Problema do mês :: Problem of the month #2

ver/see Problema do mês Problem of the month

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Resolução :: Solution

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Enunciado do Problema

Admita que n=1,2,3,\dots . Seja x\ge 0  um número  real, \dbinom{x}{0}=1 e  \dbinom{x}{n}=\dfrac{x\left( x-1\right) \cdots\left( x-n+1\right) }{n!}. Deduza a identidade \dbinom{x}{n}+\dbinom{x}{n-1}=\dbinom{x+1}{n}.

  • 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 n=1,2,3,\dots . Let x\ge 0 be a real  number, \dbinom{x}{0}=1 and \dbinom{x}{n}=\dfrac{x\left( x-1\right) \cdots\left( x-n+1\right) }{n!}.  Derive the identity  \dbinom{x}{n}+\dbinom{x}{n-1}=\dbinom{x+1}{n}.

  • The deadline for submitting solutions is September 9, 2009 either via e-mail  acltavares@sapo.pt or comment box.

Sobre Américo Tavares

eng. electrotécnico reformado / retired electrical engineer
Esta entrada foi publicada em Matemática, Math, Problem, Problem Of The Month, Problema do mês, Problemas com as etiquetas , , , , . ligação permanente.

4 respostas a Problema do mês :: Problem of the month #2

  1. PB diz:

    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

  2. Anónimo diz:

    The proof by PB is nice.

  3. MathOMan diz:

    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.

  4. Américo Tavares diz:

    Recebi por mail a seguinte prova de antonio girao, 21.07.09:

    \dbinom{x}{n}=\dfrac{x!}{\left( x-n\right) !n!}

    \dbinom{x}{n-1}=\dfrac{x!}{\left( x-n+1\right) !\left( n-1\right) }

    \dfrac{x!}{\left( x-n\right) !n!}+\dfrac{x!}{\left( x-n+1\right) !\left( n-1\right) !} =\dfrac{x!}{\left( x-n\right) !n!}+\dfrac{x!n}{\left( x-n+1\right) !\left( n-1\right) !n} =\dfrac{x!\left( x-n+1\right) }{\left( x-n\right) !n!\left( x-n+1\right) }+\dfrac{x!n}{\left( x-n+1\right) !\left( n-1\right) n}

    =\dfrac{x!\left( x-n+1\right) }{\left( x-n+1\right) !n!}+\dfrac{x!n}{\left( x-n+1\right) !n!}=\dfrac{x!\left[ \left( x-n+1\right) +n\right] }{\left( x-n+1\right) !n!} =\dfrac{x!\left( x+1\right) }{\left( x-n+1\right) !n!}=\dfrac{\left( x+1\right) !}{\left( x+1-n\right) !n!}=\dbinom{x+1}{n}

    Nota minha: o leitor usou a notação

    \dfrac{x!}{\left( x-n\right) !}=x\left( x-1\right) \cdots \left( x-n+1\right)

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