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According to many definitions I've seen the relative error is defined by

$$E = \frac{x-x_0}{x}$$ where $x$ is the "true" value. But some people use instead $$\frac{x-x_0}{x_0}. $$ Is this incorrect?

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1 Answer 1

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Let $\Delta x$ be the absolute error. Then $\Delta x=x_0-x$ where $x$ is true value and $x_0$ is measured value (sometimes with the absolute value taken). Relative error $\delta x$ is defined by $$\delta x=\dfrac{\Delta x}{x}=\dfrac{x_0-x}{x}$$

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  • $\begingroup$ So when people divide by $x_0$, they are doing it wrong? I myself wouldn't divide by $x_0$ but I cannot figure out why some people do it without motivating it, e.g. that "in this case it doesn't make any difference if we bla bla bla" $\endgroup$ Dec 2, 2014 at 21:46
  • $\begingroup$ If $\Delta x\to0$ or $x\to\infty$ there is no difference. For example: if $x=10^5$ and $x_0=10^5+1$ there is very small difference between first and second formula, but $\dfrac{x_0-x}{x}$ is only correct formula. $\endgroup$
    – user164524
    Dec 2, 2014 at 21:53

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