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Retaining the three terms in the series, estimate the remaining series using "Big Oh" notation with the best integer value possible, as $x\to 0$. The series is

$$\ln (\tan (x)) =\ln(x)+ \frac{x^2}{3} + \frac{7x^4}{90} + \frac{62x^6}{2835}+..., \quad \left( 0< \left|x\right| <\frac{\pi}{2} \right)$$

I was thinking of proving

$$f(x)=\left| \ln (\tan (x)) - \left(\ln(x)+ \frac{x^2}{3} + \frac{7x^4}{90} \right) \right| \leq C \left|x^4\right|$$

where $g(x)=x^4$.

Then i'd calculate this

$$ \left| \frac{f(x)}{g(x)} \right| = \left| \frac{\ln (\tan (x)) - \left( \ln(x)+ \frac{x^2}{3} + \frac{7x^4}{90} \right)} {x^4} \right|$$

I think the easiest way would be using this method where you choose $k=1$ and $x>1$, but i'm not sure how to do it with this expression as there are fractions, the natural logarithm and even tangens.

The problem, here b): enter image description here

Stranqer95
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  • It should be $O(x^6)$. For the best possible constant, use the representation of the remainder of the Taylor series. There will be a derivative, evaluated at some point $\xi\in (-\pi/2;\pi/2)$ and compute the maximum value of the absolute value of this derivative. – Svetoslav Sep 06 '15 at 14:15
  • There was 2 problem the first one was a) Retaining three terms in the series, estimate the remaining series using little oh-notation with the best integer value possible, $x \to 0$. The second one is b) Repeat the problem using big Oh-notation. That's why I assumed that I should only use the first three terms i.e. not including $x^6$. I just uploaded the problem. – Stranqer95 Sep 06 '15 at 14:21
  • But see that in the second problem, the 4-th term is $x^6$. Therefore it will be $O(x^6)$ – Svetoslav Sep 06 '15 at 14:26
  • I see. But I'm not sure how to use the method you described. Is it possible to solve the problem with the method i described? – Stranqer95 Sep 06 '15 at 14:41
  • The method you linked tells you how to show big O when $x\rightarrow \infty$. Taking the absolute value of the difference of $f$ and the first three terms is actually finding the reminder of the Taylor series, as I told you before. So you can use the Mean-value form of the remainder (see wikipedia). – Svetoslav Sep 06 '15 at 14:49
  • I'm not quite sure, but is this right $\frac{\ln(\tan(x)^{(6)} (\xi (x))}{720} \leq \frac{1}{720}$ ? – Stranqer95 Sep 06 '15 at 15:20

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