Let $f: \mathbb{R} \to \mathbb{R}$ (I guess it's supposed to be $\mathbb{R}^2 \to \mathbb{R}$) be defined by $f(x,y) := y e^{4x^2}$ Find the relative condition number of with respect to the 1-norm.
What I've done
Since $f$ is continuously differentiable, we can use the characterisation: $$ \kappa_1(f) := \frac{\| f'(x,y) \|_1 \| (x,y) \|_1}{\| f(x,y) \|_1} $$ Now, $f'(x,y) = e^{4x^2} \begin{pmatrix}8xy & 1\end{pmatrix}$. For matrices $A$, we have $\| A \|_1 := \max_{i \in \{1, \ldots, n\}} \sum_{j = 1}^{m} | a_{i,j} |$ and for vectors $v$ $\| v \|_1 = \sum_{I = 1}^{n} |v_i|$, therefore I got $$ \kappa_1(f) = \frac{\left(e^{4x^2} \cdot \max(8 |xy|, 1)\right) \cdot \max(|x|, |y|)}{e^{4x^2} |y|} = \frac{\max(8 |xy|, 1) \cdot \max(|x|, |y|)}{|y|}, $$ but the answer key suggests $\kappa_1(f) = \max(1, 8x^2)$. What is my mistake?
I also don't think that the result from the answer key can be correct since it is independent of $y$: i.e. for $x = 1$ and $y = 2$ my term gives 16, but the correct answer according to answer key is 8.
The next question is: find $\kappa_1\left(\frac{1}{2}, \frac{1}{2}\right)$ for $g(x,y) := y^2 e^{2x}$. Sing my method from above I get $$ \kappa_1(g) = \frac{2}{y^2} \max(|y|, y^2), \max( |x|, |y|), $$ which yields the correct result for $x = y = \frac{1}{2}$.