推 ADAH33:感謝^^真是太厲害了^^ 59.104.60.147 03/02 11:16
※ 引述《ADAH33 (逐漸消失的生命)》之銘言：
: let f : R -> R be cont
: f(x) = 1 + (0到1積分) kf(x-ky) dy + (0到x-k積分) f(y)dy
: where k > 0
: find f(x)
∫(0->1) kf(x-ky) dy Let u=x-ky, du=-kdy, y=0 ==> u=x, y=1 ==> u=x-k
=∫(x->x-k) -f(u)du
=∫(x-k->x) f(u)du
=∫(x-k->x) f(y)dy
f(x) = ∫(0->1) kf(x-ky) dy + ∫(0->x-k) f(y)dy
= 1 + ∫(x-k->x) f(y)dy+∫(0->x-k) f(y)dy
= 1 + ∫(0->x) f(y)dy
By the fundamental theorem of calculus, we have
f'(x) = f(x).
==> f(x) = Ae^x. ( f(0)=A )
And f(x) = 1 + ∫(0->x) f(y)dy ==> f(0) = 1 = A
f(x) = e^x
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