具體數學第二版第四章習題（5)

61 假設$\frac{\hat{m}}{\hat{n}}$是$\frac{m^{'}}{n^{'}}$，如今證實$\frac{\hat{m}}{\hat{n}}=\frac{m^{''}}{n^{''}}$

$\hat{m}\perp \hat{n},\frac{\hat{m}}{\hat{n}}>\frac{m^{'}}{n^{'}},N=\frac{n+N}{n^{'}}n^{'}-n\geq \hat{n}>(\frac{n+N}{n^{'}}-1)n^{'}-n=N-n^{'}>0\rightarrow \frac{\hat{m}}{\hat{n}}\geq \frac{m^{''}}{n^{''}}$(這裏爲何能夠推出來這個結論)。

而後若是等於號不知足，那麼$n^{''}=(\hat{m}n^{'}-m^{'}\hat{n})n^{''}=n^{'}(\hat{m}n^{''}-m^{''}\hat{n})+\hat{n}(m^{''}n^{'}-m^{'}n^{''})\geq n^{'}+\hat{n}>N$出現矛盾

62 $e=2^{-1}+(2^{-2}+2^{-3}-2^{-6}-2^{-7})+(2^{-12}+2^{-13}-2^{-20}-2^{-21})+..=2^{-1}+\sum_{k\geq 0}(2^{-(2k^{2}+6k+3)}-2^{-(4k^{2}+10k+7)})$

63(1) 首先，假設$n=4$時，不存在$a,b,c$知足$a^{4}+b^{4}\neq c^{4}$

若是$n>4$且$n$不是素數使得$a^{n}+b^{n}=c^{d}$，因爲存在一個因子$d$使得$n=kd$，那麼$a^{n}+b^{n}=c^{n}\rightarrow  (a^{\frac{n}{d}})^{d}+(b^{\frac{n}{d}})^{d}=(c^{\frac{n}{d}})^{d}$。因此最小的必定是素數。

(2) 設$a+b=x\rightarrow \left (\frac{c^{p}}{x}=\frac{a^{p}+b^{p}}{x}=\frac{a^{p}+(x-a)^{p}}{x}  \right )\equiv pa^{p-1}(mod(x))$.。另外最小的反例知足$a\perp x$.

若是$p$不能整除$x$,那麼$Gcd(x,\frac{c^{p}}{x})=Gcd(x,pa^{p-1})=1$，因此必定存在$m$使得$x=m^{p}$

若是$p$能夠整除$x$，那麼$\frac{c^{p}}{x}$能被$p$整除，可是不能被$p^{2}$整除，因此有$x=p^{p-1}m^{p}$

64 生成數列的代碼

#include <iostream>
#include <sstream>
#include <string>
#include <vector>

// The max number of sign of Pn
constexpr int kMaxSignNumber = 300;

// The max N to generate
constexpr int kMaxN = 20;

class Node {
 public:
  void Add(const std::string &s) {
    if (Size() + 1 <= kMaxSignNumber) {
      ops.emplace_back(s);
    }
  }

  void Add(int x, int y) { Add(std::to_string(x) + "/" + std::to_string(y)); }

  std::string Get(int x) {
    --x;
    if (0 <= x && x < Size()) {
      return ops[x];
    }
    return "";
  }

  std::string ToString() const {
    std::stringstream ss;
    for (size_t i = 0; i < ops.size(); ++i) {
      if (i % 2 == 0) {
        ss << Transform(ops[i]);
      } else {
        ss << ops[i];
      }
    }
    return ss.str();
  }

 private:
  std::string Transform(const std::string &num) const {
    auto p = num.find_first_of('/');
    return "\\frac{" + num.substr(0, p) + "}{" + num.substr(p + 1) + "}";
  }

  int Size() const { return static_cast<int>(ops.size()); }

  std::vector<std::string> ops;
};

Node a[kMaxN + 1];

int main() {
  for (int i = 0; i < kMaxSignNumber / 2; ++i) {
    a[1].Add(i, 1);
    a[1].Add("<");
  }
  for (int N = 1; N < kMaxN; ++N) {
    for (int k = 1; k <= kMaxSignNumber; ++k) {
      for (int j = (k - 1) * N + 1; j < k * N; ++j) {
        a[N + 1].Add(a[N].Get(j));
      }
      if (k * N % 2 == 1) {
        a[N + 1].Add(k - 1, N + 1);
        a[N + 1].Add("=");
      } else {
        a[N + 1].Add(a[N].Get(k * N));
        a[N + 1].Add(k - 1, N + 1);
      }
      a[N + 1].Add(a[N].Get(k * N));
    }
  }
  for (int i = 1; i <= kMaxN; ++i) {
    std::cout << i << " : " << a[i].ToString() << "\n";
  }
  return 0;
}

1 : $\frac{0}{1}<\frac{1}{1}<\frac{2}{1}<\frac{3}{1}<\frac{4}{1}<\frac{5}{1}<\frac{6}{1}<\frac{7}{1}<\frac{8}{1}<\frac{9}{1}<\frac{10}{1}<\frac{11}{1}<\frac{12}{1}<\frac{13}{1}<\frac{14}{1}<\frac{15}{1}<\frac{16}{1}<\frac{17}{1}<\frac{18}{1}<\frac{19}{1}<\frac{20}{1}<\frac{21}{1}<\frac{22}{1}<\frac{23}{1}<\frac{24}{1}<\frac{25}{1}<\frac{26}{1}<\frac{27}{1}<\frac{28}{1}<\frac{29}{1}<\frac{30}{1}<\frac{31}{1}<\frac{32}{1}<\frac{33}{1}<\frac{34}{1}<\frac{35}{1}<\frac{36}{1}<\frac{37}{1}<\frac{38}{1}<\frac{39}{1}$
2 : $\frac{0}{2}=\frac{0}{1}<\frac{1}{2}<\frac{2}{2}=\frac{1}{1}<\frac{3}{2}<\frac{4}{2}=\frac{2}{1}<\frac{5}{2}<\frac{6}{2}=\frac{3}{1}<\frac{7}{2}<\frac{8}{2}=\frac{4}{1}<\frac{9}{2}<\frac{10}{2}=\frac{5}{1}<\frac{11}{2}<\frac{12}{2}=\frac{6}{1}<\frac{13}{2}<\frac{14}{2}=\frac{7}{1}<\frac{15}{2}<\frac{16}{2}=\frac{8}{1}<\frac{17}{2}<\frac{18}{2}=\frac{9}{1}<\frac{19}{2}<\frac{20}{2}=\frac{10}{1}<\frac{21}{2}<\frac{22}{2}=\frac{11}{1}<\frac{23}{2}<\frac{24}{2}=\frac{12}{1}<\frac{25}{2}<\frac{26}{2}$
3 : $\frac{0}{2}=\frac{0}{3}=\frac{0}{1}<\frac{1}{3}<\frac{1}{2}<\frac{2}{3}<\frac{2}{2}=\frac{3}{3}=\frac{1}{1}<\frac{4}{3}<\frac{3}{2}<\frac{5}{3}<\frac{4}{2}=\frac{6}{3}=\frac{2}{1}<\frac{7}{3}<\frac{5}{2}<\frac{8}{3}<\frac{6}{2}=\frac{9}{3}=\frac{3}{1}<\frac{10}{3}<\frac{7}{2}<\frac{11}{3}<\frac{8}{2}=\frac{12}{3}=\frac{4}{1}<\frac{13}{3}<\frac{9}{2}<\frac{14}{3}<\frac{10}{2}=\frac{15}{3}=\frac{5}{1}<\frac{16}{3}<\frac{11}{2}<\frac{17}{3}<\frac{12}{2}=\frac{18}{3}=\frac{6}{1}<\frac{19}{3}$
4 : $\frac{0}{2}=\frac{0}{4}=\frac{0}{3}=\frac{0}{1}<\frac{1}{4}<\frac{1}{3}<\frac{2}{4}=\frac{1}{2}<\frac{2}{3}<\frac{3}{4}<\frac{2}{2}=\frac{4}{4}=\frac{3}{3}=\frac{1}{1}<\frac{5}{4}<\frac{4}{3}<\frac{6}{4}=\frac{3}{2}<\frac{5}{3}<\frac{7}{4}<\frac{4}{2}=\frac{8}{4}=\frac{6}{3}=\frac{2}{1}<\frac{9}{4}<\frac{7}{3}<\frac{10}{4}=\frac{5}{2}<\frac{8}{3}<\frac{11}{4}<\frac{6}{2}=\frac{12}{4}=\frac{9}{3}=\frac{3}{1}<\frac{13}{4}<\frac{10}{3}<\frac{14}{4}=\frac{7}{2}<\frac{11}{3}<\frac{15}{4}$
5 : $\frac{0}{2}=\frac{0}{4}=\frac{0}{5}=\frac{0}{3}=\frac{0}{1}<\frac{1}{5}<\frac{1}{4}<\frac{1}{3}<\frac{2}{5}<\frac{2}{4}=\frac{1}{2}<\frac{3}{5}<\frac{2}{3}<\frac{3}{4}<\frac{4}{5}<\frac{2}{2}=\frac{4}{4}=\frac{5}{5}=\frac{3}{3}=\frac{1}{1}<\frac{6}{5}<\frac{5}{4}<\frac{4}{3}<\frac{7}{5}<\frac{6}{4}=\frac{3}{2}<\frac{8}{5}<\frac{5}{3}<\frac{7}{4}<\frac{9}{5}<\frac{4}{2}=\frac{8}{4}=\frac{10}{5}=\frac{6}{3}=\frac{2}{1}<\frac{11}{5}<\frac{9}{4}<\frac{7}{3}<\frac{12}{5}<\frac{10}{4}$
6 : $\frac{0}{2}=\frac{0}{4}=\frac{0}{6}=\frac{0}{5}=\frac{0}{3}=\frac{0}{1}<\frac{1}{6}<\frac{1}{5}<\frac{1}{4}<\frac{2}{6}=\frac{1}{3}<\frac{2}{5}<\frac{2}{4}=\frac{3}{6}=\frac{1}{2}<\frac{3}{5}<\frac{4}{6}=\frac{2}{3}<\frac{3}{4}<\frac{4}{5}<\frac{5}{6}<\frac{2}{2}=\frac{4}{4}=\frac{6}{6}=\frac{5}{5}=\frac{3}{3}=\frac{1}{1}<\frac{7}{6}<\frac{6}{5}<\frac{5}{4}<\frac{8}{6}=\frac{4}{3}<\frac{7}{5}<\frac{6}{4}=\frac{9}{6}=\frac{3}{2}<\frac{8}{5}<\frac{10}{6}=\frac{5}{3}<\frac{7}{4}$
7 : $\frac{0}{2}=\frac{0}{4}=\frac{0}{6}=\frac{0}{7}=\frac{0}{5}=\frac{0}{3}=\frac{0}{1}<\frac{1}{7}<\frac{1}{6}<\frac{1}{5}<\frac{1}{4}<\frac{2}{7}<\frac{2}{6}=\frac{1}{3}<\frac{2}{5}<\frac{3}{7}<\frac{2}{4}=\frac{3}{6}=\frac{1}{2}<\frac{4}{7}<\frac{3}{5}<\frac{4}{6}=\frac{2}{3}<\frac{5}{7}<\frac{3}{4}<\frac{4}{5}<\frac{5}{6}<\frac{6}{7}<\frac{2}{2}=\frac{4}{4}=\frac{6}{6}=\frac{7}{7}=\frac{5}{5}=\frac{3}{3}=\frac{1}{1}<\frac{8}{7}<\frac{7}{6}<\frac{6}{5}<\frac{5}{4}<\frac{9}{7}$
8 : $\frac{0}{2}=\frac{0}{4}=\frac{0}{6}=\frac{0}{8}=\frac{0}{7}=\frac{0}{5}=\frac{0}{3}=\frac{0}{1}<\frac{1}{8}<\frac{1}{7}<\frac{1}{6}<\frac{1}{5}<\frac{2}{8}=\frac{1}{4}<\frac{2}{7}<\frac{2}{6}=\frac{1}{3}<\frac{3}{8}<\frac{2}{5}<\frac{3}{7}<\frac{2}{4}=\frac{4}{8}=\frac{3}{6}=\frac{1}{2}<\frac{4}{7}<\frac{3}{5}<\frac{5}{8}<\frac{4}{6}=\frac{2}{3}<\frac{5}{7}<\frac{6}{8}=\frac{3}{4}<\frac{4}{5}<\frac{5}{6}<\frac{6}{7}<\frac{7}{8}<\frac{2}{2}=\frac{4}{4}=\frac{6}{6}=\frac{8}{8}$
9 : $\frac{0}{2}=\frac{0}{4}=\frac{0}{6}=\frac{0}{8}=\frac{0}{9}=\frac{0}{7}=\frac{0}{5}=\frac{0}{3}=\frac{0}{1}<\frac{1}{9}<\frac{1}{8}<\frac{1}{7}<\frac{1}{6}<\frac{1}{5}<\frac{2}{9}<\frac{2}{8}=\frac{1}{4}<\frac{2}{7}<\frac{2}{6}=\frac{3}{9}=\frac{1}{3}<\frac{3}{8}<\frac{2}{5}<\frac{3}{7}<\frac{4}{9}<\frac{2}{4}=\frac{4}{8}=\frac{3}{6}=\frac{1}{2}<\frac{5}{9}<\frac{4}{7}<\frac{3}{5}<\frac{5}{8}<\frac{4}{6}=\frac{6}{9}=\frac{2}{3}<\frac{5}{7}<\frac{6}{8}=\frac{3}{4}<\frac{7}{9}$

10 :$ \frac{0}{2}=\frac{0}{4}=\frac{0}{6}=\frac{0}{8}=\frac{0}{10}=\frac{0}{9}=\frac{0}{7}=\frac{0}{5}=\frac{0}{3}=\frac{0}{1}<\frac{1}{10}<\frac{1}{9}<\frac{1}{8}<\frac{1}{7}<\frac{1}{6}<\frac{2}{10}=\frac{1}{5}<\frac{2}{9}<\frac{2}{8}=\frac{1}{4}<\frac{2}{7}<\frac{3}{10}<\frac{2}{6}=\frac{3}{9}=\frac{1}{3}<\frac{3}{8}<\frac{4}{10}=\frac{2}{5}<\frac{3}{7}<\frac{4}{9}<\frac{2}{4}=\frac{4}{8}=\frac{5}{10}=\frac{3}{6}=\frac{1}{2}<\frac{5}{9}<\frac{4}{7}<\frac{6}{10}=\frac{3}{5}<\frac{5}{8}$
11 : $\frac{0}{2}=\frac{0}{4}=\frac{0}{6}=\frac{0}{8}=\frac{0}{10}=\frac{0}{11}=\frac{0}{9}=\frac{0}{7}=\frac{0}{5}=\frac{0}{3}=\frac{0}{1}<\frac{1}{11}<\frac{1}{10}<\frac{1}{9}<\frac{1}{8}<\frac{1}{7}<\frac{1}{6}<\frac{2}{11}<\frac{2}{10}=\frac{1}{5}<\frac{2}{9}<\frac{2}{8}=\frac{1}{4}<\frac{3}{11}<\frac{2}{7}<\frac{3}{10}<\frac{2}{6}=\frac{3}{9}=\frac{1}{3}<\frac{4}{11}<\frac{3}{8}<\frac{4}{10}=\frac{2}{5}<\frac{3}{7}<\frac{4}{9}<\frac{5}{11}<\frac{2}{4}=\frac{4}{8}=\frac{5}{10}=\frac{3}{6}=\frac{1}{2}<\frac{6}{11}<\frac{5}{9}<\frac{4}{7}<\frac{6}{10}=\frac{3}{5}<\frac{5}{8}<\frac{7}{11}<\frac{4}{6}=\frac{6}{9}=\frac{2}{3}<\frac{7}{10}<\frac{5}{7}<\frac{8}{11}<\frac{6}{8}=\frac{3}{4}<\frac{7}{9}<\frac{8}{10}=\frac{4}{5}<\frac{9}{11}<\frac{5}{6}<\frac{6}{7}<\frac{7}{8}<\frac{8}{9}<\frac{9}{10}<\frac{10}{11}<\frac{2}{2}=\frac{4}{4}=\frac{6}{6}=\frac{8}{8}=\frac{10}{10}=\frac{11}{11}=\frac{9}{9}=\frac{7}{7}=\frac{5}{5}$
12 : $\frac{0}{2}=\frac{0}{4}=\frac{0}{6}=\frac{0}{8}=\frac{0}{10}=\frac{0}{12}=\frac{0}{11}=\frac{0}{9}=\frac{0}{7}=\frac{0}{5}=\frac{0}{3}=\frac{0}{1}<\frac{1}{12}<\frac{1}{11}<\frac{1}{10}<\frac{1}{9}<\frac{1}{8}<\frac{1}{7}<\frac{2}{12}=\frac{1}{6}<\frac{2}{11}<\frac{2}{10}=\frac{1}{5}<\frac{2}{9}<\frac{2}{8}=\frac{3}{12}=\frac{1}{4}<\frac{3}{11}<\frac{2}{7}<\frac{3}{10}<\frac{2}{6}=\frac{4}{12}=\frac{3}{9}=\frac{1}{3}<\frac{4}{11}<\frac{3}{8}<\frac{4}{10}=\frac{2}{5}<\frac{5}{12}<\frac{3}{7}$

首先能夠發現，相等的數字出現的規律爲$\frac{2m}{2n},\frac{4m}{4n},\frac{6m}{6n},..,\frac{rm}{rn},...\frac{5m}{5n},\frac{3m}{3n},\frac{m}{n}$

假設$P_{N}$是正確的，來證實$P_{N+1}$是正確的。證實兩部分：

（1）若是$kN$是奇數，那麼$\frac{k-1}{N+1}=P_{N,kN}$

（2）若是$kN$是偶數，那麼$P_{N,kN-1}P_{N,kN}\frac{k-1}{N+1}P_{N,kN}P_{N,kN+1}$，其中$P_{N,kN}$爲比較符號。

首先計算$P_{N}$中小於$\frac{k-1}{N+1}$的數字個數。

$\sum_{n=1}^{N}\sum_{m}[0\leq \frac{m}{n}<\frac{k-1}{N+1}]=\sum_{n=1}^{N}\left \lceil \frac{(k-1)n}{N+1} \right \rceil=\sum_{n=1}^{N}\left \lfloor \frac{(k-1)n+N}{N+1} \right \rfloor=\frac{(k-2)N}{2}+\frac{d-1}{2}+d\left \lfloor \frac{N}{d} \right \rfloor$

最後一步由第三章的公式3.32獲得。其中$d=Gcd(k-1,n+1)\rightarrow N\equiv d-1(mod(d))\rightarrow d\left \lfloor \frac{N}{d} \right \rfloor=N-(d-1)$，因此總的個數爲$\frac{kN-d+1}{2}$

 其中，$P_{N}$中與$\frac{k-1}{N+1}$相等且在$P_{N+1}$中在$\frac{k-1}{N+1}$以前的個數爲$\frac{1}{2}(d-1-[d=2q])$

(1)若是$kN$是奇數，那麼$d$爲偶數，那麼$\frac{k-1}{N+1}$在$P_{N}$中位於$\frac{kN-d+1}{2}+\frac{1}{2}(d-1-1)=\frac{kN-1}{2}$個數字以後，因此$\frac{k-1}{N+1}$前面有$kN-1$個字符，因此$\frac{k-1}{N+1}=P_{N,kN}$

(2)若是$kN$是偶數，則$d$爲奇數，則$\frac{k-1}{N+1}$在$P_{N}$中位於$\frac{kN}{2}$個數字以後。若是$d=1$那麼$P_{N}$中沒有數字等於$\frac{k-1}{N+1}$,此時$P_{N,kN}$是小於號。若是$d$爲偶數，那麼$\frac{k-1}{N+1}$在兩個相等的元素中間，此時$P_{N,kN}$爲等於號。

65 多是

66 應該是

67 總的趨勢是若是第一個數和第$n$個數獲得的$f(1,n)=\frac{a_{n}}{Gcd(a_{1},a_{n})}$不是最大的話，說明他們中間的位置不多，那麼中間的某兩個元素$i,j$獲得的$Gcd(a_{i},a_{j})$會很小，從而使得$f(i,j)\geq n$

68  可能存在

69 應該成立

70 由這裏第24題的結論有，等式成立當且僅當$\upsilon _{3}(n)=\upsilon _{2}(n)$.$\upsilon _{p}(n)$表示$n$的$p$進制中各位數字之和。1和6知足，$\upsilon _{3}(1)=\upsilon _{2}(1)=1,\upsilon _{3}(6)=\upsilon _{2}(6)=2$.因此應該有不少

71 看起來並很少

72 $a=-1$時全部素數都知足。任意的$a$應該都有無窮個$n$知足

73 應該是

74 大約是$p(1-e^{-1})$