Exponential rule for propagation of errors

Question

I'm struggling to follow what feels like it should be a simple method. I often see that for if you want to find the error of an exponential, it follows:

if $y = x^{n}$ we have $\frac{\Delta y}{y} = n\frac{\Delta x}{x}$.

Where I'm getting confused, is that it seems like you can get $\frac{\Delta y}{y} > 1$ quite easily in situations when the answer clearly can't be negative.

For example, setting $x=2, \Delta x=0.5, n=5, $ gives $\frac{\Delta y}{y} = 1.25 $ and $y = 32 $. So from my understanding, we get $y = 32 \pm 40$ using this approach. I don't see how negative values can be within the uncertainty in this case. Similarly, if I were to actually modify $x$ by $0.5$ (i.e. $1.5 \le x \le 2.5$) then $7.6 \le y \le 98$.

I'm trying to understand what I'm missing. It seems like the larger $n$ is, the less accurate it becomes. Is it related to this method being a simplified version?

Answer 1

It's not a simplified version, it's a linearized version:

$$ y = x^n $$ means:

$$ y(x) \approx y(x_0) + (x-x_0)\frac{dy}{dx}_{|_{x=x_0}} + \frac 1 2 (x-x_0)^2\frac{d^2y}{dx^2}_{|_{x=x_0}}+ \frac 1 6 (x-x_0)^3\frac{d^3y}{dx^3}_{|_{x=x_0}} + \ldots$$

With $\Delta x \equiv (x-x_0)$:

$$ y(x) - y_0 = \Delta y \approx (nx_0^{n-1})\Delta x + \frac 1 2 (n(n-1)x_0^{n-2})\Delta x^2 + \ldots$$

In error analysis, it's customary to keep just the linear term. If your errors on $x$ are so large that that is not a good approximation, then getting non-physical values of $y$ can be expected.

Note that:

$$ \big(\frac{2.5} 2 \big)^5 \approx 3 $$

so the error bar is 3x the value. Not good for a linear approximation in a monomial, and of course, the larger $n$, the worse it gets.

There are a few options to proceed:

1: Say, "My measurement is terrible. Call it and "order-of-magnitude" measurement".

2: Use a computer to Monte Carlo a gaussian error ($\sigma = \frac 1 2$) input, and calculate the output. (This works, but is not necessary for such a simple function)

3: Define new variables: $Y=\ln y$ and $X=\ln x$, and fit a line to $Y = nX$. Then, all results will be in the correct domain.