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By searching the Inverse Symbolic Calculator, we appear to be able to make the following conjecture about a real root to the equation:

$$\sum\limits_{k=0}^d \frac{x^k a_{k+1}}{k!}=0 \tag{1}$$

Let the lower triangular matrix $A$ be:

$$A=\binom{n-1}{k-1} a_{n-k+1} \tag{2}$$

where $n=1,2,3,4,5,...N$ and $k=1,2,3,4,5,...N$, with $N>>d$, and where the parentheses is the binomial function.

Calculate the matrix inverse $$B=A^{-1} \tag{3}$$, consider the first column of matrix $B$:

$$b_n=B(n,1) \tag{4}$$

and take the limiting ratio:

$$x=\lim_{n\to \infty } \, \frac{(n-1) b_{n-1}}{b_n} \tag{5}$$

Under what conditions for the coefficients:

$$a_1,...,a_{d} \tag{6}$$

is the limiting ratio $x$ in $(5)$ a real root solution to $(1)$

$$\sum\limits_{k=0}^d \frac{x^k a_{k+1}}{k!}=0$$ ?

Is the conjecture true at all?

I apologize for not letting the index of $a$ begin with $0$ instead of $1$.

Here is the Mathematica program for the conjecture:

Clear[A, B, a, b, x];
a = {1, 3, 5, 8, 5, 41, 39, 57, 53, 47, 13, 19, 0, 0, 0, 0, 0, 0, 0, 
   0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
   0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
   0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
   0, 0, 0, 0, 0, 0, 0, 0, 0};
nn = Length[a];
d = Max[Flatten[Position[Sign[Abs[a]], 1]]]
A = Table[
   Table[If[n >= k, Binomial[n - 1, k - 1]*a[[n - k + 1]], 0], {k, 1, 
     Length[a]}], {n, 1, Length[a]}];
b = Inverse[A][[All, 1]];
x = N[Table[(n - 1)*b[[n - 1]]/b[[n]], {n, nn - 8, nn - 1}], 30]
Sum[a[[k + 1]]/k!*x^k, {k, 0, d}]
Clear[x];
Sum[a[[k + 1]]/k!*x^k, {k, 0, d}]

The limiting ratio output from the program is: $$x=-0.474390307209018254579812222047$$

and this appears to be a solution to:

$$1+3 x+\frac{5 x^2}{2}+\frac{4 x^3}{3}+\frac{5 x^4}{24}+\frac{41 x^5}{120}+\frac{13 x^6}{240}+\frac{19 x^7}{1680}+\frac{53 x^8}{40320}+\frac{47 x^9}{362880}+\frac{13 x^{10}}{3628800}+\frac{19 x^{11}}{39916800}=0$$

In the program one needs to have $N$ much bigger than $d$ in order to see the conjecture in the output. Therefore there are a lot of trailing zeros in the coefficients $a$. You can add more zeros manually to the vector $a$ if you want to.

I don't know how to tag this question properly.

A related proof by joriki:
https://math.stackexchange.com/a/60385/8530
OEIS entry:
https://oeis.org/A167196
Related limiting ratio:
https://oeis.org/A132049


OEIS searches:
https://oeis.org/A322262
https://oeis.org/A006153

A much more conventional formulation of what I am doing in Mathematica:

Clear[x, b];
polynomial = (1 + 2*x + x^2/2! + x^3/3! + x^4/4! + x^5/5!);
digits = 100;
b = With[{nn = 200}, 
   CoefficientList[Series[1/polynomial, {x, 0, nn}], 
     x] Range[0, nn]!] ;
nn = Length[b] - 10;
x = N[Table[(n - 1)*b[[n - 1]]/b[[n]], {n, nn - 8, nn - 1}], digits]
polynomial

Thanks to Harvey Dale in the OEIS for how to write the part:

b = With[{nn = 200}, 
   CoefficientList[Series[1/polynomial, {x, 0, nn}], 
     x] Range[0, nn]!] ;

And thereby the polynomial need not have factorials:

Clear[x, b];
polynomial = (1 - 2 x + 3*x^2 - 5 x^3 + 7 x^4 - 11 x^5);
digits = 100;
b = With[{nn = 4000}, 
   CoefficientList[Series[1/polynomial, {x, 0, nn}], 
     x] Range[0, nn]!] ;
nn = Length[b] - 10;
x = N[Table[(n - 1)*b[[n - 1]]/b[[n]], {n, nn - 8, nn - 1}], digits]
polynomial

The following program uses the method in the question by first Taylor expanding the Riemann zeta function at $zi$ and real part equal to $0$, and then adding trailing zeros to the vector $a$. Expanding at real part equal to $1$ gives a similar plot.

(*start*)
start = 10;
end = 30;
Monitor[list = Table[zz = 10;
   d = 20;
   a = Flatten[{CoefficientList[
        Normal[Series[Zeta[x + z*N[I, d]], {x, 0, zz}]], x]*
       Range[0, zz]!, Range[d]*0}];
   nn = Length[a];
   A = Table[
     Table[If[n >= k, Binomial[n - 1, k - 1]*a[[n - k + 1]], 0], {k, 
       1, Length[a]}], {n, 1, Length[a]}];
   Quiet[b = Inverse[A][[All, 1]]];
   z*I + N[(nn - 1)*b[[nn - 1]]/b[[nn]], 40], {z, start, end, 1/10}], 
 z*10]
ListLinePlot[Re[list], PlotRange -> {-1, 3}, DataRange -> {start, end}]
ListLinePlot[Im[list], DataRange -> {start, end}]
(*end*)

The result is an approximation to the Riemann zeta zeros, where the plot of the real part stays around $\frac{1}{2}$ except at singularities at Gram points:

Riemann zeta zeros through Taylor polynomial approximation and its matrix inverse

The heights of the steps in the staircase in the second plot are at imaginary parts of the Riemann zeta zeros.

The plots below are the same as above but from $z=10$ to $z=60$:

real and imaginary part of Riemann zeta zeros 10 to 60

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(Extended comments in reply to Matt's comment to me.)

The answer to your MO question was provided in the MSE question couched in terms of polynomials expressed as truncated power series, or ordinary generating functions (o.g.f.s) and their reciprocals. Here you use truncated Taylor series, or exponential generating functions (e.g.f.s). The series for the reciprocals can be found several parallel ways. Three are 1) by recursion relations (as in the MSE-Q) related to regular or binomial convolution of the coefficients of the generating function with those of its reciprocal, generated by the product of the pair of g.f.s, which by construction is unity; 2) for e.g.f.s, through evaluation of the signed face partition polynomials of the of permutahedra (A133314), and, for o.g.f.s, by the signed, refined Pascal partition polynomials (A263633), encoding the Newton identity relating the symmetric elementary polynomials to complete symmetric homogeneous polynomials; and 3) by inverting a Pascal matrix diagonally multiplied by the Taylor coefficients as you do here. All three methods are explained in A133314 (see also my blog post "Skipping over Dimensions, Juggling Zeros in the Matrix," and others, in the companion OEIS entry.

Whenever you are looking at multiplicative inversion of g.f.s, Appell sequences are lurking nearby.

Interesting paper in your comment. if you throw in some binomial coefficients and normalize, the polynomial in eqn. 1 in the paper (any polynomial) can be re-expressed in umbral notation as $(e.+x)^n=E_n(x)=s(x)$ with lower order polynomials in the related Appell sequence obtained by differentiation or by noting $E_k(x)=(x+e.)^{k}$ for $k \leq n$. The paper seeks to construct zeros of this polynomial equation.

Eqn. 2 in the paper can be couched in terms of the umbral operational calculus. If you let $\frac{m_{n-k+1}}{m_n}=\frac{b_{k-1}}{(k-1)!}$ for $1 \leq k \leq n$ and $b_k=0$ for $k \geq n-1$ , then the eqn. becomes

$$e^{b.D_s}x(s)=x(b.+s)=x(B.(s)) = \frac{-m_{n+1}}{m_n}$$

where $D_s=d/ds$ and $B.(s)^n=B_n(s)= (b.+s)$. Clearly, $e^{b.D_x}$ is the diff op that generates the Appell polynomials when acting on $x^n$.

So, the analysis in the paper also seems closely related to Appell sequences, so perhaps there is something to gain from this perspective, but I don't have much time to pursue it at the moment.

In addition, a solution about $x=0$ to eqn. 1 in the paper can be approached through compositional inversion of the o.g.f. $s(x) = a./(1-a.x)$ through A133437 or A134264, but I'm not sure yet how this might relate to your results.

As I believe you already know, this Appell approach ties in with your interest in the Riemann zeta function through investigations of Jensen and later Polya (probably generating the initial interest in what became known as Appell polynomials). The Riemann hypothesis is true if and only if all the Appell polynomials formed from the Taylor series coefficients of the shifted, rotated, entire Landau-Riemann xi function $\xi(1/2+it)$ are hyberbolic, i.e., have only real zeros.

(Titchmarsh also investigated the zeros of the partial sums of Taylor series, no doubt due to the RH.)

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  • $\begingroup$ log(2) = A052882/A000670, Pi/4 = n*A001586(n-1)/A001586(n) as n-->infinity, and this oeis.org/A167196 were my starting points. I noticed that the coefficients of your list partition transform turned up in one of the series expansions. Also I have a typo in the oeis in the limiting ratio for square roots. $\endgroup$ – Mats Granvik Aug 2 at 10:49
  • $\begingroup$ I posted a new question: mathoverflow.net/a/368534/25104 $\endgroup$ – Mats Granvik Aug 7 at 4:37

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