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Suppose $f:\mathbb{R}\rightarrow\mathbb{R}$ is a $C^{2}$ function. Is true that for every $x<h$, there exists $\theta\in(x,x+h)$, such that $$f(x+h)-f(x)=f'(x)h+\frac{1}{2}f''(\theta)h^2$$

Im studying optmization and the author uses this fact in $\mathbb{R}^{n}$, so i think that if i can prove this in $\mathbb{R}$, hence the adaptation is easy, but is that true?

Thanks

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This looks like Taylor's Theorem to me. See en.wikipedia.org/wiki/Taylor%27s_Theorem –  user35959 Oct 23 '12 at 19:09
    
But in Taylor theorem we have the rest. What happened with the rest in this case? –  Tomás Oct 23 '12 at 19:11
    
@Tomás This is Taylor's theorem, indeed. The "remainder" is the term $$\frac 1 2 f''(\xi)h^2$$ Note that that is actually expanding" around $x$, that is $$f(h+x)=f(x)+f'(x)h+R_{x,2}(h)$$ –  Pedro Tamaroff Oct 23 '12 at 19:13
    
@PeterTamaroff, thank you. –  Tomás Oct 23 '12 at 19:13
    
@Tomás Taylor's theorem is distinct from Taylor series. The term $\frac{1}{2}f''(\theta)h^2$ encapsulates the error term in the Taylor polynomial. –  EuYu Oct 23 '12 at 19:15

1 Answer 1

up vote 2 down vote accepted

This is just the Lagrange form of the remainder for Taylor's Theorem.

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So easy, thanks. –  Tomás Oct 23 '12 at 19:13

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