非常简单的傅里叶级数展开 常用傅里叶级数展开式怎么证明
\u5085\u91cc\u53f6\u7ea7\u6570\u5c55\u5f00\u5982\u56fe\u6240\u793a\uff1a
\u8bc1\u660e\uff1a\u6839\u636e\u5085\u91cc\u53f6\u7ea7\u6570\u7684\u5b9a\u4e49\uff0c\u82e5\u5c06f(x)\u5c55\u5f00\u6210\u4f59\u5f26\u7ea7\u6570\uff0c\u5219f(x)=(a0)/2+\u2211ancosnx\uff0c\u5176\u4e2d\uff0can=(2/\u03c0)\u222b(0,\u03c0)f(x)cosnxdx\uff0cn=0,1,2\uff0c\u2026\uff0c\u221e\u3002\u672c\u9898\u4e2d\uff0cf(x)=sinx\uff0c\u5219an=(2/\u03c0)\u222b(0,\u03c0)sinxcosnxdx\u3002 \u2234a0=(2/\u03c0)\u222b(0,\u03c0)sinxdx=(-2/\u03c0)cosx\u4e28(x=0,\u03c0)=4/\u03c0\uff0ca1=\u222b(0,\u03c0)sinxcosxdx=0\uff0c\u800cn\u22600,1\u65f6\uff0c\u222b(0,\u03c0)sinxcosnxdx=(1/2)\u222b(0,\u03c0)[sin(n+1)x-sin(n-1)x]dx=(1/2){1/(n+1)-[(-1)^(n+1)]-1/(n-1)+[(-1)^(n+1)]/(n+1)}\u3002\u663e\u7136\uff0cn=2k+1\u65f6\uff0can=0\u3001n=2k\u65f6\uff0can=(-4/\u03c0)/[(2k+1)(2k-1)](k=1,2,\u2026\u2026\u221e\uff09\uff0c \u2234sinx=2/\u03c0+\u2211a2kcos2kx=2/\u03c0-(4/\u03c0)\u2211(cos2kx)/[(2k+1)(2k-1)]\uff0c\u5373\u2211(cos2nx)/[(2n+1)(2n-1)]=1/2-(\u03c0/4)sinx(n=1,2,\u2026\u2026\uff0c\u221e\uff09\u3002\u4f9b\u53c2\u8003\u3002
因为∫axcosnxdx=ax/n*sin(nx)-a/n∫sin(nx)dx=ax/n*sin(nx)+a/n²*cos(nx)+C∫axsinnxdx=-ax/n*cos(nx)+a/n∫cos(nx)dx=a/n²*sin(nx)-ax/n*cos(nx)+C
所以an=∫(-π到π)axcosnxdx=0
bn=∫(-π到π)axsinnxdx=-2aπ/n*cos(nπ)
故若n为奇数，则bn=2aπ/n
若n为偶数，则bn=-2aπ/n
所以函数f(x)的傅里叶级数为
f(x)=2aπ*sinx-2aπ/2*sin2x+2aπ/3*sin3x-2aπ/4*sin4x+……
绛旓細鎵浠ュ嚱鏁癴(x)鐨鍌呴噷鍙剁骇鏁涓 f(x)=2a蟺*sinx-2a蟺/2*sin2x+2a蟺/3*sin3x-2a蟺/4*sin4x+鈥︹
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