f(5)是个常数,那就按照常数的傅立叶变换
L[1] = 0
所以L[f(5)] = 0
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F(w) = ℱ[f(t)] dF(w)/dw = (-i)ℱ[tf(t)] ℱ[tf(t)] = i dF(w)/dw ℱ[tf(3t)]=(1/3)ℱ[3tf(3t)]=(1/9) i[dF(w/3)/dw] 2. 令x=1-t, ℱ[(1-t)f(1-t)] = -e^(-iw) i dF(-w)/dw 很多中间过程受字符限制打不出来
F(w) = ℱ[f(t)]
dF(w)/dw = (-i)ℱ[tf(t)]
ℱ[tf(t)] = i dF(w)/dw
ℱ[tf(3t)]=(1/3)ℱ[3tf(3t)]=(1/9) i[dF(w/3)/dw]
2. 令x=1-t,
ℱ[(1-t)f(1-t)]
= -e^(-iw) i dF(-w)/dw
很多中间过程受字符限制打不出来