Tuesday, July 26, 2016

calculus - Maclaurin polynomial of tan(x)



The method used to find the Maclaurin polynomial of sin(x), cos(x), and $e^x$ requires finding several derivatives of the function. However, you can only take a couple derivatives of tan(x) before it becomes unbearable to calculate.




Is there a relatively easy way to find the Maclaurin polynomial of tan(x)?



I considered using tan(x)=sin(x)/cos(x) somehow, but I couldn't figure out how.


Answer



Long division of series.



$$ \matrix{ & x + \frac{x^3}{3} + \frac{2 x^5}{15} + \dots
\cr 1 - \frac{x^2}{2} + \frac{x^4}{24} + \ldots & ) \overline{x - \frac{x^3}{6} + \frac{x^5}{120} + \dots}\cr
& x - \frac{x^3}{2} + \frac{x^5}{24} + \dots\cr & --------\cr
&

\frac{x^3}{3} - \frac{x^5}{30} + \dots\cr
& \frac{x^3}{3} - \frac{x^5}{6} + \dots\cr
& ------\cr
&\frac{2 x^5}{15} + \dots \cr
&\frac{2 x^5}{15} + \dots \cr
& ----}$$


No comments:

Post a Comment

analysis - Injection, making bijection

I have injection $f \colon A \rightarrow B$ and I want to get bijection. Can I just resting codomain to $f(A)$? I know that every function i...