ver/see Problema do mês Problem of the month
Enunciado do Problema
Admita que . Seja um número real, e . Deduza a identidade .
- O prazo limite para apresentação das resoluções é 9.09.2009, quer via email acltavares@sapo.pt ou comentando no blogue.
Problem Statement
Suppose that . Let be a real number, and . Derive the identity .
- The deadline for submitting solutions is September 9, 2009 either via e-mail acltavares@sapo.pt or comment box.
Hello,
Let’s write the identity we want to proove like this : P(x)=Q(x).
P(x) and Q(x) are obviously polynomial functions.
It is well known that the identity is true if x is an positive integer.
Hence P(n)=Q(n) for all positive integers n.
We can now use this fact : if two polynomials P(x) and Q(x) have the same values for x in a infinite set then P(x)=Q(x) for all x.
Pierre
The proof by PB is nice.
By multiplying both sides with n! the equation becomes
X (X – 1) … (X – n + 1) + n X (X – 1) … (X – n + 2) = (X + 1) X (X – 1) … (X – n + 2)
Both sides are polynomials of degree n with highest coefficient 1 and same roots, i.e., 0 , …, n – 2 (evident) and -1 (easy check). Hence they are the same.
Recebi por mail a seguinte prova de antonio girao, 21.07.09:
Nota minha: o leitor usou a notação