Machine Floating Point Theorem Completely stuck on this floating point question.
Let $x \in \mathbb{R}$ have the following floating point representation:
$$
x = (-1)^s[0.a_1a_2\dots a_ta_{t+1}\dots]\cdot \beta^e
$$
[Where $\beta$ is the base]
Define the floating point round off  to $t$ significant figures to be:
$$
fl(x) = (-1)^s[0.a_1a_2\dots\tilde{a}_t] , \quad \tilde{a}_t = \begin{cases} a_t  &\text{if}\; a_{t + 1} < \beta /2 \\ a_t + 1 &\text{if}\; a_{t + 1} \geq \beta /2 \end{cases}
$$
And the following flop computation for $x, y \in \mathbb{R}$ to be:
$$
x \ominus y := fl(fl(x) - fl(y))
$$
Given that the relative error follows:
$$
\frac{|x - fl(x)|}{|x|} \leq u \left[= \frac{1}{2}\beta^{1-t} \right]
$$
Show that [triangle inequality may help]:
$$
\frac{x \ominus y - (x - y)}{|x - y|} \leq u(2 + u)\frac{|x| + |y|}{|x - y|}
$$
Any ideas?
 A: I am stuck with a question very similar to yours, but I can answer yours.
\begin{equation*}
 \left\vert \frac{x \ominus y - (x - y)}{x - y} \right\vert \leq 
 \left\vert \frac{x \ominus y - (fl(x) - fl(y))}{x - y} \right\vert + \left\vert \frac{fl(x) - fl(y) - (x - y)}{x - y}\right\vert.
\end{equation*}
Let's find an upper bound for the first addend.
\begin{equation*}
 \left\vert \frac{x \ominus y - (fl(x) - fl(y))}{x - y}\right\vert =
 \left\vert \frac{fl(fl(x) - fl(y)) - (fl(x) - fl(y))}{x - y}\right\vert =
 \left\vert \frac{(fl(x) - fl(y))(1 + \delta) - (fl(x) - fl(y))}{x - y}\right\vert =
 \left\vert \frac{(fl(x) - fl(y))\delta}{x - y}\right\vert \leq
 \left\vert \frac{(fl(x) - fl(y))}{x - y}\right\vert u \leq
 \left\vert \frac{fl(x) - x}{x}x + x + \frac{fl(y) - y}{y}y + y\right\vert \frac{u}{\left\vert x - y\right\vert} \leq
 \left(\left\vert \frac{fl(x) - x}{x}\right\vert \left\vert x\right\vert + \left\vert x\right\vert + \left\vert\frac{fl(y) - y}{y}\right\vert \left\vert y\right\vert + \left\vert y\right\vert\right) \frac{u}{\left\vert x - y\right\vert} \leq
 (\left\vert x\right\vert + \left\vert y\right\vert)(u + 1)\frac{u}{\left\vert x - y\right\vert}.
\end{equation*}
Let's find an upper bound for the second addend.
\begin{equation*}
 \left\vert \frac{fl(x) - fl(y) - (x - y)}{x - y}\right\vert \leq
 \left\vert \frac{fl(x) - x}{x - y}\right\vert + \left\vert \frac{fl(y) - y}{x - y}\right\vert =
 \left\vert \frac{fl(x) - x}{x}\frac{x}{x - y}\right\vert + \left\vert \frac{fl(y) - y}{y}\frac{y}{x - y}\right\vert \leq
 u\frac{\left\vert x\right\vert + \left\vert y\right\vert}{\left\vert x - y\right\vert}.
\end{equation*}
Putting all together we find
\begin{equation*}
 \left\vert \frac{x \ominus y - (fl(x) - fl(y))}{x - y} \right\vert + \left\vert \frac{fl(x) - fl(y) - (x - y)}{x - y}\right\vert \leq 
 (\left\vert x\right\vert + \left\vert y\right\vert)(u + 1)\frac{u}{\left\vert x - y\right\vert} + u\frac{\left\vert x\right\vert + \left\vert y\right\vert}{\left\vert x - y\right\vert} =
 \frac{\left\vert x\right\vert + \left\vert y\right\vert}{\left\vert x - y\right\vert}((u+1)u + u) =
 \frac{\left\vert x\right\vert + \left\vert y\right\vert}{\left\vert x - y\right\vert}((u+2)u).
\end{equation*}
So your problem is solved.
This is mine, very linked to yours Relative error of machine summation.
