How do you find the Taylor polynomial of degree 10 of the function #arctan(x^3)# at a = 0?

Question

Christopher Allers · Accepted Answer

#arctan x^3 = x^3-x^9/3+x^15/5+o(x^15)#

Start from the sum of the geometric series:

#sum_(n=0)^oo q^n = 1/(1-q)#

for #abs q <1#.

Let #q = -x^2#:

#1/(1+x^2) = sum_(n=0)^oo (-x^2)^n = sum_(n=0)^oo (-1)^nx^(2n)#

converging for #abs x <1#.

In the interval of convergence we can integrate term by term:

#int_0^t dx/(1+x^2) = sum_(n=0)^oo (-1)^n int_0^t x^(2n)dx#

#arctan t = sum_(n=0)^oo (-1)^n t^(2n+1)/(2n+1)#

Let now #t=x^3#:

#arctan x^3 = sum_(n=0)^oo (-1)^n (x^3)^(2n+1)/(2n+1)#

#arctan x^3 = sum_(n=0)^oo (-1)^n x^(6n+3)/(2n+1)#

and truncating at the third term:

#arctan x^3 = x^3-x^9/3+x^15/5+o(x^15)#

Emma Aldridge · Answer

undefined To find the Taylor polynomial of degree 10 of the function arctan(x^3) at $a = 0$, we can use the Taylor series expansion formula for arctan(x). The Taylor series expansion for arctan(x) is given by:

$$ \arctan(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \frac{x^9}{9} - \frac{x^{11}}{11} + \ldots $$

Substituting $x^3$ for $x$ in the series, we get:

$$ \arctan(x^3) = x^3 - \frac{(x^3)^3}{3} + \frac{(x^3)^5}{5} - \frac{(x^3)^7}{7} + \frac{(x^3)^9}{9} - \frac{(x^3)^{11}}{11} + \ldots $$

$$ = x^3 - \frac{x^9}{3} + \frac{x^{15}}{5} - \frac{x^{21}}{7} + \frac{x^{27}}{9} - \frac{x^{33}}{11} + \ldots $$

The Taylor polynomial of degree 10 of the function arctan(x^3) at $a = 0$ is obtained by considering the terms up to $x^{10}$ in the series expansion. Thus, the Taylor polynomial of degree 10 is:

$$ P_{10}(x) = x^3 - \frac{x^9}{3} + \frac{x^{15}}{5} - \frac{x^{21}}{7} + \frac{x^{27}}{9} $$