(1)<\/strong>\u3000\\(f(x)\\) \u304c\u6700\u5c0f\u3068\u306a\u308b \\(x\\) \u306e\u5024 \\(s _ p\\) \u3092\u6c42\u3081, \\(y = f(x)\\) \u306e\u30b0\u30e9\u30d5\u3092\u63cf\u3051.<\/p><\/li>\r\n (2)<\/strong>\u3000\r\n\\[\r\ng(t) = \\displaystyle\\int _ t^{t+1} f(x) e^{t-x} \\, dx\r\n\\]\r\n\u3068\u304a\u304f. \\(g(t)\\) \u304c\u6700\u5c0f\u3068\u306a\u308b \\(t\\) \u306e\u5024 \\(t _ p\\) \u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n (3)<\/strong>\u3000\\(0 \\lt p \\leqq 1\\) \u306e\u3068\u304d,\r\n\\[\r\n1+\\dfrac{p}{2} \\leqq \\dfrac{e^p-1}{p} \\leqq 1+\\dfrac{p}{2}+p^2\r\n\\]\r\n\u304c\u6210\u7acb\u3059\u308b\u3053\u3068\u3092\u7528\u3044\u3066, \u53f3\u5074\u304b\u3089\u306e\u6975\u9650 \\(\\displaystyle\\lim _ {p \\rightarrow +0} (t _ p -s _ p)\\) \u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n<\/ol>\r\n (1)<\/strong><\/p>\r\n \\(f(x) = e^x \\left( e^{px} -1 \\right)\\) \u3088\u308a\r\n\\[\\begin{align}\r\nf'(x) & = (p+1) e^{(p+1)x} -e^x = e^x \\left\\{ (p+1)e^{px} -1 \\right\\} , \\\\\r\nf''(x) & = (p+1)^2 e^{(p+1)x} -e^x = e^x \\left\\{ (p+1)^2e^{px} -1 \\right\\}\r\n\\end{align}\\]\r\n\\(f'(x) =0\\) \u3092\u89e3\u304f\u3068\r\n\\[\\begin{align}\r\ne^{px} & = \\dfrac{1}{p+1} \\\\\r\n\\text{\u2234} \\quad x & = -\\dfrac{1}{p} \\log (p+1)\r\n\\end{align}\\]\r\n\\(f''(x) =0\\) \u3092\u89e3\u304f\u3068\r\n\\[\\begin{align}\r\ne^{px} & = \\dfrac{1}{(p+1)^2} \\\\\r\n\\text{\u2234} \\quad x & = -\\dfrac{2}{p} \\log (p+1)\r\n\\end{align}\\]\r\n\u3055\u3089\u306b\r\n\\[\r\n\\displaystyle\\lim _ {x \\rightarrow -\\infty} f(x) = 0 , \\ \\displaystyle\\lim _ {x \\rightarrow \\infty} f(x) = +\\infty\r\n\\]\r\n\u306a\u306e\u3067, \\(f(x)\\) \u306e\u5897\u6e1b\u3068\u51f9\u51f8\u306f\u4e0b\u8868\u306e\u3088\u3046\u306b\u306a\u308b.\r\n\\[\r\n\\begin{array}{c|ccccccc} x & ( -\\infty ) & \\cdots & -\\dfrac{2 \\log (p+1)}{p} & \\cdots & -\\dfrac{\\log (p+1)}{p} & \\cdots & ( \\infty ) \\\\ \\hline f'(x) & & - & & + & 0 & + & \\\\ \\hline f''(x) & & - & 0 & + & & + & \\\\ \\hline f(x) & (0) & \\searrow ( \\cap ) & & \\searrow ( \\cup ) & \\text{\u6700\u5c0f} & \\nearrow ( \\cup ) & ( \\infty ) \\end{array}\r\n\\]\r\n\u3088\u3063\u3066\r\n\\[\r\ns _ p =\\underline{-\\dfrac{1}{p} \\log (p+1)}\r\n\\]\r\n\u307e\u305f\r\n\\[\\begin{align}\r\nf \\left( -\\dfrac{1}{p} \\log (p+1) \\right) & = \\left( \\dfrac{1}{p+1} \\right)^{\\frac{1}{p}} \\left( \\dfrac{1}{p+1} -1 \\right) \\\\\r\n& = -\\dfrac{p}{(p+1)^{\\frac{p+1}{p}}} , \\\\\r\nf \\left( -\\dfrac{2}{p} \\log (p+1) \\right) & = \\left( \\dfrac{1}{p+1} \\right)^{\\frac{2}{p}} \\left\\{ \\dfrac{1}{(p+1)^2} -1 \\right\\} \\\\\r\n& = -\\dfrac{p(p+2)}{(p+1)^{\\frac{2(p+1)}{p}}}\r\n\\end{align}\\]\r\n\u306a\u306e\u3067, \u30b0\u30e9\u30d5\u306e\u6982\u5f62\u306f\u4e0b\u56f3\u306e\u3088\u3046\u306b\u306a\u308b.<\/p>\r\n (2)<\/strong><\/p>\r\n \\[\\begin{align}\r\ng(t) & = \\displaystyle\\int _ t^{t+1} e^{t-x} \\, dx = e^t \\displaystyle\\int _ t^{t+1} \\left( e^px -1 \\right) \\, dx \\\\\r\n& = e^t \\left[ \\dfrac{e^{px}}{p} -x \\right] _ t^{t+1} \\\\\r\n& = e^t \\left\\{ \\dfrac{e^{p(t+1)}}{p} -t-1 -\\dfrac{e^{pt}}{p} +t \\right\\} \\\\\r\n& = e^t \\left\\{ \\dfrac{e^{pt} \\left( e^p -1 \\right)}{p} -1 \\right\\} \\\\\r\n& = \\dfrac{e^p -1}{p} e^{(p+1)t} -e^t\r\n\\end{align}\\]\r\n\u3057\u305f\u304c\u3063\u3066\r\n\\[\\begin{align}\r\ng'(t) & = \\dfrac{(p+1) \\left( e^p -1 \\right)}{p} e^{(p+1)t} -e^t \\\\\r\n& = e^t \\left\\{ \\dfrac{(p+1) \\left( e^p -1 \\right)}{p} e^{pt} -1 \\right\\}\r\n\\end{align}\\]\r\n\\(g'(t) = 0\\) \u3092\u89e3\u304f\u3068\r\n\\[\\begin{align}\r\ne^{pt} & = \\dfrac{p}{(p+1) \\left( e^p -1 \\right)} \\\\\r\n\\text{\u2234} \\quad x & = -\\dfrac{1}{p} \\log \\dfrac{(p+1) \\left( e^p -1 \\right)}{p}\r\n\\end{align}\\]\r\n\u3057\u305f\u304c\u3063\u3066, \\(f(x)\\) \u306e\u5897\u6e1b\u306f\u4e0b\u8868\u306e\u3088\u3046\u306b\u306a\u308b.\r\n\\[\r\n\\begin{array}{c|ccc} x & \\cdots & -\\dfrac{1}{p} \\log \\dfrac{(p+1) \\left( e^p -1 \\right)}{p} & \\cdots \\\\ \\hline f'(x) & - & 0 & + \\\\ \\hline f(x) & \\searrow & \\text{\u6700\u5c0f} & \\nearrow \\end{array}\r\n\\]\r\n\u3088\u3063\u3066\r\n\\[\r\nt _ p = \\underline{-\\dfrac{1}{p} \\log \\dfrac{(p+1) \\left( e^p -1 \\right)}{p}}\r\n\\]\r\n (3)<\/strong><\/p>\r\n \\[\\begin{align}\r\nt _ p -s _ p & = -\\dfrac{1}{p} \\left\\{ \\log \\dfrac{(p+1) \\left( e^p -1 \\right)}{p} -\\log (p+1) \\right\\} \\\\\r\n& = -\\dfrac{1}{p} \\log \\dfrac{e^p -1}{p} \\\\\r\n& = -\\log \\left( \\dfrac{e^p -1}{p} \\right)^{\\frac{1}{p}} \\quad ... [1]\r\n\\end{align}\\]\r\n\\(p \\gt 0\\) \u306a\u306e\u3067, \u4e0e\u5f0f\u3088\u308a\r\n\\[\r\n\\left( 1 +\\dfrac{p}{2} \\right)^{\\frac{1}{p}} \\leqq \\left( \\dfrac{e^p -1}{p} \\right)^{\\frac{1}{p}} \\leqq \\left( 1 +\\dfrac{p}{2} +p^2 \\right)^{\\frac{1}{p}}\r\n\\]\r\n\u3053\u306e\u4e0d\u7b49\u5f0f\u306b\u3064\u3044\u3066, \\(p \\rightarrow +0\\) \u306e\u3068\u304d\u3092\u8003\u3048\u308b\u3068\r\n\\[\\begin{align}\r\n\\left( 1 +\\dfrac{p}{2} \\right)^{\\frac{1}{p}} & = \\left( 1+\\dfrac{p}{2} \\right)^{\\frac{1}{2} \\cdot \\frac{2}{p}} \\\\\r\n& \\rightarrow e^{\\frac{1}{2}} , \\\\\r\n\\left( 1 +\\dfrac{p}{2} +p^2 \\right)^{\\frac{1}{p}} & = \\left\\{ 1 +\\dfrac{p(2p+1)}{2} \\right\\}^{\\frac{1}{p}} \\\\\r\n& = \\left\\{ 1+\\dfrac{p(2p+1)}{2} \\right\\}^{\\frac{2p+1}{2} \\cdot \\frac{2}{p(2p+1)}} \\\\\r\n& \\rightarrow e^{\\frac{1}{2}}\r\n\\end{align}\\]\r\n\u3057\u305f\u304c\u3063\u3066, \u306f\u3055\u307f\u3046\u3061\u306e\u539f\u7406\u3088\u308a\r\n\\[\r\n\\displaystyle\\lim _ {p \\rightarrow +0} \\left( \\dfrac{e^p -1}{p} \\right)^{\\frac{1}{p}} = e^{\\frac{1}{2}}\r\n\\]\r\n\u3088\u3063\u3066, [1] \u3068\u3042\u308f\u305b\u3066\r\n\\[\r\n\\displaystyle\\lim _ {p \\rightarrow +0} \\left( t _ p -s _ p \\right) = -\\log e^{\\frac{1}{2}} = \\underline{-\\dfrac{1}{2}}\r\n\\]\r\n","protected":false},"excerpt":{"rendered":"\u5b9f\u6570 \\(p \\gt 0\\) \u306b\u5bfe\u3057\u3066, \\[ f(x) = e^{(p+1)x} -e^x \\] \u3068\u304a\u304f. \u4ee5\u4e0b\u306e\u554f\u306b\u7b54\u3048\u3088. (1)\u3000\\(f(x)\\) \u304c\u6700\u5c0f\u3068\u306a\u308b \\(x\\) \u306e\u5024 \\(s _ p\\) \u3092\u6c42\u3081, \\ … \u7d9a\u304d\u3092\u8aad\u3080
\r\n\r\n\u3010 \u89e3 \u7b54 \u3011<\/h4>\r\n
![\"waseda_r_200905_01\"](\"\/\/www.roundown.net\/nyushi\/wp-content\/uploads\/waseda_r_200905_01.png\")
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