# Why is propagation of uncertainties quadratic rather than linear? [duplicate]

1) Up until now, during practical work sessions, I always used these formulas for uncertainty propagation:

if $C = A+B$ or $C = A-B$ $$\Delta C = \Delta A + \Delta B$$ if $C = AB$ or $C = \frac{A}{B}$ $$\frac{\Delta C}{C} = \frac{\Delta A}{A} + \frac{\Delta B}{B}$$ if $C = A^m$ $$\frac{\Delta C}{C} = |m|\frac{\Delta A}{A}$$

These formulas are derived from the expression for the differential of a function of multiple variables:

if $C = f(A,B)$ $$dC = \frac{\partial C}{\partial A}dA + \frac{\partial C}{\partial B}dB \Rightarrow \Delta C = \frac{\partial C}{\partial A}\Delta A + \frac{\partial C}{\partial B}\Delta B$$

and I think that makes sense because the value C is just a function of two variables that happen to be measurements.

2) But this morning, I was told that this is wrong and I should actually use these instead:

if $C = A+B$ or $C = A-B$ $$(\Delta C)^2 = (\Delta A)^2 + (\Delta B)^2$$ if $C = AB$ or $C = \frac{A}{B}$ $$\left(\frac{\Delta C}{C}\right)^2 = \left(\frac{\Delta A}{A}\right)^2 + \left(\frac{\Delta B}{B}\right)^2$$

which are apparently derived from a general formula:

$$(\Delta C)^2 = \left(\frac{\partial C}{\partial x_1}\Delta x_1\right)^2 + \left(\frac{\partial C}{\partial x_2}\Delta x_2\right)^2 + \hspace{0.3cm}...$$

or

$$\Delta C = \sqrt{\left(\frac{\partial C}{\partial x_1}\Delta x_1\right)^2 + \left(\frac{\partial C}{\partial x_2}\Delta x_2\right)^2 + \hspace{0.3cm}...}$$

3) So:

• Why is this formula better?
• Where does it come from?
• What does it actually represent? (Do I recognize the shape of a norm in that last formula?)
• What is wrong with the formula of the differential?

I'm asking a lot, but the propagation of uncertainty always confused the ship out of me.

• "What is wrong with the formula of the differential?" Why would you think it is right in the first place. The errors are standard deviations, not differentials. Also, Wikipedia has the general approach, where the formula under "Simplification" is what physicists usually use, assuming independent variables. – ACuriousMind Mar 17 '15 at 16:46
• Also, this a essentially a duplicate of physics.stackexchange.com/q/59628/50583, since Chris White's answer there derives the general formula. – ACuriousMind Mar 17 '15 at 16:48
• "Why would you think it is right in the first place. The errors are standard deviations, not differentials." --> For one thing, I've been using it for 2 years and it always seemed pretty accurate. Also, because I trusted (wrongfully ?) the formulas that I had been given by the teacher. – mwa1 Mar 17 '15 at 17:01
• Alas questions like this are becoming more common. Someone out there keeps teaching students the wrong formulas. – user10851 Mar 17 '15 at 17:13
• This question seems a little more comprehensive than the other one, so I'm not entirely convinced it is a duplicate. (Determination of duplicates should be made based on the question, not the answers.) But it may be. Regardless it is a very well posed question. – David Z Mar 18 '15 at 5:17

I think that the easiest way to understand this is in the formula for addition. If you consider your quantity $C$ which depends on $A$ and $B$ such that $C=A+B$, then the formula $\Delta C = \Delta A + \Delta B$ overestimates the error values. You can visualize this as a rectangle with $A$ on the x-axis and $B$ on the y-axis with the area enclosed being $\Delta C$. However, the likelihood of having errors in the corners, i.e. the extreme values for both $A$ and $B$ are very small. Therefore, instead of using a rectangle to represent the error, we instead use an ellipse with $\Delta C^2 = \Delta A^2 + \Delta B^2$. This gets rid of the extremely unlikely error values. 