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发表于 24-12-2011 02:50 PM
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发表于 24-12-2011 02:56 PM
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发表于 25-12-2011 05:22 PM
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Chapter2 polynomial
x^2 + px +q = 0 ,q not equal to 0,
roots are a(alpha) and B(beta)
one root of x^2 + p'x +q = 0 is ka,
show the other root is B/k.
p' 是什么啊? |
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发表于 25-12-2011 07:50 PM
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Chapter2 polynomial
x^2 + px +q = 0 ,q not equal to 0,
roots are a(alpha) and B(beta)
one ro ...
huatiang 发表于 25-12-2011 05:22 PM
就把它当成另外一个unknown咯。。。 |
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发表于 25-12-2011 09:50 PM
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回复 2864# Allmaths
p' not equal to p, but p(p') = 1,是这样吗? |
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发表于 25-12-2011 09:54 PM
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回复 Allmaths
p' not equal to p, but p(p') = 1,是这样吗?
huatiang 发表于 25-12-2011 09:50 PM
做么有这样的东西出来的? |
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发表于 26-12-2011 01:10 AM
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做么有这样的东西出来的?
Allmaths 发表于 25-12-2011 09:54 PM
果然是宝刀未老啊。。。。 |
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发表于 26-12-2011 01:49 AM
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果然是宝刀未老啊。。。。
Log 发表于 26-12-2011 01:10 AM
老大来了。。。小弟就不算什么了。。。 |
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发表于 21-2-2012 04:16 PM
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回复 2868# Allmaths
我有paper2的问题想请教一下。去paper2讨论区,帮帮忙,谢谢。 |
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发表于 21-2-2012 04:18 PM
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回复 2867# Log
Allmaths的老大一定很强咯,得空帮帮忙一下吧
我有paper2的问题,去paper2讨论专区教教我吧,谢谢。 |
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发表于 13-3-2012 06:11 PM
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given y= sin x / x^2, how to prove x^2 d2y/dx2 + 4 dy/dx +(x^2 + 2)y= 0? |
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发表于 13-3-2012 07:46 PM
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given y= sin x / x^2, how to prove x^2 d2y/dx2 + 4 dy/dx +(x^2 + 2)y= 0?
happy=] 发表于 13-3-2012 06:11 PM
题目有误。。应该是4x(dy/dx)
y=(sin x)/x^2
x^2y=sin x
x^2(dy/dx)+2xy=cos x
x^2(d^2y/dx^2)+2x(dy/dx)+2x(dy/dx)+2y= -sin x
x^2(d^2y/dx^2)+4x(dy/dx)+2y= -x^2y
x^2(d^2y/dx^2)+4x(dy/dx)+2y+x^2y=0
x^2(d^2y/dx^2)+4x(dy/dx)+(x^2+2)y=0 |
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发表于 13-3-2012 09:35 PM
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回复 2872# Allmaths
thx...maybe the textbook print wrong d... |
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发表于 14-3-2012 07:49 PM
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发表于 15-3-2012 05:40 PM
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這裡還有沒有人啊
誰幫忙看一下這題
算不到不甘心呃
鉛字是答案
dawn29 发表于 14-3-2012 07:49 PM
∫1/(1+x^2)^2 dx
Let x=tan u,
dx=sec^2 (u) du
∫1/(1+x^2)^2 dx=∫[sec^2 (u)]/[sec^2 (u)]^2 du
=∫du/sec^2 (u)
=∫cos^2 (u) du
=(1/2) ∫[cos (2u)+1]du
=[sin (2u)]/4 + u/2 +C
u=tan^-1 x,
∫1/(1+x^2)^2 dx=[sin (2(tan^-1 (x))]/4 + [tan^-1 (x)]/2 +C
=(1/2)[x/(1+x^2)+tan^-1 (x)]+C |
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发表于 15-3-2012 10:52 PM
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发表于 15-3-2012 10:56 PM
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发表于 16-3-2012 01:08 AM
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假期沒人可以問啊啊啊
dawn29 发表于 15-3-2012 10:56 PM
(1) Let u=2x+1,
x=(u-1)/2
du/2=dx
substitute进去就可以了
(2) Let x=2 cos u
dx=-2 sin (u) du
when x=1, u=pi/3
when x=2, u=0
∫ dx/[x^2(4-x^2)^(1/2)] from 1 to 2= ∫ [-2sin (u) du]/{[4cos^2 (u)][4-4cos^2 (u)]^(1/2)} from pi/3 to 0
=(-1/4) ∫ sec^2 (u) du
=(-1/4)[tan (0)-tan(pi/3)]
=[(3)^(1/2)]/4
(3)Let u=2x-1
x=(u+1)/2
du/2=dx
substitute进去就可以了
(4)Let u=x^2
du/2=x dx
substitute进去就可以了 |
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发表于 16-3-2012 09:44 PM
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A right circular cone of base radius r and height h has a total surface area S and volume V. Show that 9V^2=r^2(S^2-2π r^2S). HOW TO SHOW???? |
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发表于 17-3-2012 03:18 AM
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A right circular cone of base radius r and height h has a total surface area S and volume V. Show th ...
happy=] 发表于 16-3-2012 09:44 PM
S=(pi)r^2+(pi)r(r^2+h^2)^(1/2) ---eq 1
V=(1/3)(pi)(r^2)(h) ---eq 2
from eq 2,
9V^2=(pi^2)(r^4)(h^2)
from eq 1,
S/[(pi)(r)]=r+(r^2+h^2)^(1/2)
h^2=S^2/[(pi)(r)]^2-2S/(pi)
Sub h^2=S^2/[(pi)(r)]^2-2S/(pi) into 9V^2=(pi^2)(r^4)(h^2)
9V^2=(pi^2)(r^4){S^2/[(pi)(r)]^2-2S/(pi)}
9V^2=r^2[S^2-2(pi)(r^2)S] |
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