[f(x)=frac{a_{0}}{2}+sum_{n=1}^{infty}left(a_{n} cos n x+b_{n} sin n x
ight)
]
不难看出其中的线性组合形式,最关键的就是解出这些系数。
一种解法
第一步先求 (a_0) :
[int_{0}^{2pi} f(x) d x=int_{0}^{2pi} frac{a_{0}}{2} d x+int_{0}^{2pi} sum_{n=1}^{infty} a_{n} cos n x d x+int_{0}^{2pi} sum_{n=1}^{infty} b_{n} sin n x d x
]
根据三角函数系基函数的正交性:
[egin{aligned}int_{0}^{2pi} sum_{n=1}^{infty} a_{n} cos n x d x+int_{0}^{2pi} sum_{n=1}^{infty} b_{n} sin n x d x &= a_{n}int_{0}^{2pi} sum_{n=1}^{infty} 1cdotcos n x d x+ b_{n} int_{0}^{2pi} sum_{n=1}^{infty}1cdotsin n x d x\&=a_n cdot0+b_ncdot 0\&=0end{aligned}
]
所以:
[int_{0}^{2pi} f(x) d x=int_{0}^{2pi} frac{a_{0}}{2} d x
]
[int_{0}^{2pi} f(x) cos m x d x=int_{0}^{2pi} frac{a_0}{2} cos m x d x+int_{0}^{2pi} sum_{n=1}^{infty} a_{n} cos n x cos m x d x+int_{0}^{2pi} sum_{n=1}^{infty}b_{n} sin nx cos m xd x
]
根据三角函数系基函数的正交性:
[int_{0}^{2pi} frac{a_0}{2} cos m x d x =0\int_{0}^{2pi} sum_{n=1}^{infty}b_{n} sin nx cos m xd x=0\int_{0}^{2pi} sum_{n=1}^{infty} a_{n} cos n x cos m x d x = int_{0}^{2pi} a_{n} cos n x cos m x d x quad m=n
]
带入得:
[int_{0}^{2pi} f(x) cos n x d x=int_{0}^{2pi} a_{n} cos^2 n x d x
]
其中 (int_{0}^{2pi} cos^2 n x d x = pi) ,可得:
[a_{n}=frac{1}{pi} int_{0}^{2pi} f(x) cos n x d x
]
第三步求 (b_n) :
与求 (a_n) 同理,先对等式两边同乘 (sin mx) ,经过计算可得:
[b_{n}=frac{1}{pi} int_{0}^{2pi} f(x) sin n x d x
]
综上,我们可以对周期为2(pi)的周期函数按照傅里叶级数展开:
[f(x)=frac{a_{0}}{2}+sum_{n=1}^{infty} a_{n} cos n x+sum_{n=1}^{infty} b_n sin n x
]
其中:
[egin{aligned}&a_{0}=frac{1}{ pi} int_{0}^{2pi} f(x) d x\&a_{n}=frac{1}{pi} int_{0}^{2pi} f(x) cos n x d x\&b_{n}=frac{1}{pi} int_{0}^{2pi} f(x) sin n x d xend{aligned}
]
[x = frac{pi}{L} t \ t = frac{L}{pi}x\f(t) = f(frac{L}{pi}x) = g(x)
]
根据 (f(t) = f(t+2L)) ,有:
[g(x) = g(x+2pi)
]
这样一来便可以通过对周期 (T = 2pi) 的函数 (g(x)) 按照傅里叶级数展开:
[g(x)=frac{a_{0}}{2}+sum_{n=1}^{infty} a_{n} cos n x+sum_{n=1}^{infty} b_n sin n x
]
其中:
[egin{aligned}&a_{0}=frac{1}{ pi} int_{0}^{2pi} g(x) d x\&a_{n}=frac{1}{pi} int_{0}^{2pi} g(x) cos n x d x\&b_{n}=frac{1}{pi} int_{0}^{2pi} g(x) sin n x d xend{aligned}
]
代入 (x = frac{pi}{L} t) 得:
[egin{aligned}f(t) &= g(x)\cos nx &= cos frac{npi}{L}t \sin nx &= sin frac{npi}{L}t\int_{0}^{2pi}dx &= int_{0}^{2L}dfrac{pi}{L}t\&=frac{pi}{L}int_{0}^{2L}dtend{aligned}
]
将上式代入函数 (g(x)) 的傅里叶级数展开可得:
[f(t)=frac{a_{0}}{2}+sum_{n=1}^{infty} a_{n} cos frac{npi}{L}t+sum_{n=1}^{infty} b_n sin frac{npi}{L}t
]
其中:
[egin{aligned}&a_{0}=frac{1}{L} int_{0}^{2pi} f(t) d t\&a_{n}=frac{1}{L} int_{0}^{2pi} f(t) cos frac{npi}{L}t\&b_{n}=frac{1}{L} int_{0}^{2pi} f(t) sin frac{npi}{L}tend{aligned}
]