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

My problem


Corrollary C

I am to use Corrollary C to find a better approximation for my problem. I'm trying to understand what exactly is going on but I'm having severe problems understanding it.

I sort of understand the error formula, what I don't get is how to choose this M and N and what exactly is means that $g(2) = 1$ and all that..

share|cite|improve this question

Suggestion: In your problem $a=2$ and $x=1.8$. Since $$|g''(t)|<1+(t-2)^2$$ you have that $$-1-(t-2)^2<g''(t)<1+(t-2)^2$$ which should give you $M,N$. But $M$ and $N$ should be constants so that the inequality holds for all $t$ between $a=2$ and $x=1.8$. I think (but please doublecheck it) that $$|g''(t)|<1+(t-2)^2<1+(1.8-2)^2=1.04$$ So your constants $M$ and $N$ are equal to $-1.04$ and $1.04$ respectively. You cannot plug in $2$ in the above inequality to obtain narrower bounds, since according to Corollary C, you need constants that work for every $t$ in the interval $(1.8, 2)$. Unfortunately $M$ and $N$ do not have the same sign, so you cannot use the better approximation stated in Corollary C. Your approximation is given by $$g(1.8)=g(2)+g'(2)(1.8-2)$$ Your error can be as big as the half the length of the interval, that is $$\frac{N-M}{2}(1.8-2)^2$$ and not $$\frac{N-M}{4}(1.8-2)^2$$ as the better approximation would have yielded.

share|cite|improve this answer
Do you mean to say that I can't find a better approximation on the problem? Is the text book wrong? – Paze Mar 9 '14 at 17:43
@Paze Of course not. I mean that for you M and N have the same sign and therefore you cannot use "the better approximation which is used when M and N have different sign". Have you read my answer (or your book) at all? – Jimmy R. Mar 9 '14 at 20:15

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.