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 December 23rd, 2016, 03:30 AM #1 Member   Joined: Mar 2016 From: Sweden Posts: 32 Thanks: 4 Taylor series I have the following problem to solve $$\lim_{x\to0} (\frac{cosx}{cos2x})^{1/x^2}$$ so I take ln and use Taylor series for cosx. For cosx I get $\mathcal{O}(x^4)$. Then I am confused when I should use the Taylor series for $ln(1-2x^2+\mathcal{O}(x^4))$. I am confused about the $\mathcal{O}(x^4)$ inside ln when using Taylor. Hopefully you understand my question, thanks
 December 23rd, 2016, 04:00 AM #2 Senior Member     Joined: Sep 2007 From: USA Posts: 349 Thanks: 67 Math Focus: Calculus Honestly, I wouldn't use Taylor series for that limit. Is it a requirement? Thanks from v8archie
December 23rd, 2016, 06:20 AM   #3
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Quote:
 Originally Posted by Compendium Honestly, I wouldn't use Taylor series for that limit. Is it a requirement?
No, it is not. But my question is not about the limit but about the Taylor series.

December 23rd, 2016, 06:44 AM   #4
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Quote:
 Originally Posted by matteamanda Then I am confused when I should use the Taylor series for $ln(1-2x^2+\mathcal{O}(x^4))$. I am confused about the $\mathcal{O}(x^4)$ inside ln when using Taylor.
\begin{align*} \ln(1+x) &= x-\tfrac12x^2+\mathcal{O}(x^3) \\ \ln \big(1-2x^2+\mathcal{O}(x^4) \big) &= \big( 1-2x^2+\mathcal{O}(x^4) \big) - \frac12 \big(1-2x^2+\mathcal{O}(x^4) \big)^2 + \mathcal{O}(x^3) \end{align*}
and then expand noting that $a\mathcal{O}(x^n)^2=\mathcal{O}(x^{2n})$ and $ax^m\mathcal{O}(x^n)=\mathcal{O}(x^{m+n})$.

Edit: and for $m \lt n$ we have $\mathcal{O}(x^m) + \mathcal{O}(x^n) = \mathcal{O}(x^{m})$.

Last edited by v8archie; December 23rd, 2016 at 06:48 AM.

January 16th, 2017, 10:03 PM   #5
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Quote:
 Originally Posted by v8archie \begin{align*} \ln(1+x) &= x-\tfrac12x^2+\mathcal{O}(x^3) \\ \ln \big(1-2x^2+\mathcal{O}(x^4) \big) &= \color{blue}{\big( 1-2x^2+\mathcal{O}(x^4) \big) - \frac12 \big(1-2x^2+\mathcal{O}(x^4) \big)^2 + \mathcal{O}(x^3)} \end{align*} and then expand noting that $a\mathcal{O}(x^n)^2=\mathcal{O}(x^{2n})$ and $ax^m\mathcal{O}(x^n)=\mathcal{O}(x^{m+n})$. Edit: and for $m \lt n$ we have $\mathcal{O}(x^m) + \mathcal{O}(x^n) = \mathcal{O}(x^{m})$.
I think the answer has a mistake because you copy but fix.

$\ln \big(1-2x^2+\mathcal{O}(x^4) \big)= \big(-2x^2+\mathcal{O}(x^4) \big)- \frac12 \big(-2x^2+\mathcal{O}(x^4) \big)^2+ \mathcal{O}(x^3).$

Last edited by skipjack; January 17th, 2017 at 10:33 AM.

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