Find the Condition Number for $w\left(a+b\right)=a^2+b^2$ This is what I got:
$$w\left(a+b\right)=a^2+b^2$$
$$ \alpha =a\left(1+E_1\right),\:\beta \:=b\left(1+E_2\right)$$
$$ \frac{\left|w\left(\alpha ,\beta \right)-w\left(a,b\right)\right|}{\left|w\left(a,b\right)\right|}=\frac{\left|a^2\left(1+E_1\right)^2+b^2\left(1+E_2\right)^2-\left(a^2+b^2\right)\right|}{\left|a^2+b^2\right|}=\frac{\left|a^2\left(2E_1+E_1^2\right)+b^2\left(2E_2+E_2^2\right)\right|}{\left|a^2+b^2\right|} $$
Let $E\::=max\left(E_1,\:E_2\right)$, then:
$$\le \frac{\left|a^2\left(2E+E^2\right)+b^2\left(2E+E^2\right)\right|}{\left|a^2+b^2\right|}=2E+E^2$$
So we got:
$$ \frac{\left|w\left(\alpha ,\beta \right)-w\left(a,b\right)\right|}{\left|w\left(a,b\right)\right|} \le  2E+E^2 $$
For $E\::=max\left(E_1,\:E_2\right)$
But I don't know what to do further on, how to actually find $cond\left(w,\:a,\:b\right)$.
I know that the basic definition is:
$$\frac{\left|\phi \left(d+\Delta d\right)-\phi \left(d\right)\right|}{\left|\phi \left(d\right)\right|}\le cond\left(\phi ,\:d\right)\:\frac{\left|\Delta d\right|}{\left|d\right|}$$
But what I got is completely independent of our "d" here (a and b). And also, in this example, how would $\frac{\left|\Delta d\right|}{\left|d\right|}$ look like?
 A: As $w$ is bivariate the inequality becomes $$\frac{|w(a+\epsilon_a,\beta+\epsilon_b)-w(a,b)|}{|w(a,b)|}\le\kappa(a,b)\frac{\|(\epsilon_a,\epsilon_b)\|}{\|(a,b)\|}$$ where $\epsilon_a,\epsilon_b>0$ and $\|\cdot\|$ is a norm on the codomain of $w$ (which is $\Bbb R_{\ge0}$).
As $\epsilon_a,\epsilon_b\to0^+$, taking the Euclidean norm yields \begin{align}\kappa(a,b)&\ge\frac{|\epsilon_a^2+\epsilon_b^2+2(a\epsilon_a+b\epsilon_b)|}{|a^2+b^2|}\sqrt{\frac{a^2+b^2}{\epsilon_a^2+\epsilon_b^2}}\end{align} as a necessary condition. Note that \begin{align}\frac{|\epsilon_a^2+\epsilon_b^2+2(a\epsilon_a+b\epsilon_b)|}{|a^2+b^2|}\sqrt{\frac{a^2+b^2}{\epsilon_a^2+\epsilon_b^2}}&=\frac1{\sqrt{a^2+b^2}}\left|\frac{\epsilon_a^2+\epsilon_b^2+2(a\epsilon_a+b\epsilon_b)}{\sqrt{\epsilon_a^2+\epsilon_b^2}}\right|\\&\le\frac1{\sqrt{a^2+b^2}}\left(\sqrt{\epsilon_a^2+\epsilon_b^2}+2\left|\frac{a\epsilon_a+b\epsilon_b}{\sqrt{\epsilon_a^2+\epsilon_b^2}}\right|\right)\\&\le2\cdot\frac{\max\{|a|,|b|\}}{\sqrt{a^2+b^2}}+E\end{align} for a small $E>0$. Therefore, choosing $$\kappa(a,b)=2\cdot\frac{\max\{|a|,|b|\}}{\sqrt{a^2+b^2}}+1$$ suffices.
It does not matter whether we use $\epsilon_a$ or $E_a:=a\epsilon_a$ as both tend to zero as $\epsilon_a\to0^+$.
