数学Paper 1讨论专区

hongji

find the sum of square of the first n terms of the positive odd number

..也是用1/6n(n+1)(n+1)?

peaceboy

回复 2401# hongji


    1^2 + 3^2 + 5^2 +.......+（2n-1）^2


sum of r^2 formula = (r/6)(r+1)(2r+1)
sub r = 2n-1
((2n-1)/6)(2n-1+1)(2(2n-1)+1) =

然后自己来

hongji

prove that ,if x is so small that terms in x^3 and higher powers may be neglected,then [√1+x/1-x]=1+x+1/2x^2.
by substituting a suitable value of x in your result,show that √11 is approximately equal to 663/200


i use ascending power of x
[(1+x)/(1-x)]^1/2  
=(1+x)^1/2 (1-x)^-1/2
=(1+[1/2]/1(x)   +  (1/2)(-1/2)]/2(x^)2   +[(1/2)(-1/2)(-3/2)]/6(x)^3


=(1+[(-1/2)]/1(-x)+[(-1/2)(-3/2)/2] (-x)^2+[(-1/2)(-3/2)(-5/2)]/6(-x)^3

=2+1/2x^2+13/16x^3..
but i don't knowx=?

whyyie

回复 2403# hongji

√1+x/1-x 是 √(1+x)/ (1-x)还是√[(1+x)/(1-x)]

hongji

回复  hongji

√1+x/1-x 是 √(1+x)/ (1-x)还是√[(1+x)/(1-x)]
whyyie 发表于 10-12-2010 11:11 PM

    is √[(1+x)/(1-x)]

√(1+x)/√(1-x)

whyyie

[(1+x)/(1-x)]^1/2 = (1+x)^(1/2) - (1-x)^(-1/2)

(1+x)^(1/2) = 1+ x/2 - (x^2)/8
(1-x)^(-1/2) = 1+ x/2 +(3x^2)/8

=>[(1+x)/(1-x)]^1/2 = 1+x + (x^2)/2

By substituting x = 1/10......

介意我问下这题目哪来的?

Allmaths

[(1+x)/(1-x)]^1/2 = (1+x)^(1/2) - (1-x)^(-1/2)

(1+x)^(1/2) = 1+ x/2 - (x^2)/8
(1-x)^(-1/2) = 1+ ...
whyyie 发表于 10-12-2010 11:30 PM

   
话说这题目在没有经验下要找suitable value of x 是比较难。。。

Minute

MATRICES
If A=(a b)
       (c d)
SHOW that adj(adj A)=A

Help please...

whyyie

回复 2408# Minute

Minute

回复 2409# whyyie


   谢谢....

Lov瑜瑜4ever

how to integrate e^-x?

Allmaths

how to integrate e^-x?
Lov瑜瑜4ever 发表于 11-12-2010 01:54 PM

用∫f '(x)e^f(x) dx= e^f(x)


   ∫e^-x dx=-∫-e^-x dx
            =-e^-x+C

Lov瑜瑜4ever

用∫f '(x)e^f(x) dx= e^f(x)


   ∫e^-x dx=-∫-e^-x dx
            =-e^-x+C
Allmaths 发表于 11-12-2010 01:58 PM

The continuous random variable X has probability density function
f(x)=e^-x , x>0
     =0 , otherwise

Find the cumulative distribution function, F(x).

Ans: F(x)=0 , x<0
            =1-e^-x , x>0

nkrealman

回复 2413# Lov瑜瑜4ever


  只需要 integrate 从 0 到 x

Lov瑜瑜4ever

回复  Lov瑜瑜4ever


  只需要 integrate 从 0 到 x
nkrealman 发表于 11-12-2010 02:21 PM

d/dx(e^-x)=-e^-x
integrate e^-x=-e^-x
那么d/dx(e^-x)=integrate e^-x吗？

Allmaths

d/dx(e^-x)=-e^-x
integrate e^-x=-e^-x
那么d/dx(e^-x)=integrate e^-x吗？
Lov瑜瑜4ever 发表于 11-12-2010 02:33 PM

    应该不算吧...

因为integrate后还有constant...

Lov瑜瑜4ever

应该不算吧...

因为integrate后还有constant...
Allmaths 发表于 11-12-2010 02:37 PM

那么我intergrate e^-mx的时候都用differentiation来做
然后再加1个C
这样也不是一样？
还是只能用integration by substitution哦？

nkrealman

回复 2417# Lov瑜瑜4ever


   不明白你说什么

Allmaths

那么我intergrate e^-mx的时候都用differentiation来做
然后再加1个C
这样也不是一样？
还是只能用i ...
Lov瑜瑜4ever 发表于 11-12-2010 02:41 PM

如果说d/dx(e^-x)=integrate e^-x, 这情况是在C=0的时候才对...

通常integrate e^x 之类的都用general formula, ∫f'(x)e^f(x) dx=e^f(x)+C


不过用substitution 也可以啦...


∫e^-mx dx


Let  u=-mx

du/dx=-m
-du/m=dx


∫e^-mx dx=∫-(e^u)/m du
               =-(e^-mx)/m+C

Lov瑜瑜4ever

很乱下...我在研究下...谢谢了2位...