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Re: A potatoe of a residu!

> What should be spelled C-I-N-C-H is the result of inputting
> Residue[1/Sinh[z],{z,I Pi}] .
> This should be the same as inputting
> Residue[1/(I Sin[-I z]),{z,I Pi}] .
> Both inputs result in the output
> Residue[Csch[z],{z, I Pi}].
> To feed Mathematica the dictionary necessary to get over the hurdle,  
> one must input
> Residue[2/(Exp[z]-Exp[-z]),{z,I Pi}].
> This results in the correct output, -1.
> Eeeeeeeeeeeeeekk!

This is caused by an oversight in the internal code for Residue.
In early versions of Mathematica, hyperbolic functions were automatically
converted to exponentials, so functions like Residue never had to
deal with hyperbolic functions directly.  When these automatic conversions
were removed, Residue was modified to include the new functions, but
a typo in the modification causes it to fail for Csch.

As you point out, Residue does know how to do the problem if Csch
is converted to something else:

In[30]:= Residue[Unevaluated[1/Sinh[z]], {z, I Pi}]

Out[30]= -1

n[37]:= res[f_, x_] := 
             Residue[Unevaluated[f], x] /. Csch[t_] :> 1/Sinh[t]

In[38]:= res[1/Sinh[z],{z,I Pi}]

Out[38]= -1

In[39]:= res[Csch[z],{z, I Pi}]

Out[39]= -1

Dave Withoff
Wolfram Research, Inc.
withoff at

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