#3189 -1/taylor(x,x,0,1) => 0

[Stavros Macrakis]

1/taylor(x,x,0,1) => 1/x+...   OK
taylor(x,x,0,1)*taylor(1/x,x,0,1) => 1+... OK

-1/taylor(x,x,0,1) => 0+...   WRONG
2/taylor(x,x,0,1) => 0+...    WRONG
x/taylor(x,x,0,1) => 0+...    WRONG
taylor(x,x,0,1)/taylor(x,x,0,1) => 0+... WRONG

[Stavros Macrakis]

The bug report read
~~~~
taylor(x,x,0,1)*taylor(1/x,x,0,1) => 1/x+... OK
~~~~
but of course the correct result is 1+..., which is what it currently returns. This was probably a typo in the bug report.

Taylor also now reports (I don't know if it did before) that it is assuming that 1/x+... is zero in the WRONG cases. I don't think that should excuse the incorrect results.