证明:
设 k1e^x+k2e^2x+k3e^3x = 0.
x分别取0,1,2,得
k1+k2+k3 =0
k1e+k2e^2+k3e^3 = 0
k1e^2+k2e^4+k3e^6 = 0
即
k1+k2+k3 =0
k1+k2e+k3e^2 = 0
k1+k2e^2+k3e^4 = 0
因为行列式 (Vandermonder行列式)
1 1 1
1 e e^2
1 e^2 e^4
=(e-1)(e^2-1)(e^2-e)
=e(e+1)(e-1)^3
≠0.
所以 k1=k2=k3=0
所以 e^x ,e^2x,e^3x线性无关.