# Error propagation for exponents

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?

• 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"? Sep 2, 2015 at 3:32
• @march the x is the independent variable and $n$ is some constant. $f_{\text{best}}$ is the best estimate. Sep 2, 2015 at 3:33
• 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? Sep 2, 2015 at 3:36
• @march no, I just put them there for clarity. Sep 2, 2015 at 3:44
• 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. Sep 2, 2015 at 3:47

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)

$$f(x) = x^n \Rightarrow \delta f(x) = \left|n x^{n-1} \right| \delta x$$
regardless of $n$ integer or not. EDIT: This is assuming the errors $\delta x$ follow Gaussian statistics as in the below comment.
• 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 $f(x)$. Sep 2, 2015 at 3:19
• 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. Sep 2, 2015 at 3:30