Calcuator forensics history question

[Garth Wilson]

Here's a repeat of part of a post of mine from several years ago (which I can't find now to put in a link, and the search here does not work well) in response to a complaint that the HP-41 calculator did not have good accuracy because alog(log(60)) showed 59.999999950:

This is something slide-rule users will have a better understanding of. It's not really any bug in the machine, just a limitation of the number of digits it uses.

Consider: ln(60) is 4.09434456222, according to my HP-71B which uses more digits than the 41 does. Using only 10 digits, e^4.094344562 is 59.9999999867, and e^4.94344563 (ie, incrementing the least-significant digit by 1) is 60.0000000467, both quite a ways off from 60 in the 10th digit. IOW, there is no 10-digit number that will give you 60.00000000 when you take the natural antilog.

You find this more in for example cos(1°), only one digit, where you get .99984769516; but using only four digits of it to convert back to the angle, .9998 gets you over 14% high, at 1.1459..., and .9999 gets you .81029..., almost 19% low. There is no 4-digit number which, when taking its ACOS, will result in 1.0° (two digits), let alone 1.000° (four digits).