<s>[INST] Polynomial long division - intermediate steps missing I'm trying to find the intermediate step:
$$\frac{1}{(x-\frac{x^2}{2}+\frac{x^3}{12}-\frac{x^4}{144}+...)^2} = \frac{1}{x^2}+\frac{1}{x}+\frac{7}{12}+\frac{19x}{72} ...$$
Is there a quick way to find these first few terms that I'm missing here? Thanks
 [/INST] The first terms of the doninator of the LHS are$$x^2-x^3+\frac{5 x^4}{12}-\frac{7 x^5}{72}+\cdots=x^2\left(1-x+\frac{5 x^2}{12}-\frac{7 x^3}{72}+\cdots\right).$$Therefore, the RHS can be written as$$\frac1{x^2}\left(b_0+b_1x+b_2x^2+b_3x^3+\cdots\right)$$and we must have$$\left(1-x+\frac{5 x^2}{12}-\frac{7 x^3}{72}+\cdots\right)\left(b_0+b_1x+b_2x^2+b_3x^3+\cdots\right)=1$$and therefore$$\left\{\begin{array}{l}b_0=1\\b_1-b_0=0\\b_2-b_1+\frac5{12}b_0=0\\b_3-b_2+\frac5{12}b_1-\frac7{72}b_0=0,\end{array}\right.$$from which we get that$$\left\{\begin{array}{l}b_0=1\\b_1=1\\b_2=\frac7{72}\\b_3=\frac{19}{72}.\end{array}\right.$$But that's all you can get from the information provided.
</s>