設(shè)f(x) = ax^2 + bx + c
f(0) = 1
則 c =1
f(x) = ax^2 + bx + 1
f(x+1) = a(x+1)^2 + b(x+1) + 1 = ax^2 + (b+2a)x + a + b + 1
f(x+1) - f(x)
= [ax^2 + (b+2a)x + a + b + 1] - [ ax^2 + bx + 1]
= 2ax + a + b
f(x+1) - f(x) = 2x 對任何x成立,則
2a = 2
a + b = 0
a = 1
b = -1
f(x) = x^2 - x + 1