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The sequence of polynomial equations $$ (a+n)x^n+(b-n)x^{n-1}+f(n)=0 $$

has real roots near 1 having an asymptotic series $$ x_n=1+\frac{y_n-a-b}{n}+O(y_n^2n^{-2}), $$ where $$ y_n=W(-e^{a+b}f(n)). $$ Here $W$ is the Lambert function.

Now, I want to apply this result to the special sequence of polynomials $$ x^n-x^{n-1}-1=0. $$ So I guess I have to choose $$ a=1-n,\qquad b=n-1,\qquad f(n)=-1. $$

Then I should have by above $$ x_n=1+\frac{y_n}{n}+O(y_n^2n^{-2}) $$ with $y_n=W(1)$.

On the other hand, I know that $$ x_n\sim1+\frac{\log n}{n}. $$

Do these two results fit together? Maybe $y_n=W(1)$ is equivalent to $y_n=\log(n)$?

Rhjg
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  • According to Wolfram Alpha, the difference $W(1) - \log(n)$ grows as $n$ increases. That difference does have zero near $n = 2$ where you could call them approximately equal, I guess. As $n$ gets large, both $\frac{W(1)}{n}$ and $\frac{\log(n)}{n}$ get squashed to 0 anyway. – Andy Walls Jan 28 '18 at 12:52
  • If you want to try something different than the Lambert function approach, you may be able to modify Glasser's approach to finding the roots of $x^n-x-t=0$, https://www.sciencedirect.com/science/article/pii/S0377042700002879 . But it might not work out so well. I'll note that your polynomial can be written as $x^{n-1}(x-1) - 1 = 0$, which is kind of interesting, in that the roots are close to the $n^{th}$ roots of unity. – Andy Walls Jan 28 '18 at 13:06
  • The Lambert function approach would be faster and simpler, since you can precompute $W(1)$. That is assuming you only want the real root near 1. – Andy Walls Jan 28 '18 at 13:46
  • I think $x_n$ is the largest positive real root, isn’t it? – Rhjg Jan 28 '18 at 15:30
  • Yes. The roots are arranged in nearly a circle in the complex plane, close to the $n^{th}$ roots of unity. The root near 1 is the largest real root. Wolfram Alpha gives you a nice plot. Here's the case for $n=8$ for example: https://www.wolframalpha.com/input/?i=x%5E7(x-1)-1%3D0 – Andy Walls Jan 28 '18 at 16:11

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