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Consider $f(z)=z^2+az+b$ we put it's root in vector $[z_1, z_2]^T \in \mathbb{C}^2$. Find condition number in maximum norm of finding the root vector when changing variable a where $a=-2, b=3$

I know in general that for function $g$ finding it's root condition number is given of formula $\displaystyle\frac{1}{g'(x^*)}$ where $g(x^*)=0$ but in our case we have vector and I don't know how to handle it. Calculating $\displaystyle \frac{1}{f'(z_1)}=\frac{1}{i \cdot \sqrt{2}} = \frac{1}{f'(z_2)}$ so taking max I get cond. number is $\displaystyle \frac{1}{i \cdot \sqrt{2}}$ ? But it's not real valued so I'm suppose I did something wrong. Thanks in advance

Watson
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Jessy
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1 Answers1

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You are supposed to compute the first order asymptotic for $$ \max_{k=1,2}\left|\frac{Δz_k}{Δa}\right|. $$ In first order, the equation $f(z)=εq(z)$ (the $ε$ to indicate a small perturbation can be written as $$ f'(z_k)Δz_k=εq(z_k)\iff \frac{Δz_k}ε=\frac{q(z_k)}{f'(z_k)} $$ thus with $q(z)=-z$ and $Δa=ε$ $$ \frac{Δz_k}{Δa} =-\frac{z_k}{f'(z_k)} =-\frac{1\pm i\sqrt2}{\pm i\sqrt2}. $$

Your cited formula is for the perturbation by constants, i.e., for the variation of $b$.

Lutz Lehmann
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  • Could you explain how did you get $f(z)=\epsilon q(z)$ ? and what $q(z)$ is ? – Jessy Apr 17 '16 at 13:23
  • I don't see why $q(z)=-z$ since we have $f(z_k)=0=\epsilon \cdot q(z_k)$ so this should imply that $q(z_k)=0$ – Jessy Apr 17 '16 at 13:45
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    You are exploring $z^2+(a+Δa)z+b=0 \iff f(z)=-Δa·z$. Or more general the implicit equation $f(z(ε))=ε·q(z(ε))$ with first derivative at $ε=0$ as $f'(z(0))·z'(0)=q(z(0))$. – Lutz Lehmann Apr 17 '16 at 13:49