The earlier submit derived the id
and mentioned in a footnote that the id holds not less than for x > 1 and y > 1. That’s true, however let’s see why the footnote is critical.
Let’s have Mathematica plot
The plot will likely be 0 the place the id above holds.
The plot is certainly flat for x > 1 and y > 1, and extra, however not in every single place.
If we mix the 2 sq. roots
and plot once more we nonetheless get a sound id for x > 1 and y > 1, however the plot adjustments.
It’s because √a √b doesn’t essentially equal √(ab) when the arguments could also be adverse.
The sq. root operate and the arccosh operate aren’t naturally single-valued features. They require department cuts to power them to be single-valued, and the 2 features require totally different department cuts. I’m going into this in some element right here.
There’s a strategy to reformulate our id in order that it holds in every single place. If we substitute
with
which is equal for z > 1, the corresponding id holds in every single place.
We are able to confirm this with the next 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.
In contrast, the code
FullSimplify[ Cosh[ArcCosh[x] + ArcCosh[y]] - x y - Sqrt[x^2 - 1] Sqrt[y^2 - 1]]
merely returns its enter with no simplification, until 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.
Associated posts
The submit Nearer take a look at an id first appeared on John D. Prepare dinner.
