Take the 2-minute tour ×
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.

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 .

share|improve this question
    
What is $\Delta(c^2)$ in this context? More general, how does $\Delta$ act on $c$? –  Pedro Tamaroff Sep 8 '12 at 19:20
    
@PeterTamaroff - I think the OP means the error estimate. –  nbubis Sep 8 '12 at 19:22
    
@nbubis Oh, right. $d(c^2)=2c dc$, maybe? –  Pedro Tamaroff Sep 8 '12 at 19:27
    
@PeterTamaroff : I edited my post –  URL87 Sep 8 '12 at 19:28
add comment

2 Answers

up vote 6 down vote accepted

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 $d(c^2) / dc = 2c$: $$\Delta(c^2) = 2c \Delta c$$

share|improve this answer
add comment

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.

share|improve this answer
    
+1 for mentioning whether $(\Delta x)^2$ is "small enough" –  Hurkyl Sep 8 '12 at 21:25
add comment

Your Answer

 
discard

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.