泰勒展开 — Taylor Expansion

泰勒展开是希望基于某区间一点x0x_0x0展开，用一组简单的幂函数xax^axa来近似一个复杂的函数f(x)f(x)f(x)在该区间的局部。公式如下：
f(x)=a0+a1(x−x0)+a2(x−x0)2+a3(x−x0)3+...f(x) = a_0 + a_1(x - x_0) + a_2(x - x_0)^2 + a_3(x - x_0)^3 + ... f(x)=a0+a1(x−x0)+a2(x−x0)2+a3(x−x0)3+...

实际效果如下图所示。蓝色的线条是sinsinsin函数，黄色线条是sinsinsin函数不同阶（即上式右侧保留的项数）在x=0x=0x=0处的泰勒展开，随着阶数升高（保留其泰勒展开的更多项），其泰勒级数在更广的区间上近似sinsinsin函数。

泰勒展开的应用场景例如：我们很难直接求得sin(1)sin(1)sin(1)的值，但我们可以通过sinsinsin的泰勒级数求得sin(1)sin(1)sin(1)的近似值，且展开项越多，精度越高。
sin(1)=0.8414709848078965taylor_expansion_of_sin_with_order_3(1)=0.8333333333333334taylor_expansion_of_sin_with_order_5(1)=0.8416666666666667taylor_expansion_of_sin_with_order_7(1)=0.841468253968254sin(1) = 0.8414709848078965 \\ taylor\_expansion\_of\_sin\_with\_order\_3(1) = 0.8333333333333334 \\ taylor\_expansion\_of\_sin\_with\_order\_5(1) = 0.8416666666666667 \\ taylor\_expansion\_of\_sin\_with\_order\_7(1) = 0.841468253968254 sin(1)=0.8414709848078965taylor_expansion_of_sin_with_order_3(1)=0.8333333333333334taylor_expansion_of_sin_with_order_5(1)=0.8416666666666667taylor_expansion_of_sin_with_order_7(1)=0.841468253968254

现在我们来求得上式中各项的系数。为求a0a_0a0，令x=x0x=x_0x=x0：
f(x0)=a0+a1(x0−x0)+a2(x0−x0)2+a3(x0−x0)3+...f(x_0) = a_0 + a_1(x_0 - x_0) + a_2(x_0 - x_0)^2 + a_3(x_0 - x_0)^3 + ... f(x0)=a0+a1(x0−x0)+a2(x0−x0)2+a3(x0−x0)3+...
因此，a0=f(x0)a_0=f(x_0)a0=f(x0)。为求a1a_1a1，我们先对原始左右求导：
ddxf(x)=a1+2a2(x−x0)+3a3(x−x0)2+...\frac{d}{dx}f(x) = a_1 + 2a_2(x - x_0) + 3a_3(x - x_0)^2 + ... dxdf(x)=a1+2a2(x−x0)+3a3(x−x0)2+...
再令x=x0x=x_0x=x0：
ddxf(x0)=a1+2a2(x0−x0)+3a3(x0−x0)2+...\frac{d}{dx}f(x_0) = a_1 + 2a_2(x_0 - x_0) + 3a_3(x_0 - x_0)^2 + ... dxdf(x0)=a1+2a2(x0−x0)+3a3(x0−x0)2+...
因此，a1=dfdx∣x=x0a_1 = \frac{df}{dx}\Big|_{x=x_0}a1=dxdf∣∣∣x=x0。为求a2a_2a2，我们先对原始左右求二阶导：
d2dx2f(x)=2a2+3a3(x−x0)+4∗3a4(x−x0)2+...\frac{d^2}{dx^2}f(x) = 2a_2 + 3a_3(x - x_0) + 4*3a_4(x - x_0)^2 + ... dx2d2f(x)=2a2+3a3(x−x0)+4∗3a4(x−x0)2+...
再令x=x0x=x_0x=x0：
d2dx2f(x0)=2a2+3a3(x0−x0)+4∗3a4(x0−x0)2+...\frac{d^2}{dx^2}f(x_0) = 2a_2 + 3a_3(x_0 - x_0) + 4*3a_4(x_0 - x_0)^2 + ... dx2d2f(x0)=2a2+3a3(x0−x0)+4∗3a4(x0−x0)2+...
因此，a2=12d2fdx2∣x=x0a_2 = \frac{1}{2}\frac{d^2f}{dx^2}\Big|_{x=x_0}a2=21dx2d2f∣∣∣x=x0。如此往复，我们得到aka_kak的计算通式：
ak=1k!dkfdxk∣x=x0a_k = \frac{1}{k!}\frac{d^kf}{dx^k}\Big|_{x=x_0} ak=k!1dxkdkf∣∣∣x=x0
最后，我们得到函数f(x)f(x)f(x)在x=x0x=x_0x=x0处展开的泰勒级数：
f(x)=f(x0)+f′(x0)Δx+12!f′′(x0)(Δx)2+13!f′′′(x0)(Δx)3+...f(x) = f(x_0) + f^{\prime}(x_0)\Delta{x} + \frac{1}{2!}f^{\prime \prime}(x_0)(\Delta{x})^2 + \frac{1}{3!}f^{\prime \prime \prime}(x_0)(\Delta{x})^3 + ... f(x)=f(x0)+f′(x0)Δx+2!1f′′(x0)(Δx)2+3!1f′′′(x0)(Δx)3+...
其中，Δx=x−x0\Delta{x} = x-x_0Δx=x−x0。

对于f(x±h)f(x\pm h)f(x±h)的泰勒展开，我们只需要将上式中的xxx替换为x±hx \pm hx±h，x0x_0x0替换为xxx：
f(x±h)=f(x)+f′(x)(±h)+12!f′′(x)(±h)2+13!f′′′(x)(±h)3+...f(x \pm h) = f(x) + f^{\prime}(x)(\pm h) + \frac{1}{2!}f^{\prime \prime}(x)(\pm h)^2 + \frac{1}{3!}f^{\prime \prime \prime}(x)(\pm h)^3 + ... f(x±h)=f(x)+f′(x)(±h)+2!1f′′(x)(±h)2+3!1f′′′(x)(±h)3+...

二元函数f(x,y)f(x, y)f(x,y)的泰勒展开

对于f(x,y)f(x, y)f(x,y)在(x0,y0)(x_0, y_0)(x0,y0)处的泰勒展开为：
f(x,y)=f(x0,y0)+fx′(x0,y0)(x−x0)+fy′(x0,y0)(y−y0)+12!fx′′(x0,y0)(x−x0)2+12!fy′′(x0,y0)(y−y0)2+12!fxy′′(x0,y0)(x−x0)(y−y0)+12!fyx′′(x0,y0)(x−x0)(y−y0)+...\begin{aligned} f(x, y) = & f(x_0, y_0) + f^{\prime}_x(x_0,y_0)(x-x_0) + f^{\prime}_y(x_0,y_0)(y-y_0) \\ & + \frac{1}{2!}f^{\prime \prime}_x(x_0,y_0)(x-x_0)^2 + \frac{1}{2!}f^{\prime \prime}_y(x_0,y_0)(y-y_0)^2 \\ & + \frac{1}{2!}f^{\prime \prime}_{xy}(x_0,y_0)(x-x_0)(y-y_0) + \frac{1}{2!}f^{\prime \prime}_{yx}(x_0,y_0)(x-x_0)(y-y_0) \\ & + ... \end{aligned} f(x,y)=f(x0,y0)+fx′(x0,y0)(x−x0)+fy′(x0,y0)(y−y0)+2!1fx′′(x0,y0)(x−x0)2+2!1fy′′(x0,y0)(y−y0)2+2!1fxy′′(x0,y0)(x−x0)(y−y0)+2!1fyx′′(x0,y0)(x−x0)(y−y0)+...

多元函数f(x1,x2,...,xn)f(x_1, x_2, ..., x_n)f(x1,x2,...,xn)的泰勒展开

对于f(x1,x2,...,xn)f(x_1, x_2, ..., x_n)f(x1,x2,...,xn)在(x1,0,x2,0,...,xn,0)(x_{1,0}, x_{2,0}, ..., x_{n,0})(x1,0,x2,0,...,xn,0)处的泰勒展开为：
f(x1,x2,...,xn)=a0+∑m=1∞[∑jkj=m∑ax1,...,xn;m(x1−x1,0)k1…(xn−xn,0)kn]f(x_1, x_2, ..., x_n) = a_0 + \sum^{\infty}_{m=1}\Big[_{\sum_{j}k_j = m} \sum a_{x_1,...,x_n;m} (x_1-x_{1,0})^{k_1}\dots(x_n-x_{n,0})^{k_n}\Big] f(x1,x2,...,xn)=a0+m=1∑∞[∑jkj=m∑ax1,...,xn;m(x1−x1,0)k1…(xn−xn,0)kn]
系数aaa的计算通式为：
a0=f(x1,0,x2,0,...,xn,0)ax1,...,xn;m=1m∂mf∂k1x1…∂knxn∣x1,0,x2,0,...,xn,0\begin{aligned} & a_0 = f(x_{1,0}, x_{2,0}, ..., x_{n,0}) \\ & a_{x_1,...,x_n;m} = \frac{1}{m}\frac{\partial^mf}{\partial^{k_1}x_1\dots\partial^{k_n}x_n} \Big|_{x_{1,0}, x_{2,0}, ..., x_{n,0}} \end{aligned} a0=f(x1,0,x2,0,...,xn,0)ax1,...,xn;m=m1∂k1x1…∂knxn∂mf∣∣∣x1,0,x2,0,...,xn,0
其中，x1,...,xn;mx_1,...,x_n;mx1,...,xn;m表示对{x1,...,xn}\{x_1,...,x_n\}{x1,...,xn}进行mmm次有放回的排列组合。