$$\cos(x) - \frac{1+ax^2}{1+bx^2} \text{ on } x \to 0$$

If $\displaystyle \cos(x) = 1 - \frac{x^2}{2} + \frac{x^4}{4!} - \cdots $

Then we should choose $a, b$ in a such way that it's Taylor series is close to this.

However, I'm not sure how to approach this. I tried to take several derivates of second term to see its value on $x\_0 = 0$, but it becomes complicated and I don't see general formula for $n$-th derivative at point zero to find $a$ and $b$.