x^3怎么展开成余弦级数,然后求Σ1/n^4的极限

对函数f(x)=x³ (0≤x≤1)进行偶延拓a₀=2*∫{0,1}f(x)dx=2*∫{0,1} x³dx=1/2v*x⁴|{0,1}=1/2an=2*∫{0,1}f(x)*cos(n*π*x)dx=2*∫{0,1} x³*cos(n*π*x)dx=2/(n*π)*∫{0,1} x³ d[sin(n*π*x)]=2/(n*π)* x³* sin(n*π*x) |{0,1}-2/(n*π)∫{0,1} sin(n*π*x) d(x³)=-6/(n*π)∫{0,1} x²*sin(n*π*x) dx=6/(n²*π²)∫{0,1} x² d[cos(n*π*x)]=6/(n²*π²) x²* cos (n*π*x) |{0,1}-6/(n²*π²)*∫{0,1} cos(n*π*x) d(x²)=6*cos (n*π)/(n²*π²)-12/(n²*π²)*∫{0,1}x* cos(n*π*x) dx=6*cos (n*π)/(n²*π²)-12/(n³*π³)*∫{0,1}xd[sin(n*π*x)]=6*cos (n*π)/(n²*π²)-12/(n³*π³)*x* sin(n*π*x) |{0,1}+12/(n³*π³)*∫{0,1} sin(n*π*x) dx=6*cos (n*π)/(n²*π²)-12/(n⁴*π⁴)*cos(n*π*x) |{0,1}=6*cos (n*π)/(n²*π²)-12/(n⁴*π⁴)*[cos(n*π)-1]=6* (-1)^n/(n²*π²)-12/(n⁴*π⁴)*[(-1)^n -1]当n=2*k-1,k=1,2,。。。时,an= -6/[(2*k-1)²*π²]+24/[(2*k-1)⁴*π⁴]当n=2*k,k=1,2,。。。时,an=6/[(2*k)²*π²]f(x)= a₀/2+∑{n=1,∞} an*cos(n*π*x)= 1/4+∑{k=1,∞}{-6/[(2*k-1)²*π²]+24/[(2*k-1)⁴*π⁴]}*cos[(2*k-1)*π*x]+∑{k=1,∞}6/[(2*k)²*π²]*cos(2*k*π*x)当-1≤x≤0时,f(x)= -x³当0≤x≤1 时,f(x)= x³取x=0,则0=1/4-6/π²*∑{k=1,∞}1/(2*k-1)² +24/π⁴∑{k=1,∞} 1/(2*k-1)⁴+6/π²*∑{k=1,∞}1/(2*k)² ①注意到∑{k=1,∞}1/ k²=π²/6 ② （可对函数f(x)=x (0≤x≤1)进行偶延拓计算得到）∑{k=1,∞}1/(2*k)² =1/4*∑{k=1,∞}1/ k²=π²/24 ③∵∑{k=1,∞}1/ k²=∑{k=1,∞}1/(2*k)²+∑{k=1,∞}1/(2*k-1)²∴∑{k=1,∞}1/(2*k-1)² =3/4*∑{k=1,∞}1/ k²=π²/8 ④由①②③④式可得∑{k=1,∞} 1/(2*k-1)⁴=π⁴/96又∵∑{n=1,∞} 1/n⁴=∑{k=1,∞}1/(2*k)⁴+∑{k=1,∞}1/(2*k-1)⁴=1/16*∑{k=1,∞}1/k⁴+∑{k=1,∞}1/(2*k-1)⁴∴∑{k=1,∞} 1/k⁴=16/15*∑{k=1,∞}1/(2*k-1)⁴=16/15*π⁴/96=π⁴/90即∑{n=1,∞} 1/n⁴=π⁴/90