T*******x
发帖数: 8565 | 1 这个题在格点上求级数，每一条coprime格点所在的射线上有一个zeta(3)。这种方法明
显有好几种方式可以扩展。我本希望有方法直接算coprime格点上的级数和，硬算。现
在这个办法有点迂回，我感觉。当然，它不依赖于Stark的结果，是直接证明。
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发帖数: 1 | 2 用这么复杂吗？
难道不是简单变形求极限。结果不是pi^2/6?......
: 这个题在格点上求级数，每一条coprime格点所在的射线上有一个zeta(3)。这种
方法明
: 显有好几种方式可以扩展。我本希望有方法直接算coprime格点上的级数和，硬
算。现
: 在这个办法有点迂回，我感觉。当然，它不依赖于Stark的结果，是直接证明。
: /2
: /2
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p*****k
发帖数: 318 | 3 off-topic question. i'm just curious, how to score sole 2pts in football? never saw it in a game. is it same as the one for 8pts?
the first question is pretty simple, only 1 is not possible (6,7 and 8 are redundant anyway).
for the general case, i think the following lemma might help:
two natural numbers p and q. if gcd(p,q)=1, i.e., p and q are coprime, then any integer n>=pq can be expressed as ap+bq, where a and b are nonnegative integers.
so let's say you if x_1 is coprime with some x_i |
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m********4
发帖数: 1837 | 4 I tried and may figure it out this kind of slow way:
Lemma 1a: If a|c, b|c, and a, b are coprime, then ab|c.
Lemma 1b: lcm(a,b)*gcd(a,b)=ab.
Proof: Let m=lcm(a,b), d=gcd(a,b). Then a|m, b|m.
=> d*(a/d)|m, d*(b/d)|m.
a/d|m/d, b/d|m/d.
Because a/d and b/d are coprime by the
definition of gcd, then applying Lemma 1 we would get
(a/d)(b/d)|m/d.
... 阅读全帖 |
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T*******x
发帖数: 8565 | 5 刚才出了个题，出简单了。再出个难点的：
let A be the set of ordered pair of positive coprime integer (p,q). For
example (1,1), (1,3), (3,1), (15,8).
Find sum of 1/(p*q*(p+q)), where (p,q) runs through set A. |
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发帖数: 1 | 6 你今天跪了几个老黑？
盹盹盹
：刚才出了个题，出简单了。再出个难点的：
：let A be the set of ordered pair of positive coprime integer (p,q). For
：example (1,1), (1,3), (3,1), (15,8).
：Find sum of 1/(p*q*(p+q)), where (p,q) runs through set A.
：☆ 发自 iPhone 买买提 1.24.10 |
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发帖数: 1 | 7 sum n,m = 1 to infinity 1/[nm(n+m)]= sum d=1 to infinity 1/d^3 sum p,q
coprime 1/[pq(p+q)]。所以只需算等式左边，再除以zeta(3)即可。记等式左边的二重
级数和为S。如把二重级数按k=m+n归类，得 S = sum k=2 to infinity 2/k^2 *(1+1/2
+...+1/(k-1))。如把二重级数按m归类，得 S = sum m=1 to infinity 1/m^2 *(1+1/2
+...+1/m)。用 2S-S 易得 S=2zeta(3)。所以题目所求的值为2。 |
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T*******x
发帖数: 8565 | 8 漂亮！
E. L. Stark 证明了 S = 2zeta(3). Wolfram 上有链接。我看到了这个结果，再用以
前推出的Harmonic Series 的公式代入，做一些变换，得到这个题目，及其结果。
但是我并不知道 E.L. Stark 的结果是怎么证明的。看来就是这么证明的。
这个证明同时给出两个不平凡的结果：
一个是 EL Stark 的结果，也就是 S = 2zeta(3)。
另一个是本题的结果，也就是这个coprime sum等于2。
两个结果都很好！
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T*******x
发帖数: 8565 | 9 差的这一点在“出个题”中补齐了。
现在我记录一下这个证明：证明Z[sqrt(-5)] is integrally closed.
1. the integral closure of Z[sqrt(-5)] in Q[sqrt(-5)]=Q(sqrt(-5)) is equal
to the integral closure of Z in Q(sqrt(-5)).
2. The integral closure of Z in Q(sqrt(-5)) is Z[sqrt(-5)].
Therefore, the integral closure of Z[sqrt(-5)] in Q(sqrt(-5)) is itself, and
thus it's integrally closed.
1比较容易，因为any element alpha integral over R=Z[sqrt(-5)] has a monic
polynomial r(x) in R[x] such that r(alpha)=0. 把r(x)的系数按照整数和带sqrt(-
5)的数分开，分成等式两边，然后... 阅读全帖 |
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p*g
发帖数: 141 | 10 那个用random(m) 来生成random(n)的有什么提示么？
比如m/n coprime？ |
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p*****k
发帖数: 318 | 13 note 11 and 60 are co-prime, so the general solution is:
X = 121 (mod 660), or 660*m+121, where m is a non-negative integer.
but i think OP actually meant X modulo all numbers from 2 to 10
results 1, i.e., X=1(mod 2520) and X=0(mod 11).
again 11 and 2520 are coprime. by Euclidean algorithm,
the general solution is: X = 25201 (mod 27720) |
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x******a
发帖数: 6336 | 14 3 and 20 are coprime... you can get any integer with 3m+20n if no other
condition on m and n than they are integers.
if m>=0, n>=0 are enforced, the largest I got is 37. |
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