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后面写错了..sorry

Posted by majia5 at 2009-09-21 17:23:37
In Reply To:A的推导全过程 Posted by:majia5 at 2009-09-21 17:02:30

ans=sigma{EE(n)} (n=2,3,...)
=C*sigma{(nanda *T*q)^n / n!*[nQ^n-nQ^(n-1)-Q^n+1]}
=C*{sigma{(nanda *T*q)^n / n!*(nQ^n)}-sigma{(nanda *T*q)^n / n!*(nQ^(n-1))}-sigma{(nanda *T*q)^n / n!*Q^n}+sigma{(nanda *T*q)^n / n!}}
=C*{(nanda *T*q*Q)*sigma{(nanda *T*q*Q)^(n-1) / (n-1)!} -(nanda *T*q)*sigma{(nanda *T)^(n-1) / (n-1)!}-sigma{(nanda *T)^n / n!}+sigma{(nanda *T*q)^n / n!}}
=C*{(nanda *T)*(e^(nanda *T)-1)-(nanda *T*q)*(e^(nanda *T)-1)-(e^(nanda *T)-1-nanda *T)+(e^(nanda *T*q)-1-nanda *T*q)}
=C*{(nanda *T*p)*(e^(nanda *T)-1)-(e^(nanda *T)-1-nanda *T)+(e^(nanda *T*q)-1-nanda *T*q)}
=exp(-nanda*T)/p*{...}
=(nanda *T)-exp(-nanda*T)*((e^(nanda *T)-1)-(e^(nanda *T*q)-1))/p
=(nanda *T)-exp(-nanda*T)*(e^(nanda *T)-e^(nanda *T*q))/p
=(nanda *T)-(1-e^(-nanda *T*p))/p
=(nanda *T)+(e^(-nanda *T*p) -1 )/p
完毕