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Re: [obm-l] Expansão de termos -proposta de problema
- To: obm-l@xxxxxxxxxxxxxx
- Subject: Re: [obm-l] Expansão de termos -proposta de problema
- From: "Marcelo Salhab Brogliato" <msbrogli@xxxxxxxxx>
- Date: Mon, 29 Oct 2007 01:42:54 -0300
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Olá Rodrigo,
são os números de Stirling (http://en.wikipedia.org/wiki/Stirling_number
).
vamos mostrar algumas coisas legais...
digamos que:
x(x-1)(x-2)...(x-n+1) = Sum {k=1..n} S[n, k].x^k
então:
x(x-1)(x-2)...(x-n+1)(x-n) = Sum{k=1..n+1} S[n+1, k].x^k
pegando a primeira e multiplicando por (x-n), temos:
x(x-1)(x-2)...(x-n+1)(x-n) = ( Sum {k=1..n} S[n, k].x^k
).(x-n)
Sum{k=1..n+1} S[n+1, k].x^k = ( Sum {k=1..n} S[n, k].x^k
).(x-n)
Sum{k=1..n+1} S[n+1, k].x^k = ( Sum {k=1..n} S[n, k].x^(k+1)
) - ( Sum {k=1..n} nS[n, k].x^k
)
Sum{k=1..n+1} S[n+1, k].x^k = ( Sum {k=2..n+1} S[n, k-1].x^k
) - ( Sum {k=1..n} nS[n, k].x^k
)
Sum{k=2..n} S[n+1, k].x^k + S[n+1, n+1].x^(n+1) + S[n+1, 1].x = ( Sum {k=2..n} (S[n, k-1] - nS[n, k]).x^k
) + S[n, n].x^(n+1) - nS[n, 1].x
portanto:
S[n+1, 1] = -nS[n, 1]
S[n+1, k] = S[n, k-1] - nS[n, k] ... k=2, 3, ..., n
S[n+1, n+1] = S[n, n]
x=x... entao: S[1,1] = 1
x(x-1) = x^2 - x
S[2, 1] = -
1.S[1, 1] = -1 ... ok!
S[2, 2] = S[1, 1] = 1 ... ok!
x(x-1)(x-2) = x^3 - 3x + 2x
S[3, 1] = -2.S[2, 1] = -2.(-1) = 2 ... ok!
S[3, 2] = S[2, 1] - 2.S[2, 2] = -1 - 2.1 = -3 ... ok!
S[3, 3] = S[2, 2] = 1 ... ok!
x(x-1)(x-2)(x-3) = x^4 - 6x^3 + 11x^2 - 6x
S[4, 1] = -3.S[3, 1] = -3.2 = -6 .... ok!
S[4, 2] = S[3, 1] - 3.S[3, 2] = 2 - 3.(-3) = 11 ... ok!
S[4, 3] = S[3, 2] - 3.S[3, 3] = -3 - 3.1 = -6 ... ok!
S[4, 4] = S[3, 3] = 1 ... ok!
e assim por diante.. :)
abraços,
Salhab
On 10/27/07, Rodrigo Renji <rodrigo.uff.math@xxxxxxxxx> wrote:
Tente encontrar uma formula para os coeficientes da potência que
aparecem na expansão de
x(x-1)(x-2). ... (x-n)
i.e
x=x
x(x-1)=x²-x
x(x-1)(x-2)=x³-3x+2x
x(x-1)(x-2)(x-3)=x^4 -6x³+11x²-6x
etc...
(a fórmula existe, é uma recorrência de duas variáveis)
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