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In relation with a probabilistic problem for concrete numbers I came upon the generalization of an idea and there’s a problem:
Let’s suppose we are throwing by special cubes with sides numbered serially by 1,2,…,m and the probability of toss of each number is the same according to the classical definition of probability). Let denote the sum of all numbers on the sides of all cubes shortly after the last toss. Which sum has the highest probability.
Let be random variables that represents the value of the toss on cubes in this order and . Then should be the value of the most probable sum on the sides of cubes (or in case that isn’t integer let’s take the closest two integers which E X is situated between). At last it’s clear that , then:
But I would like to solve it more classically and exactly, so first I need to solve this:
Let be some natural numbers such that , then the esencial question is how many (orderly) somes does there exist such that each x_i is a natural number from 1 to maximally and:
.
Could anybody give me any analytic formula (depending on ) for number of all convenient some (or any hint).
Obviously, the number of all convenient somes is also the number of all adic combinations with reprise of elements such that each element can appear minimally once and maximally times.
I’ve tried to use the principle of inclusion and exclusion, but it doesn’t give me the right result.
Thank you….
Q5:
Lemma:
Prove of lemma:
(Induction) Cases n = 2, 3: Obvious by calculation. Suppose cases n = k – 2, k – 1 are true. When n = k,
Thus the lemma is true. Thus
Thus .
(It doesn’t need to be written in “expanded form”, does it? )
Q8:
Suppose x = a + bi, y = b + ci, z = c + ai. Hence: a = Re(x) b = Re(y) c = Re(z) And the original statement becomes:
Assume . Then:
.
Notice that since a,b,c > 0, we have .
Moreover, the cosine values do not change if we scale the whole system, so let’s assume x, y, z
Q5:
Lemma:
Prove of lemma:
(Induction) Cases n = 2, 3: Obvious by calculation. Suppose cases n = k – 2, k – 1 are true. When n = k,
Thus the lemma is true. Thus
Thus .
(It doesn’t need to be written in “expanded form”, does it? )
Q8:
Suppose x = a + bi, y = b + ci, z = c + ai. Hence: a = Re(x) b = Re(y) c = Re(z) And the original statement becomes:
Assume . Then:
.
Notice that since a,b,c > 0, we have .
Moreover, the cosine values do not change if we scale the whole system, so let’s assume x, y, z
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數學白癡 ≧▽≦ Joined: 26 Apr 2004Posts: 11 




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Fernando wrote: 
設平面上有3個單位向量a,b,c (1)若a與b垂直,則計算向量c(c dot a)a (c dot b)b 的長度 (2)若a與b垂直,則 c 是否有可能同時與a及b垂直? (3)證明:平面上不可能有3個互相垂直的非零向量? p.s:所有的a b c上面皆有一個向量的符號,而c dot a是指內積的意思 順帶問幾個觀念: 1.向量a dot 向量b 和向量a*向量b有什麼不同?
2.零向量的意義?是否會與所有向量垂直or what? 
先答觀念: 1)
是向量的點積 (Dot product), 是 a 投射到 b 上的長度再乘以 b 的長度, 即 , 其中 是 a和b之夾角.
而 是向量的叉積 (Cross product), 其垂直於 a 和 b, 且長度為 a、b 所構成的平行四邊形. 即 , 其中 為一垂直於 a 和 b 的單位向量.
點積和叉積 最大的分別是 點積 會產生一 純量 (Scalar), 而叉積會產生向量 (Vector) (*正式來說是偽向量 Pseudovector, 因為那個單位向量有2個選法)
2)
零向量 (Zero vector or Null vector、0) 是長度為 0 的向量, 故亦沒有方向.
—————————————————————————— 回到問題:
已知 a = b = c = 1.
(1) 因 a 與 b 垂直, 故有 . 另外還有 . 所以:
(2) 不能. 但若 a,b,c 在三維空間中則可.
(3) 假設平面上存在4點 A,B,C,O 使得 , 且定義 , 那麼由 可知 a 和 c 必定在同一線上 (因為 ), 與 矛盾.