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 $x = y = \frac{1}{2}$.

  • $\begingroup$ There is no mistake. Your answer to the problem, as posed, is correct. Their answer corresponds to the case where you measure the relative change in $(x,y)$ as the maximum of the relative change in $x$ and the relative change in $y$. It is also a meaningful way to think of the sensitivity but it has little to do with what is formally asked in the formulation of the problem. $\endgroup$ – fedja May 13 at 15:33
  • $\begingroup$ @fedja Can you please explain ( / expand on) what their answer corresponds to? I couldn't follow :/ Please feel free to formulate this as an answer instead of a comment, too. $\endgroup$ – Viktor Glombik May 13 at 19:50
  • $\begingroup$ I mean that their idea is, apparently, to measure the relative change in $(x,y)$ as $\frac{|\delta x|}{|x|}+\frac{|\delta y|}{|y|}$ instead of $\frac{\|(\delta x,\delta y)\|_1}{\|(x,y)\|_1}$ $\endgroup$ – fedja May 13 at 20:24
  • $\begingroup$ As a quick note: You used the $\infty$-norm instead of the $1$-norm in your calculations (for instance, $\|f'(x,y)\|_1=\|e^{4x^2}(8xy, \, 1)\|_1=e^{4x^2}(8\vert xy\vert + 1)$.) However, using the correct norm I still didn't get the same answer as your answer key... $\endgroup$ – Maximilian Janisch May 16 at 19:14

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