首页数学和编程微积分

# 三角函数的正交性

参考视频

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给出一个周期函数，能否表示为傅里叶级数，而傅里叶级数中的每一个参数是怎么来的？
之后系列第一个问题的回答是:

# 三角函数系

{0,1,sin⁡x,cos⁡x,sin⁡2x,cos⁡2x,…,sin⁡nx,cos⁡nx}⇓{sin⁡0x,cos⁡0x,sin⁡2x,cos⁡2x,…,sin⁡nx,cos⁡nx}\{0, 1, \sin x, \cos x, \sin 2x, \cos 2x, \ldots, \sin nx, \cos nx \} \\ \Downarrow \\ \{ \sin 0x, \cos 0x, \sin 2x, \cos 2x, \ldots, \sin nx, \cos nx \}

# 正交

∫−ππsin⁡nxcos⁡mx dx=0(n≠m)\int_{-\pi}^{\pi} \sin nx \cos mx \, dx = 0 \quad (n \neq m)

注：当 n=mn = m 时，上式仍然为零。

∫−ππcos⁡nxcos⁡mx dx=0(n≠m)\int_{-\pi}^{\pi} \cos nx \cos mx \, dx = 0 \quad (n \neq m)

注：当 n=mn = m 时，上式为 π\pi。

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a⃗=(1,2,5)b⃗=(1,2,−1)a⃗⋅b⃗=1⋅1+2⋅2−5⋅1=0\vec{a} = (1,2,5) \\ \vec{b} = (1,2,-1) \\ \vec{a} \cdot \vec{b} = 1 \cdot 1 + 2 \cdot 2 - 5 \cdot 1 = 0

推广得

a⃗=(a1,a2,a3,…,an)b⃗=(b1,b2,…,bn)a⃗⋅b⃗=a1b1+a2b2+…+anbn=∑i=1naibi=0\vec{a} = (a_1, a_2, a_3, \ldots, a_n) \\ \vec{b} = (b_1, b_2, \ldots, b_n) \\ \vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + \ldots + a_n b_n = \sum_{i=1}^{n} a_i b_i = 0

继续推广

a=f(x)b=g(x)a = f(x) \\ b = g(x)

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加和变成取积分，得到

a⋅b=∫x0x1f(x)g(x)dx=0a \cdot b = \int_{x_0}^{x_1} f(x) g(x) dx = 0

这样也称为正交

# 周期为 2π2\pi 的函数展开为傅里叶级数

参考视频

T=2πf(x)=f(x+2π)T = 2\pi f(x) = f(x + 2\pi)

展开得

f(x)=∑n=0∞ancos⁡nx+∑n=0∞bnsin⁡nxf(x) = \sum_{n=0}^{\infty} a_n \cos nx + \sum_{n=0}^{\infty} b_n \sin nx

而教科书上一般是：

f(x)=a02+∑n=1∞(ancos⁡nx+bnsin⁡nx)f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n \cos nx + b_n \sin nx)

两者是等价得，证明如下：

f(x)=∑n=0∞ancos⁡nx+∑n=0∞bnsin⁡nx=a0cos0x+∑n=1∞ancos⁡nx+b0sin0x+∑n=1∞bnsin⁡nx=a0+∑n=1∞ancos⁡nx+∑n=1∞bnsin⁡nxf(x) = \sum_{n=0}^{\infty} a_n \cos nx + \sum_{n=0}^{\infty} b_n \sin nx = a_0cos0x + \sum_{n=1}^{\infty} a_n \cos nx + b_0sin0x + \sum_{n=1}^{\infty} b_n \sin nx = a_0 + \sum_{n=1}^{\infty} a_n \cos nx + \sum_{n=1}^{\infty} b_n \sin nx

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∫−ππf(x)dx=a0∫−ππdx=a0x∣−ππ=2π⇒a0=12π∫−ππf(x)dx\int_{-\pi}^{\pi}f(x)dx=a_0\int_{-\pi}^{\pi}dx=a_0x|_{-\pi}^{\pi}=2\pi \\ \Rightarrow a_0=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)dx

如果在推导过程中把a0a_0 变成a02\frac{a_0}{2}，那么最终得到

a0=1π∫−ππf(x)dxa_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x)dx

这样看起来更舒服，而且和下面系数ana_n 的数学表达形式一致

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∫−ππf(x)cosmxdx=∫−ππ∑n=1∞ancosnxcosmxdx\int_{-\pi}^{\pi}f(x)cosmxdx= \int_{-\pi}^{\pi}\sum_{n=1}^{\infty}a_n cosnx cosmx dx

由于三角函数系得正交性，只有 m=n 时，右边得式子才不为零，即

∫−ππf(x)cosmxdx=∫−ππancosnxcosnxdx=an∫−ππcos2nxdx=anπ⇒an=1π∫−ππf(x)cosnxdx\int_{-\pi}^{\pi}f(x)cosmxdx= \int_{-\pi}^{\pi}a_n cosnx cosnx dx = a_n \int_{-\pi}^{\pi}cos^2nx dx = a_n\pi \Rightarrow a_n = \frac{1}{\pi}\int_{-\pi}^{\pi}f(x)cosnxdx

同理得：

bn=1π∫−ππf(x)sin⁡nxdxb_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f(x) \sin nx dx

总结

T=2πf(x)=f(x+2π)T = 2\pi f(x) = f(x + 2\pi)

展开得

f(x)=a02+∑n=1∞(ancos⁡nx+bnsin⁡nx)f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n \cos nx + b_n \sin nx)

其中

a0=12π∫−ππf(x)dxan=1π∫−ππf(x)cosnxdxbn=1π∫−ππf(x)sin⁡nxdxa_0 =\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)dx \\ a_n = \frac{1}{\pi}\int_{-\pi}^{\pi}f(x)cosnxdx \\ b_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f(x) \sin nx dx

# 周期为 2L2L 的函数展开为傅里叶级数

参考视频

f(t)=f(t+2L)f(t) = f(t + 2L)

换元

x=πLt⇒t=Lπxx = \frac{\pi}{L} t \quad \Rightarrow \quad t = \frac{L}{\pi} x

tt xx
2L2L 2π2\pi
4L4L 4π4\pi
00 00

函数 g(x)g(x) 表达为 f(t)f(t) 后，周期T=2πT=2\pi，我们得到以下等式：

f(t)=f(Lπx)⇔g(x)f(t) = f\left(\frac{L}{\pi} x\right) \Leftrightarrow g(x)

其中 g(x)g(x) 的周期 T=2πT=2\pi 为：

g(x)=g(x+2π)g(x) = g(x + 2\pi)

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g(x)=a02+∑n=1∞(ancos⁡(nx)+bnsin⁡(nx))a0=1π∫−ππg(x)dxan=1π∫−ππg(x)cos⁡(nx)dxbn=1π∫−ππg(x)sin⁡(nx)dxg(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left( a_n \cos(nx) + b_n \sin(nx) \right) \\ a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} g(x) dx \\ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} g(x) \cos(nx) dx \\ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} g(x) \sin(nx) dx \\

x=πLtcos⁡(nx)=cos⁡(nπLt)sin⁡(nx)=sin⁡(nπLt)g(x)=f(t)∫−ππdx=∫−LLd(πLt)x = \frac{\pi}{L} t\cos(nx) = \cos\left(n \frac{\pi}{L} t\right)\sin(nx) = \sin\left(n \frac{\pi}{L} t\right)g(x) = f(t)\int_{-\pi}^{\pi} dx = \int_{-L}^{L} d\left( \frac{\pi}{L} t \right)

xt−π−LπL\begin{array}{|c|c|} \hline x & t \\ \hline -\pi & -L \\ \hline \pi & L \\ \hline \end{array}

1π∫−ππdx=1π⋅πL∫−LLdt=1L∫−LLdt\frac{1}{\pi} \int_{-\pi}^{\pi} dx = \frac{1}{\pi} \cdot \frac{\pi}{L} \int_{-L}^{L} dt = \frac{1}{L} \int_{-L}^{L} dt

g(x)=a02+∑n=1∞(ancos⁡(nx)+bnsin⁡(nx))⇒f(t)=a02+∑n=1∞(ancos⁡(nπLt)+bnsin⁡(nπLt))g(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left( a_n \cos(nx) + b_n \sin(nx) \right) \\ \Rightarrow f(t) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left( a_n \cos\left(n \frac{\pi}{L} t\right) + b_n \sin\left(n \frac{\pi}{L} t\right) \right)

a0=1π∫−ππg(x) dx⇒a0=1L∫−LLf(t) dtan=1π∫−ππg(x)cos⁡(nx) dx⇒an=1L∫−LLf(t)cos⁡(nπLt) dtbn=1π∫−ππg(x)sin⁡(nx) dx⇒bn=1L∫−LLf(t)sin⁡(nπLt) dta_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} g(x) \, dx \Rightarrow a_0 = \frac{1}{L} \int_{-L}^{L} f(t) \, dt \\ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} g(x) \cos(nx) \, dx \Rightarrow a_n = \frac{1}{L} \int_{-L}^{L} f(t) \cos\left(n \frac{\pi}{L} t\right) \, dt \\ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} g(x) \sin(nx) \, dx \Rightarrow b_n = \frac{1}{L} \int_{-L}^{L} f(t) \sin\left(n \frac{\pi}{L} t\right) \, dt

工程中：时间 tt 从 00 开始，周期为 T=2LT = 2L，角频率为 ω=πL=2πT\omega = \frac{\pi}{L} = \frac{2\pi}{T}

∫−LLdt→∫02Ldt→∫0Tdt \int_{-L}^{L} dt \rightarrow \int_{0}^{2L} dt \rightarrow \int_{0}^{T} dt

f(t)=a02+∑n=1∞(ancos⁡(nωt)+bnsin⁡(nωt)) f(t) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left( a_n \cos(n \omega t) + b_n \sin(n \omega t) \right)

a0=2T∫0Tf(t) dt a_0 = \frac{2}{T} \int_{0}^{T} f(t) \, dt

an=2T∫0Tf(t)cos⁡(nωt) dt a_n = \frac{2}{T} \int_{0}^{T} f(t) \cos(n \omega t) \, dt

bn=2T∫0Tf(t)sin⁡(nωt) dt b_n = \frac{2}{T} \int_{0}^{T} f(t) \sin(n \omega t) \, dt

但是当 T→∞T \rightarrow \infty 时，f(t)f(t) 就不再为周期函数。

# 傅里叶级数的复数形式

参考视频

  • 角频率：

    ω=πL=2πT\omega = \frac{\pi}{L} = \frac{2\pi}{T}

  • 积分关系：

    ∫−LLdt→∫02Ldt→∫0Tdt\int_{-L}^{L} dt \rightarrow \int_{0}^{2L} dt \rightarrow \int_{0}^{T} dt

  • 函数表示：

    f(t)=a02+∑n=1∞(ancos⁡(nωt)+bnsin⁡(nωt))f(t) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left( a_n \cos(n \omega t) + b_n \sin(n \omega t) \right)

  • 系数计算：

    a0=2T∫0Tf(t) dta_0 = \frac{2}{T} \int_{0}^{T} f(t) \, dt

    an=2T∫0Tf(t)cos⁡(nωt) dta_n = \frac{2}{T} \int_{0}^{T} f(t) \cos(n \omega t) \, dt

    bn=2T∫0Tf(t)sin⁡(nωt) dtb_n = \frac{2}{T} \int_{0}^{T} f(t) \sin(n \omega t) \, dt

欧拉公式

eiθ=cos⁡θ+isin⁡θ⇓cos⁡θ=12(eiθ+e−iθ)sin⁡θ=−12i(eiθ−e−iθ)e^{i\theta} = \cos\theta + i\sin\theta \\ \Downarrow \\ \cos\theta = \frac{1}{2} \left( e^{i\theta} + e^{-i\theta} \right) \\ \sin\theta = -\frac{1}{2i} \left( e^{i\theta} - e^{-i\theta} \right)

  • 将函数f(t)f(t) 重新表示为复数形式：

    f(t)=a02+∑n=1∞[an12(einωt+e−inωt)−12ibn(einωt−e−inωt)]f(t) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left[ a_n \frac{1}{2}\left( e^{i n \omega t} + e^{-i n \omega t} \right) - \frac{1}{2i} b_n \left( e^{i n \omega t} - e^{-i n \omega t} \right) \right]

  • 进一步整理得到：

    f(t)=a02+∑n=1∞[an−ibn2einωt+an+ibn2e−inωt]f(t) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left[\frac{a_n - i b_n}{2} e^{i n \omega t} + \frac{a_n + i b_n}{2} e^{-i n \omega t}\right]

  • 最终表示为：

    f(t)=∑−∞∞Cneinωtf(t) = \sum_{-\infty}^{\infty} C_n e^{i n \omega t}

系数CnC_n

Cn={a02,n=0an−ibn2,n=1,2,3,…a−n+ib−n2,n=−1,−2,−3,…C_n = \begin{cases} \frac{a_0}{2}, & n = 0 \\ \frac{a_n - i b_n}{2}, & n = 1, 2, 3, \ldots \\ \frac{a_{-n} + i b_{-n}}{2}, & n = -1, -2, -3, \ldots \end{cases}

  • n=0n = 0:

    C0=a02=1T∫0Tf(t) dtC_0 = \frac{a_0}{2} = \frac{1}{T} \int_{0}^{T} f(t) \, dt

  • n=1,2,3,…n = 1, 2, 3, \ldots:

    Cn=12(2T∫0Tf(t)cos⁡(nωt) dt−i2T∫0Tf(t)sin⁡(nωt) dt)C_n = \frac{1}{2} \left( \frac{2}{T} \int_{0}^{T} f(t) \cos(n \omega t) \, dt - i \frac{2}{T} \int_{0}^{T} f(t) \sin(n \omega t) \, dt \right)

    =1T∫0Tf(t)(cos⁡(nωt)−isin⁡(nωt))dt= \frac{1}{T} \int_{0}^{T} f(t) \left( \cos(n \omega t) - i \sin(n \omega t) \right) dt

  • n=−1,−2,−3,…n = -1, -2, -3, \ldots:

    Cn=12[2T∫0Tf(t)cos⁡(−nωt) dt+i2T∫0Tf(t)sin⁡(−nωt) dt]C_n = \frac{1}{2} \left[ \frac{2}{T} \int_{0}^{T} f(t) \cos(-n \omega t) \, dt + i \frac{2}{T} \int_{0}^{T} f(t) \sin(-n \omega t) \, dt \right]

    =1T∫0Tf(t)[cos⁡(−nωt)+isin⁡(−nωt)]dt= \frac{1}{T} \int_{0}^{T} f(t) \left[ \cos(-n \omega t) + i \sin(-n \omega t) \right] dt

f(t)=f(t+T)f(t) = f(t + T)

f(t)=∑−∞∞Cneinωtf(t) = \sum_{-\infty}^{\infty} C_n e^{i n \omega t}

Cn=1T∫0Tf(t)e−inωt dtC_n = \frac{1}{T} \int_{0}^{T} f(t) e^{-i n \omega t} \, dt