# Closed form series for reciprocal cubic function

consider a cubic of the form f(x)=$$x^3-2x+z$$

Is it possible to derive a power series of coefficients for the function $$x^y/f(x)$$, for some $$y=0,1,2...$$ that does not require the use of Faà di Bruno's formula, or does not require multiple iterations of some recursive scheme to get the x^n coefficients .

for example:

The 6th coefficient is $$((32 - 12 z^2))/z^6$$

So how does one derive such functions of z without having to use a cumbersome recursive scheme ?

It is possible to derive such a closed-form expression for the coefficients of $$\frac{3x^2-2}{x^3-2x+z}$$

$$\displaystyle -z^{-m}2^{m+1} + \sum_{n=0} \frac{z^{2-m}z^{2n} (1+m)}{2^{3n-m+2}(n+1)} \binom{3n+1-m}{n}$$

for $$3n+1-m<0$$

EDIT:

The closest I got for the $$\frac{1}{f(x)}$$ case, based on symmetric polynomials,

is $$(f(z))^2=4z(a^{-2m-1}+b^{-2m-1}+c^{-2m-1})-4(-z)^{-m}((a^{m+2}+b^{m+2}+c^{m+2}))+-2(-z)^{1-m}((a^{m-1}+b^{m-1}+c^{m-1}))+((a^{-2m+2}+b^{-2m+2}+c^{-2m+2}))$$

then...

$$f(z)/\sqrt{(32-27z^2)}$$

is a perfect square , which gives the coefficient

The $$a,b,c$$ are roots of $$f(x)$$, which are functions of z and have a finite expressions as a sum of binomials via the Lagrange inversion theorem.

So for $$m=3$$

https://www.wolframalpha.com/input/?i=%288z%287z%5E4-56z%5E2%2B64%29%2Fz%5E7-8%28z%5E2-2%29%2Fz%5E4-40%2Fz%5E2-8%2Fz%5E2%29%2F%2832-27z%5E2%29

the square root of which is the third coefficient of

• If you do a Partial Fraction Decomposition on your rational function, you get its power series expansion as a linear combination of binomial series of x. Dec 6, 2021 at 18:43
• I apologize I was not aware of your edit. Feel free to roll back mine or ask me to do so. Dec 6, 2021 at 19:09

The reciprocal quadratic is trivial, cause you can just apply the binomial theorem to the roots. This does not work for higher degree ones.

For the reciprocal of the cubic, the P(z) terms are:

$$\sum_{n=0}^{} \frac{z^{2n-m}2^{m-1}}{2^{3n}} \binom{3n-m}{n}$$ for $$3n-m<0$$

This is a single summation that does not require multiple sums or a iterative/recursion process.

Consider the following partial fractional decomposition of the roots:

$$(b-c)/(x-a)+(c-a)/(x-b)+(a-b)/(x-c)=\frac{g}{x^3-2x+z}$$

note:

$$g=((b - a) (c - a) (b - c))=+-\sqrt{32-27z^2}$$

the Taylor expansion of the above fractions has the $$x^{m-1}$$ coefficients:

$$\frac{a^{-m}(b-c)+c^{-m}(a-b)+b^{-m}(c-a)}{b/a-c/a-a/b+a+b/c+a/c-b/c}$$

The proof requires having to show that a certain cancellation process occurs in the expression above. The a,b,c's are infinite series as function of z for the roots. This reduces to the finite binomial series above.

Through the use of a hypergeometric identity, the $$\sqrt{32-27z^2}$$ terms factors out the numerator and denominator. This is necessary for simplifying the expression.

example: find the 13th term

which shown above is

$$(x^{12} (z^8 - 160 z^6 + 1792 z^4 - 5120 z^2 + 4096))/z^{13}$$

which is the same as the formula:

https://www.wolframalpha.com/input?i=+sum_%28n%3D0%29%5E4+++z%5E%282n%29%2F%282%5E%283n%2B1-m%29%29*C%283n-m%2Cn%29%2Cm%3D13

general case:

$$\sum_{n=0}^{} \frac{z^{(k-1)n-m} p^{m-1}}{p^{kn}} \binom{kn-m}{n}$$ for $$kn-m<0$$

as the z coefficients for $$\frac{1}{x^k-px+z}$$