How to solve the equation $1-2x-x^t=0$

Question

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

$$P_n(x)=1-\sum_{k=1}^{n}x^k=0$$

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

$$P_n(x)=2-\frac{1-x^{n+1}}{1-x}$$

So we get

$$P_n(x)=\frac{1-2x+x^{n+1}}{1-x}$$

Finally, we have the equation

$$1-2x+x^{n+1}=0$$

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.

Thanks.

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.

Answer 1

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

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

Answer 2

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.