{"id":302,"date":"2022-04-30T21:07:10","date_gmt":"2022-04-30T13:07:10","guid":{"rendered":"https:\/\/qwq.cafe\/?p=302"},"modified":"2022-09-14T20:07:11","modified_gmt":"2022-09-14T12:07:11","slug":"%e6%af%8f%e6%97%a5%e4%b8%80%e9%a2%98qwq%ef%bc%88%e6%95%b0%e5%ad%a6%e5%88%86%e6%9e%90%ef%bc%89","status":"publish","type":"post","link":"https:\/\/qwq.cafe\/?p=302","title":{"rendered":"\u4ebf\u65e5\u4e00\u9898qwq(\u6570\u5b66\u5206\u6790)(\u5df2\u505c\u66f4)"},"content":{"rendered":"

(\u8fd9\u91cc\u6536\u96c6\u4e86\u672c\u4eba\u505a\u8fc7\u7684\u4e00\u4e9b\u6bd4\u8f83\u5de7\u5999\u7684\u6570\u5b66\u5206\u6790\u9898\u76ee\uff0c\u4e89\u53d6\u505a\u5230\u6bcf\u65e5\u4e00\u9898\u3002\u5e94\u8be5\u7528\u4e0d\u4e86\u591a\u4e45\u5c31\u53d8\u6210\u4ebf\u65e5\u4e00\u9898\u4e86\uff08\u7b11\uff09<\/del><\/p>\n

<\/p>\n

\u5f53\u7136\u8fd9\u91cc\u7684\u505a\u6cd5\u4e0d\u4e00\u5b9a\u662f\u6700\u5de7\u5999\u7684\uff0c\u751a\u81f3\u53ef\u80fd\u662f\u9519\u8bef\u7684\uff0c\u5982\u679c\u6709\u66f4\u5de7\u5999\u7684\u505a\u6cd5\u6216\u8005\u53d1\u73b0\u9519\u8bef\u6b22\u8fce\u6307\u51fa\u3002<\/p>\n

\u53e6\u5916\u4e5f\u6b22\u8fce\u8bc4\u8bba\u4f60\u89c9\u5f97\u6bd4\u8f83\u5de7\u5999\u7684\u9898\u76ee\u3002<\/p>\n

2022.04.26<\/h2>\n
\n

\u6c42\u8bc1 $\\sum^{+\\infty}_{n=1}\\frac{1}{n^2}=\\frac{\\pi^2}{6}$.<\/p>\n<\/blockquote>\n

Proof.<\/p>\n

$\\arcsin$ \u7684 Maclaurin \u5c55\u5f00\u5f0f\u4e3a $ \\arcsin x=x+\\sum^{+\\infty}_{k=1}\\frac{(2k-1)!!}{(2k)!!}\\frac{x^{2k+1}}{2k+1} $ .<\/p>\n

\u4ee4 $ t=\\arcsin x $ \u5f97 $ t=\\sin t+\\sum^{+\\infty}_{k=1}\\frac{(2k-1)!!}{(2k)!!}\\frac{\\sin^{2k+1} t}{2k+1} $ .<\/p>\n

\u7531\u4e8e $ |\\frac{(2k-1)!!}{(2k)!!}\\frac{\\sin^{2k+1} t}{2k+1}|\\le\\frac{(2k-1)!!}{(2k)!!(2k+1)} $ \u4e14
\n$$\\begin{align}
\n&\\lim_{k\\to+\\infty}k(\\frac{(2k-1)!!}{(2k)!!(2k+1)}\\frac{(2k+2)!!(2k+3)}{(2k+1)!!}-1)\\\\
\n=&\\lim_{k\\to+\\infty}k(\\frac{(2k+2)(2k+3)}{(2k+1)^2}-1)\\\\
\n=&\\lim_{k\\to+\\infty}k\\cdot\\frac{6k+5}{4k^2+4k+1}\\\\
\n=&\\frac 3 2
\n\\end{align}
\n$$
\n\u7531 Raabe \u5224\u522b\u6cd5\u77e5 $ \\sum_{k=1}^{+\\infty}\\frac{(2k-1)!!}{(2k)!!(2k+1)}$ \u6536\u655b\uff0c\u518d\u7531\u4f18\u7ea7\u6570\u5224\u522b\u6cd5\u77e5$ \\sum^{+\\infty}_{k=1}\\frac{(2k-1)!!}{(2k)!!}\\frac{\\sin^{2k+1} t}{2k+1} $ \u5728 $\\mathbb{R}$ \u4e0a\u4e00\u81f4\u6536\u655b\u3002<\/p>\n

\u6240\u4ee5 $ \\int_0^{\\frac {\\pi} 2} t\\mathrm{d}t=\\int_0^{\\frac {\\pi} 2}\\sin t\\mathrm{d}t+\\sum^{+\\infty}_{k=1}\\int_0^{\\frac {\\pi} 2}\\frac{(2k-1)!!}{(2k)!!}\\frac{\\sin^{2k+1} t}{(2k+1)^2}\\mathrm{d}t\\ (1)$<\/p>\n

\u4ee4 $ I_k=\\int_0^{\\frac \\pi 2}\\sin^{2k+1} t\\ \\mathrm{d}t $<\/p>\n

\u5219\u6709
\n$$\\begin{align}
\nI_k&=\\int_0^{\\frac \\pi 2}\\sin^{2k+1} t\\ \\mathrm{d}t\\\\
\n&=\\int_0^{\\frac \\pi 2}(-\\cos t)'\\sin^{2k} t\\ \\mathrm{d}t\\\\
\n&={(-\\cos t\\sin^{2k}t)}\\bigg|^{t=\\frac \\pi 2}_{t=0}+\\int_0^\\frac{\\pi}2\\cos t\\cdot2k\\sin^{2k-1} t\\cdot\\cos t\\ \\mathrm{d}t\\\\
\n&=2k\\int_0^{\\frac\\pi 2}(1-\\sin^2t)\\sin^{2k-1}t\\ \\mathrm{d}t\\\\
\n&=2kI_{k-1}-2kI_k\\\\
\n\\Rightarrow I_k&=\\frac{2k}{2k+1}I_{k-1}
\n\\end{align}
\n$$
\n\u5176\u4e2d $ I_0=1 $ \uff0c\u6240\u4ee5 $ I_k=\\frac{(2k)!!}{(2k+1)!!}$.<\/p>\n

\u6240\u4ee5\u7ed3\u5408\u5f0f(1)\u5f97 $ \\frac {\\pi^2} 8=\\sum_{k=0}^{+\\infty}\\frac{1}{(2k+1)^2} $ .
\n$$\\sum^{+\\infty}_{n=1}\\frac{1}{n^2}=\\sum_{k=0}^{+\\infty}\\frac{1}{(2k+1)^2}+\\sum_{k=1}^{+\\infty}\\frac{1}{(2k)^2}=\\frac{\\pi^2} 8+\\frac 1 4\\sum^{+\\infty}_{n=1}\\frac{1}{n^2}
\n$$
\n\u6240\u4ee5 $\\sum^{+\\infty}_{n=1}\\frac{1}{n^2}=\\frac{\\pi^2}{6}\\ \\square$<\/p>\n

2022.04.30<\/h2>\n
\n

\u6c42 $ \\lim_{x\\to 0} \\frac{\\int_x^{2x}\\sin \\frac 1 t\\mathrm{d} t}{x} $.<\/p>\n<\/blockquote>\n

\u89e3\uff1a<\/p>\n

\u56e0\u4e3a $ (\\cos \\frac{1}{t})'= \\frac{\\sin \\frac{1} t}{t^2}$<\/p>\n

\u6240\u4ee5 $ \\int_{x}^{2x}\\sin\\frac 1 t\\mathrm{d}t= \\int_{x}^{2x}t^2(\\cos\\frac 1 t)'\\mathrm{d}t=t^2\\cos\\frac 1 t\\bigg|_{t=x}^{t=2x}-\\int_{x}^{2x}2t\\cos\\frac 1 t\\mathrm{d}t$
\n$$\\lim_{x\\to0}\\frac{\\int_x^{2x}\\sin \\frac 1 t\\mathrm{d} t}{x} =\\lim_{x\\to0}x\\cos\\frac{1}{x}+4x\\cos\\frac{1}{2x}-\\frac{\\int_{x}^{2x}2t\\cos\\frac 1 t\\mathrm{d}t}{x}=\\lim_{x\\to 0}-\\frac{\\int_{x}^{2x}2t\\cos\\frac 1 t\\mathrm{d}t}{x}
\n$$
\n\u518d\u7531 L’Hospital \u6cd5\u5219\u77e5
\n$$\\lim_{x\\to 0}-\\frac{\\int_{x}^{2x}2t\\cos\\frac 1 t\\mathrm{d}t}{x}=\\lim_{x\\to 0}-\\frac{4x\\cos\\frac 1 {2x}-2x\\cos \\frac 1 x}{1}=0\\ \\square
\n$$<\/p>\n

2022.05.01<\/h2>\n
\n

\u5bf9 $(0,1)$ \u533a\u95f4\u4e2d\u7684\u5c0f\u6570 $x=0.x_1x_2x_3…$ \uff0c\u5b9a\u4e49 $ f(x)=0.x_10x_20x_3… $ .<\/p>\n

(1). \u8bc1\u660e $ f\\in R[0,1] $.<\/p>\n<\/blockquote>\n

Proof.<\/p>\n

\u56e0\u4e3a $\\forall x,y\\in(0,1)(x<y)$ \u6613\u5f97 $ f(x)<f(y) $ \uff0c\u6240\u4ee5 $ f $ \u5728 $ (0,1) $ \u4e0a\u5355\u8c03\u9012\u589e\uff0c\u6240\u4ee5 $ f $ \u7684\u95f4\u65ad\u70b9\u96c6\u81f3\u591a\u53ef\u6570\uff0c\u662f\u96f6\u6d4b\u96c6\u3002<\/p>\n

\u53c8\u56e0\u4e3a $ f$ \u6709\u754c\uff0c\u7531 Lebesgue \u53ef\u79ef\u51c6\u5219\u77e5 $ f\\in R[0,1]\\ \\square$ <\/p>\n

\n

(2). \u6c42 $ f $ \u7684\u95f4\u65ad\u70b9\u96c6 $ D_f $.<\/p>\n<\/blockquote>\n

\u89e3\uff1a<\/p>\n

\u5148\u8bc1\u660e\u5982\u679c $ x $ \u662f $ (0,1) $ \u4e0a\u7684\u6709\u9650\u5c0f\u6570\uff0c\u5219 $ x\\in D_f $.<\/p>\n

\u4ee4 $ x=0.x_1x_2\\dots x_n $\uff0c\u5176\u4e2d $ n $ \u4e3a $ x $ \u7684\u5c0f\u6570\u4f4d\u6570\u3002<\/p>\n

\u4ee4 $ a_k=0.x_1x_2…x_{n-1}(x_n-1)\\underbrace{99\\dots9}_{k}$ \uff0c\u5219\u6709 $ \\lim_{k\\to+\\infty}a_k=x $<\/p>\n

$\\forall k\\in \\mathbb{N}, f(a_k)=0.x_10x_20\\dots x_{n-1}0(x_n-1)0909\\dots <0.x_10x_20\\dots x_{n-1}0(x_n-1)1$ <\/p>\n

\u6240\u4ee5 $ f(a_k)\\nrightarrow f(x)\\ (k\\to+\\infty) $ \uff0c\u6240\u4ee5 $ x \\in D_f$.<\/p>\n

\u518d\u8bc1\u660e\u5982\u679c $ x $ \u662f $ (0,1) $ \u4e0a\u7684\u65e0\u7a77\u5c0f\u6570\uff0c\u5219 $ x\\in D_f^c $.<\/p>\n

\u4ee4 $ x=0.x_1x_2x_2\\dots $ <\/p>\n

$ \\forall n\\in \\mathbb{N}_+,\\ \\exists \\delta=\\min\\{x-0.x_1x_2…x_n,0.x_1x_2…x_n-x+10^{-n}\\} $<\/p>\n

s.t. $\\forall y=0.y_1y_2y_3 \\dots \\in \\mathring{U}_\\delta(x)$ \uff0c\u6ee1\u8db3 $ y_i=x_i(\\forall i\\in\\{1,2,…,n\\}) $ <\/p>\n

\u5219 $ f(x) $ \u4e0e $ f(y) $ \u7684\u5c0f\u6570\u90e8\u5206\u524d $ 2n $ \u4f4d\u76f8\u540c\uff0c\u6240\u4ee5 $|f(x)-f(y)|<\\frac{1}{10^n} $ <\/p>\n

\u7531\u6781\u9650\u7b49\u4ef7\u5b9a\u4e49\u77e5 $ \\lim_{y\\to x} f(y)=f(x) $ \uff0c\u6240\u4ee5 $ x\\in D_f^c $ .<\/p>\n

\u7efc\u4e0a\u6240\u8ff0 $ D_f=\\{x=0.x_1x_2x_2\\dots\\in(0,1)|\\exists N\\in\\mathbb{N},\\forall n>N,x_n=0\\}\\ \\square$ <\/p>\n

2022.05.02<\/h2>\n
\n

\u8bc1\u660e\u5eb7\u6258\u5c14-\u4f2f\u6069\u65af\u5766-\u65bd\u7f57\u5fb7\u5b9a\u7406\u3002<\/p>\n

\u5373<\/p>\n

\u5982\u679c\u96c6\u5408 $A$ \u548c\u96c6\u5408 $ B $ \u6ee1\u8db3 $ |A|\\le|B| $ \u4e14 $ |A|\\ge|B| $ \uff0c\u5219\u6709 $ |A|=|B| $.<\/p>\n

\u5176\u4e2d $ |A| $ \u8868\u793a\u96c6\u5408 $ A $ \u7684\u52bf\u3002<\/p>\n<\/blockquote>\n

Proof.<\/p>\n

\u56e0\u4e3a $ |A|\\le|B| $ \uff0c\u6240\u4ee5 $ \\exists $ \u5355\u5c04 $ \\phi:A\\rightarrow B $. <\/p>\n

\u56e0\u4e3a $ |A|\\ge|B| $ \uff0c\u6240\u4ee5 $ \\exists $ \u5355\u5c04 $ \\psi:B\\rightarrow A $.<\/p>\n

\u56e0\u4e3a $ \\phi,\\psi $ \u5747\u4e3a\u5355\u5c04\uff0c\u6240\u4ee5 $ \\phi\\circ\\psi $ \u4e5f\u4e3a\u5355\u5c04\uff0c\u540c\u65f6\u56e0\u4e3a $ \\phi $ \u4e3a\u5355\u5c04\uff0c\u6240\u4ee5 $ |A|=|\\phi(A)| $ \uff0c\u6240\u4ee5\u8981\u8bc1 $ |A|=|B| $ \uff0c\u53ea\u9700\u8bc1 $ |\\phi(A)|=|B| $. <\/p>\n

\u4ee4 $ X=B,Y=\\phi(A),Z=\\phi(\\psi(B)) $ \uff0c\u5219\u95ee\u9898\u8f6c\u5316\u4e3a\uff1a\u5df2\u77e5 $ X\\supset Y \\supset Z $ \uff0c\u4e14\u5b58\u5728\u4ece $ X $ \u5230 $ Z $ \u7684\u53cc\u5c04\uff08\u6ee1\u5c04\u662f\u56e0\u4e3a $Z$ \u4e3a $ \\phi \\circ\\psi $ \u7684\u8c61\u96c6\uff09\uff0c\u6c42\u8bc1 $ |X|=|Y| $.<\/p>\n

\u4ee4 $ f=\\phi\\circ\\psi $ \uff0c\u5b9a\u4e49 $ f^n=f\\circ f^{n-1} $ \uff0c\u5176\u4e2d $ f^0=I_X $ \uff08$ I_X $ \u8868\u793a $ X $ \u4e0a\u7684\u6052\u7b49\u6620\u5c04\uff09<\/p>\n

\u5b9a\u4e49 $ g:X\\rightarrow Y, x\\mapsto g(x)=\\begin{cases}f(x),&\\exists n\\in \\mathbb{N},\\ \\text{s.t.}\\ x\\in f^{n}(X)\\textbackslash f^{n}(Y),\\\\x,& \\text{otherwise};\\end{cases} $ <\/p>\n

\u63a5\u4e0b\u6765\u53ea\u9700\u8bc1\u660e $ g $ \u662f\u53cc\u5c04\uff0c\u5219\u6709 $ |X|=|Y| $.<\/p>\n

(1). \u8bc1\u660e $ g $ \u662f\u5355\u5c04\u3002<\/p>\n

\u82e5 $g(x_1)=g(x_2)$ \uff0c<\/p>\n

<1>. $ \\exists n,m\\in\\mathbb N,\\ \\text{s.t.}\\ x_1\\in f^n(X)\\textbackslash f^n(Y),x_2\\in f^m(X)\\textbackslash f^m(Y)$<\/p>\n

$ f(x_1)=f(x_2)\\Rightarrow x_1=x_2 $.<\/p>\n

<2>. $ \\nexists n,m\\in\\mathbb N,\\ \\text{s.t.}\\ x_1\\in f^n(X)\\textbackslash f^n(Y),x_2\\in f^m(X)\\textbackslash f^m(Y) $<\/p>\n

$g(x_2)=g(x_1)=x_1\\land g(x_2)=x_2\\Rightarrow x_1=x_2$<\/p>\n

<3>. $ \\exists n\\in\\mathbb N,\\ \\text{s.t.}\\ x_1\\in f^n(X)\\textbackslash f^n(Y)$ \u4e14 $ \\nexists m\\in\\mathbb N,\\ \\text{s.t.}\\ x_2\\in f^m(X)\\textbackslash f^m(Y)$ <\/p>\n

\u5219\u6709 $x_2=f(x_1)\\land x_1\\in f^n(X)\\textbackslash f^n(Y)\\Rightarrow x_2\\in f(f^n(X)\\textbackslash f^n(Y))=f^{n+1}(X)\\textbackslash f^{n+1}(Y)$ \u4ea7\u751f\u77db\u76fe\uff0c\u6240\u4ee5\u8fd9\u79cd\u60c5\u51b5\u4e0d\u5b58\u5728\u3002<\/p>\n

(2). \u8bc1\u660e $ g $ \u662f\u6ee1\u5c04\u3002<\/p>\n

$\\forall y\\in Y,$<\/p>\n

<1>. \u82e5 $ \\nexists n\\in \\mathbb{N},\\ \\text{s.t.}\\ y\\in f^n(X)\\textbackslash f^n(Y)$ <\/p>\n

\u5219\u6709 $ y\\in X\\land g(y)=y $.<\/p>\n

<2>. \u82e5 $ \\exists n\\in \\mathbb{N},\\ \\text{s.t.}\\ y\\in f^n(X)\\textbackslash f^n(Y)$ <\/p>\n

\u5219\u6709 $ y\\in Y\\Rightarrow y\\notin X\\textbackslash Y\\Rightarrow n\\ge 1 $.<\/p>\n

$f^{-1}(y)\\in f^{-1}(f^n(X)\\textbackslash f^n(Y))=f^{n-1}(X)\\textbackslash f^{n-1}(Y)\\land (n-1)\\in \\mathbb{N}$.<\/p>\n

\u6240\u4ee5 $ f^{-1}(y)\\in X \\land g(f^{-1}(y))=y $.<\/p>\n

\u7531(1)(2)\u77e5 $g$ \u4e3a\u6ee1\u5c04\u3002$\\square$<\/p>\n","protected":false},"excerpt":{"rendered":"

(\u8fd9\u91cc\u6536\u96c6\u4e86\u672c\u4eba\u505a\u8fc7\u7684\u4e00\u4e9b\u6bd4\u8f83\u5de7\u5999\u7684\u6570\u5b66\u5206\u6790\u9898\u76ee\uff0c\u4e89\u53d6\u505a\u5230\u6bcf\u65e5\u4e00\u9898\u3002\u5e94\u8be5\u7528\u4e0d\u4e86\u591a\u4e45\u5c31\u53d8\u6210\u4ebf\u65e5\u4e00\u9898\u4e86\uff08\u7b11\uff09<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[8],"class_list":["post-302","post","type-post","status-publish","format-standard","hentry","category-6","tag-8"],"_links":{"self":[{"href":"https:\/\/qwq.cafe\/index.php?rest_route=\/wp\/v2\/posts\/302","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/qwq.cafe\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/qwq.cafe\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/qwq.cafe\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/qwq.cafe\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=302"}],"version-history":[{"count":18,"href":"https:\/\/qwq.cafe\/index.php?rest_route=\/wp\/v2\/posts\/302\/revisions"}],"predecessor-version":[{"id":331,"href":"https:\/\/qwq.cafe\/index.php?rest_route=\/wp\/v2\/posts\/302\/revisions\/331"}],"wp:attachment":[{"href":"https:\/\/qwq.cafe\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=302"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/qwq.cafe\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=302"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/qwq.cafe\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=302"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}