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Let $\tilde{f} $ be some algorithm: we have: $$ \| f(x) - \tilde{f}(x) \| = \| f(x) - f(\tilde{x}) \| \leq \|f'(x) \| \|x - \tilde{x} \|$$

I'm curious on the last step, how did they get the inequality? It kind of looks like the mean value theorem but f' is dependent on x and not on some variable between x and $\tilde{x}$ so I don't think it's that

thanks

DH.
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    This does not make sense. Define $f(x)=x^2$. Define $x=0$, $\tilde{x}=1$. Then $|f(x)-f(\tilde{x})|=1$, but $|f'(x)||x-\tilde{x}| = 0$. – Michael May 07 '15 at 02:26

1 Answers1

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Indeed, the vector version of the mean value theorem says $$ \|f(x)-f(\bar x)\|\le \sup_{z\in [x,\bar x]}\|f'(z)\|·\|x-\bar x\| $$ where $[x,\bar x]$ is the line segment between the points.

Perhaps the intended meaning was that this holds to first order, i.e., $$ \|f(x)-f(\bar x)\|\le \|f'(x)\|·\|x-\bar x\|+O(\|x-\bar x\|^2). $$ In the numerical error analysis, one often only considers the first order terms of the error, as when the relative difference $\|x-\bar x\|\sim\mu·\|x\|$ is a small (or even medium large) multiple of the constant of machine precision, the second order effects fall far below machine precision, thus do not influence the floating point result.

Lutz Lehmann
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