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对一个恒等式的深入审视Closer look at an identity

johndcook.com·2026-05-19 节选正文

文章回顾并验证了一个先前提出的恒等式,重点分析了其在 x > 1 且 y > 1 条件下成立的原因,同时解释了为何需要特别注明这一限制条件。通过使用 Mathematica 绘制函数图像,作者展示了当 x ≤ 1 或 y ≤ 1 时图像不再平坦,说明恒等式在这些区域不成立。这表明原恒等式的有效性依赖于变量的取值范围,突显了数学证明中边界条件的重要性。

John

The previous post derived the identity

and said in a footnote that the identity holds at least for x > 1 and y > 1. That’s true, but let’s see why the footnote is necessary.

Let’s have Mathematica plot

The plot will be 0 where the identity above holds.

The plot is indeed flat for x > 1 and y > 1, and more, but not everywhere.

If we combine the two square roots

and plot again we still get a valid identity for x > 1 and y > 1, but the plot changes.

This is because √a √b does not necessarily equal √(ab) when the arguments may be negative.

The square root function and the arccosh function are not naturally single-valued functions. They require branch cuts to force them to be single-valued, and the two functions require different branch cuts. I go into this in some detail here.

There is a way to reformulate our identity so that it holds everywhere. If we replace

with

which is equivalent for z > 1, the corresponding identity holds everywhere.

We can verify this with the following Mathematica code.

f[z_] := Exp[(1/2) (Log[z - 1 ] + Log[z + 1])]
FullSimplify[Cosh[ArcCosh[x] + ArcCosh[y]] - x y - f[x] f[y]]

This returns 0.

By contrast, the code

FullSimplify[
 Cosh[ArcCosh[x] + ArcCosh[y]] - x y - Sqrt[x^2 - 1] Sqrt[y^2 - 1]]

simply returns its input with no simplification, unless we add restrictions on x and y. The code

FullSimplify[
 Cosh[ArcCosh[x] + ArcCosh[y]] - x y - Sqrt[x^2 - 1] Sqrt[y^2 - 1], 
 Assumptions -> {x > -1 && y > -1}]

does return 0.

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