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My problem

Theorem

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..

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1 Answer 1

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.

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Do you mean to say that I can't find a better approximation on the problem? Is the text book wrong? –  Paze Mar 9 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? –  Stef Mar 9 at 20:15

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