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[函数] 函数方程$f(x)=-f(x^{-1})$

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hbghlyj Posted at 2020-6-30 21:54:16 |Read mode
Last edited by hbghlyj at 2024-12-13 15:05:00求所有连续函数$f:\mathbf{R}^+\to\mathbf{R}$满足$f(x)=-f(x^{-1}),\forall x\in\mathbf R^+$

解为\(f(x)=g(\ln(x))\)
其中$g:\mathbf{R}\to\mathbf{R}$是任意连续奇函数

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 Author| hbghlyj Posted at 2024-12-13 22:58:18
但是有一个答案是\[f(x)=\sum\limits_{k=1}^{\infty} c_k(x^k-x^{-k})\ldots\ldots(2)\]其中$c_k(k=±1,±2,\cdots)$是任意实数
这个答案是否包含所有的解?
例如
$f(x)=\ln x$
$f(x)=\arctan x-\frac\pi4$
它们能否表示为(2)的形式?

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2025-4-21 21:56 GMT+8

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