题目

Problem

Given that

\cosh y = x \qquad \text{and} \qquad y < 0,

use the definition of $\cosh y$ in terms of exponential functions to prove that

y = \ln\left(x - \sqrt{x^2 - 1}\right).

(6)

题目中文翻译

已知

\cosh y = x \qquad \text{且} \qquad y < 0

利用 $\cosh y$ 的指数定义证明

y = \ln\left(x - \sqrt{x^2 - 1}\right)

解答

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按题目要求写出 $\cosh y$ 的指数定义，乘以 $e^y$ 后得到关于 $e^y$ 的二次方程。两个根互为倒数；条件 $y<0$ 意味着 $0<e^y<1$ ，因此必须选择较小的根。

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Using the exponential definition of $\cosh y$ ,

x=\frac{e^y+e^{-y}}{2}.

Multiplying by $2e^y$ gives

e^{2y}-2xe^y+1=0.

Treating this as a quadratic in $e^y$ ,

\begin{align*} e^y =&\,\frac{2x\pm\sqrt{4x^2-4}}{2} \\ =&\,x\pm\sqrt{x^2-1}. \end{align*}

Since $y<0$ , $0<e^y<1$ . Also $x=\cosh y>1$ , so

x+\sqrt{x^2-1}>1.

Therefore, the smaller root must be selected:

e^y=x-\sqrt{x^2-1}.

Taking natural logarithms,

\boxed{y=\ln\bigl(x-\sqrt{x^2-1}\bigr)}.

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从待证结果出发，先求 $e^y$ ，再把其倒数有理化得到 $e^{-y}$ 。两者相加后恰好为 $2x$ ，从而恢复 $\cosh y=x$ ；同时检查对数真数位于 $(0,1)$ ，保证 $y<0$ 。

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Let

y=\ln\bigl(x-\sqrt{x^2-1}\bigr).

Then

e^y=x-\sqrt{x^2-1}.

Rationalising its reciprocal,

\begin{align*} e^{-y} =&\,\frac1{x-\sqrt{x^2-1}} \\ =&\,x+\sqrt{x^2-1}, \end{align*}

because

x^2-(x^2-1)=1.

Therefore,

\begin{align*} \cosh y =&\,\frac{e^y+e^{-y}}2 \\ =&\,\frac{x-\sqrt{x^2-1} +x+\sqrt{x^2-1}}2 \\ =&\,x. \end{align*}

Moreover, for $x>1$ ,

0<x-\sqrt{x^2-1}<1,

so $y<0$ . Hence the stated expression is the required negative solution.