# Propagation of errors

I was going through a problem in finding error in measurement of focal length, if object and image distances are given with appropriate errors. I found two methods to find the error at different places and want to know which one is more accurate and why?

One method is based on the fact that fractional errors add up and second is pure differential. I wrote down both the solutions and posted them below. • Try again. Assume errors in $u$ and $v$ are uncorrelated, which means: $(df)^2 = (f_udu)^2 + (f_vdv)^2$ – JEB Mar 26 '18 at 23:06

The two approaches give identical results, but neither of the things you've written is correct for error propagation.

In your approach on the left you've done the differential calculus correctly.

In your approach on the right, you have a sign error on the term like $1/(u+v)$:

\begin{align} \mathrm df &= \frac{\partial f}{\partial u} \mathrm du + \frac{\partial f}{\partial v} \mathrm dv \\ &= \left( \frac v{u+v} - \frac{uv}{(u+v)^2} \right) \mathrm du + \left( \frac u{u+v} - \frac{uv}{(u+v)^2} \right) \mathrm dv \\ &= f\left( \frac 1u - \frac1{u+v} \right)\mathrm du + f\left( \frac 1v - \frac1{u+v} \right)\mathrm dv \\&= f^2 \frac{\mathrm du}{u^2} + f^2 \frac{\mathrm dv}{v^2} \end{align}

and in your initial question you used something other than $f=20/3\rm\,cm$. This is good way to check that you haven't (or have) made a silly mistake: the results should be algebraically identical.

However, when propagating uncertainties, you should add independent uncertainties in quadrature, not linearly. Most every text on error analysis discusses this; here's two.

• I have corrected the part about using wrong value, but still don't understand your point about sign error. If you point me to a good text, I will go through it but can't really see how both methods will give same result. – Atinder Mar 27 '18 at 5:10
• @Atinder Edited. – rob Mar 27 '18 at 14:20
• @JEB It's true that fractional errors always grow when multiplying or dividing, and it's true that absolute errors always grow while adding and subtracting. However it's possible for the fractional error to shrink when additions are involved. (The canonical example is averaging, where the fractional uncertainty on an average is smaller than the fractional uncertainty on any individual measurement.) That's the case here: the correct analysis gives $\mathrm df/f \approx 0.7\% < \mathrm du/u$. – rob Mar 27 '18 at 18:01