The Museum of HP Calculators

HP Forum Archive 13

 Round off (beautiful...)Message #1 Posted by Tizedes Csaba on 13 Aug 2003, 1:40 a.m. In the first math exam we got this example: What is the limit of this serie: (the calculator use not was allowable!) an=(1+1/(2^n+1))^(2^(n+1)) where n->infinity (n is integer) the limit is e^2 (~7.39), but look, what happen if we try to calculate it! Grab your 48, and plot it: 0 40 XRNG 0 10 YRNG 'n' INDEP ERASE DRAW You'll see, the value of function about 37.5-37.8 will be 1. (not 7.39) That happen, when 1+1/(2^n+1)=1 for 12 digit, the limit will be 1. Where? Its simple calculable: 1/(2^n+1)<5e-12 -> n>37.54 Csaba

 Re: Round off (beautiful...)Message #2 Posted by hugh on 13 Aug 2003, 3:28 p.m.,in response to message #1 by Tizedes Csaba like you say, let k=2^n+1, then your problem is written (((1+1/k)^k)/(1+1/k))^2 ~ e^2 as k->inf. the difficulty for the calculator is (1+x/k)^k -> exp(x) as k->inf because you have here a classic 1+epsilon problem. the 1+eps problem is a defect of floating point numbers as real approximations. floats (inc calculator BCD) are only dense around zero. with a single floating point, 1+x-1 = x, only where abs(x) > 10^(-d) where d is the number of internal working digits. when IEEE floating point was invented (1983??), it was a big step forward in standardisation. i believe its time for calculator makers to move forward again onto a fundamentally new way of approximating reals.

Go back to the main exhibit hall