0
$\begingroup$

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?

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

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 $4\%$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.