根據(jù)題意,f(x)=|x|為周期為2π的函數(shù),而且因?yàn)閒(x)=f(-x),所以f(x)為偶函數(shù).f(x)可展開為傅里葉級(jí)數(shù):
f(x)=a0+ ∑(n=1→∞)(ancosnωx+ bnsinnωx)
上式中:ω=2π/2π=1,系數(shù)a0、an、bn由下式?jīng)Q定:
a0=(1/2π)∫(-π,π)f(x)dx
=(1/2π)∫(-π,π)|x|dx
=(1/2π)×2×∫(0,π)xdx
=π/2
an=(1/π)∫(-π,π)|x|cosnωxdx
=(2/π)∫(0,π)xcosnxdx
=(2/nπ)∫(0,π)xdsinnx
=(2/nπ)[xsinnx(0,π)-∫(0,π)sinnxdx]
=(2/nπ)[(1/n)cosnx(0,π)]
=2((-1)^n-1)/(πn^2)
bn=(1/π)∫(-π,π)|x|sinnωxdx
由于|x|sinnωx是奇函數(shù),所以bn=0,所以:f(x)=π/2+ ∑(n=1→∞)2((-1)^n-1)/(πn^2)cosnx,
由上面可見,當(dāng)n為偶數(shù)時(shí),an=0,所以
f(x)可寫作如下形式:
f(x)=π/2+ ∑(n=1→∞)(-4)/[π(2n-1)^2]cosnx,即
|x|=π/2-(4/π)∑(n=1→∞)cosnx/(2n-1)^2