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漸化式の考え方とは？
- 漸化式の考え方とは、数列の次の項を前の項やそれ以前の項で表現する方法です。
- 具体的には、与えられた式や条件を使って数列の項を計算していくことで、その数列の一般項や性質を求めることができます。
- この文章では、漸化式の考え方を用いて数列の項を置き換える方法について説明されています。具体的な置き換え方については、式や条件に基づいて新しい数列を定義することで実現されています。
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- カードトリック問題の意味について
こんにちは。前に質問させてもらい、 内容として 「21 枚のカードがあり、絵が見えるように上にして、三枚ずつ七行に並べていく。七行・三列の配列で考える。 そして、観客に一枚を選ばせ、三列のうちのどの列なのか言わせる。カードは手渡さないで、指示された列が真ん 中になるよう各列のカードを集めたあと、また繰り返す。このイカサマを 3ラウンド繰り返せば、イカサマ師は観 客の選んだ札を言い当てられる。」 ということで、理解できました。そして、そのあとの英文のないようで、数式が出てきているのですが、 どのような意味なのかが、読みとれません＞＜ その内容は、 The convergence of the process We consider a pack with PQ cards which will be arranged in an array of P columns, each of Q cards or, equivalently, Q rows, each of P cards. The case described above is with P=3 and Q=7. We shall only analyse the case when P and Q are odd and greater than 1 (leaving the other cases to the interested reader), and we write P=2p+1 and Q=2q+1, where p>=1 and q>=1. We are interested in the central card in the pack; this has exactly 1/2(P×Q-1) cards above and below it and, labelling the cards 1,2,3,... from the top of the pack, the central card is labelled C, where C=1/2(P×Q+1)=p×Q+q+1=q×P+p+1. Suppose now that the chosen card lies in the Xn th position in the pack after n rounds have been played. During the play in the next round the first Xn cards will be spread out over the first few rows, and the chosen card will occur in the r th row, where r is the smallest integer greater than or equal to Xn/P. We write <Y> for the smallest integer m satisfying m>=Y. It follows that, after reforming the pack by ensuring that the column containing the chosen card is the central column, the position Xn+1 of the chosen card will satisfy the recurrence relation Xn+1=pQ+<Xn/P>. This is a recurrence relation, but it is non-linear, and we do not know the initial value X0 of the chosen card in the pack; all we know is that 1=<X0=<PQ. Nevertheless, this information is sufficient to give us a complete solution of our problem, and we shall see that the dynamics of the trick is reminiscent of the convergence of iterates of a function towards an attracting fixed point of the function. We begin by noting that there is a symmetry in the process, so that we may confine our attention to those cases where the chosen card lies in the top half of the pack; thus we may assume that 1=<X0=<C. An induction argument now shows that, for all n, Xn=<C. For suppose that Xn=<C; then Xn+1=pQ+<Xn/P>=<pQ+<C/P> =pQ+<(qP+p+1)/P> =pQ+q+1=C This argument also shows that if Xn=C, then Xn+1=C. Observe next that X1>pQ. We shall now show that there is some integer N with pO<X1<X2<....<Xn-1<Xn=C=Xn+1=Xn+2=.... ・・・(1) いつも質問させてもらい、本当に助かってます。時間がある方、アドバイスお願いします＞＜
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- カードトリック問題について
こんにちは。カードについて書いてあることはわかるのですが、内容がよくわかりません。アドバイスお願いします＞＜ The card trick Consider the following card trick. The trickster has 21 cards in a pack; the audience chooses one card (leaving it in the pack) and the pack is shuffled. The trickster deals out the cards face up in a row of three, then in a second row of three, and so on, the cards finally making an array of seven rows of three cards each(or three columns of seven cards each). The audience states which column the chosen card is in, the trickster collects up the columns to reform the pack ensuring that the chosen column is placed between the other two. This process (which we call a round) is repeated twice more. The trickster can now tell the audience which card was chosen, for it will be the card in the centre of the pack with ten cards before it and ten cards after it. Of course, the trickster can add the usual embellishments to deflect the audience's attention from the crucial issues. Why does this trick work? What will happen if we take a different size pack? We shall analyse the trick for a pack of any size, explain why it works, and show how many rounds are needed. Unexpectedly perhaps(but only at first), we shall see that if the trick were to be carried out with an array of, for example, 999×999 cards, the chosen card would find its way to the centre of the pack in just two rounds. The convergence of the process We consider a pack with PQ cards which will be arranged in an array of P columns, each of Q cards or, equivalently, Q rows, each of P cards. The case described above is with P=3 and Q=7. We shall only analyse the case when P and Q are odd and greater than 1 (leaving the other cases to the interested reader), and we write P=2p+1 and Q=2q+1, where p>=1 and q>=1. We are interested in the central card in the pack; this has exactly 1/2(P×Q-1) cards above and below it and, labelling the cards 1,2,3,... from the top of the pack, the central card is labelled C, where C=1/2(P×Q+1)=p×Q+q+1=q×P+p+1. Suppose now that the chosen card lies in the Xn th position in the pack after n rounds have been played. During the play in the next round the first Xn cards will be spread out over the first few rows, and the chosen card will occur in the r th row, where r is the smallest integer greater than or equal to Xn/P. We write <Y> for the smallest integer m satisfying m>=Y. It follows that, after reforming the pack by ensuring that the column containing the chosen card is the central column, the position Xn+1 of the chosen card will satisfy the recurrence relation Xn+1=pQ+<Xn/P>. This is a recurrence relation, but it is non-linear, and we do not know the initial value X0 of the chosen card in the pack; all we know is that 1=<X0=<PQ. Nevertheless, this information is sufficient to give us a complete solution of our problem, and we shall see that the dynamics of the trick is reminiscent of the convergence of iterates of a function towards an attracting fixed point of the function. We begin by noting that there is a symmetry in the process, so that we may confine our attention to those cases where the chosen card lies in the top half of the pack; thus we may assume that 1=<X0=<C. An induction argument now shows that, for all n, Xn=<C. For suppose that Xn=<C; then Xn+1=pQ+<Xn/P>=<pQ+<C/P> =pQ+<(qP+p+1)/P> =pQ+q+1=C This argument also shows that if Xn=C, then Xn+1=C. Observe next that X1>pQ. We shall now show that there is some integer N with pO<X1<X2<....<Xn-1<Xn=C=Xn+1=Xn+2=.... ・・・(1)
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- カードトリック問題についての文章について
こんにちは。何回も質問してすみません＞＜ カードトリックについて、次のような内容であることが今までの質問でわかりました。 「21 枚のカードを、絵が見えるように上にして、三枚ずつ七行に並べていく。七行・三列の行列にする。そして、観客に一枚を選ばせ、三列のうちのどの列なのか言わせる。カードは手渡さない。指示された列 が真ん中になるよう各列のカードを集めたあと、また繰り返す。これを３ラウンド繰り返せば、観客の選んだ札を言い当てられる。さらに、一般化して、P*Q 枚の札を P 列(columns)、Q 行(rows)に配列した場合(上の例では、P=3 and Q=7)を考える。P, Q が奇数の場合、P=2p+1 および Q=2q+1 (p>=1 and q>=1)とする。(上の例では、p=1 and q=3)重ねたカードの上から順番に 1, 2, 3, ... と番号付けすると、真ん中のカードの上下には (P*Q-1)/2 枚ずつあり(上の例では10枚ずつ)真ん中のカードの番号(C)は C = (P*Q+1)/2 = p*Q+q+1 = q*P+p+1 である。(上の例では C=11)このトリックを第 n ラウンドした後、選ばれたカードが重ねたカードの上から X(n) 番目だったとする。次の(n+1)ラウンドでそれが第 r 行に現れるとすれば、その r は X(n)/P 以上の整数で最小の値のものである。m>=Y の最小整数を <Y> と書けば、選ばれたカードが真ん中の列になるようカードを重ねなおしたあと、選ばれた札の番号 X(n+1) は次のような関係を満たす。 X(n+1) = p*Q+<X(n)/P> 問題の対称性から、選ばれたカードが重ねたカードの上半分にあるだけに注目すればよい。そこで 1=<X(0)=<C と仮定する。上記の帰納法により、すべての n について X(n)=<C である。そうすると、 X(n+1) = p*Q+<X(n)/P> = <p*Q+<C/P> = p*Q+<(q*P+p+1)/P> = p*Q+q+1=C 同様にして、Xn=C のとき X(n+1) = C である。また X(1)>p*Q の場合は、 pO<X(1)<X(2)<....<X(N-1)<X(N)=C=X(N+1)=X(N+2)=....(不動点 !) となる整数 N があることを示せる。(上の例では、第3ラウンドでカードを行列に配列したとき、選ばれたカードは必ず第4行にあり、観客にどの列にあるかを指示させたら、どのカードかがわかる。もう一度カードを重ねて行列に配りなおせば、選ばれたカードはど真ん中にくる。さらに繰り返すと、いつもど真ん中にくる。)」という内容でした。この続きに、下のような文章があったのですが、どのようにつながっているのかが、わかりません（泣） わからない文章は、 [The number of rounds] In the case of the trick as first described, P=3,Q=7 and C=11. In this case, X1>=8, X2>=7＋<8/3>=10, 11>=X3=7＋<10/3>=11. Thus (as stated in the first section) the chosen card is in the central position after three rounds.We turn now to discuss the number of rounds needed for the pack of general size. First, as X1>pQ we deduce that C>=X2>=pQ＋<(pQ＋1)/P>. Now P>=Q implies that pQ>=qP which, in turn, implies that C>=X2>=pQ＋<(qP＋1)/P>=pQ＋q＋1=C. Thus if P>=Q (and, in particular, if P=Q) then the chosen card is already in the central position after the second round regardless of the size of P and Q. A little thought should now show that, in retrospect, this is rather obviously so. The situation when Q>P is more complicated, and we shall show that the chosen card is always in the cantral position by the n th round provided that n>=1＋logQ/logP.・・・(2) We shall also given an example to show that in some cases this lower bound is indeed the smallest n for which the chosen card is centrally placed, so the inequality (2) cannot be improved upon. If P>=Q, then (2) simply says that Xn=C as soon as n=2, a fact we have already observed above. Note also that (2) shows that, for a fixed P, the number of rounds needed tends to ＋∞ as Q does. We begin with the example to show that (2) cannot be improved upon. Example Consider the trick with P=3, and Q=3^N, so that C=2/1(3^(N＋1)＋1), and let the selected card be the first in the psck. Then X1=pQ＋<1/3>=3^N＋1 X2=3^N＋<3/1(3^N＋1)>=3^N＋3^(N-1)＋1, ・ ・ です。 質問１：いきなりlogがでできているところで、何でか？ 質問２：∞の文章の意味 特にこの２つが疑問です＞＜ アドバイスお願いします＞＜
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It is rare that vs it is rare for O to V 以下のニュアンスの違いがわかる方はいらっしゃるでしょうか。 It is very rare that a nation has been/is heavily influenced by its historical and cultural characteristics like the US It is very rare for a nation to have been influenced by its historical and cultural characteristics to the extent [that] the US has. 語尾のlike US と to the extent ... 部分は無視してください。 主眼は、that と for to の違いにあります。 どうぞよろしくお願いします。
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海外からのメールで、何となくは分かるのですが、上手く訳せず返信に困っています どなたか、訳していただけないでしょうか？よろしくお願いします。 We will inquire for the letter and hope that it still turns up. Don't worry-in case it is lost,we will ask for a refund and return your payment. Please allow 2 weeks for the inquiry or that the letter still arrives. 以上になります。
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