Tomi's Challenges - #4

Let's tidy up this expression first. $\newline$

$f(x)=-\frac{6^{11}}{990}x^{11}+\frac{6^{9}}{9}x^{9}-8\cdot6^{7}x^{7}+336\cdot6^{5}x^{5}-6720\cdot6^{3}x^{3}+8!\cdot6x$ $\newline$ $f(x)=-\frac{1}{990}(6x)^{11}+\frac{1}{9}(6x)^{9}-8\cdot(6x)^{7}+336\cdot(6x)^{5}-6720\cdot(6x)^{3}+8!\cdot(6x)$

Notice that $\frac{1}{990}=\frac{1}{11\cdot10\cdot9}=\frac{8!}{11!}, \frac{1}{9}=\frac{8!}{9!}, 8=\frac{8!}{7!}, 336=8\cdot7\cdot6=\frac{8!}{5!}, 6720=4\cdot5\cdot336=4\cdot5\cdot6\cdot7\cdot8=\frac{8!}{3!}$ .

$f(x)=-\frac{8!}{11!}(6x)^{11}+\frac{8!}{9!}(6x)^{9}-\frac{8!}{7!}\cdot(6x)^{7}+\frac{8!}{5!}\cdot(6x)^{5}-\frac{8!}{3!}\cdot(6x)^{3}+8!\cdot(6x)$

$f(x)=8!\cdot(-\frac{1}{11!}(6x)^{11}+\frac{1}{9!}(6x)^{9}-\frac{1}{7!}\cdot(6x)^{7}+\frac{1}{5!}\cdot(6x)^{5}-\frac{1}{3!}\cdot(6x)^{3}+(6x))$ .

This expression is approximately equal to the following, using the Taylor series for sine.

$f(x)\approx 8!sin(6x)$

Note that this only holds for intervals such as $[-1/4, 1]$ , because for bigger or smaller values this polynomial explodes into infinity. The maximum value of this function is $8!$ or $40320$ .