1
$\begingroup$

Consider the set of machine numbers $M(10, 2, 0)$. (The "zero-length" for the exponent is to be understood such that there is only the sign ± and 0 available for the exponent. We interpret "+" as "+1" and "−" as "−1". The available exponential factors are thus $10^{+1}, 10^0, 10^{−1}$.)

Perform the addition \begin{equation} 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{12}\end{equation} first from left to right \begin{equation} (...((1+\frac{1}{2})+\frac{1}{3})+...)+\frac{1}{12}\end{equation} and then from right to left \begin{equation} 1+(...+(\frac{1}{10}+(\frac{1}{11}+\frac{1}{12}))...)\end{equation} in the set $M(10,2,0).$

Start by first mapping the set of real numbers $1,\frac{1}{2},\frac{1}{3},...,\frac{1}{12}\in \mathbb{R}$ onto $M(10,2,0)$ via the "float operator" $fl$. For simplicity, we assume that $fl$ works by chopping off the digits for which there is not enough storage in $M(10, 2, 0).$

Which summation order gives the more accurate result, when compared to the results of the same calculation performed in $\mathbb{R}$?

So I'm studying ODEs and this exercise just came up. I can't figure out how to do it correctly. My idea is that we convert all the fractions into exponential notation, i.e. $fl(\frac{1}{6})=1.66\cdot 10^{-1}$. but I'm not sure how to compute the following:

first from left to right \begin{equation} (...((1+\frac{1}{2})+\frac{1}{3})+...)+\frac{1}{12}\end{equation} and then from right to left \begin{equation} 1+(...+(\frac{1}{10}+(\frac{1}{11}+\frac{1}{12}))...)\end{equation} in the set $M(10,2,0).$

Should I convert all the fractions and then add them up or should I do it like this:

\begin{equation} fl(fl(...fl(1+\frac{1}{2})...)) \end{equation}

EDIT: I just tried it again. Let's consider from right to left, then I have \begin{equation} fl(\frac{1}{11}+\frac{1}{12}) \end{equation} Now I continue to \begin{equation} fl(fl(\frac{1}{11}+\frac{1}{12})+\frac{1}{10}) \end{equation} And \begin{equation} fl(fl(fl(\frac{1}{11}+\frac{1}{12})+\frac{1}{10})+\frac{1}{9}) \end{equation} If I keep going I end with 3.101

If I do from left to right by the same method, I get 3.1. Is it the correct way to do it?

$\endgroup$
  • 1
    $\begingroup$ I think both. All the fractions themselves, and all the intermediate results are stored/computed by the machine so should be in the form that the machine handles. $\endgroup$ – Jaap Scherphuis Oct 3 '18 at 13:50
1
$\begingroup$

The point of this exercise is to show how the inherent finiteness of floating-point representations affects calculations. Therefore the second approach is correct - i.e convert all fractions to floating point, then perform the indicated calculations -in floating point also-.

I leave it to you to figure out why the second approach is more accurate.

$\endgroup$
  • $\begingroup$ Yes thank you. The problem is that I don't know how to convert it correct. I.e. $fl(1/3)=3.33\cdot 10^{-1}$ or $fl(1/3)=0.33\cdot 10^{0}$ how about $fl(1/7)=0.14\cdot 10^0$ can I do $fl(1/7)=1.42\cdot 10^{-1}$ $\endgroup$ – Joey Adams Oct 3 '18 at 15:37
  • $\begingroup$ So I did both way with method 2 after converting each fraction. The problem is that I end up with the same answer. I.e. 3.1 now. $\endgroup$ – Joey Adams Oct 3 '18 at 15:44
  • $\begingroup$ Now I got 3.101 when going from right to left. If I go from left to right I got 3.1, so a little difference. I just computed 1/11+1/12 and then took the fl(1/11+1/12) and added 1/10, such that fl(fl(1/11+1/12)+1/10) and so on... $\endgroup$ – Joey Adams Oct 3 '18 at 16:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.