Example 1

Computation of the errors for the product of real numbers

Let $r_1,r_2$ two real numbers, and assume that $r_1^0,r_2^0$ are known rational approximations for $r_1$ and $r_2$, of accuracies $\delta_1$ and $\delta_2$ respectively.

We can write:

$$\vert r_1r_2-r_1^0r_2^0\vert = \vert r_1(r_2-r_2^0)+r_2^0(r_1^0-r_1)\vert \le \vert r_1\vert\,\vert r_2-r_2^0\vert+\vert r_2^0\vert\,\vert r_1-r_1^0\vert$$

$$\le (\vert r_1^0\vert+\delta_1)\delta_2+\vert r_2^0\vert\delta_1.$$

Thus, $\vert r_1^0\vert\delta_2+\delta_1\delta_2+\vert r_2^0\vert\delta_1$ is a suitable choice for the accuracy of $r_1^0r_2^0$.

Let now $\varepsilon>0$. We are looking for $\tilde \delta_1,\tilde\delta_2>0$ and $\tilde r_1^0,\tilde r_2^0$, approximations of accuracies $\tilde \delta_1,\tilde\delta_2>0$ for $r_1$ and $r_2$, so that:

$$\vert r_1r_2-\tilde r_1^0\tilde r_2^0\vert\le\varepsilon.$$

We mean to make use of the known approximations, $r_1^0,r_2^0$.

We can write $\vert r_1r_2-\tilde r_1^0\tilde r_2^0\vert\le \vert r_1\vert\tilde\delta_2 +\vert\tilde r_2^0\vert\tilde\delta_1$. We know that $\vert r_1\vert\le \vert r_1^0\vert+\delta_1$, and since $r_2^0$ and $\tilde r_2^0$ are approximations of $r_2$, we have $\vert\tilde r_2^0\vert\le \vert r_2^0\vert+\delta_2+\tilde\delta_2$; therefore:

$$\vert r_1r_2-\tilde r_1^0\tilde r_2^0\vert\le (\vert r_1^0\vert+\... ...tilde\delta_2 + (\vert r_2^0\vert+\delta_2+\tilde\delta_2)\vert\tilde\delta_1.$$

If we impose that each term in the right member of the above inequality be equal to $\varepsilon/2$, we can extract suitable values for $\tilde\delta_1$ and $\tilde\delta_2$:

$$\tilde\delta_2 = \frac{\varepsilon}{2(\vert r_1^0\vert+\delta_1)}... ...lde\delta_1 = \frac{\varepsilon}{2(\vert r_2^0\vert+\delta_2+\tilde\delta_2)}.$$