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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?

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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
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.

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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

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