2.13.tex

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Suppose that all numbers are positive.  Note $c_1$ and $p_1$ the
center and percentage tolerance of the interval $I_1$ and $c_2$ and
$p_2$ for the interval $I_2$.

We have,
\[ I_1 = [c_1(1-p_1), c_1(1+p_1)]\ \mbox{and}\ 
I_2 = [c_2(1-p_2), c_2(1+p_2)].\]
We then deduce the product $P$ of the two intervals
\[ P = [c_1c_2(1-p_1)(1-p_2), c_1c_2(1+p_1)(1+p_2)].\]
And finally the percentage tolerance $p$of $P$,
\begin{eqnarray*}
p &=& \frac{(1+p_1)(1+p_2) - (1-p_1)(1-p_2)}
{(1+p_1)(1+p_2) + (1-p_1)(1-p_2)} \\
&=& \frac{p_1+p_2}{1+p_1p_2}
\end{eqnarray*}
Under the assumption of small percentage tolerances, we then deduce
that
\[ p \simeq p_1 + p_2.\]
\end{document}