{"id":211,"date":"2018-07-04T09:19:31","date_gmt":"2018-07-04T08:19:31","guid":{"rendered":"https:\/\/donder-adventures.com\/wordpress\/?p=211"},"modified":"2018-07-04T09:19:31","modified_gmt":"2018-07-04T08:19:31","slug":"%e7%bd%aa%e8%88%87%e7%ad%8f","status":"publish","type":"post","link":"https:\/\/donder-adventures.com\/wordpress\/%e7%bd%aa%e8%88%87%e7%ad%8f\/","title":{"rendered":"\u7f6a\u8207\u7b4f"},"content":{"rendered":"

\u524d\u5e7e\u65e5\uff0c\u6536\u5230\u4e00\u9053\u4f86\u81ea\u5f8c\u8f29\u7684\u4e2d\u4e94 M2 \u6578\u5b78\u8a66\u984c\u3002
\n\u554f\u984c\u51fa\u5f97\u4e0d\u932f\uff0c\u9084\u8a18\u5f97 M2 \u8ab2\u7a0b\u5167\u5bb9\u7684\u670b\u53cb\u8acb\u52d9\u5fc5\u8a66\u4e00\u4e0b\u3002<\/p>\n

\uff08\u7528\u96fb\u8166\u770b \"\LaTeX\" \u8f03\u597d\u3002\uff09<\/p>\n

\u554f\u984c<\/h2>\n
    \n
  1. Let \"a\" be any positive constant and \"f(x)\" be any function such that \"\displaystyle\int_0^a f(x) \;\text{d}x\" exists. Show that\n

      <\/span>   <\/span>\"\[\int_0^a f(x) \;\text{d}x = \int_0^a f(a - x) \;\text{d}x\text{.}\]\"<\/p>\n<\/li>\n

  2. It is given that \"\displaystyle\int_0^\pi \ln(1 + 2k\cos x + k^2) \;\text{d}x\" exists for all real values of \"k\". Let\n

      <\/span>   <\/span>\"\[I = \int_0^\pi \ln(1 + 2b\cos x + b^2) \;\text{d}x\]\"<\/p>\n

    and<\/p>\n

      <\/span>   <\/span>\"\[\displaystyle J = \int_0^\pi \ln(1 - 2b\cos x + b^2) \;\text{d}x\]\"<\/p>\n

    , where \"b\" is any constant.<\/p>\n

      \n
    1. Using the result of (a), show that \"I = J\".<\/li>\n
    2. Show that \"\displaystyle I + J = \int_0^\pi \ln(1 - 2b^2\cos x + b^4) \;\text{d}x\".<\/li>\n<\/ol>\n<\/li>\n
    3. Evaluate \"\displaystyle\int_0^\pi \ln(\sin x) \;\text{d}x\".<\/li>\n<\/ol>\n

      ———\u6211\u662f\u5206\u9694\u7dda———<\/p>\n

       <\/p>\n

       <\/p>\n

       <\/p>\n

       <\/p>\n

       <\/p>\n

       <\/p>\n

       <\/p>\n

       <\/p>\n

       <\/p>\n

       <\/p>\n

       <\/p>\n

       <\/p>\n

       <\/p>\n

       <\/p>\n

       <\/p>\n

       <\/p>\n

       <\/p>\n

      ———\u6211\u662f\u5206\u9694\u7dda———<\/p>\n

      \u7b54\u6848<\/h2>\n

      (a) \u548c (b)(i) \u57fa\u672c\u4e0a\u662f\u9001\u5206\u984c\u3002\u4e0d\u61c2\u7684\u540c\u5b78\u61c9\u9762\u58c1\u601d\u904e\u3002<\/p>\n

        \n
      1. Let \"x = a - y\". Then, \"\text{d}y = -\text{d}x\".\n

          <\/span>   <\/span>\"<\/p>\n<\/li>\n

      2. \u5f9e (b)(ii) \u958b\u59cb\u5c31\u96e3\u89e3\u4e86\u3002\n
          \n
        1. By (a),\n

            <\/span>   <\/span>\"\begin{align*}<\/p>\n

          Hence \"I = J\".<\/li>\n

        2. Before proving this, let’s prove a lemma:\n

            <\/span>   <\/span>\"<\/p>\n

          Now we can move on.<\/p>\n

            <\/span>   <\/span>\"\begin{align*}<\/p>\n<\/li>\n<\/ol>\n<\/li>\n

        3. From (b), we can deduce that:\n

            <\/span>   <\/span>\"\[2\int_0^\pi \ln(1 -2b\cos x + b^2) \;\text{d}x = \int_0^{\pi}\ln(1-2b^2\cos x + b^4) \;\text{d}x\]\"<\/p>\n

          Set \"b = 1\":<\/p>\n

            <\/span>   <\/span>\"\begin{align*}<\/p>\n<\/li>\n<\/ol>\n

          \u611f\u60f3<\/h2>\n

          (b)(ii): \u505a\u5230 \"\cos 2x\" \u7684\u6642\u5019\uff0c\u6211\u4e00\u958b\u59cb\u9084\u4ee5\u70ba\u662f\u51fa\u932f\u984c\u3002\u82e5\u679c\u4e2d\u4e86\u9019\u9677\u9631\uff0c\u5927\u6982\u6703\u5931\u53bb\u4e00\u534a\u5206\u6578\u5427\u3002\u54c8\u54c8\u54c8\u22ef\u22ef
          \n(c): \u4e00\u822c\u4e2d\u5b78\u751f\u61c9\u8a72\u5f88\u96e3\u770b\u51fa (b) \u548c (c) \u4e4b\u9593\u7684\u95dc\u4fc2\u3002\uff08\u807d\u8aaa\u8003\u8a66\u4e2d\u53ea\u6709\u4e00\u4eba\u770b\u5f97\u5230\uff09\u51fa\u984c\u7684\u5230\u5e95\u7d93\u6b77\u4e86\u751a\u9ebc\uff0c\u624d\u60f3\u5f97\u51fa\u9019\u7a2e\u554f\u984c\u5462\uff1f<\/p>\n

          \u9019\u984c\u51fa\u5f97\u975e\u5e38\u597d\u3002<\/p>\n

          \u4ee5\u524d\u6211\u5728\u6bcd\u6821\u8003\u8a66\u7684\u6642\u5019\uff0c\u8001\u5e2b\u5011\u5f88\u559c\u6b61\u73a9 out syllabus \u7684\u904a\u6232\u3002\u521d\u4e2d\u5316\u5b78\u51fa reactivity of metals\uff0c\u9ad8\u4e2d\u6578\u5b78\u51fa t-method (t formula)\uff0cmatrix diagonalization\u22ef\u22ef\u70ba\u4e86\u5206\u51fa\u9ad8\u4e0b\u800c\u73a9 out syllabus\uff0c\u8b8a\u76f8\u9f13\u52f5\u5b78\u751f\u5b78\u7fd2 DSE \u4ee5\u5916\u7684\u77e5\u8b58\u3002\u5c0d\u4e00\u5fc3\u60f3\u5165\u5927\u5b78\u7684\u5b78\u751f\u800c\u8a00\uff0c\u6b64\u8209\u53ea\u662f\u6d6a\u8cbb\u6642\u9593\u3002\u5728\u6709\u9650\u7684\u6642\u9593\u4e2d\uff0c\u8003\u751f\u4e0d\u6703\u80fd\u5920\u61c9\u4ed8\u5230\u7121\u9650\u7684\u8003\u8a66\u7bc4\u570d\u3002<\/p>\n

          \u53ea\u8981\u80af\u82b1\u9ede\u5fc3\u601d\uff0c\u5373\u4f7f\u662f in syllabus \u7684\u554f\u984c\u4e5f\u53ef\u4ee5\u51fa\u5f97\u975e\u5e38\u96e3\u3002\u9019\u984c\u5c31\u662f\u4f8b\u5b50\u3002\u4e0d\u904e\u76f4\u6284 AL \u554f\u984c\u6bd4\u60f3\u9019\u4e9b\u8f15\u9b06\u767e\u500d\u5427\u3002<\/p>\n

          \u6b64\u5916\uff0c\u984c\u76ee\u4e5f\u5de7\u5999\u5730\u904b\u7528\u4e86 \"\sin\" \u548c \"\log\" \u7684\u7279\u6027\u3002\u96e3\u5ea6\u9ad8\u4e4b\u9918\u4e5f\u8003\u616e\u4e86\u7f8e\u611f\uff0c\u975e\u5e38\u597d\u3002<\/p>\n

          \u6a19\u984c\u54a9\u4e8b\uff1f<\/h2>\n

          \u7f6a\u00a0\u2192 sin \u2192 sine<\/p>\n

          \u7b4f \u2192 \u6728\u7b4f\u00a0\u2192 log \u2192 logarithm<\/p>\n

          \u721b gag \u569f\u35ce\u3002\u7b11\u5413\u5566\u3002<\/p>\n

          \"\"
          \u6771\u5f35\u897f\u671b\u4e3b\u6301\u6559\u4e2d\u98a8\u60a3\u8005\u5982\u4f55\u5f37\u984f\u6b61\u7b11<\/figcaption><\/figure>\n","protected":false},"excerpt":{"rendered":"

          \u524d\u5e7e\u65e5\uff0c\u6536\u5230\u4e00\u9053\u4f86\u81ea\u5f8c\u8f29\u7684\u4e2d\u4e94 M2 \u6578\u5b78\u8a66\u984c\u3002 \u554f\u984c\u51fa\u5f97\u4e0d\u932f\uff0c\u9084\u8a18\u5f97 M2 \u8ab2\u7a0b\u5167\u5bb9\u7684\u670b\u53cb\u8acb\u52d9\u5fc5\u8a66\u4e00\u4e0b\u3002 \uff08 … \u95b1\u8b80\u66f4\u591a<\/a><\/p>\n","protected":false},"author":1,"featured_media":261,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"footnotes":""},"categories":[35],"tags":[37],"class_list":["post-211","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-35","tag-m2"],"_links":{"self":[{"href":"https:\/\/donder-adventures.com\/wordpress\/wp-json\/wp\/v2\/posts\/211","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/donder-adventures.com\/wordpress\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/donder-adventures.com\/wordpress\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/donder-adventures.com\/wordpress\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/donder-adventures.com\/wordpress\/wp-json\/wp\/v2\/comments?post=211"}],"version-history":[{"count":106,"href":"https:\/\/donder-adventures.com\/wordpress\/wp-json\/wp\/v2\/posts\/211\/revisions"}],"predecessor-version":[{"id":326,"href":"https:\/\/donder-adventures.com\/wordpress\/wp-json\/wp\/v2\/posts\/211\/revisions\/326"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/donder-adventures.com\/wordpress\/wp-json\/wp\/v2\/media\/261"}],"wp:attachment":[{"href":"https:\/\/donder-adventures.com\/wordpress\/wp-json\/wp\/v2\/media?parent=211"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/donder-adventures.com\/wordpress\/wp-json\/wp\/v2\/categories?post=211"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/donder-adventures.com\/wordpress\/wp-json\/wp\/v2\/tags?post=211"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}