On Mon, Jul 15, 2019 at 4:26 PM Bill Gosper <billgosper@gmail.com> wrote:

In[95]:= $Version

Out[95]= "12.0.0 for Mac OS X x86 (64-bit) (February 25, 2019)"

In[76]:= InverseFunction@InverseFunction // FunctionExpand

Out[76]= InverseFunction[InverseFunction]

Shouldn't this just be Identity? E.g.,

In[81]:= InverseFunction@InverseFunction@Tan

Out[81]= Tan

But In[82]:= (InverseFunction@InverseFunction)@Tan

Out[82]= InverseFunction[InverseFunction][Tan]

(Hang onto your hat:) In[83]:= %[\[Pi]/4]

Out[83]= -2 (24 \[Pi] + ArcTan[4/\[Pi] + Sqrt[16 + \[Pi]^2]/\[Pi]])

!!!

After consulting the American Psychiatric Association's *Diagnostic and Statistical Manual of Mental Disorders* (*DSM–5*), Julian says: This isn't a bug. InverseFunction[InverseFunction] shouldn't be Identity; it's the inverse of the operator which inverts a function, i.e. it takes f^-1 to f, i.e. it's InverseFunction—except not, because it has to handle non-uniqueness somewhat differently. Thus In[165]:= FullSimplify[Tan[InverseFunction[InverseFunction][Tan][π/4]]] Out[165]= π/4 What it chooses to do is weird, and I don't know why Plot[InverseFunction[InverseFunction][Tan][x] - ArcTan[x], {x, -10, 10}] looks how it does, but it's not strictly wrong. ------ Try this! It might be a portal to another reality! —rwg AKA (ries)

Out[90]= -49 \[Pi] + ArcCos[4/Sqrt[16 + \[Pi]^2]]

In[91]:= FullSimplify[% - %83]

Out[91]= -(\[Pi]/2) - ArcTan[4/\[Pi]] + 2 ArcTan[(4 + Sqrt[16 + \[Pi]^2])/\[Pi]]

Another bug!

In[92]:= Sign@%

During evaluation of In[92]:= N::meprec: Internal precision limit $MaxExtraPrecision = 50.` reached while evaluating -(\[Pi]/2)-ArcTan[4/\[Pi]]+2 ArcTan[(4+Sqrt[16+Power[<<2>>]])/\[Pi]].

Out[92]= Sign[-(\[Pi]/2) - ArcTan[4/\[Pi]] + 2 ArcTan[(4 + Sqrt[16 + \[Pi]^2])/\[Pi]]] —Bill