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Proof: 0 == I Pi

Anonymous User
Anonymous User
Posted 6 years ago

Consider the following code:

In[34]:= mathematicaDefinitionOfArcCoth[x_] := 
 1/2 (-Log[x - 1] + Log[x + 1])

In[35]:= zero[x_] := 
 ArcCoth[x] - mathematicaDefinitionOfArcCoth[x]

In[36]:= zero[-1/2] // N

Out[36]= 0. + 3.14159 I

The definition of mathematicaDefinitionOfArcCoth is taken from the docs, section "Background & Context".

POSTED BY: Anonymous User
6 Replies
Posted 6 years ago

I wasn't sure what you were trying to do. I couldn't find your definition for mathematicaDefinitionOfArcCoth at the link you posted. The definition that I used was under Properties and Relations:

TrigToExp[ArcCoth[z]]

Out = -(1/2) Log[1-1/z]+1/2 Log[1+1/z]

But, something appears to be wrong with ArcCoth[-1/2] when N is applied or when using inexact arguments. It stays unevaluated with exact arguments outside of the domain.You should submit a bug report.

POSTED BY: David G
Anonymous User
Anonymous User
Posted 6 years ago

mathematicaDefinitionOfArcCoth comes from ArcCoth, Properties & Relations, see the final sentence and then back up two sentences.

A music-software company gives me free products when I beta-test for them, so I have to give my bug-report-submitting time to the other company.

POSTED BY: Anonymous User

Your mathematicaDefinitionOfArcCoth has a branch cut discontinuity in the complex plane, and -1/2 lies precisely along the cut. You can expect the numerical algorithms to give inconsistent results. It woud be nice of Mathematica to issue a warning in such cases, though.

POSTED BY: Gianluca Gorni
Anonymous User
Anonymous User
Posted 6 years ago

You can expect the numerical algorithms to give inconsistent results.

Mathematica gives the same false result even if you omit the "N"; the issue here is not a matter of any numerical algorithm.

The issue is rather a lack of coordination and communication between the developer who implemented ArcCoth and the person who authored the "Background & Context" section in the docs. As far as I have found, a fix would require no change to the implementation of ArcCoth; a fix would require only a correction to the documentation. I foresee no fix ever happening, though, because the consensus among the community is that wrong is better when wrong is what Mr. Wolfram chooses.

POSTED BY: Anonymous User
Posted 6 years ago

Try:

FullSimplify[-ArcCoth[z] - 1/2 Log[1 - 1/z] + 1/2 Log[1 + 1/z]]

The result is 0.

POSTED BY: David G
Anonymous User
Anonymous User
Posted 6 years ago

Thank you. If "Back & Context" said what you said, there would be no issue. Instead, "Back & Context" shows a different expression, one unequal to ArcCoth. Along the discontinuity, the discrepancy makes all the difference in the world.

POSTED BY: Anonymous User
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