{"id":2373,"date":"2022-02-26T11:16:39","date_gmt":"2022-02-26T02:16:39","guid":{"rendered":"https:\/\/home.hirosaki-u.ac.jp\/relativity\/?page_id=2373"},"modified":"2024-07-08T17:36:48","modified_gmt":"2024-07-08T08:36:48","slug":"%e5%91%a8%e6%9c%9f-2pi-%e3%81%ae%e3%83%95%e3%83%bc%e3%83%aa%e3%82%a8%e7%b4%9a%e6%95%b0%e5%b1%95%e9%96%8b%e3%81%ae%e4%be%8b","status":"publish","type":"page","link":"https:\/\/home.hirosaki-u.ac.jp\/relativity\/%e7%90%86%e5%b7%a5%e7%b3%bb%e3%81%ae%e6%95%b0%e5%ad%a6c\/%e3%83%95%e3%83%bc%e3%83%aa%e3%82%a8%e8%a7%a3%e6%9e%90\/%e3%83%95%e3%83%bc%e3%83%aa%e3%82%a8%e7%b4%9a%e6%95%b0\/%e5%91%a8%e6%9c%9f-2pi-%e3%81%ae%e3%83%95%e3%83%bc%e3%83%aa%e3%82%a8%e7%b4%9a%e6%95%b0%e5%b1%95%e9%96%8b%e3%81%ae%e4%be%8b\/","title":{"rendered":"\u5468\u671f $2\\pi$ \u306e\u30d5\u30fc\u30ea\u30a8\u7d1a\u6570\u5c55\u958b\u306e\u4f8b"},"content":{"rendered":"

\\( -\\pi \\le x \\le \\pi \\) \u3067\u5b9a\u7fa9\u3055\u308c\u305f\u95a2\u6570 \\( f(x) = x^2 \\) \u306e\u30d5\u30fc\u30ea\u30a8\u7d1a\u6570\u5c55\u958b\u3084\u3063\u3066\u3093\u304b\u3044\uff01<\/p>\n

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\"\"<\/p>\n

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\u533a\u9593 \\( -\\pi \\le x \\le \\pi \\) \u3067\u5b9a\u7fa9\u3055\u308c\u305f\u95a2\u6570 \\( f(x) = x^2 \\) \u3092\uff0c\u533a\u9593\u5916\u3067\u306f\u5468\u671f $2\\pi$ \u306e\u5468\u671f\u95a2\u6570\u3068\u307f\u306a\u3057\u3066…<\/p>\n

\"\"<\/p>\n

\u30d5\u30fc\u30ea\u30a8\u7d1a\u6570\u5c55\u958b<\/h3>\n

$$ f(x) = \\frac{a_0}{2} + \\sum_{n=1}^{\\infty} \\bigl( a_n \\cos n x + b_n \\sin nx \\bigr) $$<\/p>\n

\u30d5\u30fc\u30ea\u30a8\u4fc2\u6570\uff1a
$$a_0 = \\frac{1}{\\pi} \\int_{-\\pi}^{\\pi} f(x) dx = \\frac{2}{\\pi} \\int_0^{\\pi} x^2 dx = \\frac{2}{\\pi} \\Bigl[ \\frac{x^3}{3} \\Bigr]_0^{\\pi} = \\frac{2}{3} \\pi^2 $$<\/p>\n

\\(n \\geq 1 \\) \u306b\u5bfe\u3057\u3066
\\begin{eqnarray}
a_n &=& \\frac{1}{\\pi} \\int_{-\\pi}^{\\pi} f(x) \\cos n x\\, dx =\u00a0\\frac{1}{\\pi} \\int_{-\\pi}^{\\pi} x^2 \\cos n x\\, dx \\\\
&=& \\frac{1}{\\pi}\\Bigl[ x^2 \\frac{1}{n} \\sin n x \\Bigr]_{-\\pi}^{\\pi} –\u00a0 \\frac{1}{\\pi}\\int_{-\\pi}^{\\pi} 2 x \\frac{1}{n} \\sin n x\\, dx \\\\
&=& \\frac{1}{\\pi}\\Bigl[ x^2 \\frac{1}{n^2} \\cos n x \\Bigr]_{-\\pi}^{\\pi}-\u00a0 \\frac{1}{\\pi}\\int_{-\\pi}^{\\pi} 2 \\frac{1}{n^2} \\cos n x\\, dx \\\\
&=& \\frac{2}{\\pi} \\times 2 \\pi \\frac{1}{n^2} \\cos n \\pi \\\\
&=& \\frac{ 4 \\cos n \\pi}{n^2} = \\frac{4\\cdot (-1)^n}{n^2}
\\end{eqnarray}
$$ b_n = \\frac{1}{\\pi} \\int_{-\\pi}^{\\pi} f(x) \\sin n x\\, dx =\\frac{1}{\\pi} \\int_{-\\pi}^{\\pi} x^2 \\sin n x\\, dx = 0$$ \uff08\u88ab\u7a4d\u5206\u95a2\u6570\u304c\u5947\u95a2\u6570\u3060\u304b\u3089\u3002\u307e\u305f\u306f \\(f(x)\\) \u304c\u5076\u95a2\u6570\u3060\u304b\u3089\uff0c\u5947\u95a2\u6570\u3067\u3042\u308b \\(\\sin n x\\) \u306e\u30d5\u30fc\u30ea\u30a8\u4fc2\u6570 \\(b_n\\) \u306f\u5168\u3066\u30bc\u30ed\uff0c\u3068\u3044\u3063\u3066\u3082\u3088\u3044\u3002\uff09<\/p>\n

\\begin{eqnarray}
\\therefore\\ f(x) &=& \\frac{a_0}{2} + a_1 \\cos x + a_2 \\cos 2 x + a_3 \\cos 3 x + a_4 \\cos 4x + \\cdots \\\\
&=& \\frac{\\pi^2}{3} -4 \\cos x + \\cos 2 x -\\frac{4}{9} \\cos 3 x + \\frac{1}{4} \\cos 4 x + \\cdots
\\end{eqnarray}<\/p>\n

\u52d5\u753b\u3067\u793a\u3059\u30d5\u30fc\u30ea\u30a8\u7d1a\u6570\u5c55\u958b \\(f(x) = x^2 \\) \u306e\u5834\u5408<\/h3>\n
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