■初等関数研究村■
■ このスレッドは過去ログ倉庫に格納されています
初等関数(しょとうかんすう、英: Elementary function)とは、 実数または複素数の1変数関数で、代数関数、指数関数、対数関数、 三角関数、逆三角関数および、それらの合成関数を作ることを 有限回繰り返して得られる関数のことである ガンマ関数、楕円関数、ベッセル関数、誤差関数などは初等関数でない 初等関数のうちで代数関数でないものを初等超越関数という 双曲線関数やその逆関数も初等関数である 初等関数の導関数はつねに初等関数になる k=26, 6854100615782599621 8 * 9 [26] : 6854100615782599680 Table[sum[C(2n-1+α,k-1),{n,1,a}],{k,1,b}] a=n(n+1)/2-1 b=n(n+1) を満たす差分追尾数列αを見つけてくれ〜(・ω・)ノ 0qc3OXjXwBt1HD7Mt228BdYw7VinFEl43Zoc9tkhSu6hmEi1WEZ6OqB3FSe3k7L0qXrj8fNHrzBhhPTzD8WhAjrDXv1k55mQkf5uiOqjKuWUGgYFZuNZUagqAX9wNZzwaH4BlgoDtLscwycAwYQ7tuMa9CoGneWU5TXTTYEhxrUUJB0qmsR19UqgNcuwuN3oX8QyXNG 43bDtYxhwTZApiRugaYbzsFfDKZmuR212sTlb3HrYDe9jytaLBgeojHWZxPkzLDqO3djDHR4YEE7wySKMFA1WfilujRqD7izkWmcUWhFiRZrxFJAByshrPMNynEJEpGtdOg7Qx1jjMcB2nGphazIOhgkUuKFgGIiMh65hqcNYbtLdSVeTIncn2bR8pUncW95wGWymgC0 J8hGG9KXTgycc8wu65xqHO4p5z5oqxLBQRZuY5NdFK6pM1UaaUzIAlBvvWS49LKCsiIbUDX0KKFIWjAdkRpo4aZkTjXtlBABqjBeDgeH64Kv8QRCkv4NklJWJU6uagXQG8uqws2ZLhzyEs04D6ycyNc9s3LAeDaywG9mQ3jlBFhHE7ba2qlDJlN9ixypMXRleRQZWryk 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□□□■■■■■■■■■■■■ □□□□■■■■■■■■■■■ □□□□□■■■■■■■■■■ □□□□□□■■■■■■■■■ □□□□□□□■■■■■■■■ □□□□□□□□■■■■■■■ □□□□□□□□□■■■■■■ □□□□□□□□□□■■■■■ □□□□□□□□□□□■■■■ □□□□□□□□□□□□■■■ □□□□□□□□□□□□□■■ □□□□□□□□□□□□□□■ 2n x 2n の正方形を 1 x 2 のドミノで埋める場合の数を考えます たとえば、2x2の正方形を1x2のドミノで埋める場合の数は、2通りです 4x4の正方形を1x2のドミノで埋める場合の数は、36通りです 一般に、n=0,1,2,3,,,,のとき、 1, 2, 36, 6728, 12988816, 258584046368,,, となり、一般項は、 Π[j=1 to n]Π[k=1 to n]{4cos^2 πj/(2n+1)+4cos^2 πk/(2n+1)} となるようなのですが、 どのようにその公式が導かれるのでしょうか? wikipedia https://en.wikipedia.org/wiki/Domino_tiling によると Temperley & Fisher (1961) and Kasteleyn (1961) によって独立に発見されたとある 多分元論文は Temperley, H. N. V.; Fisher, Michael E. (1961), "Dimer problem in statistical mechanics-an exact result", Philosophical Magazine, 6 (68): 1061-1063, doi:10.1080/14786436108243366 Kasteleyn, P. W. (1961), "The statistics of dimers on a lattice. I. The number of dimer arrangements on a quadratic lattice", Physica, 27 (12): 1209-1225, Bibcode:1961Phy....27.1209K, doi:10.1016/0031-8914(61)90063-5. 原論文読むのが早い これに証明載ってるかも https://inis.iaea.org/collection/NCLCollectionStore/_Public/38/098/38098203.pdf?r=1& ;r=1 Section2 A famous result of Kasteleyn [8] and Temperley and Fisher [18] counts the number of domino tilings of a chessboard (or any other rectangular region). In this section we explain Kasteleyn's proof. 「ドミノによるタイル張り」(京大・理) 36p. http://www.ms.u-tokyo.ac.jp/ ~kazushi/proceedings/domino.pdf 「長方形領域のドミノタイル張りについて」(青学大・理工) 17p. http://www.gem.aoyama.ac.jp/ ~kyo/sotsuken/2010/fujino_sotsuron_2010.pdf ■平面充填(へいめんじゅうてん) 平面内を有限種類の平面図形(タイル)で隙間なく敷き詰める操作である 敷き詰めたタイルからなる平面全体を平面充填形という 平面敷き詰め、タイル貼り、タイリング (tiling) 、テセレーション (tessellation) ともいう ただし「平面」を明言しない場合は、曲面充填や、 場合によっては2次元以外の空間の充填を含む 広義のテセレーション等については、空間充填を参照 平面充填は広義の空間充填の一種で、2次元ユークリッド空間の 充填である 多面体は多角形による球面充填(曲面充填の一種)と 考えることができる そのため、多角形による平面充填は多面体と共通点が多く、 便宜上多面体に含めて論じられることもある ルジンの問題(Luzin - のもんだい)とは、 正方形に関してニコライ・ルジン (Nikolai Luzin) が考えた問題である 「任意の正方形を、2個以上の全て異なる大きさの正方形に分割できるか」 という問題であり、ルジンはこの問題の解は存在しないと予想したが、 その後幾つかの例が発見された 2, 4, 6, 7, 8, 9, 11, 15, 16, 17, 18, 19, 24, 25, 27, 29, 33, 35, 37, 42, 50 の 計21枚の正方形 Table[C(0,n-2 mod18)+3C(0,n-4)+3C(1,n-7)+7C(0,n-11)+C(1,n-16)+C(1,n-18),{n,1,27}] {0, 1, 0, 3, 0, 0, 3, 3, 0, 0, 7, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0} こういう数列を簡単に作る方法は? Table[{1-n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)(n-7)(n-8)(n-9)(n-10)(n-11)(n-12)/13!}/4,{n,0,13}] Table[(1-C(0,n-13))/4,{n,0,13}] 同じ出力で遥かに式を短くできる 56を2進法表記で桁をリストアップし, リスト長が8になるようにリストの左側にゼロを足し加える: In[3]:=IntegerDigits[56, 2, 8] Out[3]={0,0,1,1,1,0,0,0} FromDigits[{1,0,1,0,0,1,0,0,0}, 2] 328 Table[2n-1,{n,1,9}]+IntegerDigits[328, 2, 9] {2, 3, 6, 7, 9, 12, 13, 15, 17} 3を法としたときの剰余: Mod[{1, 2, 3, 4, 5, 6, 7}, 3] {1,2,0,1,2,0,1} 2進値リストからもとの数を再生する: IntegerDigits[56, 2, 8]; FromDigits[%, 2] a_n=1/4((-1)^n-(1+2i)(-i)^n-(1-2i)i^n+9) 1, 3, 3, 2, 1, 3, 3, 2, 1, 3, 3, 2, 1, 3, 3, 2, 1, 3, 3, 2, 『与えられた数より小さい素数の個数について』 Chu-Vandermonde identity n個のものからk個取り出す場合の数と k個取り残す場合の数は等しい C(n,k)=C(n,n-k) ■平方完成 y=ax^2-(-a+2)x-a-a+2 =a(x^2-(-a+2)x/a)-a-a+2 =a{(x-(-a+2)/(2a))^2-(-a+2)^2/(4a^2)}-a-a+2 =a(x-(-a+2)/(2a))^2-(-a+2)^2/(4a)-a-a+2 =a(x-(-a+2)/(2a))^2-(-a+2)^2/(4a)-2a+2 =a(x-(-a+2)/(2a))^2-(a^2-4a+4)/(4a)-2a+2 =a(x-(-a+2)/(2a))^2-(a^2-4a+4)/(4a)-(8a^2)/(4a)+(8a)/(4a) =a(x-(-a+2)/(2a))^2-(a^2-4a+4+8a^2-8a)/(4a) =a(x-(-a+2)/(2a))^2-(9a^2-12a+4)/(4a) トランプの束がある 2〜10までの数字が描かれたカードが各スートに1枚ずつと、 ジョーカーのカードが24枚ある 全てを混ぜて無作為に切り直して12枚のカードを無作為に引いたとき その12枚のカードのうちジョーカー以外にいずれも違う数字が 書かれている確率はいくらか Sum[choose(24,k)*choose(9,12-k)*4^(12-k),{k,3,12}]/(choose(60,12)) Sum[C(24,k)C(9,12-k)4^(12-k),{k,3,12}]/(C(60,12)) 出力 7371811052/66636135475 FromDigits[{1,0,1,0,0,1,0,0}, 2] 164 ガンマ関数とベータ関数 https://lecture.ecc.u-tokyo.ac.jp/ ~nkiyono/2006/miya-gamma.pdf 第一種の合流型超幾何関数(クンマー) 1F1[a; b; z] = 1+Σ[k=1, ∞] {a(a+1)・・・・(a+k-1)/b(b+1)・・・・(b+k-1)} z^k/k! 1F1[-n; -2n; z] = {n!/(2n)!} Σ[k=0, n] {(2n-k)!/(n-k)!k!} z^k ブリストル大学の数学者Andrew Booker氏が、 33を3つの立方数の合計で表すこと、すなわち 33=x^3+y^3+z^3という方程式の解を求めることに成功した (8866128975287528)^3+(-8778405442862239)^3+(-2736111468807040)^3=33 https://fabcross.jp/news/2019/20190507_33.html Table[choose(17,k-1)+choose(15,k-1)+choose(13,k-1)+choose(11,k-1)+choose(10,k-1)+choose(8,k-1)+choose(5,k-1)+choose(4,k-1)+choose(1,k-1),{k,1,20}] chooseを一つにした式に変形できますか? 三つならできた 短軸有利☆ Table[sum[C(2n-1+C(0,n-2)+C(1,n-4),k-1),{n,1,9}],{k,1,20}] Table[C(0,n-2 mod4),{n,1,10}] {0, 1, 0, 0, 0, 1, 0, 0, 0, 1} 長軸有利☆ Table[C(9,k-1)+C(7,k-1)+C(6,k-1)+C(3,k-1)+C(2,k-1),{k,1,12}] Table[sum[C(2n-1+C(0,n-1)+C(0,n-3),k-1),{n,1,5}],{k,1,12}] Table[sum[C(2n-1+C(0,3mod n),k-1),{n,1,5}],{k,1,12}] {5, 27, 76, 140, 176, 153, 92, 37, 9, 1, 0, 0} 同じ出力で式が短くなってゆく Table[C(0,2mod n),{n,1,10}] {1, 1, 0, 0, 0, 0, 0, 0, 0, 0} Table[C(0,3mod n),{n,1,10}] {1, 0, 1, 0, 0, 0, 0, 0, 0, 0} Table[C(0,4mod n),{n,1,10}] {1, 1, 0, 1, 0, 0, 0, 0, 0, 0} Table[C(0,5mod n),{n,1,10}] {1, 0, 0, 0, 1, 0, 0, 0, 0, 0} Table[C(0,6mod n),{n,1,10}] {1, 1, 1, 0, 0, 1, 0, 0, 0, 0} Table[C(0,7mod n),{n,1,10}] {1, 0, 0, 0, 0, 0, 1, 0, 0, 0} Table[C(0,8mod n),{n,1,10}] {1, 1, 0, 1, 0, 0, 0, 1, 0, 0} Table[C(0,9mod n),{n,1,10}] {1, 0, 1, 0, 0, 0, 0, 0, 1, 0} 【即時】金券五百円分とすかいらーく券を即ゲット https://pbs.twimg.com/media/D9F0S6KUcAAk_1s.jpg 1. スマホでたいむばんくを入手 iOS https://t.co/ik17bynKNT Android https://t.co/uxTzFEk2ee 2. 会員登録を済ませる 3. マイページへ移動する 4. 紹介コード → 入力する [Rirz Tu](空白抜き) 今なら更に16日23:59までの登録で倍額の600円を入手可 両方ゲットしてもおつりが来ます 数分で終えられるのでぜひお試し下さい 👀 Rock54: Caution(BBR-MD5:b73a9cd27f0065c395082e3925dacf01) 短軸有利☆ Table[C(9,k-1)+C(7,k-1)+C(5,k-1)+C(4,k-1)+C(1,k-1),{k,1,12}] Table[sum[C(2n-1+C(0,n-2),k-1),{n,1,5}],{k,1,12}] {5, 26, 73, 133, 167, 148, 91, 37, 9, 1, 0, 0} k-1を一つにして式を短縮 0qc3OXjXwBt1HD7Mt228BdYw7VinFEl43Zoc9tkhSu6hmEi1WEZ6OqB3FSe3k7L0qXrj8fNHrzBhhPTzD8WhAjrDXv1k55mQkf5uiOqjKuWUGgYFZuNZUagqAX9wNZzwaH4BlgoDtLscwycAwYQ7tuMa9CoGneWU5TXTTYEhxrUUJB0qmsR19UqgNcuwuN3oX8QyXNG 43bDtYxhwTZApiRugaYbzsFfDKZmuR212sTlb3HrYDe9jytaLBgeojHWZxPkzLDqO3djDHR4YEE7wySKMFA1WfilujRqD7izkWmcUWhFiRZrxFJAByshrPMNynEJEpGtdOg7Qx1jjMcB2nGphazIOhgkUuKFgGIiMh65hqcNYbtLdSVeTIncn2bR8pUncW95wGWymgC0 J8hGG9KXTgycc8wu65xqHO4p5z5oqxLBQRZuY5NdFK6pM1UaaUzIAlBvvWS49LKCsiIbUDX0KKFIWjAdkRpo4aZkTjXtlBABqjBeDgeH64Kv8QRCkv4NklJWJU6uagXQG8uqws2ZLhzyEs04D6ycyNc9s3LAeDaywG9mQ3jlBFhHE7ba2qlDJlN9ixypMXRleRQZWryk 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0qc3OXjXwBt1HD7Mt228BdYw7VinFEl43Zoc9tkhSu6hmEi1WEZ6OqB3FSe3k7L0qXrj8fNHrzBhhPTzD8WhAjrDXv1k55mQkf5uiOqjKuWUGgYFZuNZUagqAX9wNZzwaH4BlgoDtLscwycAwYQ7tuMa9CoGneWU5TXTTYEhxrUUJB0qmsR19UqgNcuwuN3oX8QyXNG 43bDtYxhwTZApiRugaYbzsFfDKZmuR212sTlb3HrYDe9jytaLBgeojHWZxPkzLDqO3djDHR4YEE7wySKMFA1WfilujRqD7izkWmcUWhFiRZrxFJAByshrPMNynEJEpGtdOg7Qx1jjMcB2nGphazIOhgkUuKFgGIiMh65hqcNYbtLdSVeTIncn2bR8pUncW95wGWymgC0 J8hGG9KXTgycc8wu65xqHO4p5z5oqxLBQRZuY5NdFK6pM1UaaUzIAlBvvWS49LKCsiIbUDX0KKFIWjAdkRpo4aZkTjXtlBABqjBeDgeH64Kv8QRCkv4NklJWJU6uagXQG8uqws2ZLhzyEs04D6ycyNc9s3LAeDaywG9mQ3jlBFhHE7ba2qlDJlN9ixypMXRleRQZWryk 5f1wRmcZzXXh9xqvyHtHqtXG06b6iQ5OhfrAUZwU8Scwkh52X7iNRot4vwSfMrmjYoGlVIvhK9djdlkiGy03ly9O6SmKKfkBYZJCK8zLNCJux0nBGVJWVe90kIjRFBTCjOfe11bfeVXfLUM9mLp0zyFrfY4a1dC7nS9pShB2iDxRGp6Vn2SlReeXnc6mqJ6KfhY9L8gR 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Kvr3idZa3OUlonClNyGyT3u2Xrtde47Cr6m4tG1j7AurlCjmUXvLPaQDQLlhymjaNIkWblKiKeVhk601XohEk7mNq3FyXjLGTJjx4csGI9MHt9vijbaaAQMFGIi8A28SQA1Ie3oELhvKeuLzK9ZYmGMEVqj4GtOgB719u1e1KHHqpfnGgwmMFMpRTjoTrEl4f9KFathh kjasjfkajkljfklajjfjksdjksjalkjflkjasjfkjasjfkjajjs2333354994998989029929050295895028902802058299202095898582982092029209029029 sgssl;slg;ld;1 ComplexExpand[(1+E^(I(1+n)Pi)+2n)/4] ■連続投稿・重複 連続投稿・コピー&ペースト 連続投稿で利用者の会話を害しているものは削除対象になります 個々の内容に違いがあっても、荒らしを目的としていると判断したものは同様です コピー&ペーストやテンプレートの存在するものは、アレンジが施してあれば 残しますが、全く変更されていない・一部のみの変更で内容の変わらないもの、 スレッドの趣旨と違うもの、不快感を与えるのが目的なもの、 などは荒らしの意図があると判断して削除対象になります ※お手数ですが削除依頼できる方お願いします<(_ _)> ■DoS攻撃(ドスこうげき)(英:Denial of Service attack) 情報セキュリティにおける可用性を侵害する攻撃手法で、 ウェブサービスを稼働しているサーバやネットワークなどの リソース(資源)に意図的に過剰な負荷をかけたり 脆弱性をついたりする事でサービスを妨害する攻撃、 サービス妨害攻撃である >>1 は関係ないスレゴミを書き込むキチガイです。対応できるかたアクセス禁止をお願いします。 【ロビーのお約束】 削除の要件(禁止されること) 荒らし依頼・ブラクラの張付け等第3者に迷惑がかかる行為 アダルト広告・勧誘・悪質な掲示板宣伝などのアドレス等張りつけ 煽り・煽りに対する返答・叩き・誹謗中傷等(差別発言等含む) コピペ・アスキーアート等必要以上の張り付け または 第3者に迷惑が掛かる行為や発言であった場合は削除対象にします ■掲示板・スレッドの趣旨とは違う投稿 レス・発言 スレッドの趣旨から外れすぎ、議論または会話が成立しないほどの 状態になった場合は削除対象になります 故意にスレッドの運営・成長を妨害していると判断した場合も同様です ■投稿目的による削除対象 レス・発言 議論を妨げる煽り、不必要に差別の意図をもった発言、 第三者を不快にする暴言や排他的馴れ合い、 同一の内容を複数行書いたもの、 過度な性的妄想・下品である、等は削除対象とします 確率空間においては, A ∈ F を事象 (event) と呼ぶ. 100!中の二進数字の桁数を求める: In[1]:=IntegerLength[100!, 2] Out[1]=525 ((-1)^n)(((-1)^n)n+n+4(-1)^n+2)/2 1 5 1 7 1 9 1 11 1 13 1 15 1 17 1 かなりエレガント☆ てめーが、糞まきちらしておいて俺は被害者だー、馬鹿乙 短軸有利☆ Table[C(9,k-1)+C(7,k-1)+C(5,k-1)+C(4,k-1)+C(1,k-1),{k,1,12}] Cの数は宝一つの時の当たり数の5 9+7+5+4+1=26は宝二個の時の当たり数になる 長軸有利☆ Table[C(9,k-1)+C(7,k-1)+C(6,k-1)+C(3,k-1)+C(2,k-1),{k,1,12}] Cの数は宝一つの時の当たり数の5 9+7+6+3+2=27は宝二個の時の当たり数になる 同様に20マスの場合は 短軸有利のCの数は宝一つの時の当たり数の9 17+15+13+11+10+8+5+4+1=84 長軸有利のCの数は宝一つの時の当たり数の9 17+15+13+12+8+7+6+3+2=83 短軸有利☆ Table[C(9,k-1)+C(7,k-1)+C(5,k-1)+C(4,k-1)+C(1,k-1),{k,1,12}] Table[sum[C(2n-1+C(0,n-2),k-1),{n,1,5}],{k,1,12}] {5, 26, 73, 133, 167, 148, 91, 37, 9, 1, 0, 0} k-1を一つにして式を短縮 合流型超幾何微分方程式 (confluent hypergeometric differential equation) ■■■■■■■■■■■ ■□□□□□□□□□■ ■□■■■■■■■□■ ■□■□□□□□■□■ ■□■□■■■□■□■ ■□■□■□□□■□■ ■□■□■■■■■□■ ■□■□□□□□□□■ ■□■■■■■■■■■ ■■■■■■ □□□□□■ □■■■□■ □■□□□■ □■■■■■ 'Let's Make a Deal' host Monty Hall dies aged 96 ITV News-2017/09/30 Monty Hall, one of the US's most popular television game show hosts, has died aged 96, his son has said. Born Monte Halperin on 25 August 1921, for nearly three decades Hall hosted 'Let's Make a Deal', the hugely successful television show that he co-created. 1-(165n-3n^2+351)/(208n-7n^2+468) (4n+9)(n-13)/(7n^2-208n-468) ・マクローリン展開 入力例:series[tan x] 合流型超幾何関数 歴史的には、18世紀に Euler が初めて超幾何微分方程式と その解の研究を手掛けた 19世紀初頭になると、J. C. F. Gauss や N. H. Abel 等によって 級数の収束性についての厳密な理論が展開され、 超幾何級数にも応用された 19世紀中葉では複素解析学が整備され、 G. F. B. Riemann などの著名な数学者によって、 複素領域で定義された線形常微分方程式の解となる 関数の大域的理論や多価関数としての構造が深く研究された https://rio2016.5ch.net/test/read.cgi/math/1540218853/161,194-198 に書いてある事がちゃんと読めれば 宝の数が何個になっても 場合わけ+多項式で記述できるのはすぐわかる 読めよ 数学板なんだから ↑ これだと宝二個の多項式しか作れない しかも偶数と奇数が分離していて美しくない 解答としては不十分 ■目からウロコ!の最短ロジックはこちら https://rio2016.5ch.net/test/read.cgi/math/1560604951/2-4 思考を小学生モードにすることにより 数式処理ソフトのSageMathなしで 偶数と奇数の分離しない回答に最短で到達! ■https://rio2016.5ch.net/test/read.cgi/math/1540218853/161 二つの関数を一つに合成する P1st (6n^3+20n^2-n-27)(n-1)/24 (奇数)……@ (6n^4+14n^3-21n^2-26n+24)/24 (偶数)……A Q1st (6n^2+10n-3)(n+1)(n-1)/24 (奇数)……B (6n^2-2n-5)(n+2)n/24 (偶数)……C 奇数[1 0 1 0 1 0 1 0 1 0 1 0 1 0]のみ出力する関数は ((-1)^(n+1)+1)/2 ……D 偶数[0 1 0 1 0 1 0 1 0 1 0 1 0 1]のみ出力する関数は ((-1)^n+1)/2 ……E @xD+AxE ((6n^3+20n^2-n-27)(n-1)/24)(((-1)^(n+1)+1)/2)+((6n^4+14n^3-21n^2-26n+24)/24)(((-1)^n+1)/2) ∴P1st ={12n^4+28n^3-42n^2-52n-3(-1)^n+51}/48 BxD+CxE ((6n^2+10n-3)(n+1)(n-1)/24)(((-1)^(n+1)+1)/2)+((6n^2-2n-5)(n+2)n/24)(((-1)^n+1)/2) ∴Q1st ={12n^4+20n^3-18n^2-20n-3(-1)^n+3}/48 >>4 と一致Match ■Obituary - John Forbes Nash, Jr. (1928 - 2015) Swarajya-2015/05/25 Nash is mostly known for his equilibrium concept called as “Nash Equilibrium”. For many years before his seminal paper, legends like von Neumann were working on the theory of games with a special focus on Zero-sum games. ComplexExpand[(1+E^(I(1+n)Pi)+2n)/4] Piはπ Table[(E^(I n Pi)(2+n+E^(I n Pi)(4+n)))/2,{n,1,56}] {1, 5, 1, 7, 1, 9, 1, 11, 1, 13, 1, 15, 1, 17, 1, 19, 1, 21, 1, 23,1, 25, 1, 27, 1, 29, 1, 31, 1, 33, 1, 35, 1, 37, 1, 39, 1, 41, 1, 43, 1, 45, 1, 47, 1, 49, 1, 51, 1, 53, 1, 55, 1, 57, 1, 59} a_n=(2n+(-1)^(n+1)+1)/4 1 1 2 2 3 3 4 4 5 5 6 6 7 7 1/4(2n+e^(iπ(n+1))+1) (1+E^(I(1+n)Pi)+2n)/4 1/4(2n+e^(i πn+i π)+1) (1+E^(I Pi+I nPi)+2n)/4 ComplexExpand[(1+E^(I(1+n)Pi)+2n)/4] 1/4(2n+e^(iπ n+iπ)+1) n/2-1/4 i sin(π n)-1/4 cos(π n)+1/4 ComplexExpand[(1+E^(I Pi+I n Pi)+2 n)/4] 1/4+n/2-Cos[n Pi]/4-(I/4) Sin[n Pi] ■スイッチング関数 Table[-C(1,n-2)+C(1,n-5)+C(1,n-9)+C(1,n-10),{n,1,10}] {0, -1, -1, 0, 1, 1, 0, 0, 1, 2} 153043438141440=4(18!!)+2(20!!)+78(24!!) 153043438141440=4(18!!)+2(20!!)+3(26!!) 53760=512(7!!) ((-1)^n-(1+2 i)(-i)^n-(1-2 i)i^n+9)/4 1, 3, 3, 2, 1, 3, 3, 2, 1, 3, 3, 2, 1, 3, 3, 2, 1, 3, 3, 2, ■ このスレッドは過去ログ倉庫に格納されています
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