Timeline for Why is $\exp(\ln(x))-x\neq0$ in floating point arithmetic?
Current License: CC BY-SA 3.0
3 events
| when toggle format | what | by | license | comment | |
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| Apr 23, 2016 at 2:23 | comment | added | EngrStudent | I was about to say "happy round-off to you". | |
| Apr 22, 2016 at 20:58 | comment | added | Charles | This is correct. The virtual representation of any number has limited precision (matlab uses double precision by default) and manipulations (e.g. exp(log(x))) are computed with finite precision. After subtracting x, there is a loss of precision. | |
| Apr 22, 2016 at 20:42 | history | answered | user1841833 | CC BY-SA 3.0 |