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Re: [obm-l] De volta às recursões em duas variáveis!!!
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- Subject: Re: [obm-l] De volta às recursões em duas variáveis!!!
- From: "saulo nilson" <saulo.nilson@xxxxxxxxx>
- Date: Tue, 4 Dec 2007 12:07:46 -0300
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A(x,1)=(n-x)A(x+1,0)-(x+1)A(x,0)=(n-x)-x-1=n-1-2x
A(x,y+1)=(n-x)A(x+1,y) -(x+1)A(x,y)
A(x,y+2)=(n-x)A(x+1,y+1)-(x+1)A(x,y+1)
somatorioA(x,k)(1,n)=(n-x)somatorioA(x+1,k)(0,n-1)-(x+1)somatorioA(x,k)(0,n-1)
A(x,0)+A(x,n)+(x+2)somatorioA(x,k)(1,n-1)=(n-x)somatorioA(x+1,k)(0,n-1)
A(n,0)+A(n,n)+(n+2)somaA(n,k)(1,n-1)=0
somaA(n,k)(1,n-1)=-(A(n,0)+A(n,n))/(n+2)
somaA(n,k)(0,n)=(A(n,0)+A(n,n))(n+1)/(n+2)
falta achar o A(n,n)
Oi pessoal...tenho o seguinte problema...
...
A: N²-->R
x,y e N
n e N (parâmetro)
A(x,y+1)=(n-x)A(x+1,y) -(x+1)A(x,y)
A(x,0)=1
A(x,y)=???
A(x,0)+A(x,1)+...+A(x,n)=S(x,n)=???
Agradeceria bastante se alguém me resolvesse este trambolho...
aliás...para o felizardo tenho várias outras dúvidas...e também, claro, porquê das mesmas...
Até! :)