f(n)>0,f(n+1)>0,
-p=α+β,q=αβ.
min{f(n),f(n+1)}在二者相等时取到最大
f(n)=f(n+1)时,n^2+pn+q=(n+1)^2+p(n+1)+q,2n+1+p=0,
-p=2n+1,α+β=2n+1,n=(α+β-1)/2,
f(n)
=n^2-(2n+1)n+q
=-n^2-n+q
=-n(n+1)+q
=-(α+β-1)(α+β+1)/4+αβ
=αβ-[(α+β)^2-1]/4
=(1-α+β)(1+α-β)/4
=[1-(α-β)^2]/4
f(n)>0,f(n+1)>0,
-p=α+β,q=αβ.
min{f(n),f(n+1)}在二者相等时取到最大
f(n)=f(n+1)时,n^2+pn+q=(n+1)^2+p(n+1)+q,2n+1+p=0,
-p=2n+1,α+β=2n+1,n=(α+β-1)/2,
f(n)
=n^2-(2n+1)n+q
=-n^2-n+q
=-n(n+1)+q
=-(α+β-1)(α+β+1)/4+αβ
=αβ-[(α+β)^2-1]/4
=(1-α+β)(1+α-β)/4
=[1-(α-β)^2]/4