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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?

Winfield Chen
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NabbKitha
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    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. – Jaap Scherphuis Oct 03 '18 at 13:50

1 Answers1

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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.

PMar
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  • 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 $fl(1/7)=1.42\cdot 10^{-1}$ – NabbKitha Oct 03 '18 at 15:37
  • 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. – NabbKitha Oct 03 '18 at 15:44
  • 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... – NabbKitha Oct 03 '18 at 16:12