BC ONLY- (C) f (x) = a 0 + a 1 x + a 2 x^2 + a 3 x^3 + · · · ; if f (0) = 1, then a 0 = 1.
f′ (x) = a 1 + 2a 2 x + 3a 3 x^2 + 4a 4 x^3 + · · · ; f′ (0) = −f (0) = −1,
so a 1 = −1. Since f′ (x) = −f (x), f (x) = −f′(x):
1 − x+ a 2 x^2 + a 3 x^3 + ... = −(−1 + 2a 2 x + 3a 3 x^2 + 4a 4 x^3 + ...) identically. Thus,