m*********y
发帖数: 1735 | 1 1)
假设第二个方程的解是x1,x2，所以第一个方程的解就是(x1+5),(x2+5)
所以，x1+x2=-7,x1*x2=3;
b=-(x1+5)+(x2+5)=-(x1+x2)-10=-3
c=(x1+5)*(x2+5)=x1*x2+5(x1+x2)+25=3-35+25=-7
2)
{27+9b+3c+d=0 (1)
64+16b+4c+d=0;(2)
125+25b+5c+d=0;(3)
}
(2)-(1)=(4)
(3)-(2)=(5)
(5)-(4)=(6)
倒回去
b=-12 c=47; d=-60;
3) 假设三个根是x1,x2,x3
(x-x1)(x-x2)(x-x3)=0
x1x2+x2x3+x1x3=-13
x1+x2+x3=3;
x1x2x3=-15;
1/x1+1/x2+1/x3=(x1x2+x2x3+x1x3)/(x1x2x3)=13/15
4)the same way as above
x1^2+x2^2+x3^2=(x1+x2+x3)^2-(x1x2+x2x3+x1x3)=15^2-66=159
5)
almost the same
first solu... 阅读全帖 |
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d*z
发帖数: 150 | 2 来自主题: Mathematics版 - 极大值问题 i)已知x1,x2,x3,x4,x5是非负实数，，而且x1+x2+x3+x4+x5=1,请问
2(x1x2+x2x3+x3x4+x4x5+x5x1)+x1x3+x2x4+x3x5+x4x1+x5x2
的最大值是多少
ii)已知x1,x2,x3,x4,x5,x6是非负实数，，而且x1+x2+x3+x4+x5+x6=1,请问
2(x1x2+x2x3+x3x4+x4x5+x5x6+x6x1)+x1x3+x2x4+x3x5+x4x6+x5x1+x6x2
的最大值是多少 |
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r*****t
发帖数: 286 | 3 ☆─────────────────────────────────────☆
mechanics (mechanics) 于 (Tue Feb 27 15:39:10 2007) 提到:
有一个string, x1x2…xn, 顺序分割, 比如说
1) (x1)(x2x3)(x4x5)x6
2) (X1x2x3)(x4x5x6)
每种分割对应一个cost, 比如说第一种分割, cost是C(x1)+C(x2x3)+C(x4x5)+C(x6)
第二种分割, cost是C(x1x2x3)+C(x4x5x6)
如何找到一种分割使cost最小
☆─────────────────────────────────────☆
Phase (amplitude) 于 (Tue Feb 27 15:46:56 2007) 提到:
interview question?
☆─────────────────────────────────────☆
kmalloc (婚后的幸福生活(儿子能背诗了)) 于 (Tue Feb 27 15:50:13 2007) |
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DQ
发帖数: 1580 | 7 两岁的娃。。。
简单的：除了衣服啥都不用带。吃的可以吃table food了，尿布你嫌重，可以到了目的
地买一包小的，wipe同样道理。vegas也不冷，带点娃的夏装换洗衣服就好了，一般按
天数+2带（实在不够洗了一天就干了）。娃洗澡的洗发水之类的。路上无聊的话带两本
书或者两个娃爱的玩具。当然stroller和car seat不算在内。
复杂的：娃的各种食物,衣服天数x2x3。。。,尿布wipe，娃洗澡的浴巾，毛巾，洗头膏
，香皂，diaper cream等等。娃睡觉时的各种物品。娃吃饭坐的booster，娃上厕所用
的小马桶。。。
总体上你就按娃的衣食住行分类看着带呗。 |
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B******1
发帖数: 9094 | 8 What is the sum of the squares of the solutions of x3-15x2+66x-80=0?
(x1)^2 + (x2)^2 + (x3)^2 = (x1 + x2 + x3)^2 - 2*(x1x2 + x1x3 + x2x3)
= 80^2 -2*(66)=6400 - 132 = 6268 |
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f*******g
发帖数: 245 | 9 how come ? I am confused by this statement. Have a calculation:
Grand parentals: X1Y vs X2X3
Father: X2(or3)Y vs Mother: X4X5
Daughter: X2(or3)X4(or5)
now, the genetical material (x chromosome) between father and his brother
has 50% identical.
The genetical material (x chromosome) between father and dauthter has 50%
identical.
Conclusion: the chance for a female to share identical X chromosome with her
uncle is 25%. |
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l****i
发帖数: 4609 | 10 母亲的基因：
X1X2
父亲的基因：
X3Y
儿子的基因
X1Y； X2Y
女儿的基因
X1X3；X2X3
为什么生的儿子智力状况会不一样，因为母亲遗传给儿子的基因还是不同的，所以查一
下祖宗八代还是有必要的。 |
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k***s
发帖数: 206 | 11 If i have a feature expansion:
{0, 1}^n -> {0, 1}^2n. for example. If n = 3, then
Theta(x1,x2,x3) = (1, x1, x2, x3, x1x2, x1x3, x2x3, x1x2x3): Every possible
"logical OR" from these 3 variables (including 1).
In this case, show K(x,y) for Theta can be computed in polynomial time: poly
(n).
K(x,y) = Theta(x) dot product Theta(y).
这里Theta(y) 指的是什么？？？？ |
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j******a
发帖数: 1599 | 12 【 以下文字转载自 Mathematics 讨论区 】
发信人: jiamajia (其实我特讨厌高丽棒子), 信区: Mathematics
标 题: 牛仁们帮我看看这个概率题
发信站: BBS 未名空间站 (Sat Mar 1 12:47:40 2008)
Using the joint characteristic function show that if X1, X2, X3, X4 are
jointly Gaussian (correlated) random variables with zero mean, then
E[X1X2X3X4] = E[X1X2]E[X3X4] + E[X1X3]E[X2X4] + E[X1X4]E[X2X3]
TIA |
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j******a
发帖数: 1599 | 13 Using the joint characteristic function show that if X1, X2, X3, X4 are
jointly Gaussian (correlated) random variables with zero mean, then
E[X1X2X3X4] = E[X1X2]E[X3X4] + E[X1X3]E[X2X4] + E[X1X4]E[X2X3]
TIA |
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k***s
发帖数: 206 | 14 If i have a feature expansion:
{0, 1}^n -> {0, 1}^2n. for example. If n = 3, then
Theta(x1,x2,x3) = (1, x1, x2, x3, x1x2, x1x3, x2x3, x1x2x3): Every possible
"logical OR" from these 3 variables (including 1).
In this case, show K(x,y) for Theta can be computed in polynomial time: poly
(n).
K(x,y) = Theta(x) dot product Theta(y).
这里Theta(y) 指的是什么？？？？ |
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j******a
发帖数: 1599 | 15 【 以下文字转载自 Mathematics 讨论区 】
发信人: jiamajia (其实我特讨厌高丽棒子), 信区: Mathematics
标 题: 牛仁们帮我看看这个概率题
发信站: BBS 未名空间站 (Sat Mar 1 12:47:40 2008)
Using the joint characteristic function show that if X1, X2, X3, X4 are
jointly Gaussian (correlated) random variables with zero mean, then
E[X1X2X3X4] = E[X1X2]E[X3X4] + E[X1X3]E[X2X4] + E[X1X4]E[X2X3]
TIA |
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s******h
发帖数: 539 | 16 直接算，然后利用independence和对称性。
Cov((X2 - X1)^2, (X2 - X3)^2)
= Cov(X2^2 - 2X2X1, X2^2 - 2X2X3)
= Var(X2^2) - 2Cov(X2^2, X2X3) - 2Cov(X2^2, X1X2) + 4Cov(X1X2, X3X2)
= Var(X2^2)
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