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# We first note that 1 2n n X k=0 (n − 2k)2pn k

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Problem 11821

(American Mathematical Monthly, Vol.122, February 2015) Proposed by F. Holland and C. Koester (Ireland).

Let p be a positive integer. Prove that

lim

n→∞

1 2nnp

n

X

k=0

(n − 2k)2pn k



=

p

Y

j=1

(2j − 1).

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

We first note that 1 2n

n

X

k=0

(n − 2k)2pn k



= 1 2n

2p

X

j=0

2p j

 n2p−j·

n

X

k=0

n k

 (−2k)j

= 1 2n

2p

X

j=0

2p j

  x2p−j (2p − j)!



enx· xj j!



(1 + e−2x)n

= x2p (2p)!

  ex+ e−x 2

n

= D2p((cosh(x))n)(0).

The Fa´a di Bruno’s formula says that

D2p(f ◦ g)(x) =X (2p)!

Q2p

j=1(kj)!(j!)kj · (Dkf )(g(x)) ·

2p

Y

j=1

((Djg)(x))kj

where k =P2p

j=1kj and the sum is taken over all 2p-tuples of nonnegative integers (k1, . . . , k2p) for whichP2p

j=1jkj= 2p.

In our case f (x) = xn and g(x) = cosh(x) and since

Dk(xn)(cosh(0)) =

k−1

Y

j=0

(n − j) and Dj(cosh)(0) =

(1 if j is even 0 if j is odd , it follows that we can take k1= k3= · · · = k2p−1= 0.

Hence k =Pp

j=1k2j, the costraint becomesPp

j=1jk2j= p, and

D2p((cosh(x))n)(0) =X (2p)!

Qp

j=1(k2j)!((2j)!)k2j ·

k−1

Y

j=0

(n − j).

The RHS is a polynomial in n of degree the maximum value of k which is equal to p. It is attained only when k2= p and k4= · · · = k2p= 0, therefore we finally find that

n→∞lim 1 2nnp

n

X

k=0

(n − 2k)2pn k



= lim

n→∞

D2p((cosh(x))n)(0)

np = (2p)!

p!(2!)p =

p

Y

j=1

(2j − 1).



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