ANALYSIS

(1) f(0)=0,
(2) f(x+x)+ f(x-x)= 2[ f(x)+f(x)] => f(2x)= 4f(x)
f(x-x)+ f(x+x) = 2[ f(x)+f(-x)] =>4f(x)= 2[f(x)+f(-x)]=> f(-x)=f(x)
(3) 設 f(1)= a
f(n+1)+f(n-1)= 2[ f(n)+ f(1)]=> f(n+1)= 2f(n)- f(n-1) + 2a (遞迴)
=> f(n)= a n^2
(4) x=y= 1/n 時, f(2/n)+f(0)= 4f(1/n)
x=2/n, y=1/n => f(3/n)+ f(1/n)= 2[ f(2/n) + f(1/n)] => f(3/n)= 9 f(1/n)
=>(by MI) f(m/n)= m^2 f(1/n) => f(1/n)= 1/n^2 f(1)
=> f(n/m)= (n/m)^2 f(1)
(5)故 f(r) = r^2 f(1) = a r^2
f(x)= lim(r->x) f(r)= lim(r->x) ar^2 = ax^2