Homework problem here, would appreciate an explanation to the answer of this question.

Problem: Find the rate of convergence of

$$ \lim\limits_{h \to 0} \frac{\sin(h)}{h} = 0 $$

The book solves this by using the expansion of the Taylor series. But only too the second polynomial. Like so:

$$ \left|\frac{\sin(h)}{h} -1 \right| = \left| \frac{h - \frac{h^3}{6}\sin(x(h))}{h} - 1 \right| = \left|-\frac{h^2}{6}\sin(x(h)) \right| = O(h^2) $$ (Where $x(h) \in (0,h)$)

But im not sure how they arbitrarily just chose to expand to the second taylor polynomial. What gave them the intuition to use the Taylor Polymnomial in the first place? More to the point, what did they choose to expand the polynomial about?

If anyone could help me that would be great. thanks!

  • $\begingroup$ You have some mistakes in what you've written after "Like so" $\endgroup$ – Omnomnomnom Sep 6 '14 at 21:12
  • $\begingroup$ Ah yeah, cos(x) probably should not be there. $\endgroup$ – user3389436 Sep 6 '14 at 21:13
  • $\begingroup$ My edit is another typical way of looking at it $\endgroup$ – Omnomnomnom Sep 6 '14 at 21:14

A brief answer to all points:

They chose to expand about $0$ because the limit was as $h \to 0$. That tends to work out best in most cases.

The key to choosing how far to expand is to simply expand "enough". In fact, you could expand further if you want to, but you'd find that the same basic steps apply. The only difference is that we'd be left with, for example, $$ \cdots = \left|-\frac 16 h^2 + \frac 1{120} h^4\right| = o(h^2) $$ In general, the Taylor polynomial centered at $x = a$ is a good way of dissecting how a function "behaves" near $a$, and is generally useful in limit problems like these.

  • $\begingroup$ thanks for the answer. Wouldn't your example converge with $O(h^4)$ since clearly the $h^4$ is larger for any h? Also that is an interesting statement about taylor polynomials. I wish my professor OR the book would have passed that along. $\endgroup$ – user3389436 Sep 6 '14 at 21:15
  • $\begingroup$ @user3389436 We should be thinking about $o$ rather than $O$, since we are concerned with behavior as $h \to 0$ rather than $h \to \infty$. $h^4$ will be progressively smaller than $h^2$ the closer $h$ gets to $0$. $\endgroup$ – Omnomnomnom Sep 6 '14 at 21:16
  • $\begingroup$ Ah, so here we are considering the convergence of the function to $0$, which in this case would be $\sin(h)$ due to the limit. In that case, $h^4$ will certainly be smaller than $h^2$ as $n \to 0$. Right? $\endgroup$ – user3389436 Sep 6 '14 at 21:20
  • $\begingroup$ Not sure what you mean by your first sentence. Yes, $h^2$ dominates $h^4$ as $h \to 0$. $\endgroup$ – Omnomnomnom Sep 6 '14 at 21:25

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