题目

x^3怎么展开成余弦级数,然后求Σ1/n^4的极限
就是傅里叶展开

提问时间：2021-01-05

答案

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}=1/2
an=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