\\(f(x)\\) \u306f\u3059\u3079\u3066\u306e\u5b9f\u6570 \\(x\\) \u306b\u304a\u3044\u3066\u5fae\u5206\u53ef\u80fd\u306a\u95a2\u6570\u3067, \u95a2\u4fc2\u5f0f\r\n\\[\r\nf(2x) = ( e^x+1 ) f(x)\r\n\\]\r\n\u3092\u307f\u305f\u3057\u3066\u3044\u308b\u3068\u3059\u308b. \u4ee5\u4e0b\u306e\u554f\u306b\u7b54\u3048\u3088.<\/p>\r\n
(1)<\/strong>\u3000\\(f(0) =0\\) \u3092\u793a\u305b.<\/p><\/li>\r\n (2)<\/strong>\u3000\\(x \\neq 0\\) \u306b\u5bfe\u3057\u3066\r\n\\[\r\n\\dfrac{f(x)}{e^x-1} = \\dfrac{f \\left( \\frac{x}{2} \\right)}{e^{\\frac{x}{2}}-1}\r\n\\]\r\n\u304c\u6210\u308a\u7acb\u3064\u3053\u3068\u3092\u793a\u305b.<\/p><\/li>\r\n (3)<\/strong>\u3000\u5fae\u5206\u306e\u5b9a\u7fa9\u3092\u7528\u3044\u3066, \\(f'(0) = \\displaystyle\\lim _ {h \\rightarrow 0} \\dfrac{f(h)}{e^h-1}\\) \u3092\u793a\u305b.<\/p><\/li>\r\n (4)<\/strong>\u3000\\(f(x) = (e^x-1) f'(0)\\) \u304c\u6210\u308a\u7acb\u3064\u3053\u3068\u3092\u793a\u305b.<\/p><\/li>\r\n<\/ol>\r\n (1)<\/strong><\/p>\r\n \u95a2\u4fc2\u5f0f\u306b \\(x=0\\) \u3092\u4ee3\u5165\u3059\u308b\u3068\r\n\\[\\begin{align}\r\nf(0) & = 2 f(0) \\\\\r\n\\text{\u2234} \\quad f(0) & = \\underline{0}\r\n\\end{align}\\]\r\n (2)<\/strong><\/p>\r\n \u95a2\u4fc2\u5f0f\u306e\u4e21\u8fba\u306b \\(e^x-1 ( \\neq 0 )\\) \u3092\u639b\u3051\u308b\u3068\r\n\\[\\begin{align}\r\n(e^x-1) f(2x) & = (e^{2x}-1) f(x) \\\\\r\n\\text{\u2234} \\quad \\dfrac{f(2x)}{e^{2x}-1} & = \\dfrac{f(x)}{e^x-1}\r\n\\end{align}\\]\r\n\\(x\\) \u3092 \\(\\dfrac{x}{2}\\) \u306b\u7f6e\u304d\u63db\u3048\u308c\u3070\r\n\\[\r\n\\dfrac{f(x)}{e^x-1} = \\dfrac{f \\left( \\frac{x}{2} \\right)}{e^{\\frac{x}{2}}-1}\r\n\\]\r\n (3)<\/strong><\/p>\r\n \\(g(x) = e^x\\) \u3068\u304a\u304f\u3068, \\(g'(x) =e^x\\) .\r\n\\[\r\ng'(0) = \\displaystyle\\lim _ {h \\rightarrow 0} \\dfrac{e^h -1}{h} = e^0 = 1\r\n\\]\r\n\u3053\u308c\u3092\u7528\u3044\u308c\u3070\r\n\\[\\begin{align}\r\nf'(0) & = \\displaystyle\\lim _ {h \\rightarrow 0} \\dfrac{f(h) -f(0)}{h} = \\displaystyle\\lim _ {h \\rightarrow 0} \\dfrac{f(h)}{h} \\cdot \\dfrac{h}{e^h-1} \\\\\r\n& = \\underline{\\displaystyle\\lim _ {h \\rightarrow 0} \\dfrac{f(h)}{e^h-1}}\r\n\\end{align}\\]\r\n (4)<\/strong><\/p>\r\n (2)<\/strong> \u306e\u7d50\u679c\u3092\u7e70\u8fd4\u3057\u7528\u3044, (3)<\/strong> \u306e\u7d50\u679c\u3092\u7528\u3044\u308c\u3070\r\n\\[\\begin{align}\r\n\\dfrac{f(x)}{e^x-1} & = \\cdots = \\displaystyle\\lim _ {n \\rightarrow \\infty} \\dfrac{f \\left( \\frac{x}{2^n} \\right)}{e^{\\frac{x}{2^n}}-1} \\\\\r\n& = \\displaystyle\\lim _ {h \\rightarrow 0} \\dfrac{f(h)}{e^h-1} = f'(0)\r\n\\end{align}\\]\r\n\u3088\u3063\u3066\r\n\\[\r\nf(x) = (e^x-1) f'(0)\r\n\\]\r\n","protected":false},"excerpt":{"rendered":"\\(f(x)\\) \u306f\u3059\u3079\u3066\u306e\u5b9f\u6570 \\(x\\) \u306b\u304a\u3044\u3066\u5fae\u5206\u53ef\u80fd\u306a\u95a2\u6570\u3067, \u95a2\u4fc2\u5f0f \\[ f(2x) = ( e^x+1 ) f(x) \\] \u3092\u307f\u305f\u3057\u3066\u3044\u308b\u3068\u3059\u308b. \u4ee5\u4e0b\u306e\u554f\u306b\u7b54\u3048\u3088. (1)\u3000\\(f(0) =0\\) \u3092\u793a … \u7d9a\u304d\u3092\u8aad\u3080
\r\n\r\n\u3010 \u89e3 \u7b54 \u3011<\/h4>\r\n