# Find the relative condition number of $f(x,y) := y e^{4x^2}$ with respect to the 1-norm.

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

• 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. – fedja May 13 at 15:33
• @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. – Viktor Glombik May 13 at 19:50
• 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 $\frac{\|(\delta x,\delta y)\|_1}{\|(x,y)\|_1}$ – fedja May 13 at 20:24
• 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... – Maximilian Janisch May 16 at 19:14