Đặt $p=2q+1$. Cmr nếu $p>3$ thì $p^{3}\mid 2^{3q}q!-(-1)^{q}(2q-1)!! $
#1
Đã gửi 13-07-2011 - 10:43
Zaraki
-
- Phó Quản lý Toán Cao cấp
-
- 4273 Bài viết
PQT
Discovery is a child’s privilege. I mean the small child, the child who is not afraid to be wrong, to look silly, to not be serious, and to act differently from everyone else. He is also not afraid that the things he is interested in are in bad taste or turn out to be different from his expectations, from what they should be, or rather he is not afraid of what they actually are. He ignores the silent and flawless consensus that is part of the air we breathe – the consensus of all the people who are, or are reputed to be, reasonable.
Grothendieck, Récoltes et Semailles (“Crops and Seeds”).
#2
Đã gửi 25-07-2011 - 11:01
Zaraki
-
- Phó Quản lý Toán Cao cấp
-
- 4273 Bài viết
PQT
$ (-1)^{q}(2q-1)!!=(2-p)(4-p)...(2q-p)=2^{q}*q![1-\dfrac{p}{2}\sigma_{1}+\dfrac{p^{2}}{4}\sigma_{2}+O(p^{3})], $
Với $ \sigma_{1}=\sum_{k=1}^{q}\dfrac{1}{k},\sigma_{2}=\sum_{1\le j<k\le q}\dfrac{1}{jk}. $
Với $ \sigma_{1}=\sum_{k=1}^{q}\dfrac{1}{k},\sigma_{2}=\sum_{1\le j<k\le q}\dfrac{1}{jk}. $
$ 2^{2q}=\sum_{k=0}^{q}C_{p}^{2k}=1-\dfrac{p}{2}\sum_{k=1}^{q}\dfrac{1}{k}\prod_{j=1}^{2k-1}(1-\dfrac{p}{j})=1-\dfrac{p}{2}\sigma_{1}+\dfrac{p^{2}}{2}\sum_{k=1}^{q}\sum_{j=1}^{2k-1}\dfrac{1}{jk} $
$ 2\sum_{k=1}^{q}}\sum_{j=1}^{2k-1}\dfrac{1}{jk}=\sum_{k=1}^{q}\sum_{i=1}^{k-1}\dfrac{1}{ik}+\sum_{k=1}^{q}\sum_{i=k+1}^{q}\dfrac{1}{k(i-\dfrac{p}{2})}=\sigma_{2}+O(p). $
Discovery is a child’s privilege. I mean the small child, the child who is not afraid to be wrong, to look silly, to not be serious, and to act differently from everyone else. He is also not afraid that the things he is interested in are in bad taste or turn out to be different from his expectations, from what they should be, or rather he is not afraid of what they actually are. He ignores the silent and flawless consensus that is part of the air we breathe – the consensus of all the people who are, or are reputed to be, reasonable.
Grothendieck, Récoltes et Semailles (“Crops and Seeds”).
#3
Đã gửi 18-08-2011 - 18:19
thangthan
-
- Thành viên
-
- 92 Bài viết
Hạ sĩ
Bạn giải bài này cho bạn hiểu đấy à.Không giải thích O(p) là gì thì làm sao người đọc bít được.$ (-1)^{q}(2q-1)!!=(2-p)(4-p)...(2q-p)=2^{q}*q![1-\dfrac{p}{2}\sigma_{1}+\dfrac{p^{2}}{4}\sigma_{2}+O(p^{3})], $
Với $ \sigma_{1}=\sum_{k=1}^{q}\dfrac{1}{k},\sigma_{2}=\sum_{1\le j<k\le q}\dfrac{1}{jk}. $$ 2^{2q}=\sum_{k=0}^{q}C_{p}^{2k}=1-\dfrac{p}{2}\sum_{k=1}^{q}\dfrac{1}{k}\prod_{j=1}^{2k-1}(1-\dfrac{p}{j})=1-\dfrac{p}{2}\sigma_{1}+\dfrac{p^{2}}{2}\sum_{k=1}^{q}\sum_{j=1}^{2k-1}\dfrac{1}{jk} $
$ 2\sum_{k=1}^{q}}\sum_{j=1}^{2k-1}\dfrac{1}{jk}=\sum_{k=1}^{q}\sum_{i=1}^{k-1}\dfrac{1}{ik}+\sum_{k=1}^{q}\sum_{i=k+1}^{q}\dfrac{1}{k(i-\dfrac{p}{2})}=\sigma_{2}+O(p). $
1 lời giải nhẹ nhàng hơn.
Ta có $2^{3q}q!(2q-1)!!=2^{2q}.2.4...2q.1.3...(2q-1)=2^{2q}(2q)!=2.4...4q$$ =2.4..(p-1)(p+1)...(2p-2)=(p-(p-2))(p-(p-4))...(p-1)(p+1)...(p+(p-2))=(p^2-(p-2)^2)...(p^2-1^2)=p^{2q}-S_1.p^{2q-2}+...+(-1)^{q-1}.S_{q-1}.p^2+(-1)^q.S_q$
Với $S_k $là tổng các tích k số từ $ 1^2,3^2,...,(p-2)^2$
Mà $S_q=1^2...(2q-1)^2=((2q-1)!!)^2$
nên $(2^{3q}q!-(2q-1)!!.(-1)^q)(2q-1)!!=p^{2q}-S_1.p^{2q-2}+...+(-1)^{q-1}.S_{q-1}.p^2$
Nhưng dễ cm $S_{q-1}$ chia hết cho p nên có dpcm.
Bài viết đã được chỉnh sửa nội dung bởi thangthan: 18-08-2011 - 18:28
- chardhdmovies yêu thích
0 người đang xem chủ đề
0 thành viên, 0 khách, 0 thành viên ẩn danh
