I'm trying to solve the equation $1-2x-x^t=0$ where $t\in\mathbb{N}:t>1$, and I really can't think about any approach that can help me out.

Actually, the original equation was


and, since the sum is a truncated geometric series, then


So we get


Finally, we have the equation


with $x \neq 0$.

Now here I'm stuck. Is there a method for solving this?

I don't really know if this approach is correct for solving the original equation. And I know that the solutions are not trivial. I don't want to use numerical methods, and I need all the solutions, including the complex ones, in terms of series or other analytic functions, if that's possible.


EDIT: I know that there's no general solution for a general equation of grade $>4$ in terms of $+ - \cdot ÷ $ and roots. This does not necessarily mean that one can't solve it though, for example through a power series.

  • $\begingroup$ For $t\le 4$ there are formulas for the solutions in terms of radicals. For $t\ge 5$ we know from Galois that there need not exist such a solution in general. Just a few minutes ago the same question was posed, with slightly different coefficients. $\endgroup$ Sep 11, 2021 at 18:15
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    $\begingroup$ For real solutions : consider the abscissas of common points to the curve $y=x^t$ and the straight line $y=2x-1$ ; you will find in general 2 (real) solutions for even values of $t$ and 3 (real) solutions for odd values of $t$. The other ones are complex... $\endgroup$
    – Jean Marie
    Sep 11, 2021 at 19:08
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    $\begingroup$ Who is the person who has proposed to close this perfectly written question attesting personal work ? $\endgroup$
    – Jean Marie
    Sep 11, 2021 at 19:12
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    $\begingroup$ This should help, if you're interested in series solutions that are expressible in terms of hypergeometric functions. $\endgroup$ Sep 11, 2021 at 20:56
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    $\begingroup$ In the particular case $t=5$ see [Bring's radical] (en.wikipedia.org/wiki/Bring_radical) $\endgroup$
    – Jean Marie
    Sep 12, 2021 at 7:15

2 Answers 2


You may use the Lagrange inversion theorem to get the analytic solution, after some serious simplification we get that the positive solution of $1-2x-x^t=0$ is given by

$$\sum_{k=0}^\infty \binom{t\,k}{k}\frac{(-1)^k}{(t-1)k+1}\,\frac{1}{2^{t\,k+1}}$$

  • $\begingroup$ Yes, exactly what I used. I prefer accepting your answer instead of mine, since I should review it. $\endgroup$ Sep 12, 2021 at 18:02
  • $\begingroup$ It shouldn't be that difficult to find the other solutions too though. $\endgroup$ Sep 12, 2021 at 18:03
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    $\begingroup$ @FrankSoll I see you also used Lagrange inversion theorem in your answer, but after some complicated transformations. I arrive at my solution applying it directly to $f(x) = x^t +2x$. $\endgroup$
    – jjagmath
    Sep 12, 2021 at 20:08
  • $\begingroup$ hm. In any case now I know how to do it. Thanks :) $\endgroup$ Sep 12, 2021 at 20:33

I managed to solve my own problem in the end, so I'm answering my own question.

Thanks to @DinosaurEgg who suggested taking a look at an interesting paper which helped me a lot.

The equation was $1-2x-x^t=0$ and multiplying both sides by $-1$ we get $x^t+2x-1=0$. Now we shall apply the substitution $x=\Omega^{1/(t-1)}$ to get:

$$\Omega^{\frac{t}{t-1}}+2\Omega^{\frac{1}{t-1}}=1$$ $$\Omega^{\frac{1}{t-1}}\left(\Omega+2\right)=1$$ $$\Omega=\Omega^{-\frac{1}{t-1}}-2$$

Now we let $\phi(\Omega)=\Omega^{-\frac{1}{t-1}}$ so

$$\Omega=\phi(\Omega)-2e^{2\pi i}$$

Lagrange’s inversion theorem states that for any function f analytic in a neighbourhood of a root the equation above.

$$f(\Omega)=f(-2e^{2\pi i})+\sum_{n=1}^{\infty}\frac{1}{n!}\frac{d^{n-1}}{d\alpha^{n-1}}\left[\phi(\alpha)f'(\alpha)\right]$$

where $\alpha=-2e^{-2\pi i}$.

Since $D_kx^p=\frac{\Gamma(p+1)x^{p-k}}{\Gamma(p-k+1)}$ if we let $f(\Omega)=x$ we get our first solution with:

$$x_1=-2e^{\frac{-2\pi i}{t-1}}+\frac{1}{t-1}\sum_{n=1}^{\infty}\frac{(-2)^ne^{2\pi i n}}{n!}\frac{\Gamma(-\frac{1}{t-1}+1)}{\Gamma(-\frac{1}{t-1}-n+1)}$$

Of course, you can then find the other roots, or simplify the formula. You can also transform the infinite sum into a finite one using Gauss' multiplication theorem, having a solution in terms of a hypergeometric function.

I may have left behind one or two mistakes writing the answer, and I'm checking, but basically, this is the approach I was looking for.


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