Expand the function $[\cos(x^3)]^\frac{-1}{2}$ into Taylor series around $x=0$ up to $O(x^6)$

Question

I need help in expansion of the following function into Taylor Series up to $O(x^6)$ $$[\cos(x^3)]^\frac{-1}{2}$$ The things I've already tried are expansion of $u^\frac{-1}{2}$ by assuming $u=cos(x^3)$ and manually calculated 4 differentials of $u^\frac{-1}{2}$ and writing their Taylor's expansion without inputing $u=0$ The Taylor expansion looks something like $$\frac{(u-a)^0}{0!}u^\frac{-1}{2} - \frac{(u-a)^1}{1!}\frac{1}{2}u^\frac{-3}{2} +\frac{(u-a)^2}{2!}\frac{3}{4}u^\frac{-5}{2}-\frac{(u-a)^3}{3!}\frac{15}{8}u^\frac{-7}{2}+\cdots$$ Initially i though of substituting $u$ as Taylor series of $\cos(x^3)$ as $$\cos(t) = \sum_{k=0}^{\infty} (-1)^k\frac{t^{2k}}{2k!}$$ by inputting $t=x^3$ so it becomes $$\cos(x^3) = \sum_{k=0}^\infty (-1)^k\frac{x^{6k}}{2k!}$$

But this makes things more complicated than normal as $u$ is in root powers and I cannot figure out what can i do to reach to a solution to this problem. Any help is appreciated.

Answer 1

$$y=\left[\cos(x^3)\right]^{-\frac{1}{2}}\implies \log(y)=-\frac{1}{2}\log\left[\cos(x^3)\right]$$ $$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^6\right)$$ $$\cos(x^3)=1-\frac{x^6}{2}+\frac{x^{12}}{24}+O\left(x^{15}\right)$$ $$\log\left[\cos(x^3)\right]=-\frac{x^6}{2}-\frac{x^{12}}{12}+O\left(x^{15}\right)$$ As, $\log(y)=-\frac{1}{2}\log\left[\cos(x^3)\right]$ $$\log(y)=\frac{x^6}{4}+\frac{x^{12}}{24}+O\left(x^{15}\right)$$ $$y=e^{\log(y)}=1+\frac{x^6}{4}+\frac{7 x^{12}}{96}+O\left(x^{15}\right)=1+\frac{x^6}{4}+O\left(x^{12}\right)$$

Answer 2

Let $y=(\cos(x^3))^{-\frac 12}$

Then $\frac {dy}{dx}=-\frac 12(\cos(x^3))^{-\frac 32}\sin(x^3)3x^2$ (using the chain rule)

Can you continue differentiating to get $\frac {d^2y}{dx^2}$, $\frac {d^3y}{dx^3}$ etc?

Answer 3

Combine $\cos(x^3) =1-x^6/2+O(x^{12}) $ with $(1-z)^{-1/2} =1+z/2+O(z^2) $ to get $1+x^6/4+O(x^{12}) $.