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I have been searching the Inverse Symbolic Calculator and then the OEIS and came up with this little program by copy pasting parts here and there:

Let $P$ be the polynomial:

$$P = a_n x^n + a_{n-1}x^{n-1} + \dotsb + a_2 x^2 + a_1 x + a_0$$

Then do the series expansion of:

$$\frac{1}{P}$$

at $x = 0$

and name the coefficients $b_1,...,b_\infty$

and take the limiting ratio:

$$x=\lim_{n\to \infty } \, \frac{b_{n-1}}{b_n}$$

For what kind of polynomials is $x$ a real root to the polynomial $P$?

Does it have to do with Lagrange inversion? I don't know Lagrange inversion.

(*Mathematica program*)
Clear[x, b];
polynomial = (1 - 2 x + 3*x^2 - 5 x^3 + 7 x^4 - 11 x^5);
digits = 100;
nn = 4000;
b = CoefficientList[Series[1/polynomial, {x, 0, nn}], x] ;
nn = Length[b];
x = N[Table[b[[n - 1]]/b[[n]], {n, nn - 8, nn - 1}], digits]
polynomial

Edit 12.7.2020:

Appears to work for zeta zeros:
(*start*)
Clear[t, b, n, k, nn, x];
"z is the parameter to vary"
z = 16
digits = 40;
polynomial = Normal[Series[Zeta[(x + z*N[I, digits])], {x, 0, 50}]];
nn = 200;
b = CoefficientList[Series[1/polynomial, {x, 0, nn}], x];
x = z*I + N[b[[nn - 1]]/b[[nn]], digits]
(*end*)

Input z = 16 gives output:
x=0.500000000000000000000000 + 14.134725141734693790457252*I

Mathematica program for plot of Imaginary and Real part of Riemann zeta zeros:

(*start apparently equivalent to Newton Raphson*)cons = 10;
ww = 400;
div = 10;
real = 0;
Monitor[TableForm[zz = Table[Clear[t, b, n, k, nn, x];
     z = N[cons + w/div, 20];
     polynomial = 
      Normal[Series[Zeta[(real + x + z*N[I, 20])], {x, 0, 10}]];
     digits = 20;
     b = With[{nn = 20}, 
       CoefficientList[Series[1/polynomial, {x, 0, nn}], x]];
     nn = Length[b] - 1;
     x = z*I + N[b[[nn - 1]]/b[[nn]], digits], {w, 0, ww}]];, w]
g1 = ListLinePlot[Flatten[Im[zz]], DataRange -> {cons, cons + ww/div}]
g2 = ListLinePlot[Flatten[Re[zz]], DataRange -> {cons, cons + ww/div},
   PlotRange -> {-2, 2}]
zz
(*end apparently equivalent to Newton Raphson*)

plus one half

From the plot above we see that the staircase takes the values of the imaginary parts of the Riemann zeta zeros, and that the lower plot takes the value $\frac{1}{2}$ which is the real part of the Riemann zeta zeros, except at what appears to be Gram points where there are singularities.

One can see that this appears to be true regardless of the value 'real' in the program, as long as 'real' is between $0$ and $1$.

The recurrence pointed out by Conrad:

Clear[f, n, k, a];
a = {1, -1, -2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
f[0] = 1;
f[n_] := -Sum[f[k]*Binomial[n, k]*a[[n - k + 1]], {k, 0, n - 1}]
Table[f[n - 1]/(n - 1)!, {n, 1, 14}]

Taken from Daniel Suteu's comment: https://oeis.org/A132096

Clear[f, n, k, a, b];
f[0] = 1;
f[n_] := -Sum[ff[k]*bin[n, k]*a[n - k + 1], {k, 0, n - 1}]
TableForm[Table[f[n - 1], {n, 1, 10}]]

$$\begin{array}{l} 1 \\ -a[2] b[1,0] \text{ff}[0] \\ -a[3] b[2,0] \text{ff}[0]-a[2] b[2,1] \text{ff}[1] \\ -a[4] b[3,0] \text{ff}[0]-a[3] b[3,1] \text{ff}[1]-a[2] b[3,2] \text{ff}[2] \\ -a[5] b[4,0] \text{ff}[0]-a[4] b[4,1] \text{ff}[1]-a[3] b[4,2] \text{ff}[2]-a[2] b[4,3] \text{ff}[3] \\ -a[6] b[5,0] \text{ff}[0]-a[5] b[5,1] \text{ff}[1]-a[4] b[5,2] \text{ff}[2]-a[3] b[5,3] \text{ff}[3]-a[2] b[5,4] \text{ff}[4] \end{array}$$

$b$ stands for binomial $a$ is the sequence of coefficients multiplied with the factorials.

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  • $\begingroup$ Posted previously on MathOverflow: mathoverflow.net/q/364186/25104 $\endgroup$ – Mats Granvik Jun 26 at 15:21
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    $\begingroup$ why does that limit exist? eg $\frac{1}{1-x^2}=\sum x^{2k}$ and the limit obviously doesn't exist here; if we know that the limit $l$ exists the result is easy to prove since assuming wlog $a_0=1$ and writing the recurrence for $b_N$ and dividing by $b_{N-1}$ and taking the limit (again assumed to exist) we get $1/l=-a_1-a_2l-...-a_nl^{n-1}$ which obviously gives $P(l)=0$ $\endgroup$ – Conrad Jun 26 at 17:04
  • $\begingroup$ Limiting ratios vs Newton Raphson iteration: pastebin.com/ixQxnpjC $\endgroup$ – Mats Granvik Jul 12 at 17:08
  • $\begingroup$ It would be interesting to do the same for the shifted normalized Landau-Riemann xi function $\xi(1/2+it)/\xi(1/2)$, an entire even function with only real zeros discovered so far. $\endgroup$ – Tom Copeland Aug 1 at 19:09
  • $\begingroup$ @TomCopeland Is that the xi function defined as: $\xi(s) = \frac{1}{2} s(s-1) \pi^{-s/2} \Gamma\left(\frac{s}{2}\right) \zeta(s)$? I can send you the code tomorrow. $\endgroup$ – Mats Granvik Aug 1 at 19:15
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Provided that $L=\lim_{n\to\infty}b_n/b_{n+1}$ exists, then this easily follows from the recurrence relation that $b_n$ satisfies.

Note that we have

$$\frac1{P(x)}=\sum_{n=0}^\infty b_kx^k\implies1=P(x)\sum_{k=0}^\infty b_kx^k$$

When expanding and collecting terms, one then gets the linear recurrence

$$0=a_nb_{k-n}+a_{n-1}b_{k-n+1}+\dots+a_1b_{k-1}+a_0b_k$$

with known solution and dominating term given by $b_k\sim Q(k)\lambda^k$ for polynomial $Q$ and $P(1/\lambda)=0$. Substituting this in, one then gets that $L=1/\lambda=\lim_{n\to\infty}b_n/b_{n+1}$.

If there are several dominating terms i.e. several roots of equal magnitude and non-zero coefficients in the expansion of $b_k$, then the limit will not converge.

| cite | improve this answer | |
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  • $\begingroup$ Why so quick to upvote? Surely someone couldn't have read this answer within 5 seconds of my posting this... $\endgroup$ – Simply Beautiful Art Jun 26 at 18:11
  • $\begingroup$ I did read the answer that quickly, or at least enough to appreciate that it was correct :) I was writing my own that was similar but you beat me to the punch. $\endgroup$ – Jair Taylor Jun 26 at 18:13
  • $\begingroup$ @JairTaylor fair enough then I suppose :p $\endgroup$ – Simply Beautiful Art Jun 26 at 18:22
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    $\begingroup$ See also oeis.org/A263633. $\endgroup$ – Tom Copeland Jun 26 at 18:32
  • $\begingroup$ @SimplyBeautifulArt Does this answer also explain why we by the same method get Riemann zeta zeros in the critical strip regardless of starting point, except at Gram points and t<10, within the critical strip? $\endgroup$ – Mats Granvik Jul 26 at 10:09

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