Say we want to know the zero of a function f, then of course we can use the iterative Newton-Raphson method, given by [itex]x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}[/itex].(adsbygoogle = window.adsbygoogle || []).push({});

But say f typically has really big values, and say you can actually, without much trouble, rewrite your problem as solving for the zero of a function g which reaches much smaller values. Is it then "better" to use [itex]x_{n+1} = x_n - \frac{g(x_n)}{g'(x_n)}[/itex] instead of [itex]x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}[/itex]? Or does it make no difference? After all, the "bigness" of f doesn't seem to influence the relative error on x (since if [.] denotes the computed value of something, then [itex][f(x_n)] = (1+\epsilon_r^{(1)})f(x_n)[/itex] and [itex][f'(x_n)] = (1+\epsilon_r^{(2)})f'(x_n)[/itex], such that [itex]\frac{[f](x_n)}{[f'](x_n)} = \frac{f(x_n)}{f'(x_n)}(1+2 \epsilon_r)[/itex], so no problem there)

On the other hand, although the method with the big f might be well conditioned(?), it might still be better to use g, since you'll avoid an overflow?

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# Is this problem ill-conditioned or not?

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