Log

用 factor formula...最快。。好像whyyie那样
如不用factor formula, 也能解，可是较长。。。