Mar 13, 2015 - GM] 13 Mar 2015. An Interesting ... March 17, 2015. Abstract ... k=0. 1 nk + n - 1 . Proof. We will evaluate the integral. â« 1. 0. (1 - x)n x 1 n dx.
arXiv:1503.04678v1 [math.GM] 13 Mar 2015
An Interesting Identity Brett Pansano Northwest Arkansas Community College March 17, 2015 Abstract This purpose of this paper is to note an interesting identity derived from an integral in Gradshteyn and Ryzhik using techniques from George Boros’(deceased) Ph.D thesis. The idenity equates a sum to a product by evaluating an integral in two different ways. A more general form of the idenity is left for further investigation.
1
Introduction
Theorem 1 Let n be non-negative integer. Then n X
(−1)k
k=0
n Y n 1 1 = nn n! . k nk + n − 1 nk + n − 1 k=0
Proof. We will evaluate the integral Z 1 n (1 − x) 1
xn
0
dx
in two different ways. We start with n
(1 − x) =
n X
k=0
n k x , (−1) k k
1
which follows directly from the binomial theorem. Dividing both sides by x n yields n n X 1 (1 − x) k n xk− n , = (−1) 1 k xn k=0
so that Z 1 n (1 − x) 0
1
xn
dx =
n X
k=0
Z 1 n X n 1 n k n k− n (−1) x (−1) dx = . k k nk + n−1 0 k
k=0
1
But the integral
R1
(1−x)n
dx can be evaluated using the fact that
1
0
xn
Z
1
xm−1 (1 − x)
n−1
dx = B (m, n)
0
where B (m, n) is the beta function. We get Z 1 n Γ (n + 1) Γ 1 − n1 (1 − x) 1 , = dx = B n + 1, 1 − 1 n Γ n + 2 − n1 xn 0 since
B(m, n) =
Γ (m) Γ (n) , Γ (m + n)
where Γ is the gamma function. We next note that 1 1 1 = n+1− Γ n+1− Γ n+2− n n n 1 1 1 n− Γ n− = n+1− n n n 1 1 1 1 = ... = n + 1 − n− n−1− n−2− n n n n 1 1 Γ 1− , ··· n− n n so that (since Γ (n + 1) = n!) Z 1 n (1 − x) dx 1 xn 0 = =
n+1−
1 n
n! n+1 Y k=1
k−
n − n1 n+1 Y
= 1
n! n − 1 − n1 n − 2 − n1 · · · 1 − n1
(nk − 1) .
k=1
n
Thus, n X
k
(−1)
k=0
n Y 1 n n nn+1 n! = nn n! = n+1 . nk + n − 1 k nk + n − 1 Y k=0 (nk − 1) k=1
so that, n X
k=0
k
(−1)
n Y nn n! 1 n 1 = n+1 . = nn n! nk + n − 1 k nk + n − 1 Y k=0 (nk − 1) k=1
2
Corollary 2 Let a be a real number. Then 1.
n X
k
(−1)
k=0
k=0
2.
n X
k=0
3.
n X
k=0
n Y 1 n 1 = nn n! nk + an − 1 k nk + an − 1
n Y n 1 1 (−1) = (an)n n! k ank + n − 1 ank + n − 1 k
k=0
n Y 1 n 1 n = (an) n! (−1) ank + n − 1 k ank + an − 1 k
k=0
Proof. This is clear. The following identity was confirmed via Mathematica 3 Y d n4 1 da nk+an−1 : (an−1)2 (n+an−1)2 (2n+an−1)2 (3n+an−1)2 by noting that k=0
d 1 1 =− 2 n. da nk + an − 1 (nk + an − 1) The following might be interesting to investigate: Conjecture 3 Let n be a natural number. Suppose f and g are differentiable functions. Then n X
k
(−1)
k=0
n Y 1 n 1 = nn n! . f (n) k + g (n) k f (n) k + g (n) k=0
References [I.S. Gradshteyn and I.M. Ryzhik, Table of Integrals, Series, and Products, Academic Press, INC., 1980]
[G. Boros and V. Moll, Irresistible Integrals, Symbolics, Analysis, and Experiments in the Evaluation of Integ
