でも、日常会話で学問的な話を持ち出して、相手を否定すると、友達いなくなるゾ。 0757132人目の素数さん2018/04/09(月) 03:26:04.32ID:9Nk3vN2i お前もともと友達いないじゃん(クソデカブーメラン) 0758132人目の素数さん2018/04/09(月) 13:03:10.86ID:DeVuYH8+ SaksとZygmundのAnalytic Functionsの情報持ってる人いる? 0759132人目の素数さん2018/04/09(月) 13:26:57.46ID:8BjdrdGM ファンクなアナルか 0760¥ ◆2VB8wsVUoo 2018/04/09(月) 15:46:58.94ID:io+q775y ¥ 0761¥ ◆2VB8wsVUoo 2018/04/09(月) 15:47:16.06ID:io+q775y ¥ 0762¥ ◆2VB8wsVUoo 2018/04/09(月) 15:47:34.74ID:io+q775y ¥ 0763¥ ◆2VB8wsVUoo 2018/04/09(月) 15:47:51.48ID:io+q775y ¥ 0764¥ ◆2VB8wsVUoo 2018/04/09(月) 15:48:11.37ID:io+q775y ¥ 0765¥ ◆2VB8wsVUoo 2018/04/09(月) 15:48:30.65ID:io+q775y ¥ 0766¥ ◆2VB8wsVUoo 2018/04/09(月) 15:48:48.06ID:io+q775y ¥ 0767¥ ◆2VB8wsVUoo 2018/04/09(月) 15:49:06.35ID:io+q775y ¥ 0768¥ ◆2VB8wsVUoo 2018/04/09(月) 15:49:24.52ID:io+q775y ¥ 0769¥ ◆2VB8wsVUoo 2018/04/09(月) 15:49:42.14ID:io+q775y ¥ 0770132人目の素数さん2018/04/10(火) 05:20:47.44ID:cgRjeD8b おまいら馬鹿だろ 0771memo2018/04/13(金) 09:11:02.74ID:p9cqkdQk Let's prove that there exists infinitely many primes. Suppose there exists only n primes. Π[j=1,n](Σ[i=0,∞]1/(p_j)^i) = (1/1+1/2+1/4+…)(1/1+1/3+1/9+…)(1/1+1/5+1/25+…)…(1/1+1/(p_n)+1/(p_n)^2+…) where p_j denotes the jth prime. By the unique factorization theorem, when we expand this, we get every reciprocal of a natural number once. So, Π[j=1,n](Σ[i=0,∞]1/(p_j)^i) = Σ[k=1,∞]1/k. Also, {1/(p_j)^i} is a geometric sequence with first term 1 and common ratio 0<1/(p_j)<1. So, Π[j=1,n](Σ[i=0,∞]1/(p_j)^i) = Π[j=1,n]1/(1-1/(p_j)) = Π[j=1,n](p_j)/((p_j)-1). We obtain Σ[k=1,∞]1/k = Π[j=1,n](p_j)/((p_j)-1). The L.H.S. is the harmonic series known to diverge, while the R.H.S. is a finite number. Contradiction. Therefore, there exists infinitely many primes.
Let's prove that the sum of the reciprocals of the primes diverges. Π[j=1,∞](p_j)/((p_j)-1) = ∞. For any x, 1+x ≦ e^x, so, Π[j=1,a](p_j)/((p_j)-1) = Π[j=1,a]1+1/((p_j)-1) ≦ Π[j=1,a]e^(1/((p_j)-1)) = e^(Σ[j=1,a]1/((p_j)-1)). Thus e^(Σ[j=1,∞]1/((p_j)-1)) = ∞ and Σ[j=1,∞]1/((p_j)-1) = ∞ follows. For j≧2, (p_j)-1 ≧ p_(j-1), and 1/((p_j)-1) ≦ 1/(p_(j-1)), so, Σ[j=1,b]1/((p_j)-1) = 1/(2-1) + Σ[j=2,b]1/((p_j)-1) ≦ 1 + Σ[j=2,b]1/(p_(j-1)) = 1 + Σ[j=1,b-1]1/(p_j). Thus Σ[j=1,∞]1/(p_j) = ∞ WWWWW. 0772¥ ◆2VB8wsVUoo 2018/04/15(日) 06:40:33.25ID:yRfavIow ¥ 0773¥ ◆2VB8wsVUoo 2018/04/15(日) 06:40:47.87ID:yRfavIow ¥ 0774¥ ◆2VB8wsVUoo 2018/04/15(日) 06:41:04.69ID:yRfavIow ¥ 0775¥ ◆2VB8wsVUoo 2018/04/15(日) 06:41:26.35ID:yRfavIow ¥ 0776¥ ◆2VB8wsVUoo 2018/04/15(日) 06:41:48.15ID:yRfavIow ¥ 0777¥ ◆2VB8wsVUoo 2018/04/15(日) 06:42:06.43ID:yRfavIow ¥ 0778¥ ◆2VB8wsVUoo 2018/04/15(日) 06:42:26.27ID:yRfavIow ¥ 0779¥ ◆2VB8wsVUoo 2018/04/15(日) 06:42:46.66ID:yRfavIow ¥ 0780¥ ◆2VB8wsVUoo 2018/04/15(日) 06:43:10.66ID:yRfavIow ¥ 0781¥ ◆2VB8wsVUoo 2018/04/15(日) 06:43:31.48ID:yRfavIow ¥ 0782132人目の素数さん2018/04/18(水) 19:40:06.76ID:G7AiDShi ウィキペディア「ジャン・デュドネ」
Let a_0 be a non-negative real number. a_0=[a_0]+{a_0}. When {a_0}≠0, a_0=[a_0]+1/(1/{a_0}).
Let a_1=1/{a_0}. a_1=[a_1]+{a_1}, a_0=[a_0]+1/([a_1]+{a_1}). When {a_1}≠0, a_0=[a_0]+1/([a_1]+1/(1/{a_1})).
If we continue replacing 1/{a_k} with a_(k+1) when {a_k}≠0, a_0 can be expressed by integers [a_n]. a_0=[a_0]+1/([a_1]+1/([a_2]+…. This fraction is called "regular continued fraction," and is often written as a_0=[a_0;a_1,a_2,…].
If a_(n+1) is large enough, a_0 can be approximated by [a_0;a_1,a_2,…,a_n]. 0828memo2018/05/12(土) 03:36:54.50ID:asoZtODm Some approximations of pi, obtained by using this method.
22/7 has been known since the ancient times. 355/113 was found by Zu Chongzhi in 480 A.D., and is commonly referred to as "Zu's ratio." Zu named 22/7 "Yuelü" and 355/113 "Milü." (2143/22)^(1/4) was found by Srinivasa Ramanujan. 0829132人目の素数さん2018/05/12(土) 11:30:08.69ID:FumkAJGr>>826 替わってやれば 0830DJgensei artchive gemmar2018/05/12(土) 11:58:09.42ID:pKtCKnP+ すげえな。狂わーん暗唱してる人たちは。 0831132人目の素数さん2018/05/12(土) 15:46:18.50ID:wJb7EV44 数というのは、距離と角度で表すものだと言う事を、今日知りました。