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Say I want to expand the function $\tan{x}$ in terms of a polynomial with remainder term of 7:th order. This means that there exists constants $c_1, c_3$ and $c_5$ such that

$$\tan{x}=c_1x+c_3x^3+c_5x^5+O(x^7).$$

Since $\sin{x}=\tan{x}\cdot \cos{x},$ the McLaurin expansions of $\sin{x}$ and $\cos{x}$ gives

$$x-\frac{x^3}{6}+\frac{x^5}{120}+O(x^7)=\left(1-\frac{x^2}{2}+\frac{x^4}{24}+O(x^6)\right)(c_1x+c_3x^3+c_5x^5+O(x^7)).$$

Question: Am I supposed to multiply through with one term at a time from each parenthesis in the RHS or are there shortcuts? My goal here is to have the RHS simplifed so that I can solve for $c_1,c_3$ and $c_5$ by identification of coefficients.

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  • $\begingroup$ If you are ok, you can accept the answer and set as solved. Thanks! $\endgroup$ – user Dec 29 '17 at 21:02
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You need to multiply term by term, th eonly shortcut is to neglect the terms which are $O(x^7)$.

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  • $\begingroup$ So, multiplying a $x^4$-term with a $x^3$-term doesn't need to be written down, but I can just bunch it into the $O(x^7)$? $\endgroup$ – Parseval Dec 29 '17 at 14:14
  • $\begingroup$ By the way, why can't I just use normal McLaurin expansion on $\tan{x}$? $\endgroup$ – Parseval Dec 29 '17 at 14:15
  • $\begingroup$ @Parseval yes of course, this is to great help in the calculation by Taylor's series! $\endgroup$ – user Dec 29 '17 at 14:15
  • $\begingroup$ @Parseval of course you can, I thuoght this was an excersise to obtain tanx series given teh series for sinx and cosx $\endgroup$ – user Dec 29 '17 at 14:17
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    $\begingroup$ @Parseval it a very powerful issue with taylor's series you can compose them and find a lot of others series and also integrate them, see for example the derivation for arctanx math.stackexchange.com/questions/29649/… $\endgroup$ – user Dec 29 '17 at 14:23

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