Ytukyg

I encountered in a text the unsupported assertion that the series expansion of

$$(1-w)^{y_1/y_2+1} + left(frac{x}{y_2z_2}right)(1-w)^{y_1/y_2} = 1$$

is

$$w = left(frac{1}{y}right)left(frac{1}{z_2}right)x
- left(frac{1}{2y^2z}right)x^2 - left(frac{1}{3y^3z^2}right)left(z_1-z_2right)x^3 - left(frac{1}{8y^4z^3}right)left(2z_1^2 - 5z_1z_2 + 2z_2^2right)x^4 + ldots
$$

for $0 le frac{x}{y_2z_2} le 1$ and given that ${ y_1, y_2, z_1, z_2 } in mathbb{N}$, $y=y_1+y_2$, $z=z_1cdot{}z_2$, $y_1z_1=y_2z_2$, and $1-w ge 0$.

The series appears to have the basic form

$$w = -sum_{i=1}^infty left(frac{x^i}{i!! cdot{} y^i z^{i-1}}right) left(??right)$$

where the term indicated by question marks lacks a recognizable pattern (${-1/z_2, 1, z_1-z_2, (2z_1-z_2)(z_1-2z_2), ldots }$).

By what approach can this series expansion be derived?

The idea seems to be to write (I changed notations to make my life easier)
$$w=sum_{k=1}^n c_k,x^k$$ and to expand
$$(1-w)^{a/b+1} + left(frac{x}{bz}right)(1-w)^{a/b} -1$$ as a Taylor series around $x=0$. Cancelling the coefficients leads to the expressions dor the $c_k$'s.

Doing it, we get
$$ 0=left(-frac{a c_1}{b}-c_1+frac{1}{b z}right)x+frac{ left((a+b) left(a
c_1^2-2 c_2 bright)-frac{2 a c_1}{z}right)}{2 b^2}x^2-frac{ left(6 c_3
b^2 z (a+b)-6 a c_2 b (c_1 z (a+b)-1)+a c_1^2 (a-b) (c_1 z
(a+b)-3)right)}{6 b^3 z}x^3+Oleft(x^4right)$$ from which, using more terms, gives the following
$$left(
begin{array}{cc}
k & c_k \
1 & frac{1}{(a+b) z} \
2 & -frac{a}{2 b (a+b)^2 z^2} \
3 & frac{a (a-b)}{3 b^2 (a+b)^3 z^3} \
4 & -frac{a (a-2 b) (2 a-b)}{8 b^3 (a+b)^4 z^4} \
5 & frac{a (a-3 b) (a-b) (3 a-b)}{15 b^4 (a+b)^5 z^5} \
6 & -frac{a (a-4 b) (2 a-3 b) (3 a-2 b) (4 a-b)}{144 b^5 (a+b)^6 z^6} \
7 & frac{a (a-5 b) (a-2 b) (a-b) (2 a-b) (5 a-b)}{70 b^6 (a+b)^7 z^7} \
8 & -frac{a (a-6 b) (2 a-5 b) (3 a-4 b) (4 a-3 b) (5 a-2 b) (6 a-b)}{5760 b^7
(a+b)^8 z^8} \
9 & frac{a (a-7 b) (3 a-5 b) (a-3 b) (5 a-3 b) (a-b) (3 a-b) (7 a-b)}{2835 b^8
(a+b)^9 z^9} \
10 & -frac{a (a-8 b) (2 a-7 b) (4 a-5 b) (5 a-4 b) (a-2 b) (7 a-2 b) (2 a-b) (8
a-b)}{44800 b^9 (a+b)^{10} z^{10}}
end{array}
right)$$

I prefer to add another answer since changing quite many things.

Let first $$a=frac {y_1}{y_2}qquad k=frac 1 {y_2,z_2}qquad text{and define} qquad
w=asum_{i=1}^n (-1)^{i-1}, c_i, left(frac{k}{a+1}right)^i x^i $$ Now, much clearer (I think) that in my previous answer, the coefficients are
$$left(
begin{array}{cc}
i & text{coefficient } c_i \
1 & frac{1}{a} \
2 & frac{1}{2} \
3 & frac{1}{3}(a-1) \
4 & frac{1}{8} (a-2) (2 a-1) \
5 & frac{1}{15} (a-3) (a-1) (3 a-1) \
6 & frac{1}{144} (a-4) (2 a-3) (3 a-2) (4 a-1) \
7 & frac{1}{70} (a-5) (a-2) (a-1) (2 a-1) (5 a-1) \
8 & frac{1}{5760} (a-6) (2 a-5) (3 a-4) (4 a-3) (5 a-2) (6 a-1)\
9 & frac{1}{2835}(a-7) (a-3) (a-1) (3 a-5) (3 a-1) (5 a-3) (7 a-1) \
10 & frac{1}{44800}(a-8) (a-2) (2 a-7) (2 a-1) (4 a-5) (5 a-4) (7 a-2) (8 a-1) \
11 &frac{1}{49896}(a-9) (a-4) (a-1) (2 a-3) (3 a-7) (3 a-2) (4 a-1) (7 a-3) (9 a-1)
end{array}
right)$$

Unfortunately, the sequence ${2,3,8,15,144,70,5760,2835,44800,49896}$ or its subsequences do not appear in $OEIS$.

You can notice two different patterns depending on the parity of $i$.