I have this question:

Consider the series

$e^{\tan(x)} = 1 + x + \dfrac{x^{2}}{2!} + \dfrac{3x^{3}}{3!} + \dfrac{9x^{4}}{4!} + \ldots $

Retaining three terms in the series, estimate the remaining series using "Little-$o$" notation with the best integer value possible, as $x\to 0$.

My question is:

What do they mean with "with the best integer value possible"? Someone who can point out the connection with little $o$ notation and a best integer vaulue possible?

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    $\begingroup$ It would greatly help matters if you included the precise question and its source. $\endgroup$ – JavaMan Sep 8 '11 at 18:46
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    $\begingroup$ Probably they're envisioning something like $... + o(x^5)$ where $5$ could be something else but is supposed to be the best integer that make the estimate work. $\endgroup$ – Henning Makholm Sep 8 '11 at 18:52
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    $\begingroup$ Three terms in the series make $1+x+\frac12x^2$ hence the question asks for the smallest possible $n$ such that $\exp(\tan(x))=1+x+\frac12x^2+o(x^n)$ when $x\to0$. $\endgroup$ – Did Sep 8 '11 at 19:45

‘Best’ here would be ‘smallest’, so as Henning said, in all likelihood you’re being asked to write $e^{\tan x} = 1 + x + \frac{x^2}{2} + o(x^n)$ as $x\to 0$ for the smallest integer $n$ that makes the statement true.

Added: As Brugerfugl points out, that should be ‘largest’, since were’ looking at $x\to 0$.

  • $\begingroup$ Well. I have now realized that it must be the biggest integer n, because the situation is x->0 and not x-> infinity When x->0 we have: x^-1 = o(x^-2) and when x->infinity x^-2 = o(x^-1) $\endgroup$ – Brugerfugl Sep 10 '11 at 17:40
  • $\begingroup$ @Brugerfugl: You’re quite right: I wasn’t thinking when I wrote that. My apologies; I’ll edit it. $\endgroup$ – Brian M. Scott Sep 10 '11 at 17:44

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