[Math] Finding the 5th order taylor polynomial of a function without brute force

approximationcalculustaylor expansion

I am currently stuck on this 5th order taylor polynomial question for almost an hour now, it's quite difficult if you brute force it but there was a hint given today by my tutor that there is an easier way to solve for the 5th taylor polynomial of this function below if you do some substitutions. Should I be substituting $x^2$ for $x$ and then find the 5th taylor polynomial and then substituting $x^2$ back in at the end or am I completely off with this suggestion?:$$f(x) = \frac{\sin(x^2)}{x^2}$$

Best Answer

Building on the hint $u=x^2$.

$\sin(u)=u-\frac{u^3}{3!}+\frac{u^5}{5!}$
$\frac{\sin(u)}{u}=1-\frac{u^2}{3!}+\frac{u^4}{5!}$

Now back substituting in $x^2$

$\frac{\sin(x^2)}{x^2}=1-\frac{x^4}{3!}+\frac{x^8}{5!}+...$

Edit Using $x=x^2$ as the substitution as op had.

$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}+...$

$\frac{\sin(x)}{x}=1-\frac{x^2}{3!}+\frac{x^4}{5!}+...$

Now back substituting $x^2$

$\frac{\sin(x^2)}{x^2}=1-\frac{(x^2)^2}{3!}+\frac{(x^2)^4}{5!}...$

It doesn't matter what you pick as your substitution. It may get confusing if you use the substitution $x=x^2$.

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