分类讨论(主要利用了有理分式的极限)
1.-1 < x < 1
x ^(2n) -> 0,x^(2n-1) -> 0
f(x) = ax^2 + bx
2.|x| > 1
f(x) = 0
3.x = 1
f(x) = f(1) = (1 + a + b) / 2
4.x = -1
f(x) = f(-1) = (-1 + a - b) / 2
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剩下的由连续性知对任意x
f(x+) = f(x-) = f(x)
取x = 1 and -1
列方程解
我比较懒,就不算了