Euler's identity in J

Question

Just discovering J language, I typed:

1+^o.*0j1

I expected the answer to be 0, but I get 0j1.22465e_16. Although this is very close to 0, I was wondering why there is a numerical error in the answer while J is supposed to be able to perform arbitrary precision arithmetic? How could I get the correct answer 0?

Answer 1

Arbitrary precision is for integers and rationals.

The constants π (1p1) and e (1x1) are still floating point numbers with precision of about 17 significant decimal digits .

The only way to produce exactly 0 is if your code is "aware" of the case. The composition ^@o. is special code for this case.

 ^@o. 0j1
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