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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

José Carlos Santos
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letslearnmath
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    Do you want more terms or an explanation for the ones you know? – lhf Dec 14 '17 at 17:53
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    Using the assistance of a computer is the quickest way. However in order for readers to be able to critique your method (I assume long hand method) you have to give a little more information about the steps you have taken to obtain your displayed result. – James Arathoon Dec 14 '17 at 18:04
  • Oh, sorry, so actually this was the solution provided us, going straight from the LHS to RHS, and I was hoping for an explanation as to how they came up with the first few terms on the RHS given those on the LHS. So an explanation for the ones I know would be most helpful. – letslearnmath Dec 14 '17 at 18:19

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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.

José Carlos Santos
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There’s another way to do this by hand, but let me deal with a slightly more general situation. You may perform a division of power series in exactly the same way you learned to divide polynomials in high school, but by writing the terms in ascending order of degree instead of descending. If, for instance, you wanted $$ \frac1{x-x^2+2x^3-7x^4+\cdots} $$ you would get the first monomial of the quotient to be $x^{-1}$, and write this up top; then multiply this by your denominator and subtract the result from $1$ to get $x-2x^2+7x^3+\cdots$; the next monomial of the quotient will thus be $1$, etc. I know that this is not much of an explanation—the process is best explained dynamically, at a blackboard.

In your particular case, I would expand the denominator, as $x^2-x^3+\frac5{12}x^4+\cdots$ and then divide this into $1$ by the method I sketched out above.

Lubin
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