三角函数 sinx, cosx 的泰勒展开推导及两个巧妙应用

  假设函数 f f f 充分光滑，即 f f f 在 x 0 x_0 x0​ 点处任意阶导数存在，取一小邻域 U ( x 0 ) \small U(x_0) U(x0​)，则 ∀   x ∈ U ( x 0 ) \small \forall\,x\in U(x_0) ∀x∈U(x0​)， f ( x ) \small f(x) f(x) 都可以展开为泰勒级数，即 f ( x ) = ∑ n = 0 ∞ f ( n ) ( x 0 ) n ! ( x − x 0 ) n = f ( x 0 ) + f ′ ( x 0 ) ( x − x 0 ) + f ′ ′ ( x 0 ) 2 ! ( x − x 0 ) 2 + ⋯ f(x)=\sum_{n=0}^{\infin}\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n=f(x_0)+f'(x_0)(x-x_0)+\frac{f''(x_0)}{2!}(x-x_0)^2+\cdots f(x)=n=0∑∞​n!f(n)(x0​)​(x−x0​)n=f(x0​)+f′(x0​)(x−x0​)+2!f′′(x0​)​(x−x0​)2+⋯考虑三角函数 sin ⁡ x , cos ⁡ x \sin x,\cos x sinx,cosx 的任意阶导函数， sin ⁡ ′ ( x ) = cos ⁡ ( x ) = sin ⁡ ( x + π 2 ) sin ⁡ ′ ′ ( x ) = cos ⁡ ′ ( x ) = sin ⁡ ′ ( x + π 2 ) = cos ⁡ ( x + π 2 ) = sin ⁡ ( x + π 2 + π 2 ) = sin ⁡ ( x + π ) \begin{aligned}\sin'(x)&=\cos(x)=\sin (x+\frac{\pi}{2})\\ \sin''(x)&=\cos'(x)=\sin'(x+\frac{\pi}{2})\\&=\cos (x+\frac{\pi}{2})=\sin (x+\frac{\pi}{2}+\frac{\pi}{2})\\&=\sin (x+\pi) \end{aligned} sin′(x)sin′′(x)​=cos(x)=sin(x+2π​)=cos′(x)=sin′(x+2π​)=cos(x+2π​)=sin(x+2π​+2π​)=sin(x+π)​假设 sin ⁡ ( n ) ( x ) = sin ⁡ ( x + n π / 2 ) \sin^{(n)}(x)=\sin(x+n\pi/2) sin(n)(x)=sin(x+nπ/2)，下面我们用 数学归纳法 来验证一下.

  首先，当 n = 0 n=0 n=0 时， sin ⁡ ( 0 ) ( x ) = sin ⁡ ( x ) \sin^{(0)}(x)=\sin(x) sin(0)(x)=sin(x)，结论显然成立.

  其次，假设当 n = k ( k ≥ 0 ) n=k(k\geq 0) n=k(k≥0) 时结论成立，即 sin ⁡ ( k ) ( x ) = sin ⁡ ( x + k π / 2 ) \displaystyle\sin^{(k)}(x)=\sin(x+k\pi/2) sin(k)(x)=sin(x+kπ/2).

  考虑 n = k + 1 n=k+1 n=k+1 的情形，此时 sin ⁡ ( k + 1 ) ( x ) = d ( sin ⁡ ( k ) ( x ) ) d x = d ( sin ⁡ ( x + k π / 2 ) ) d x = cos ⁡ ( x + k π / 2 ) = sin ⁡ ( x + k π / 2 + π / 2 ) = sin ⁡ ( x + ( k + 1 ) π / 2 ) \begin{aligned} \sin^{(k+1)}(x)&=\frac{d(\sin^{(k)}(x))}{dx}\\&=\frac{d(\sin(x+k\pi/2))}{dx}\\&=\cos(x+k\pi/2)\\&=\sin(x+k\pi/2+\pi/2)\\&=\sin(x+(k+1)\pi/2) \end{aligned} sin(k+1)(x)​=dxd(sin(k)(x))​=dxd(sin(x+kπ/2))​=cos(x+kπ/2)=sin(x+kπ/2+π/2)=sin(x+(k+1)π/2)​  所以，当 n = k + 1 n=k+1 n=k+1 时，结论照样成立.

  由归纳法原理， ∀   n ∈ N ,   sin ⁡ ( n ) ( x ) = sin ⁡ ( x + n π / 2 ) \forall\,n\in N,\,\sin^{(n)}(x)=\sin(x+n\pi/2) ∀n∈N,sin(n)(x)=sin(x+nπ/2).

则 cos ⁡ ( n ) ( x ) = sin ⁡ ( n + 1 ) ( x ) = sin ⁡ ( x + n π / 2 + π / 2 ) = cos ⁡ ( x + n π / 2 ) \cos^{(n)}(x)=\sin^{(n+1)}(x)=\sin(x+n\pi/2+\pi/2)=\cos(x+n\pi/2) cos(n)(x)=sin(n+1)(x)=sin(x+nπ/2+π/2)=cos(x+nπ/2).

  特别地，取 x = 0 x=0 x=0，

sin ⁡ ( 2 k ) ( 0 ) = sin ⁡ ( k π ) = 0 sin ⁡ ( 2 k + 1 ) ( 0 ) = sin ⁡ ( k π + π / 2 ) = ( − 1 ) k cos ⁡ ( 2 k ) ( 0 ) = cos ⁡ ( k π ) = ( − 1 ) k cos ⁡ ( 2 k + 1 ) ( 0 ) = cos ⁡ ( k π + π / 2 ) = 0 ( k = 0 , 1 , 2 , ⋯   ) \begin{aligned} &\sin^{(2k)}(0)=\sin(k\pi)=0 \\ &\sin^{(2k+1)}(0)=\sin(k\pi+\pi/2)=(-1)^{k} \\ \\ &\cos^{(2k)}(0)=\cos(k\pi)=(-1)^{k} \\ &\cos^{(2k+1)}(0)=\cos(k\pi+\pi/2)=0 \\ &(k=0,1,2,\cdots) \end{aligned} ​sin(2k)(0)=sin(kπ)=0sin(2k+1)(0)=sin(kπ+π/2)=(−1)kcos(2k)(0)=cos(kπ)=(−1)kcos(2k+1)(0)=cos(kπ+π/2)=0(k=0,1,2,⋯)​  取 x 0 = 0 x_0=0 x0​=0 的某一小邻域，将 sin ⁡ x , cos ⁡ x \sin x,\cos x sinx,cosx 泰勒展开 sin ⁡ x = ∑ n = 0 ∞ sin ⁡ ( n ) ( 0 ) n ! x n = ∑ k = 0 ∞ sin ⁡ ( 2 k ) ( 0 ) ( 2 k ) ! x 2 k + ∑ k = 0 ∞ sin ⁡ ( 2 k + 1 ) ( 0 ) ( 2 k + 1 ) ! x 2 k + 1 = ∑ k = 0 ∞ 0 ( 2 k ) ! x 2 k + ∑ k = 0 ∞ ( − 1 ) k ( 2 k + 1 ) ! x 2 k + 1 = ∑ k = 0 ∞ ( − 1 ) k ( 2 k + 1 ) ! x 2 k + 1 = x − 1 3 ! x 3 + 1 5 ! x 5 − 1 7 ! x 7 + ⋯ cos ⁡ x = ∑ n = 0 ∞ cos ⁡ ( n ) ( 0 ) n ! x n = ∑ k = 0 ∞ cos ⁡ ( 2 k ) ( 0 ) ( 2 k ) ! x 2 k + ∑ k = 0 ∞ cos ⁡ ( 2 k + 1 ) ( 0 ) ( 2 k + 1 ) ! x 2 k + 1 = ∑ k = 0 ∞ ( − 1 ) k ( 2 k ) ! x 2 k + ∑ k = 0 ∞ 0 ( 2 k + 1 ) ! x 2 k + 1 = ∑ k = 0 ∞ ( − 1 ) k ( 2 k ) ! x 2 k = 1 − 1 2 ! x 2 + 1 4 ! x 4 − 1 6 ! x 6 + ⋯ \begin{aligned} \sin x&=\sum_{n=0}^{\infin}\frac{\sin^{(n)}(0)}{n!}x^n\\ &=\sum_{k=0}^{\infin}\frac{\sin^{(2k)}(0)}{(2k)!}x^{2k}+\sum_{k=0}^{\infin}\frac{\sin^{(2k+1)}(0)}{(2k+1)!}x^{2k+1}\\ &=\sum_{k=0}^{\infin}\frac{0}{(2k)!}x^{2k}+\sum_{k=0}^{\infin}\frac{(-1)^k}{(2k+1)!}x^{2k+1}\\ &=\sum_{k=0}^{\infin}\frac{(-1)^k}{(2k+1)!}x^{2k+1}\\ &=x-\frac{1}{3!}x^3+\frac{1}{5!}x^5-\frac{1}{7!}x^7+\cdots\\ \\ \cos x&=\sum_{n=0}^{\infin}\frac{\cos^{(n)}(0)}{n!}x^n\\ &=\sum_{k=0}^{\infin}\frac{\cos^{(2k)}(0)}{(2k)!}x^{2k}+\sum_{k=0}^{\infin}\frac{\cos^{(2k+1)}(0)}{(2k+1)!}x^{2k+1}\\ &=\sum_{k=0}^{\infin}\frac{(-1)^{k}}{(2k)!}x^{2k}+\sum_{k=0}^{\infin}\frac{0}{(2k+1)!}x^{2k+1}\\ &=\sum_{k=0}^{\infin}\frac{(-1)^{k}}{(2k)!}x^{2k}\\ &=1-\frac{1}{2!}x^2+\frac{1}{4!}x^4-\frac{1}{6!}x^6+\cdots \end{aligned} sinxcosx​=n=0∑∞​n!sin(n)(0)​xn=k=0∑∞​(2k)!sin(2k)(0)​x2k+k=0∑∞​(2k+1)!sin(2k+1)(0)​x2k+1=k=0∑∞​(2k)!0​x2k+k=0∑∞​(2k+1)!(−1)k​x2k+1=k=0∑∞​(2k+1)!(−1)k​x2k+1=x−3!1​x3+5!1​x5−7!1​x7+⋯=n=0∑∞​n!cos(n)(0)​xn=k=0∑∞​(2k)!cos(2k)(0)​x2k+k=0∑∞​(2k+1)!cos(2k+1)(0)​x2k+1=k=0∑∞​(2k)!(−1)k​x2k+k=0∑∞​(2k+1)!0​x2k+1=k=0∑∞​(2k)!(−1)k​x2k=1−2!1​x2+4!1​x4−6!1​x6+⋯​则 sin ⁡ x = ∑ k = 0 ∞ ( − 1 ) k ( 2 k + 1 ) ! x 2 k + 1 = x − 1 3 ! x 3 + 1 5 ! x 5 − 1 7 ! x 7 + ⋯ cos ⁡ x = ∑ k = 0 ∞ ( − 1 ) k ( 2 k ) ! x 2 k = 1 − 1 2 ! x 2 + 1 4 ! x 4 − 1 6 ! x 6 + ⋯ \begin{aligned} &\sin x=\sum_{k=0}^{\infin}\frac{(-1)^k}{(2k+1)!}x^{2k+1} =x-\frac{1}{3!}x^3+\frac{1}{5!}x^5-\frac{1}{7!}x^7+\cdots \\ &\cos x=\sum_{k=0}^{\infin}\frac{(-1)^{k}}{(2k)!}x^{2k} =1-\frac{1}{2!}x^2+\frac{1}{4!}x^4-\frac{1}{6!}x^6+\cdots \end{aligned} ​sinx=k=0∑∞​(2k+1)!(−1)k​x2k+1=x−3!1​x3+5!1​x5−7!1​x7+⋯cosx=k=0∑∞​(2k)!(−1)k​x2k=1−2!1​x2+4!1​x4−6!1​x6+⋯​下面打算用这两个式子证些好玩儿的东西.

e i θ = cos ⁡ θ + i sin ⁡ θ e^{i\theta}=\cos\theta+i\sin\theta eiθ=cosθ+isinθ  首先将 e x e^x ex 在 x 0 = 0 x_0=0 x0​=0 处进行泰勒展开，已知 ( e x ) ′ = e x (e^x)'=e^x (ex)′=ex，则 ( e x ) ( n ) ∣ x = 0 = e x ∣ x = 0 = 1 (e^x)^{(n)}|_{x=0}=e^x|_{x=0}=1 (ex)(n)∣x=0​=ex∣x=0​=1，所以 e x = ∑ n = 0 ∞ ( e x ) ( n ) ∣ x = 0 n ! x n = ∑ n = 0 ∞ 1 n ! x n = 1 + x + 1 2 ! x 2 + 1 3 ! x 3 + ⋯ e^x=\sum_{n=0}^{\infin}\frac{(e^x)^{(n)}|_{x=0}}{n!}x^n=\sum_{n=0}^{\infin}\frac{1}{n!}x^n=1+x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+\cdots ex=n=0∑∞​n!(ex)(n)∣x=0​​xn=n=0∑∞​n!1​xn=1+x+2!1​x2+3!1​x3+⋯将 x x x 用 i θ i\theta iθ 代替，得 e i θ = 1 + i θ − 1 2 ! θ 2 − i 3 ! θ 3 + 1 4 ! θ 4 + i 5 ! θ 5 − 1 6 ! θ 6 − i 7 ! θ 7 + ⋯ = ( 1 − 1 2 ! θ 2 + 1 4 ! θ 4 − 1 6 θ 6 + ⋯   ) + i ( θ − 1 3 ! θ 3 + 1 5 ! θ 5 − 1 7 ! θ 7 + ⋯   ) \begin{aligned} e^{i\theta}&=1+i\theta-\frac{1}{2!}\theta^2-\frac{i}{3!}\theta^3+\frac{1}{4!}\theta^4+\frac{i}{5!}\theta^5-\frac{1}{6!}\theta^6-\frac{i}{7!}\theta^7+\cdots\\ &=(1-\frac{1}{2!}\theta^2+\frac{1}{4!}\theta^4-\frac{1}{6}\theta^6+\cdots)+i(\theta-\frac{1}{3!}\theta^3+\frac{1}{5!}\theta^5-\frac{1}{7!}\theta^7+\cdots) \end{aligned} eiθ​=1+iθ−2!1​θ2−3!i​θ3+4!1​θ4+5!i​θ5−6!1​θ6−7!i​θ7+⋯=(1−2!1​θ2+4!1​θ4−61​θ6+⋯)+i(θ−3!1​θ3+5!1​θ5−7!1​θ7+⋯)​将 1 − 1 2 ! θ 2 + 1 4 ! θ 4 − 1 6 θ 6 + ⋯ = cos ⁡ θ θ − 1 3 ! θ 3 + 1 5 ! θ 5 − 1 7 ! θ 7 + ⋯ = sin ⁡ θ \begin{aligned} 1-\frac{1}{2!}\theta^2+\frac{1}{4!}\theta^4-\frac{1}{6}\theta^6+\cdots=\cos\theta\\ \theta-\frac{1}{3!}\theta^3+\frac{1}{5!}\theta^5-\frac{1}{7!}\theta^7+\cdots=\sin\theta \end{aligned} 1−2!1​θ2+4!1​θ4−61​θ6+⋯=cosθθ−3!1​θ3+5!1​θ5−7!1​θ7+⋯=sinθ​代入，得 e i θ = cos ⁡ θ + i sin ⁡ θ e^{i\theta}=\cos\theta+i\sin\theta eiθ=cosθ+isinθ特别地，考虑 θ = π \theta=\pi θ=π 时， e i π + 1 = 0 e^{i\pi}+1=0 eiπ+1=0这是一个奇妙的公式，同时包含了数学中的 5 个重要的常数： e , i , π , 1 , 0 e,i,\pi,1,0 e,i,π,1,0.

另外， e i θ e^{i\theta} eiθ 的模长为1，常用于积分换元. 证明一下:

∑ n = 1 ∞ 1 n 2 = 1 + 1 2 2 + 1 3 2 + 1 4 2 + ⋯ = π 2 6 \sum_{n=1}^{\infin}\frac{1}{n^2}=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots=\frac{\pi^2}{6} n=1∑∞​n21​=1+221​+321​+421​+⋯=6π2​  虽然不叫欧拉公式，但这种解法同样是欧拉给出来的，下面让我们欣赏下欧拉的伟大思路.

  考虑函数 f ( x ) = sin ⁡ x x ( x ≠ 0 ) ,    lim ⁡ x → 0 f ( x ) = 1 f(x)=\frac{\sin x}{x}(x\neq 0),\,\,\lim_{x\to 0}f(x)=1 f(x)=xsinx​(x​=0),x→0lim​f(x)=1补充定义 f ( 0 ) = 1 f(0)=1 f(0)=1.

  因为 f ( x ) = sin ⁡ x / x f(x)=\sin x/x f(x)=sinx/x，所以其零点为 k π , k = ± 1 , ± 2 , ± 3 , ⋯ k\pi,k=\pm1,\pm2,\pm3,\cdots kπ,k=±1,±2,±3,⋯

  将 f ( x ) f(x) f(x) 看作一个多项式，可以通过零点表示，即 f ( x ) = c 1 ( x − π ) ( x + π ) ( x − 2 π ) ( x + 2 π ) ( x − 3 π ) ( x + 3 π ) ⋯ = c 1 ( x 2 − π 2 ) ( x 2 − 4 π 2 ) ( x 2 − 9 π 2 ) ⋯ \begin{aligned} f(x)&=c_1(x-\pi)(x+\pi)(x-2\pi)(x+2\pi)(x-3\pi)(x+3\pi)\cdots\\ &=c_1(x^2-\pi^2)(x^2-4\pi^2)(x^2-9\pi^2)\cdots \end{aligned} f(x)​=c1​(x−π)(x+π)(x−2π)(x+2π)(x−3π)(x+3π)⋯=c1​(x2−π2)(x2−4π2)(x2−9π2)⋯​将 sin ⁡ x = ∑ k = 0 ∞ ( − 1 ) k ( 2 k + 1 ) ! x 2 k + 1 = x − 1 3 ! x 3 + 1 5 ! x 5 − 1 7 ! x 7 + ⋯ \sin x=\sum_{k=0}^{\infin}\frac{(-1)^k}{(2k+1)!}x^{2k+1} =x-\frac{1}{3!}x^3+\frac{1}{5!}x^5-\frac{1}{7!}x^7+\cdots sinx=k=0∑∞​(2k+1)!(−1)k​x2k+1=x−3!1​x3+5!1​x5−7!1​x7+⋯代入 f ( x ) = sin ⁡ x / x f(x)=\sin x/x f(x)=sinx/x，得 f ( x ) = 1 − 1 3 ! x 2 + 1 5 ! x 4 − 1 7 ! x 6 + ⋯ f(x)=1-\frac{1}{3!}x^2+\frac{1}{5!}x^4-\frac{1}{7!}x^6+\cdots f(x)=1−3!1​x2+5!1​x4−7!1​x6+⋯将这两个式子放在一起 1 − 1 3 ! x 2 + 1 5 ! x 4 − 1 7 ! x 6 + ⋯ = c 1 ( x 2 − π 2 ) ( x 2 − 4 π 2 ) ( x 2 − 9 π 2 ) ⋯ 1-\frac{1}{3!}x^2+\frac{1}{5!}x^4-\frac{1}{7!}x^6+\cdots=c_1(x^2-\pi^2)(x^2-4\pi^2)(x^2-9\pi^2)\cdots 1−3!1​x2+5!1​x4−7!1​x6+⋯=c1​(x2−π2)(x2−4π2)(x2−9π2)⋯由两端常数项相等，得 1 = c 1 ( − π 2 ) ( − 4 π 2 ) ( − 9 π 2 ) ⋯ c 1 = 1 ( − π 2 ) ( − 4 π 2 ) ( − 9 π 2 ) ⋯ 1=c_1(-\pi^2)(-4\pi^2)(-9\pi^2)\cdots\\ c_1=\frac{1}{(-\pi^2)(-4\pi^2)(-9\pi^2)\cdots} 1=c1​(−π2)(−4π2)(−9π2)⋯c1​=(−π2)(−4π2)(−9π2)⋯1​代入原式，得 1 − 1 3 ! x 2 + 1 5 ! x 4 − 1 7 ! x 6 + ⋯ = ( 1 − x 2 π 2 ) ( 1 − x 2 4 π 2 ) ( 1 − x 2 9 π 2 ) ⋯ 1-\frac{1}{3!}x^2+\frac{1}{5!}x^4-\frac{1}{7!}x^6+\cdots=(1-\frac{x^2}{\pi^2})(1-\frac{x^2}{4\pi^2})(1-\frac{x^2}{9\pi^2})\cdots 1−3!1​x2+5!1​x4−7!1​x6+⋯=(1−π2x2​)(1−4π2x2​)(1−9π2x2​)⋯   考虑两端 x 2 x^2 x2 的系数，有人可能会问右端 x 2 x^2 x2 的系数是怎么求出来的？没错，我也想问.

  我的理解是这样的：利用组合的思想，右端的 x 2 x^2 x2 项只能这样得出：第 k k k 个括号里的 − x 2 k 2 π 2 \displaystyle-\frac{x^2}{k^2\pi^2} −k2π2x2​ 与其他括号中的 1 1 1 相乘，然后将这无穷多项相加.

  所以右端 x 2 x^2 x2 的系数为 − 1 π 2 − 1 4 π 2 − 1 9 π 2 − ⋯ = − 1 π 2 ∑ n = 1 ∞ 1 n 2 -\frac{1}{\pi^2}-\frac{1}{4\pi^2}-\frac{1}{9\pi^2}-\cdots=-\frac{1}{\pi^2}\sum_{n=1}^{\infin}\frac{1}{n^2} −π21​−4π21​−9π21​−⋯=−π21​n=1∑∞​n21​而左端 x 2 x^2 x2 的系数为 − 1 3 ! \displaystyle -\frac{1}{3!} −3!1​，所以 − 1 3 ! = − 1 π 2 ∑ n = 1 ∞ 1 n 2 π 2 6 = ∑ n = 1 ∞ 1 n 2 = 1 + 1 2 2 + 1 3 2 + 1 4 2 + ⋯ \begin{aligned} -\frac{1}{3!}&=-\frac{1}{\pi^2}\sum_{n=1}^{\infin}\frac{1}{n^2}\\ \frac{\pi^2}{6}&=\sum_{n=1}^{\infin}\frac{1}{n^2}=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots \end{aligned} −3!1​6π2​​=−π21​n=1∑∞​n21​=n=1∑∞​n21​=1+221​+321​+421​+⋯​Okay, 证明完毕 ！

如有纰漏，还请指正！