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A well known infinite series for the solution of the equation $x^p-x+t=0$ for $p>1$ is: $$x=\sum_{k=0}^{\infty}\binom{pk}{k}\frac{t^{(p-1)k+1}}{(p-1)k+1}$$ Which is well defined in the region $0\le t\le (p-1)p^{-p/(p-1)}$. This series when $t=0$ returns $$x=0$$ But the original equation $$x^p-x=0$$ Has another real root $$x=1$$ I wonder if there's a series expansion for $x(t)$ that takes $x(0)=1$ and is defined in the region $0\le t\le (p-1)p^{-p/(p-1)}$ and for some other region $t<0$.

I asked this question because I'm trying to solve for this probability question.

What I've tried: $$x=t+x^p$$ $$x-1=t+(x-1+1)^p-1$$ Using Lagrange reversion: $$x-1=t+\sum_{k=1}^{\infty}\frac{1}{k!}\frac{d^{k-1}}{dz^{k-1}}\left((z+1)^p-1\right)^k|_{z=t}$$ $$x=t+1+\sum_{k=1}^{\infty}\frac{1}{k!}\frac{d^{k-1}}{dz^{k-1}}\left(z^p-1\right)^k|_{z=t+1}$$ However things get complicated because I have to consider when $p$ is rational, and when $p$ is irrational.

Anyone? Any idea?

Thinh Dinh
  • 8,233

1 Answers1

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Let $x^{p-1}=w$ and apply the reversion theorem:

$$x^p-x+t=0\iff x^{p-1}-1=-tx^{-1}\iff w=1-t w^{-\frac1{p-1}}\\\implies x=w^\frac1{p-1}=1+\sum_{n=1}^\infty\frac{(-t)^n}{n!}\left.\frac{d^{n-1}}{dz^{n-1}}\left(\left(z^\frac1{p-1}\right)’z^{-\frac n{p-1}}\right)\right|_1$$

Then, differentiating and evaluating at $z=1$ gives the following using factorial power $x^{(v)}$:

$$\boxed{x^p-x+t=0\implies x=1+\frac1{p-1}\sum_{n=1}^\infty\frac{(-t)^n}{n!}\left(\frac{1-n}{p-1}-1\right)^{(n-1)}=1-\frac1{p-1}\sum_{n=1}^\infty\frac{t^n\Gamma\left(\frac{n-1}{p-1}+n\right)}{n!\Gamma\left(\frac{n-1}{p-1}+1\right)},|t|\le\left|(p-1)p^\frac p{1-p}\right|}$$

as shown here.

Тyma Gaidash
  • 13,576