TWO SOLUTIONS OF ONE PROBLEM OF B. S. POPOV

Kostadin G. Trenčevski and Ice B. Risteski

In this paper are given two solutions of the problem of Prof. B. S. Popov [1].

1. The Problem of Prof. B. S. Popov. In [1], Prof. Popov formulates the following problem.

Prove that the determinant D_{n} = |a_{ij}|, where

\begin{matrix}a_{ij} = 0, (j = i + s, s > 1); a_{ij} = 1, (j = i + 1); \\ a_{ij} = a, (i = j) and a_{ij} = 2i − 2, (j = i − 1), \end{matrix}

satisfies the following identities:

\begin{aligned}(i) D_{n} & = ∑_{k=0}^{⌊n/2⌋}(−1)^{k} · \frac{n!}{k!} · \frac{a^{n−2k}}{(n − 2k)!}, \\ (ii) D_{n}^{2} & = 2^{n} · n! · ∑_{k=0}^{n} \binom{n}{k} · \frac{D_{2k}}{2^{k} · k!}. \end{aligned}

2. First Solution of the Problem. The given determinant is

D_{n} = | \begin{matrix}a & 1 & 0 & ⋯ & 0 & 0 & 0 \\ 2 & a & 1 & ⋯ & 0 & 0 & 0 \\ 0 & 4 & a & ⋯ & 0 & 0 & 0 \\ · & · & · & ⋯ & · & · & · \\ 0 & 0 & 0 & ⋯ & a & 1 & 0 \\ 0 & 0 & 0 & ⋯ & 2n − 4 & a & 1 \\ 0 & 0 & 0 & ⋯ & 0 & 2n − 2 & a \end{matrix} | .

Developing it by the last column, we obtain the recurrent formula

(1)  D_{n} = aD_{n−1} − (2n − 2)D_{n−2}.

Note that D_{1} = a and D_{2} = a^{2} − 2, and putting D_{0} = 1 and D_{−1} = 0, we obtain that (1) holds for n ≥ 1. The proof of (i) and (ii) is by induction of n, and (i) will not be used in the proof of (ii).

\underline{Proof of (i)}. It is easy to verify that (i) holds for n = 1 and n = 2. Assume that (i) holds for the numbers n − 1 and n − 2, (n > 2). Then

D_{n} = aD_{n−1} − (2n − 2)D_{n−2} = a · ∑_{k=0}^{⌊(n−1)/2⌋}(−1)^{k} · \frac{(n − 1)!}{k!} · \frac{a^{n−2k−1}}{(n − 2k − 1)!}

− (2n − 2) · ∑_{k=0}^{⌊(n−2)/2⌋}(−1)^{k} · \frac{(n − 2)!}{k!} · \frac{a^{n−2k−2}}{(n − 2k − 2)!}

= ∑_{k=0}^{⌊(n−1)/2⌋}(−1)^{k} · \frac{(n − 1)!}{k!} · \frac{a^{n−2k}}{(n − 2k − 1)!}

− 2 · ∑_{k=0}^{⌊(n−2)/2⌋}(−1)^{k} · \frac{(n − 1)!}{k!} · \frac{a^{n−2k−2}}{(n − 2k − 2)!}

= ∑_{k=0}^{⌊(n−1)/2⌋}(−1)^{k} · \frac{(n − 1)!}{k!} · \frac{a^{n−2k}}{(n − 2k − 1)!}

+ 2 · ∑_{k=1}^{⌊n/2⌋}(−1)^{k−1} · \frac{(n − 1)!}{(k − 1)!} · \frac{a^{n−2k}}{(n − 2k)!}

= ∑_{k=0}^{⌊n/2⌋}(−1)^{k} · \frac{(n − 1)!}{k!} · \frac{a^{n−2k}}{(n − 2k)!}[(n − 2k) + 2k]

= ∑_{k=0}^{⌊n/2⌋}(−1)^{k} · \frac{n!}{k!} · \frac{a^{n−2k}}{(n − 2k)!},

and (i) is proved.

\underline{Proof of (ii)}. In order to prove (ii) by induction of n, D_{n}^{2} should be expressed by D_{n−1}^{2}, D_{n−2}^{2}, … analogously as in (1). Indeed, we will prove the following identity.

(2) D_{n}^{2} = (a^{2} −2n+2)D_{n−1}^{2} −(2n−2)(a^{2} −2n+2)D_{n−2}^{2} +(2n−2)(2n−4)^{2}D_{n−3}^{2},

for n ≥ 4.

The identity (1) implies

D_{n}^{2} = a^{2}D_{n−1}^{2} + 4(n − 1)^{2}D_{n−2}^{2} − 2(2n − 2)aD_{n−1}D_{n−2}.

On the other side,

aD_{n−1}D_{n−2} = D_{n−1}[D_{n−1} + (2n − 4)D_{n−3}] = D_{n−1}^{2} + (2n − 4)D_{n−1}D_{n−3}.

The identity

D_{n−1} + (2n − 4)D_{n−3} = aD_{n−2}

implies

\begin{matrix}D_{n−1}^{2} + (2n − 4)^{2}D_{n−3}^{2} + 2(2n − 4)D_{n−1}D_{n−3} = a^{2}D_{n−2}^{2}, \\ (2n − 4)D_{n−1}D_{n−3} = \frac{1}{2}[a^{2}D_{n−2}^{2} − D_{n−1}^{2} − (2n − 4)^{2}D_{n−3}^{2}]. \end{matrix}

Hence,

aD_{n−1}D_{n−2} = D_{n−1}^{2} + \frac{1}{2}[a^{2}D_{n−2}^{2} − D_{n−1}^{2} − (2n − 4)^{2}D_{n−3}^{2}]

= \frac{1}{2}[a^{2}D_{n−2}^{2} + D_{n−1}^{2} − (2n − 4)^{2}D_{n−3}^{2}],

D_{n}^{2} = a^{2}D_{n−1}^{2} + 4(n − 1)^{2}D_{n−2}^{2} − (2n − 2)[a^{2}D_{n−2}^{2} + D_{n−1}^{2} − (2n − 4)^{2}D_{n−3}^{2}] = (a^{2} − 2n + 2)D_{n−1}^{2} − (2n − 2)(a^{2} − 2n + 2)D_{n−2}^{2} + (2n − 2)(2n − 4)^{2}D_{n−3}^{2} and (2) is proved. We should also prove the following identity.

(3)  a^{2}D_{n−2} = D_{n} + (4n − 6)D_{n−2} + (2n − 4)(2n − 6)D_{n−4},

for n ≥ 5. Indeed,

D_{n} = aD_{n−1} − (2n − 2)D_{n−2} = a[aD_{n−2} − (2n − 4)D_{n−3}] − (2n − 2)D_{n−2}

= a^{2}D_{n−2} − (2n − 2)D_{n−2} − (2n − 4)aD_{n−3}
= a^{2}D_{n−2} − (2n − 2)D_{n−2} − (2n − 4)[D_{n−2} + (2n − 6)D_{n−4}]
= a^{2}D_{n−2} − (4n − 6)D_{n−2} − (2n − 4)(2n − 6)D_{n−4}

and (3) is proved.

The identity (3) implies the following identity.

\begin{aligned}D_{2k}(a^{2} − 2n + 2) & = D_{2(k+1)} + (8k + 2)D_{2k} + 4k(4k − 2)D_{2(k−1)} − (2n − 2)D_{2k} \\ & = D_{2(k+1)} + (8k + 4 − 2n)D_{2k} + 4k(4k − 2)D_{2(k−1)}. \end{aligned}

Using this identity and (3), we are able to prove (ii).

It is easy to verify (ii) for n = 1, 2, 3, 4. Assume that (ii) holds for the numbers n − 1, n − 2 and n − 3, (n ≥ 5). Let us denote

A_{j}^{i} = \binom{i}{j} · 2^{i−j} · \frac{i!}{j!}

for i ∈ ℕ and 0 ≤ j ≤ i. By algebraic transformations, the following identity

A_{k−1}^{n−1} + (8k + 4 − 2n)A_{k}^{n−1} + 4(k + 1)(4k + 2)A_{k+1}^{n−1}
−(2n − 2)A_{k−1}^{n−2} − (2n − 2)(8k + 4 − 2n)A_{k}^{n−2} − (2n − 2)4(k + 1)(4k + 2)A_{k+1}^{n−2}
+ (2n − 2)(2n − 4)^{2}A_{k}^{n−3} = A_{k}^{n}

can be proved. Indeed, multiplying it by

\frac{[(k + 1)!]^{2} · (n − k)!}{[(n − 2)!]^{2}} · 2^{k−n}

one obtains polynomial equality of n and k, and by direct calculation it can be verified. Using this identity, the inductive assumptions and the identity (2), we obtain

D_{n}^{2} = (a^{2} − 2n + 2)D_{n−1}^{2} − (2n − 2)(a^{2} − 2n + 2)D_{n−2}^{2} + (2n − 2)(2n − 4)^{2}D_{n−3}^{2}

= ∑_{k=0}^{n−1} A_{k}^{n−1}D_{2k}(a^{2} − 2n + 2) − ∑_{k=0}^{n−2} A_{k}^{n−2}D_{2k}(a^{2} − 2n + 2)(2n − 2)

+ ∑_{k=0}^{n−3} A_{k}^{n−3}D_{2k}(2n − 2)(2n − 4)^{2}

= ∑_{k=0}^{n−1} A_{k}^{n−1}[D_{2(k+1)} + (8k + 4 − 2n)D_{2k} + 4k(4k − 2)D_{2(k−1)}]

− ∑_{k=0}^{n−2} A_{k}^{n−2}(2n − 2)[D_{2(k+1)} + (8k + 4 − 2n)D_{2k} + 4k(4k − 2)D_{2(k−1)}]

+ ∑_{k=0}^{n−3} A_{k}^{n−3}(2n − 2)(2n − 4)^{2}D_{2k}

= ∑_{k=0}^{n} D_{2k}[A_{k−1}^{n−1} + (8k + 4 − 2n)A_{k}^{n−1} + 4(k − 1)(4k + 2)A_{k+1}^{n−1}

− (2n − 2)A_{k−1}^{n−2} − (2n − 2)(8k + 4 − 2n)A_{k}^{n−2} − (2n − 2)4(k + 1)(4k + 2)A_{k+1}^{n−2}

+ A_{k}^{n−3}(2n − 2)(2n − 4)^{2}] = ∑_{k=0}^{n} A_{k}^{n} · D_{2k}.

Hence (ii) is proved.

3. Second Solution of the Problem. Now we will prove the identities (i) and (ii) using the properties of the Hermite polynomials.

(i) In the theory of the special functions [2], the Hermite polynomial

(4)  H_{n}(x) = ∑_{k=0}^{⌊n/2⌋}(−1)^{k} \frac{n!(2x)^{n−2k}}{k!(n − 2k)!},

of nth degree satisfies the following equality

(5)  H_{n}(x) = 2xH_{n−1}(x) − (2n − 2)H_{n−2}(x).

Since (4) is uniquely determined by (5) with the initial conditions H_{1}(x) = 2x and H_{2}(x) = 4x^{2} − 2, by putting x = a/2 from (1), (4) and (5) we obtain

D_{n} = H_{n}(a/2) = ∑_{k=0}^{⌊n/2⌋}(−1)^{k} \frac{n!a^{n−2k}}{k!(n − 2k)!},

and (i) is proved.

(ii) Starting from the known identity of E. Feldheim [3],

(6)  H_{m}(x)H_{n}(x) = 2^{n}n! ∑_{k=0}^{n} \binom{m}{n − k}\frac{1}{2^{k}k!}H_{m−n+2k}(x), (m ≥ n)

for product of Hermite polynomials, for m = n we obtain

(7)  H_{n}^{2}(x) = 2^{n}n! ∑_{k=0}^{n} \binom{n}{n − k}\frac{1}{2^{k}k!}H_{2k}(x).

Since D_{n} satisifies the recurrent formula (1), and also (5) for x = a/2, then from (7) it follows

D_{n}^{2} = H_{n}^{2}(a/2) = 2^{n}n! ∑_{k=0}^{n} \binom{n}{k}\frac{1}{2^{k}k!}D_{2k},

and (ii) is proved.

\underline{Remark}. There are more forms for the product of the Hermite polynomials [4,5,6,7,8,9], and hence, different proofs for (ii). For example using the equation

H_{m}(x)H_{n}(x) = ∑_{r=0}^{n} 2^{r}r!\binom{m}{r}\binom{n}{r}H_{m+n−2r}(x), (m ≥ n)

of G. N. Watson [4], for m = n we obtain

(8)  D_{n}^{2} = H_{n}^{2}(a/2) = ∑_{r=0}^{n} 2^{r}r!\binom{n}{r}^{2}D_{2n−2r}.

Exchanging n − r by k in the left side of (8), we obtain (ii).

\underline{References}

1. B. S. Popov, “Problem 5,” Bull. Soc. Math. Phys. Macedoine, 2 (1951), 40.

2. D. S. Mitrinović, Uvod u Specijalne Funkcije, Gradjevinjska knjiga, Beograd, 1972, 59–74.

3. E. Feldheim, “Quelques Nouvelles Relations pour les Polynomes d’Hermite,” J. London Math. Soc., 13 (1938), 22–29.

4. G. N. Watson, “A Note on the Polynomials of Hermite and Laguerre,” J. London Math. Soc., 13 (1938), 29–32.

5. V. R. Thiruvenkata Char, “Note on Some Formulae Involving the Laguerre and Hermitian Polynomials and Bessel Functions,” Proc. Indian Acad. Sci., 10 (1939), 229–234.

6. J. Geronimus, “On the Polynomials of Legendre and Hermite,” Tôhoku Math. J., 34 (1931), 295–296.

7. N. Nielsen, “Recherches sur les Polynomes d’Hermite,” Danske Vidensk. Selsk. Mat. Fys. Medd. I, 6 (1918), 80.

8. L. Carlitz, “A Note on Hermite Polynomials,” Amer. Math. Monthly, 62 (1955), 646–647.

9. L. Carlitz, “Problem 68–7,” SIAM Review, 10 (1968), 224.

Kostadin G. Trenčevski Institute of Mathematics P. O. Box 162 91000 Skopje, Macedonia email: kostatre@iunona.pmf.ukim.edu.mk

Ice B. Risteski Bratfordska 2/3–9 91000 Skopje, Macedonia

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