<s>[INST] Unexpected (incorrect) solution to Lagrange Inversion solution to $x^4 - x^3 - x^2 - x - 1 = 0$ about the solution near $x = 2$ I am developing generalized hypergeometric solutions for a set of such polynomials.  With this example we can write $x^4 - x^3 - x^2 - x - 1 = \frac{x^5 - 2 x^4 + 1}{x - 1}$.  
Lagrange Inversion setup:  Let $f \left({x}\right) = 2\, {x}^{4} - {x}^{5} = 1 = z$.  Solve for $x$ about the point ${x}_{0} = 0$.  Thus $f \left({{x}_{0}}\right) = 0$.  Expanding via Newton's rule we have
$$
\left[{\frac{w - {x}_{0}}
{f \left({w}\right) - f \left({{x}_{0}}\right)}}\right]^{n} =
\left[{\frac{w}{2\, {w}^{4} - {w}^{5}}}\right]^{n} =
\sum_{k = 0}^{\infty} \left({\frac{1}{2}}\right)^{n + k} \binom{n + k - 1}{k} {w}^{k - 3 n}
$$
The $\left({n - 1}\right)$th derivative of ${w}^{k - 3 n}$ is 
$$
\frac{{d}^{n - 1}}{d\, {w}^{n - 1}}\, {w}^{k - 3 n} = 
\left({n - 1}\right)! \binom{k - 3 n}{n - 1} {w}^{k - 4 n + 1}.
$$
Thus,
$$
\frac{{d}^{n - 1}}{d\, {w}^{n - 1}} \left[{\frac{w - {x}_{0}} {f \left({w}\right) - f \left({{x}_{0}}\right)}}\right]^{n} =
\left({n - 1}\right)! \sum_{k = 0}^{\infty} \left({\frac{1}{2}}\right)^{n + k} \binom{k - 3 n}{n - 1} \binom{n + k - 1}{k} {w}^{k - 4\, n + 1}
$$
then we find that only the terms where $k = - 1 + 4 n$ survive or 
$$
\frac{\left({z - {x}_{0}}\right)^{n}}{n!} \lim_{w \rightarrow 0} \frac{{d}^{n - 1}}{d\, {w}^{n - 1}} \left[{\frac{w - {x}_{0}} {f \left({w}\right) - f \left({{x}_{0}}\right)}}\right]^{n} =
\frac{1}{n} \frac{2}{{2}^{5 n}} \binom{5 n - 2}{4 n - 1}.
$$
Together we have
$$
x = 2 \sum_{n = 1}^{\infty} \frac{1}{n} \frac{1}{{2}^{5 n}} \binom{5 n - 2}{4 n - 1}.
$$
Hence, the final solution
$$
x = \frac{8}{5} - \frac{8}{5}\,
{}_4{F}_{3} \left({\left\{{- \frac{1}{5}, \frac{1}{5}, \frac{2}{5}, \frac{3}{5}}\right\}, \left\{{\frac{1}{4}, \frac{1}{2}, \frac{3}{4}}\right\}, \frac{3125}{8192}}\right) = 0.724380245...
$$
However the correct solution is 
$$
x = \frac{2}{5} + \frac{8}{5}\,
{}_4{F}_{3} \left({\left\{{- \frac{1}{5}, \frac{1}{5}, \frac{2}{5}, \frac{3}{5}}\right\}, \left\{{\frac{1}{4}, \frac{1}{2}, \frac{3}{4}}\right\}, \frac{3125}{8192}}\right) = 1.92756197...
$$
Q  Why am I getting this incorrect solution?  This solution is not any of the roots.
 [/INST] You've chosen $w = 2 x^4 - x^5$, so either the inverse function is given by a Puiseux series in powers of $w^{1/4}$ or the center of the expansion is not $(0, 0)$. In the first case,
$$a_k = \lim_{x \to 0^+} \frac 1 {k!} \frac {d^{k - 1}} {d x^{k - 1}}
 \left( \frac x {(2 x^4 - x^5)^{1/4}} \right)^{\!k} =
\frac {(-1)^{k - 1} \, 2^{1 - 5 k/4}} {k!}
 \left( 2 -\frac {5 k} 4 \right)_{\!k - 1}, \\
x = \sum_{k \geq 1} a_k w^{k/4}.$$
This gives the four roots which behave like the branches of $(w/2)^{1/4}$ at zero. In the second case,
$$a_k = \lim_{x \to 2} \frac 1 {k!} \frac {d^{k - 1}} {d x^{k - 1}}
 \left( \frac {x - 2} {2 x^4 - x^5} \right)^{\!k} =
\frac {(-1)^k \, 2^{1 - 5 k}} {k!} (2 - 5 k)_{k - 1}, \\
x = 2 + \sum_{k \geq 1} a_k w^k,$$
which is regular at zero.
</s>