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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

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

for $0 \le \frac{x}{y_2z_2} \le 1$ and given that $\{ y_1, y_2, z_1, z_2 \} \in \mathbb{N}$, $y=y_1+y_2$, $z=z_1\cdot{}z_2$, $y_1z_1=y_2z_2$, 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?

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  • $\begingroup$ In my second answer, notice that $\frac k{a+1}=\frac 1{z_2(y_1+y_2)}$ $\endgroup$ Nov 24 '18 at 16:55
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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 $$ 0=\left(-\frac{a c_1}{b}-c_1+\frac{1}{b z}\right)x+\frac{ \left((a+b) \left(a c_1^2-2 c_2 b\right)-\frac{2 a c_1}{z}\right)}{2 b^2}x^2-\frac{ \left(6 c_3 b^2 z (a+b)-6 a c_2 b (c_1 z (a+b)-1)+a c_1^2 (a-b) (c_1 z (a+b)-3)\right)}{6 b^3 z}x^3+O\left(x^4\right)$$ from which, using more terms, gives the following $$\left( \begin{array}{cc} k & c_k \\ 1 & \frac{1}{(a+b) z} \\ 2 & -\frac{a}{2 b (a+b)^2 z^2} \\ 3 & \frac{a (a-b)}{3 b^2 (a+b)^3 z^3} \\ 4 & -\frac{a (a-2 b) (2 a-b)}{8 b^3 (a+b)^4 z^4} \\ 5 & \frac{a (a-3 b) (a-b) (3 a-b)}{15 b^4 (a+b)^5 z^5} \\ 6 & -\frac{a (a-4 b) (2 a-3 b) (3 a-2 b) (4 a-b)}{144 b^5 (a+b)^6 z^6} \\ 7 & \frac{a (a-5 b) (a-2 b) (a-b) (2 a-b) (5 a-b)}{70 b^6 (a+b)^7 z^7} \\ 8 & -\frac{a (a-6 b) (2 a-5 b) (3 a-4 b) (4 a-3 b) (5 a-2 b) (6 a-b)}{5760 b^7 (a+b)^8 z^8} \\ 9 & \frac{a (a-7 b) (3 a-5 b) (a-3 b) (5 a-3 b) (a-b) (3 a-b) (7 a-b)}{2835 b^8 (a+b)^9 z^9} \\ 10 & -\frac{a (a-8 b) (2 a-7 b) (4 a-5 b) (5 a-4 b) (a-2 b) (7 a-2 b) (2 a-b) (8 a-b)}{44800 b^9 (a+b)^{10} z^{10}} \end{array} \right)$$

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  • $\begingroup$ Impressive. I'm not sure that I completely follow your logic, so I proceeded by considering the $1-w$ equation as an implicit function of $x$ such that $w=f(x)$. I then began to construct a Taylor series by implicit differentiation about the point $x=0$. Doing so gave for the $k=1$ term $\frac{f^{(1)}(0)}{1!}x = 1/(z(a+b))$, which matches your $c_1$ term; $f^{(1)}(x)=df/dx$ was found to be $\frac{1}{bz*(a/b + 1 + ax/(b^2z(1-w)))}$. I was not able to relate this directly to your expression $-ac_1/b - c_1 +1/(bz)$. Would you care to elaborate on how you obtained that intermediate? Many thanks. $\endgroup$
    – user001
    Nov 24 '18 at 8:24
  • $\begingroup$ @user001. Just replace $w$ by the series and use the binomial expansion. By the way, you are very welcome ! Cheers. $\endgroup$ Nov 24 '18 at 8:29
  • $\begingroup$ When you say to replace $w$ by the series and then to use the binomial expansion, do you mean to apply the generalized binomial expansion ($(1+x)^{\alpha} = \sum_{k=0}^\infty {\alpha \choose k} x^k$) to $(1 - \sum_{k=1}^n c_k x^k)^{a/b+1} + \frac{x}{bz}(1-\sum_{k=1}^n c_k x^k)^{a/b} - 1 = 0$? Merci beaucoup! $\endgroup$
    – user001
    Nov 24 '18 at 9:20
  • $\begingroup$ @user001. This is the way ! $\endgroup$ Nov 24 '18 at 9:22
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I prefer to add another answer since changing quite many things.

Let first $$a=\frac {y_1}{y_2}\qquad k=\frac 1 {y_2\,z_2}\qquad \text{and define} \qquad w=a\sum_{i=1}^n (-1)^{i-1}\, c_i\, \left(\frac{k}{a+1}\right)^i x^i $$ 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 $i$.

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