Take the first order taylor approximation: We say $f$ is differentiable at $x_0 \in S^o$ if $\exists \space \nabla f(x_0)$ and a function $R: \mathbb{R}^n \to \mathbb{R}$ such that

$f(x) = f(x_0) + \langle \nabla f(x_0), x-x_0 \rangle + R(x-x_0)$

I don't understand what the point of the Remainder term is? Is it just trying to say there is a degree of error involved in the taylor approximation (which is the first two terms if I am not mistaken?)

What does this limit: $\lim_{x \to x_0} \frac{R(x-x_0)}{\| x-x_0 \|}=0$ convey in terms of information? If this implies that $R(x-x_0)$ grows slower than $\|x-x_0\|$ why is that useful?

  • $\begingroup$ A Taylor series represents a function as an infinite sum. Obviously, in real terms we can't write down the infinite sum, so there is a point at which we must truncate the series. If we truncate the series, there must be an error, as our truncated series only approximates the solution, instead of being equivalent. $\endgroup$ – Mattos Feb 6 '15 at 7:23

The behavior of the error is the only thing that really lets us know that it's a good approximation.

I can, for example, write $x^3=5x\sin(x)+xe^x+R(x)$, where $R(x) = x^3 -5x\sin(x)-xe^x$, and call $R$ the error. Is this useful in any way?

  • $\begingroup$ Yes, I can see the practicality of it, but what exactly then does the limit mean to me? $\endgroup$ – stromae Feb 6 '15 at 7:49
  • $\begingroup$ @stromae well, for one thing, the limit ensures that you can place upper and lower bounds on a function. This can be useful when dealing with really complicated functions, because you can just approximate them by a polynomial to get an idea of the possible behaviours. $\endgroup$ – hasnohat Feb 6 '15 at 8:07

Any function approximates another. For instance, here are two approximations of $f$:

$$f(x) \approx f(x_0),$$ $$f(x) \approx f(x_0) + \langle \nabla f(x_0), x-x_0 \rangle.$$

If we don't know how close we approximate $f$, these formulas are of little use. The remainder answers this question.

In addition, if you had a remainder that is as large as the some term in the development (which generalizes to higher orders), it would be useless to consider this term. So we require that the remainder decreases faster (when you get closer to $x_0$) than the smallest term.

  • $\begingroup$ I think I understand, but am a bit confused at what your last paragraph is trying to say. "remainder that is as large as the some term in the development" ? $\endgroup$ – stromae Feb 6 '15 at 7:48
  • $\begingroup$ If we established $f\approx3+0.2$ and evaluate that the remainder is $0.9$, why worry about the term $0.2$ which improves the accuracy by little ? $\endgroup$ – Yves Daoust Feb 6 '15 at 7:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.