Have to prove the backwards stability of the following equations:
$$x+y$$ $$x,y \in \mathbb{R}$$ with the norm $$\mid \mid \cdot \mid \mid_{\infty} $$
$$x \cdot y$$ $$x,y \in \mathbb{R}$$ with the norm $$\mid \mid \cdot \mid \mid_{\infty} $$
Tried to look up the for the backwards stabiltiy which should be $\frac{|\bar{x}-x|}{|x|} \leq O\left(\epsilon_{\text {machine}}\right)$
Simply don't know how to apply it. Some advice would be good.
Thanks
Let $\hat{x}$ and $\hat{y}$ denote the floating point representation of $x$ and $y$. If $x$ and $y$ are in the representable range, then $$\hat{x} = x(1+\delta_x), \quad |\delta_x| \leq u$$ and $$\hat{y} = y(1 + \delta_y), \quad |\delta_y| \leq u$$ where $u$ us the unit roundoff. If $\hat{x} + \hat{y}$ does not overflow, then the computed value of the sum $s = x+y$ satisfies $$ \hat{s} = (\hat{x} + \hat{y})(1 + \delta_s), \quad |\delta_s| \leq u.$$ It follows that $$ \hat{s} = x(1+ \delta_x)(1 + \delta_s) + y(1+\delta_y)(1+\delta_s) $$ We see that the computed sum is the exact sum of two numbers $\tilde{x}$ and $\tilde{y}$ given by $$ \tilde{x} = x(1+ \delta_x)(1 + \delta_s)$$ and $$ \tilde{y} = y(1+\delta_y)(1+\delta_s).$$ We have $$ \left|\frac{x - \tilde{x}}{x}\right| \leq 2u + u^2 = O(u)$$ and similarly for $y$. We conclude that the addition of real numbers in the representational range is backward stable.
I have to stress that there are good reasons for assuming that the input arguments are exact, i.e, we would normally have $\delta_x = \delta_y = 0$.