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Let us define $\alpha=x^{2}-y^{2}$ and $\beta=2xy$ for $x,y\in\mathbb{R}$. In order to calculate the $\alpha$ and $\beta$, we use the following algorithm:

$$p:=x-y;\quad q:=x+y;\quad \alpha=p\cdot q;\quad \beta=2\cdot x\cdot y$$

I wanted to prove that the above algorithm is numerically stable. I understand that numerical stability means that an error in the input will not significantly shift the produced output. Thus, intutively, I believe I should insert an $\varepsilon$ error term in the input and see how the algorithm behaves with it. However, I do not see how this could be properly done. I would be grateful for ideas.

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  • $\begingroup$ Stability doe not make a lot of sense here as the procedure is not iterative. So, there is no blow-up or divergence. Are you asking for sensitivity w.r.t parameters x and y? $\endgroup$ – dexter04 Feb 1 '13 at 11:41
  • $\begingroup$ Hm... that's certainly a valid point. I guess you are right, sensitivity would indeed make more sense here. $\endgroup$ – Johnny Westerling Feb 1 '13 at 11:58
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What you can try to show, is how your algorithm behaves, if you have a certain error in your input.

Therefore, let $\tilde x = x +\epsilon$ and $\tilde y = y +\delta$ be your perturbed input. From that we get

$$ \tilde p = \tilde x - \tilde y ,\quad \tilde q = \tilde x + \tilde y $$ And with that \begin{align} \tilde \alpha &= \tilde p \cdot \tilde q = ((x+\epsilon)-(y+\delta))\cdot ((x+\epsilon)+(y+\delta)) \\& = ...= x^2-y^2+2\epsilon x-2 \delta y + \epsilon^2-\delta^2 \\&= \alpha +2\epsilon x-2 \delta y + \epsilon^2-\delta^2 \end{align} So you can see, that your algorithm for $\alpha$ is stable, in the sense that $\tilde \alpha \rightarrow \alpha$ if $\delta,\epsilon \rightarrow 0$. For $\beta$ you obtain $\tilde \beta = \beta+ 2(\epsilon y +\delta x +\epsilon \delta) $, which is also stable in the sense above. Maybe instead of stability you mean well-conditioned?

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  • $\begingroup$ Nope, it was about stability, but it seems to me that your answer makes a good enough argument (especially in the light of dexter04's comment). Thanks! $\endgroup$ – Johnny Westerling Feb 1 '13 at 15:01

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