Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?


share|cite|improve this question
This looks like Taylor's Theorem to me. See – 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
up vote 2 down vote accepted

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

share|cite|improve this answer

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.