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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.

Francesco Sollazzi
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  • 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. – Dietrich Burde Sep 11 '21 at 18:15
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    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... – Jean Marie Sep 11 '21 at 19:08
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    Who is the person who has proposed to close this perfectly written question attesting personal work ? – Jean Marie Sep 11 '21 at 19:12
  • @DietrichBurde yeah, I know that. Not all equations don't have general solutions, though, and I was seeking help because of this. – Francesco Sollazzi Sep 11 '21 at 19:54
  • @JeanMarie thanks, I will try this tomorrow and I'll keep you updated. – Francesco Sollazzi Sep 11 '21 at 19:55
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    This should help, if you're interested in series solutions that are expressible in terms of hypergeometric functions. – K. Grammatikos Sep 11 '21 at 20:56
  • What do you mean by "solve"? As you point out in the question, for polynomials of degree 5 or higher, there is no general formula for solutions in terms of radicals. Are you looking for solutions in terms of some other family of functions? numerical solutions (e.g. those found via an approximation technique, such as Newton-Raphson)? What kind of solution are you looking for? – Xander Henderson Sep 11 '21 at 22:17
  • @JeanMarie There are reasons to close questions other than "this is a do-my-homework question which shows no effort". The close voter is Dietrich Burde, who voted to close the question as a duplicate, which is indicated in the very first comment under the question. This close reason has nothing to do with how well the question is written or whether or not it "shows work". – Xander Henderson Sep 11 '21 at 22:20
  • Is there a more general problem that led to the question involving $P_n$? [I note, for example, that it is often possible to exactly count the number roots of a polynomial within a fixed interval or region of the complex plane without anything resembling "exact formulas" for the roots] – leslie townes Sep 11 '21 at 22:20
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    In the particular case $t=5$ see [Bring's radical] (https://en.wikipedia.org/wiki/Bring_radical) – Jean Marie Sep 12 '21 at 07:15
  • @Xander Henderson 1) Thanks for clarifying, I didn't thought about this reason 2) But the question addressed by Dietrich Burde are very general : I disagree with the fact that this one is a duplicate of that one. – Jean Marie Sep 12 '21 at 07:23
  • @XanderHanderson I'm looking for analytic solutions. I think that using hypergeometric functions as DinosaurEgg pointed out would be perfect. – Francesco Sollazzi Sep 12 '21 at 07:26
  • @DinosaurEgg thanks to you I managed to solve my problem. I wrote the answer below. – Francesco Sollazzi Sep 12 '21 at 13:06
  • @FrankSoll You may want to take a look at my answer. – jjagmath Sep 12 '21 at 13:28
  • A question having a lateral connection with yours : https://math.stackexchange.com/q/161530 – Jean Marie Sep 14 '21 at 15:34

2 Answers2

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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

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

jjagmath
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4

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.

Francesco Sollazzi
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