对于αa[n+1]+βa[n]+γa[n-1]=0
设u,v,使得a[n+1]-ua[n]=v(a[n]-ua[n-1]) (这个式子可以看成等比)
展开后有:a[n+1]-(u+v)a[n]+uva[n-1]=0,所以u+v=-β/α,uv=γ/α
所以u,v是αx^2+βx+γ=0的两根 这个式子被称为特征方程
对于a[n+1]-ua[n]=v(a[n]-ua[n-1])
有:a[n+1]-ua[n]=(v^(n-1))(a[2]-ua[1]),所以a[n+1]=ua[n]+(v^(n-1))(a[2]-ua[1])
u≠v时,a[n+1]-[v^n/(u(v-u))](a[2]-ua[1])=u(a[n]-[v^(n-1)/(u(v-u))](a[2]-ua[1]))
所以a[n+1]-[v^n/(u(v-u))](a[2]-ua[1])=(u^n)(a[1]-(a[2]-ua[1])/(u(v-u)))
所以a[n+1]=[v^n/(u(v-u))](a[2]-ua[1])+(u^n)(a[1]-(a[2]-ua[1])/(u(v-u)))
所以a[n+1]=[v^(n-1)/(u(v-u))](a[2]-ua[1])+(u^(n-1))(a[1]-(a[2]-ua[1])/(u(v-u)))
此时A=a[1]-(a[2]-ua[1])/(u(v-u)),B=(a[2]-ua[1])/(u(v-u)), (x1=u,x2=v,下同)
u=v时,a[n+1]=ua[n]+(u^(n-1))(a[2]-ua[1])
所以a[n+1]-(a[2]-ua[1])(n+1)u^(n-1)=u(a[n]-(a[2]-ua[1])nu^(n-2))
所以a[n+1]-(a[2]-ua[1])(n+1)u^(n-1)=(u^n)(a[1]-(a[2]-ua[1])u^(-1))=(u^n)(2a[1]-a[2]/u)
所以a[n+1]=(a[2]-ua[1])(n+1)u^(n-1)=(u^n)(2a[1]-a[2]/u)
所以a[n]=[(a[2]-ua[1])/u]nu^(n-1)=(u^(n-1))(2a[1]-a[2]/u)
此时A=(2-u)a[1],B=(a[2]-ua[1])
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