# What is the reasoning behind the rules for compounding errors? [duplicate]

There has to be a reason behind why we add fractional errors when the involved quantities are being multiplied or divided, or why, when converting units, do we have to divide the uncertainty with the number that divides the value.

• Would Cross Validated be a better home for this question? Oct 28 '20 at 13:34
• Possible duplicates: physics.stackexchange.com/q/170886/2451 , physics.stackexchange.com/q/59628/2451 and links therein. Oct 28 '20 at 13:37
• Also, if possible, I think you should have a look at John R. Taylor's book on Error Analysis. Oct 28 '20 at 13:44
• Another "easy to read" book is Bevington's "Data Reduction and Error Analysis for the Physical Sciences. Oct 28 '20 at 13:58

Consider: y = (1/2)g$$t^2$$ Then dy = (gt)dt + (1/2)($$t^2$$)dg . If we let the differentials represent finite errors then: Δy/y = 2Δt/t + Δg/g.
To understand where these rules come from, it is helpful to deal with absolute errors rather than proportional errors. For some $$f(x,y)$$, with $$x,y$$ independent, the combination of errors formula is:
$$\sigma_f^2=\left({\partial f \over \partial x}\right)^2 \sigma_x^2+ \left({\partial f \over \partial y}\right)^2 \sigma_y^2$$
You've probably met this, or can look it up: if you want to prove it, this is done by approximating $$f=f_0+{\partial f \over \partial x}(x-x_0)+ {\partial f \over \partial y}(y-y_0)$$, and inserting it into $$<(f-f_0)^2>$$ (or $$E((f-f_0)^2)$$ in some notations. Notice that you don't have to assume the distribution is normal (Gaussian).
Then if it happens that $$f=xy$$ or $$f=x/y$$ the result $${\sigma_f^2 \over f^2}= {\sigma_x^2 \over x^2}+ {\sigma_y^2 \over y^2}$$ drops out in both cases. Which is pretty, and useful. But the combination of errors formula is more fundamental than they are.