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I read a solution of find relative error for $c^2 = a^2 + b^2 -2ab\cdot\cos(\alpha) $ and it written there the equation $\Delta(c^2)=2c(\Delta c) $ .can someone explain how to developing this equation ?

Edit : definition of $\Delta: \Delta(x) = |(x-x^*)| $ when $x^*$ is is the $x$ with the error .

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    @PeterTamaroff : I edited my post2012-09-08

2 Answers 2

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In general, the error can be computed by using the derivative as a first order approximation: $\Delta f(x) = \frac{\Delta f(x)}{\Delta x}\Delta x \simeq \frac{df(x)}{dx}\Delta x$ In your case, because $d(c^2) / dc = 2c$: $\Delta(c^2) = 2c \Delta c$

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Let $f : \mathbb{R} \to \mathbb{R}$ be defined by $f (x) = x^2$. Then, we have that

$f (x + \Delta x ) = (x + \Delta x)^2 = x^2 + 2 x \Delta x + (\Delta x)^2$

Think of $f$ as a black-box that takes values of $x$ and spits out $f (x)$. For a given $x$, we obtain $f (x)$. What happens if we perturb the input? If the input is $x + \Delta x$ then the output will be $f (x + \Delta x )$. The perturbation in the output is thus

$f (x + \Delta x ) - f (x) = 2 x \Delta x + (\Delta x)^2$

Note that the magnitude of the perturbation in the output depends on the input value $x$. If $\Delta x$ is "small enough", then the perturbation in the output can be given by its first-order approximation

$f (x + \Delta x ) - f (x) \approx 2 x \Delta x$

However, if $\Delta x$ is not "small enough", the $(\Delta x)^2$ term will have to be included.

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    +1 for mentioning whether $(\Delta x)^2$ is "small enough"2012-09-08