I am trying to figure out how $(x/y)$ in floating point arithmetic

$fl(fl(x) / fl(y))$ where $fl(x) = x(1-\delta_1)$, $fl(y) = y(1-\delta_2)$, $fl = (1-\delta_3)$

I have:

$= x/y \cdot ((1-\delta_1)/(1-\delta_2))(1-\delta_3)$

after arithmetic

$= x/y \cdot (\delta_3 \delta_1 - \delta_1 - \delta_3 + 1)/(1-\delta_2)$

Not sure how to fully write the rest I believe i have to shorten them using the info that $\delta < \epsilon$ and $\delta^2 \leq \delta$


I don't know exactly where you're trying to get to with this, but I think you probably want to Taylor expand the division:

$$\frac{1}{1 - \delta_2} = 1 + \delta_2 + \delta_2^2 + \delta_2^3 + \cdots$$

Then, for example, if $\epsilon$ is the machine epsilon and $\delta_2 > 0$:

$$1 + \delta_2 \le \frac{1}{1 - \delta_2} \le 1 + \delta_2 + \epsilon$$

That is quite a tight bound, when you think about it. (Exercise: Prove that this is true if $\delta_2 \le 0$, too.)

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  • $\begingroup$ This is from my textbook: gyazo.com/5eaa64253c3971345903f5a607ee3994 for multiplication. I thought the division version would be similar $\endgroup$ – ss sss Oct 25 '19 at 4:33
  • $\begingroup$ Sometimes you need to manipulate some inequalities. $\endgroup$ – Pseudonym Oct 25 '19 at 5:28

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