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HP Forum Archive 13

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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


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.

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