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I was told that since $x^n = x\cdot x\cdot x \cdot \ldots \cdot x$ that the error is $$\delta f = f_{\text{best}} |n|\lbrace \frac {\delta_x}{f_{\text{best}}} \rbrace$$ where $f$ is a function of $(x)$. What about $n\notin \mathbb Z, n \in \mathbb Q$? This type of proof will not work. What's the proof for that special case?

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    $\begingroup$ What is $f_{\textrm{best}}$? When you say "error propagation of exponents", do you mean that the exponent is the independent variable here, or do you really mean "error propagation with power functions"? $\endgroup$
    – march
    Sep 2, 2015 at 3:32
  • $\begingroup$ @march the x is the independent variable and $n$ is some constant. $f_{\text{best}}$ is the best estimate. $\endgroup$ Sep 2, 2015 at 3:33
  • $\begingroup$ Okay. Your title is a little misleading (hence my question). In addition, don't the $f_{\textrm{best}}$'s cancel out in your case? Is the $\{ \}$ some special notation? $\endgroup$
    – march
    Sep 2, 2015 at 3:36
  • $\begingroup$ @march no, I just put them there for clarity. $\endgroup$ Sep 2, 2015 at 3:44
  • $\begingroup$ Which means that the $f_{\textrm{best}}$'s cancel out, leading to an expression that I think is incorrect for $\delta f$. In fact, your expression is then not dimensionally correct. If the $f_{\textrm{best}}$ in the denominator is actually $x_{\textrm{best}}$, then your expression matches the expression in the answer below. $\endgroup$
    – march
    Sep 2, 2015 at 3:47

1 Answer 1

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For error propagation for one variable, it is best to use

$$\delta f(x) = \left|\frac{d f(x)}{d x}\right| \delta x$$

which is to say that the uncertainty in the function should be weighted with the derivative(or how sensitive the function is to changing the variable)

Now for your example,

$$f(x) = x^n \Rightarrow \delta f(x) = \left|n x^{n-1} \right| \delta x$$

regardless of $n$ integer or not. EDIT: This is assuming the errors $\delta x$ follow Gaussian statistics as in the below comment.

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  • $\begingroup$ Any explanation of this ratios rule? $\endgroup$
    – DanielSank
    Sep 2, 2015 at 2:54
  • $\begingroup$ That is a good question. I suppose I've always used this without any explanation of why. I'm looking at that now... This seems to only come up if $f(x) = A x$, with $A$ constant, but certainly should make sense for any $f(x)$. $\endgroup$
    – John M
    Sep 2, 2015 at 3:19
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    $\begingroup$ I'm just going to change this answer. $\delta f/f = \delta x/x$ is a good approximation, but perhaps there is a case where it isn't good. Most sources point to using the above formula. $\endgroup$
    – John M
    Sep 2, 2015 at 3:30
  • $\begingroup$ Can you explain how this matches the form given by the OP? $\endgroup$
    – march
    Sep 2, 2015 at 3:32
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    $\begingroup$ The way to do full blown error propagation is to compute the transformations of the statistical distributions involved. These simple transformations of "uncertainty" are usually derived by assuming the errors are Gaussian and tracking how the standard deviation transforms during data analysis. $\endgroup$
    – DanielSank
    Sep 2, 2015 at 3:40

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