Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In my course we have stated and used the error propagation formula:


But it was presented with no proof and I wonder if you can help me understand the formula holds?

share|cite|improve this question
You don't recognize the first few terms of the Taylor expansion? :) – J. M. Jul 19 '12 at 5:28
@J.M. OK but to me the first two terms of Taylor expansion is f(a)+f'(a)(x-a)/2. I'm working on how the two common Taylor expansions are equivalent. – Programmer 400 Jul 19 '12 at 5:36
@NickRosencrantz: The $2$ at the bottom shouldn't be there. The next term is $\frac{f''(a)}{2!}(x-a)^2$. – André Nicolas Jul 19 '12 at 6:06
@AndréNicolas Yes, exactly. I was wrong from memory. – Programmer 400 Jul 19 '12 at 7:35

1 Answer 1

up vote 2 down vote accepted

Possibly the best way to understand it is via the mean value theorem: $$f(x)-f(x_0)=f'(c)\cdot(x-x_0)$$ for some $c$ between $x_0$ and $x$. If $f'$ is continuous, $f'(c)$ can be expected to be close to $f'(x)$ or $f'(x_0)$ when $\lvert x-x_0\rvert$ is small enough.

share|cite|improve this answer
Thank you. It's somewhat more info for me to go on than just the Taylor expansion again. – Programmer 400 Jul 19 '12 at 15:55

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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