# Error propagation and relative error

Say I have measured the value

A = 50 +- 2

and from this I am calculating the value:

B(A) = A^2

From what I understand I calculate the new error using error propagation to be

delta_B = dB/dA * delta_A = 2*A*delta_A = 2*50*2 = 200

A has a therefore relative error of 2/50 = 4%, but B has a relative error of 200/2500 = 8%.

Shouldn't the relative error stay the same? It seems illogical, that a simple arithmetic calculation would change the relative error of the value.

Or am I doing something very wrong?

• The answer is right there in the factor of "2" in your formula for delta_B. It came from differentiating $A^2$. – whuber Jul 26 '15 at 21:13

## 1 Answer

Your analysis is correct. As a check, for example, try calculating $(50+2)^2=2704$, which gives a relative error of $(2704-2500)/2500=8.16\%$ for the output, approximately double that of the $4\%$ relative error of the input.

One way of looking at it is like this: if we multiply two numbers together, forming a product $XY$, where $X$ has a relative error of $4\%$ while $Y$ has no error, then the result would have a relative error of $4\%$. However, if both $X$ and $Y$ have a relative error of $4\%$, then these errors will combine to give approximately double the relative error, i.e. $8\%$. And that is what is happening here, as you are calculating the product $A^2= AA$, where each factor $A$ has a relative error of $4\%$.

• Oh, yes that makes a lot of sense suddenly (becomes quite obvious why the relative errors change when pointed out). Great job! – Jens Jul 26 '15 at 21:06