1
$\begingroup$

I've generated a set of univariate polynomials ($b=1,2,\ldots$) in $v$ of degree $b-1$. The constant term and the coefficient of $v^{b-1}$ is simply $H_b$, the $b$-th harmonic number.

The coefficients of $v$ and of $v^{b-2}$ are both \begin{equation} \left(b^2+1\right) H_b-b (b+1) = b^2 H_b-b^2+H_b-b, \end{equation} while the coefficients of $v^2$ and $v^{b-3}$ are both \begin{equation} \frac{1}{8} \left(2 \left(((b-2) b+5) b^2+4\right) H_b-b (b+1) (b (3 b-5)+10)\right)= \end{equation} \begin{equation} \frac{b^4 H_b}{4}-\frac{3 b^4}{8}-\frac{b^3 H_b}{2}+\frac{b^3}{4}+\frac{5 b^2 H_b}{4}-\frac{5 b^2}{8}+H_b-\frac{5 b}{4}. \end{equation} Additionally, the coefficients of $v^3$ and $v^{b-4}$ are both \begin{equation} \frac{1}{216} \left(6 \left((b (b ((b-6) b+22)-30)+49) b^2+36\right) H_b-b (b+1) (b (b (b (11 b-59)+193)-223)+294)\right)= \end{equation} \begin{equation} \frac{b^6 H_b}{36}-\frac{11 b^6}{216}-\frac{b^5 H_b}{6}+\frac{2 b^5}{9}+\frac{11 b^4 H_b}{18}-\frac{67 b^4}{108}-\frac{5 b^3 H_b}{6}+\frac{5 b^3}{36}+\frac{49 b^2 H_b}{36}-\frac{71 b^2}{216}+H_b-\frac{49 b}{36}. \end{equation} Further still, the coefficients of $v^4$ and $v^{b-5}$ are both (for $b>3$) \begin{equation} \frac{(b-3)^2 (b-2)^2 (b-1)^2 b^2 \left(12 H_{b-4}-25\right)+192 \left((b (b ((b-6) b+22)-30)+49) b^2+36\right) H_b-32 (b+1) (b (b (b (11 b-59)+193)-223)+294) b}{6912}= \end{equation} \begin{equation} \frac{1}{576} b^8 H_{b-4}-\frac{25 b^8}{6912}-\frac{1}{48} b^7 H_{b-4}+\frac{25 b^7}{576}+\frac{29}{288} b^6 H_{b-4}+\frac{b^6 H_b}{36}-\frac{901 b^6}{3456}-\frac{1}{4} b^5 H_{b-4}-\frac{b^5 H_b}{6}+\frac{107 b^5}{144}+\frac{193}{576} b^4 H_{b-4}+\frac{11 b^4 H_b}{18}-\frac{9113 b^4}{6912}-\frac{11}{48} b^3 H_{b-4}-\frac{5 b^3 H_b}{6}+\frac{355 b^3}{576}+\frac{1}{16} b^2 H_{b-4}+\frac{49 b^2 H_b}{36}-\frac{793 b^2}{1728}+H_b-\frac{49 b}{36}. \end{equation} Note the appearance now of $H_{b-4}$. (The coefficients are zero for $b=1,2,3,4$. Subtracting the expression for the previous coefficients for $v^3$ and $v^{b-4}$ aided our computation.)

These last coefficients (for $v^4$ and $v^{b-5}$) are equal to those for $v^3$ and $v^{b-4}$ with the addition of simply \begin{equation} \frac{(b-3)^2 (b-2)^2 (b-1)^2 b^2 \left(12 H_{b-4}-25\right)}{6912}. \end{equation}

Similarly, the coefficients for $v^5$ and $v^{b-6}$ are equal to those for the immediately previous pair (for $v^4$ and $v^{b-5}$) with the addition of \begin{equation} \frac{(b-4)^2 (b-3)^2 (b-2)^2 (b-1)^2 b^2 \left(60 H_{b-5}-137\right)}{864000}. \end{equation}

(So, a clear pattern for the generation of the next higher-order coefficients from the immediately preceding [lower] ones appears to be emerging.)

Next, in this emerging series, we have \begin{equation} \frac{(b-5)^2 (b-4)^2 (b-3)^2 (b-2)^2 (b-1)^2 b^2 \left(20 H_{b-6}-49\right)}{10368000}, \end{equation} and, then, \begin{equation} \frac{(b-6)^2 (b-5)^2 (b-4)^2 (b-3)^2 (b-2)^2 (b-1)^2 b^2 \left(140 H_{b-7}-363\right)}{3556224000}, \end{equation} followed by \begin{equation} \frac{(b-7)^2 (b-6)^2 (b-5)^2 (b-4)^2 (b-3)^2 (b-2)^2 (b-1)^2 b^2 \left(280 H_{b-8}-761\right)}{455196672000}, \end{equation} and also \begin{equation} \frac{(b-8)^2 (b-7)^2 (b-6)^2 (b-5)^2 (b-4)^2 (b-3)^2 (b-2)^2 (b-1)^2 b^2 \left(2520 H_{b-9}-7129\right)}{331838373888000}, \end{equation} and next, \begin{equation} \frac{(b-9)^2 (b-8)^2 (b-7)^2 (b-6)^2 (b-5)^2 (b-4)^2 (b-3)^2 (b-2)^2 (b-1)^2 b^2 \left(2520 H_{b-10}-7381\right)}{33183837388800000}. \end{equation} (The commonalities between the last two formulas are certainly intriguing--and will require futher verification.)

Then, next \begin{equation} \frac{(b-10)^2 (b-9)^2 (b-8)^2 (b-7)^2 (b-6)^2 (b-5)^2 (b-4)^2 (b-3)^2 (b-2)^2 (b-1)^2 b^2 \left(27720 H_{b-11}-83711\right)}{44167687564492800000}. \end{equation}

Does one know of any identifiable class of polynomials, in which these might lie? Alternatively, can one identify a generating mechanism for these coefficients?

These polynomials have arisen in my attempt to find a formula for the function $v(b,\mu)$ described in the first ("partial/half") answer to https://mathoverflow.net/questions/322958/compute-the-two-fold-partial-integral-where-the-three-fold-full-integral-is-kno

$\endgroup$
0
$\begingroup$

You can check out the OEIS for some insight. For instance, two sequences appearing in your expressions are related to Harmonic numbers (no surprise there):

25, 137, 49, 363, 761, 7129, 7381, 83711, 86021 are the numerators of Harmonic Numbers $H_{3},H_4,...$

12, 60, 20, 140, 280 are the denominators.

This means you should factor by the denominators (12,60, etc) and you have the difference of two harmonic numbers in the rightmost term. Then lead with the denominator.


Also, don't hesitate to read the comments in OEIS below each sequence, you can find connections to other sequences or problems, and though OEIS does not give you a formal proof (as the next terms of your polynomials may be 0,-1,$\pi,\pi,\pi$), these comments can certainly help.

$\endgroup$
0
$\begingroup$

The terms occurring in the first two specific differences of successive coefficients in the statement of the question, \begin{equation} \frac{\left(12 H_{b-4}-25\right)}{6912}, \end{equation} and \begin{equation} \frac{ \left(60 H_{b-5}-137\right)}{864000} \end{equation} are given by substituting $i=5,6$ into \begin{equation} \frac{\psi ^{(0)}(b-i+2)-\psi ^{(0)}(i)}{\Gamma (i)^2}, \end{equation} where the polygamma and gamma functions are indicated.

This pattern continues to hold $(i=7,\ldots,12)$ for the subsequent differences of successive coefficients presented in the question, and for those for $i>12$, as well.

The additional multiplicative factors in these coefficient difference expressions are given by \begin{equation} b^2 \left((1-b)_{i-2}\right){}^2, \end{equation} where the Pochhammer symbol is indicated.

For $i=5, 6$, we have $(1-b)^2 (2-b)^2 (3-b)^2 b^2$ and $(1-b)^2 (2-b)^2 (3-b)^2 (4-b)^2 b^2$, further agreeing with the expressions given in the question.

So, assembling these two pieces of information, we have \begin{equation} \frac{b^2 \left((1-b)_{i-2}\right){}^2 (\psi ^{(0)}(b-i+2)-\psi ^{(0)}(i))}{\Gamma (i)^2}, \end{equation} for the complete difference expressions given in the question.

Then, we can then obtain the coefficients themselves and employ them in a general formula for the $b$-th polynomial ($b>1$), \begin{equation} \sum _{k=2}^{b-1} v^{k-1} \sum _{i=1}^k \frac{b^2 \left((1-b)_{i-2}\right){}^2 (\psi ^{(0)}(b-i+2)-\psi ^{(0)}(i))}{\Gamma (i)^2}+H_b \left(v^{b-1}+1\right)= \end{equation} \begin{equation} H_b \left(\sum _{k=2}^{b-1} v^{k-1} \sum _{i=1}^k \frac{b^2 \left((1-b)_{i-2}\right){}^2 (\psi ^{(0)}(b-i+2)-\psi ^{(0)}(i))}{H_b \Gamma (i)^2}+v^{b-1}+1\right). \end{equation} For even $b$, we have \begin{equation} \sum _{k=2}^{\frac{b}{2}} \left(v^{b-k}+v^{k-1}\right) \sum _{i=1}^k \frac{b^2 \left((1-b)_{i-2}\right){}^2 (\psi ^{(0)}(b-i+2)-\psi ^{(0)}(i))}{\Gamma (i)^2}+H_b \left(v^{b-1}+1\right). \end{equation}

Now, can equivalent ("closed", hypergeometric?) formulas for these polynomials without embedded summations be found? This would presumably facilitate our analysis of the function $v(b,\mu)$ in the original motivating question https://mathoverflow.net/questions/322958/compute-the-two-fold-partial-integral-where-the-three-fold-full-integral-is-kno . Its companion function there, $w(b,\mu)$, is, in fact, now known to be hypergeometric in nature.

Let us note that of possible interest in this quest, are the identities \begin{equation} \frac{b^2 \left((1-b)_{i-2}\right){}^2}{\Gamma (i)^2}=\binom{b}{i-1}^2= \frac{b^2 \binom{b-1}{i-2}^2}{(i-1)^2}, \end{equation} as well as that \begin{equation} (\psi ^{(0)}(b-i+2)-\psi ^{(0)}(i)) =\left(H_{b-i+1}-H_{i-1}\right). \end{equation}

Let us also observe that a more concise form of the $b$-th polynomial (successfully now producing 1 for $b=1$) is \begin{equation} \sum _{k=1}^b v^{k-1} \sum _{i=0}^{k-1} \binom{b}{i}^2 (\psi ^{(0)}(b-i+1)-\psi ^{(0)}(i+1)). \end{equation}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.