化简ln ln 怎么化简?或者证明S = k[ ln (150!)- ln (N1!) - ln (N2!)- ln (N3!)- ln (N4!)- ln (N5!)]N1∈[0,150]N2∈[0,150]N3∈[0,150]N4∈[0,150]N5∈[0,150]N1 + N2 + N3 + N4 + N5 = 150;中S极值是N1=N2=N
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![化简ln ln 怎么化简?或者证明S = k[ ln (150!)- ln (N1!) - ln (N2!)- ln (N3!)- ln (N4!)- ln (N5!)]N1∈[0,150]N2∈[0,150]N3∈[0,150]N4∈[0,150]N5∈[0,150]N1 + N2 + N3 + N4 + N5 = 150;中S极值是N1=N2=N](/uploads/image/z/9303718-22-8.jpg?t=%E5%8C%96%E7%AE%80ln+ln+%E6%80%8E%E4%B9%88%E5%8C%96%E7%AE%80%3F%E6%88%96%E8%80%85%E8%AF%81%E6%98%8ES+%3D+k%5B+ln+%EF%BC%88150%21%EF%BC%89%EF%BC%8D+ln+%EF%BC%88N1%21%EF%BC%89+%EF%BC%8D+ln+%EF%BC%88N2%21%EF%BC%89%EF%BC%8D+ln+%EF%BC%88N3%21%EF%BC%89%EF%BC%8D+ln+%EF%BC%88N4%21%EF%BC%89%EF%BC%8D+ln+%EF%BC%88N5%21%EF%BC%89%5DN1%E2%88%88%5B0%2C150%5DN2%E2%88%88%5B0%2C150%5DN3%E2%88%88%5B0%2C150%5DN4%E2%88%88%5B0%2C150%5DN5%E2%88%88%5B0%2C150%5DN1+%2B+N2+%2B+N3+%2B+N4+%2B+N5+%3D+150%EF%BC%9B%E4%B8%ADS%E6%9E%81%E5%80%BC%E6%98%AFN1%3DN2%3DN)
化简ln ln 怎么化简?或者证明S = k[ ln (150!)- ln (N1!) - ln (N2!)- ln (N3!)- ln (N4!)- ln (N5!)]N1∈[0,150]N2∈[0,150]N3∈[0,150]N4∈[0,150]N5∈[0,150]N1 + N2 + N3 + N4 + N5 = 150;中S极值是N1=N2=N
化简ln
ln 怎么化简?
或者证明
S = k[ ln (150!)- ln (N1!) - ln (N2!)- ln (N3!)- ln (N4!)- ln (N5!)]
N1∈[0,150]
N2∈[0,150]
N3∈[0,150]
N4∈[0,150]
N5∈[0,150]
N1 + N2 + N3 + N4 + N5 = 150;
中S极值是N1=N2=N3=N4=N5=30时取得~
化简ln ln 怎么化简?或者证明S = k[ ln (150!)- ln (N1!) - ln (N2!)- ln (N3!)- ln (N4!)- ln (N5!)]N1∈[0,150]N2∈[0,150]N3∈[0,150]N4∈[0,150]N5∈[0,150]N1 + N2 + N3 + N4 + N5 = 150;中S极值是N1=N2=N
ln(x)跟x的单调性是一致的,x大的时候,ln(x)的值也大,所以上面的题目跟下面这个题目是一样的:
N是m的倍数,N/m=k,n1+n2+...+nm=N
N!/(n1!n2!...nm!)在(n1=n2=...=nm=N/m)时最大,也就是n1!*n2!...*nm!在(n1=n2=...=nm=N/m)时最小.把n1,n2...nm写成k+i和k-j的形式(k=N/m),再把n1!*n2!...*nm!写成(k!)的m次方乘以一个分式,就可以看出,当(n1=n2=...=nm=N/m)时n1!*n2!...*nm!最小.
加个注:分式的形式为:
(k+1)...(k+i1)...(k+1)...(k+ia)/(k-1)...(k-j1+1)...(k-1)...(k-jb+1)
其中,i1+i2+...+ia=j1+j2+...+jb,所以分子分母的项数一样多,可以看到,当i1=...=ia=j1=...=jb=0时,这个式子最小,等于1.